Model-Based Measurement of Name Concentration Risk in Credit Portfolios

Abstract

This chapter deals with the measurement of name concentrations. This type of concentration risk occurs if the weight of single credits in the portfolio does not converge to zero; thus, the individual risk component cannot be completely diversified. The main research questions on name concentrations that are considered in this chapter are:

• In which cases are the assumptions of the ASRF framework critical concerning the credit portfolio size?

• In which cases are currently discussed adjustments for the VaR-measurement able to overcome the shortcomings of the ASRF model?

Concerning the first question, it is analyzed how many credits are at least necessary implying the neglect of undiversified individual risk not to be problematic. Since there exist analytical formulas – the so-called granularity adjustment – which approximate these risks, it is further determined in which cases these formulas are able to lead to desired results.

Appendix

4.1.1 Alternative Derivation of the First-Order Granularity Adjustment

With reference to Wilde (2001), the granularity adjustment will be derived as an approximation of the difference $\Delta q$ between the true VaR of a granular portfolio ${q^{(n)}}$ and the approximation ${q^{(\infty )}}$ that results if infinite granularity is assumed to hold:

$$ \Delta q = q_\alpha^{(n)} - q_\alpha^{(\infty )}. $$

(4.104)

Instead of determining the add-on $\Delta q$ directly, it will be analyzed how much the confidence level α will be overestimated or the probability $p: = 1 - \alpha$ of exceeding the VaR will be underestimated if the portfolio is assumed to be infinitely granular. Thus, the probability

$$ \Delta p = {p^{(\infty )}} - p = \alpha - {\alpha^{(\infty )}} $$

(4.105)

refers to the overestimation of the confidence level if only the systematic loss is considered. Here, α is the specified “target” confidence level, and by definition also the probability that the systematic loss will not exceed $q_\alpha^{(\infty )}$:

$$ 1 - p = \alpha : = b{P}\left( {\tilde{L} \leq q_\alpha^{(n)}} \right) = b{P}\left( {b{E}\left[ {\tilde{L}|\tilde{x}} \right] \leq q_\alpha^{(\infty )}} \right). $$

(4.106)

By contrast, ${\alpha^{(\infty )}}$ is the actual confidence level if the VaR is approximated by the ASRF model:

$$ 1 - {p^{(\infty )}} = {\alpha^{(\infty )}}: = b{P}\left( {\tilde{L} \leq q_\alpha^{(\infty )}} \right). $$

(4.107)

Subsequent to the derivation of $\Delta p$, the result will be transformed into a shift of the loss quantile $\Delta q$.

Analogous to Appendix 2.8.3, the unconditional probability ${p^{(\infty )}}$ can be expressed in terms of the conditional probability. Then, the substitution $y: = q_\alpha^{(\infty )} + t$ is performed to center the integration at $q_\alpha^{(\infty )}$:

$$ \begin{array} {c} p + \Delta p = b{P}\left( {\tilde{L} \geq q_\alpha^{(\infty )}} \right) = \int\limits_{y = - \infty }^\infty {b{P}\left( {\tilde{L} \geq q_\alpha^{(\infty )}|\tilde{Y} = y} \right)} \,{f_Y}(y)dy \\ = \int\limits_{t = - \infty }^\infty {b{P}\left( {\tilde{L} \geq q_\alpha^{(\infty )}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right)} \,{f_Y}\left( {q_\alpha^{(\infty )} + t} \right)dt, \\ \end{array} $$

(4.108)

with the shorter notation $\tilde{Y}: = b{E}\left( {\tilde{L}|\tilde{x}} \right)$ for the conditional expectation. According to (4.106), the probability p can be written as

$$ p = b{P}\left( {\tilde{Y} \geq q_\alpha^{(\infty )}} \right) = \int\limits_{y = q_\alpha^{(\infty )}}^\infty {{f_Y}(y)\,dy} = \int\limits_{t = 0}^\infty {{f_Y}\left( {q_\alpha^{(\infty )} + t} \right)\,dt} $$

(4.109)

using the substitution $y: = q_\alpha^{(\infty )} + t$ again, so that $t(y = q_\alpha^{(\infty )}) = 0$ and $t(y = \infty ) = \infty$. Hence, (4.108) can be expressed as

$$ \begin{array} {c} \Delta p = \int\limits_{t = - \infty }^\infty {b{P}\left( {\tilde{L} \geq q_\alpha^{(\infty )}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right)} \,{f_Y}\left( {q_\alpha^{(\infty )} + t} \right)dt - \int\limits_{t = 0}^\infty {{f_Y}\left( {q_\alpha^{(\infty )} + t} \right)\,dt} \\ = \int\limits_{t = - \infty }^0 {b{P}\left( {\tilde{L} \geq q_\alpha^{(\infty )}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right)} \,{f_Y}\left( {q_\alpha^{(\infty )} + t} \right)dt \\ + \int\limits_{t = 0}^\infty {\left[ {b{P}\left( {\tilde{L} \geq q_\alpha^{(\infty )}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right) - 1} \right]{f_Y}\left( {q_\alpha^{(\infty )} + t} \right)dt} . \\ \end{array} $$

(4.110)

The following transformations are performed for simplification of the integrand in order to solve the integral. A realization of the systematic loss implies a realization of the systematic factor. As the credit loss events are assumed to be independent for a realization of the systematic factor, the conditional credit losses follow a binomial distribution, which can be approximated by a normal distribution for a sufficient number of credits. This leads to

$$ \begin{array} {c} b{P}\left( {\tilde{L} \geq q_\alpha^{(\infty )}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right) = 1 - b{P}\left( {\tilde{L} < q_\alpha^{(\infty )}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right) \\ \approx 1 - \Phi \left( {\frac{{q_\alpha^{(\infty )} - b{E}\left( {\tilde{L}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right)}}{{\sqrt {{b{V}\left( {\tilde{L}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right)}} }}} \right). \\ \end{array} $$

(4.111)

As $b{E}(\tilde{L}) = b{E}(b{E}(\tilde{L}|\tilde{x})) = b{E}(\tilde{Y})$, which is due to the law of iterated expectation, the conditional expectation of (4.111) equals

$$ b{E}\left( {\tilde{L}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right) = b{E}\left( {\tilde{Y}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right) = q_\alpha^{(\infty )} + t. $$

(4.112)

With the symmetry $1 - \Phi ( - x) = \Phi (x)$ and defining ${\sigma^2}(y): = b{V}(\tilde{L}|\tilde{Y} = y)$, (4.111) results in

$$ b{P}\left( {\tilde{L} \geq q_\alpha^{(\infty )}|\tilde{Y} = q_\alpha^{(\infty )} + t} \right) \approx 1 - \Phi \left( {\frac{{q_\alpha^{(\infty )} - q_\alpha^{(\infty )} - t}}{{\sigma \left( {q_\alpha^{(\infty )} + t} \right)}}} \right) = \Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )} + t} \right)}}} \right), $$

(4.113)

so that (4.110) can be written as

$$ \Delta p = \int\limits_{t = - \infty }^0 {\Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )} + t} \right)}}} \right)} \,{f_Y}\left( {q_\alpha^{(\infty )} + t} \right)dt + \int\limits_{t = 0}^\infty {\left[ {\Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )} + t} \right)}}} \right) - 1} \right]{f_Y}\left( {q_\alpha^{(\infty )} + t} \right)dt} . $$

(4.114)

Subsequently, several linear approximations will be performed relying on the assumption that the loss quantile of the granular portfolio is close to the systematic loss quantile and the linearizations lead to minor errors. Linearizing the density function at $q_\alpha^{(\infty )}$ leads to

$$ {f_Y}\left( {q_\alpha^{(\infty )} + t} \right) \approx {f_Y}\left( {q_\alpha^{(\infty )}} \right) + t \cdot {\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}. $$

(4.115)

The argument of the normal distribution can be approximated as

$$ \begin{array} {c} t \cdot \left( {\frac{1}{{\sigma \left( {q_\alpha^{(\infty )} + t} \right)}}} \right) \approx t \cdot \left( {\frac{1}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}} + t \cdot {{\left[ {\frac{d}{{dt}}\frac{1}{{\sigma \left( {q_\alpha^{(\infty )} + t} \right)}}} \right]}_{t = 0}}} \right) \\ = t \cdot \left( {\frac{1}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}} + t \cdot {{\left[ { - \frac{1}{{{\sigma^2}\left( {q_\alpha^{(\infty )} + t} \right)}}\frac{d}{{dt}}\sigma \left( {q_\alpha^{(\infty )} + t} \right)} \right]}_{t = 0}}} \right) \\ = \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}} - \frac{{{t^2}}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}{{\left[ {\frac{d}{{dt}}\sigma \left( {q_\alpha^{(\infty )} + t} \right)} \right]}_{t = 0}}} \right). \\ \end{array} $$

(4.116)

With the substitution $y: = q_\alpha^{(\infty )} + t$, so $dy/dt = 1$ and $y(t = 0) = q_\alpha^{(\infty )}$, the derivative of the conditional standard deviation can be rewritten as

$$ \frac{d}{{dt}}\sigma {\left. {\left( {q_\alpha^{(\infty )} + t} \right)} \right|_{t = 0}} = \frac{d}{{dy}}\sigma {\left. {(y)} \right|_{y = q_\alpha^{(\infty )}}}. $$

(4.117)

Inserting (4.115)–(4.117) in (4.114) leads to

$$ \begin{array} {c} \Delta p = \left( {\int\limits_{t = - \infty }^0 {\Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}} - \frac{{{t^2}}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}{{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}} \right)} \cdot \left[ {{f_Y}\left( {q_\alpha^{(\infty )}} \right) + t \cdot {{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}} \right]dt} \right) \\ - \left( { - \int\limits_{t = 0}^\infty {\left[ {\Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}} - \frac{{{t^2}}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}{{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}} \right) - 1} \right] \cdot \left[ {{f_Y}\left( {q_\alpha^{(\infty )}} \right) + t \cdot {{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}} \right]dt} } \right) \\ = :\Delta {p_1} - \Delta {p_2}. \\ \end{array} $$

(4.118)

When the substitution $t: = - t$ for the term $\Delta {p_2}$ is performed and the symmetry of the normal distribution $\Phi ( - x) - 1 = - \Phi (x)$ is used, both terms $\Delta {p_1}$ and $\Delta {p_2}$ are identical except for the algebraic signs:

$$ \begin{array} {c} \Delta {p_2} = - \int\limits_{t = 0}^{ - \infty } {\left[ {\Phi \left( { - \left[ {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}} + \frac{{{t^2}}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}{{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}} \right]} \right) - 1} \right]} \\ \cdot \left[ {{f_Y}\left( {q_\alpha^{(\infty )}} \right) - t \cdot {{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}} \right] \cdot \left( { - 1} \right)dt \\ = \int\limits_{t = - \infty }^0 {\Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}} + \frac{{{t^2}}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}{{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}} \right) \cdot \left( {{f_Y}\left( {q_\alpha^{(\infty )}} \right) - t \cdot {{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}} \right)dt} . \\ \end{array} $$

(4.119)

A linearization of the normal distributions in $\Delta {p_1}$ and $\Delta {p_2}$ results in

$$ \begin{array} {c} \Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}} \mp \frac{{{t^2}}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}{{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}} \right) \\ \approx \Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}}} \right) \mp \frac{{{t^2}}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}{\left. {\frac{{d\Phi (y)}}{{dy}}} \right|_{y = \frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}}}} \\ = \Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}}} \right) \mp \frac{{{t^2}}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}\varphi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}}} \right). \\ \end{array} $$

(4.120)

Using this approximation, the terms $\Delta {p_1}$ and $\Delta {p_2}$ from (4.118) can be written as

$$ \begin{array} {c} \Delta {p_{1,2}} \approx \int\limits_{t = - \infty }^0 {\underbrace {\Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}}} \right)}_{ = :{\beta_0}} \cdot \left[ {\underbrace {{f_Y}\left( {q_\alpha^{(\infty )}} \right)}_{ = :{\gamma_0}}\pm \underbrace {t{{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}}_{ = :{\gamma_1}}} \right]dt} \\ \mp \int\limits_{t = - \infty }^0 {\underbrace {\frac{{{t^2}}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}{{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}\varphi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}}} \right)}_{ = :{\beta_1}} \cdot \left[ {\underbrace {{f_Y}\left( {q_\alpha^{(\infty )}} \right)}_{ = :{\gamma_0}}\pm \underbrace {t{{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}}_{ = :{\gamma_1}}} \right]dt} . \\ \end{array} $$

(4.121)

The summands ${\beta_0},\;{\gamma_0}$ are the points around which the linearizations have been performed. The summands ${\beta_1},\;{\gamma_1}$ have resulted from the first-order approximations. Using this notation, the shift in probability $\Delta p$ of (4.118) can notably be simplified to

$$ \begin{array} {c} \Delta p \approx \Delta {p_1} - \Delta {p_2} \\ \approx \int\limits_{t = - \infty }^0 {{\beta_0}\left( {{\gamma_0} + {\gamma_1}} \right)} - {\beta_1}\left( {{\gamma_0} + {\gamma_1}} \right)dt - \int\limits_{t = - \infty }^0 {{\beta_0}\left( {{\gamma_0} - {\gamma_1}} \right)} + {\beta_1}\left( {{\gamma_0} - {\gamma_1}} \right)dt \\ = \int\limits_{t = - \infty }^0 {2{\beta_0}{\gamma_1} - 2{\beta_1}{\gamma_0}dt} . \\ \end{array} $$

(4.122)

Fortunately, both integrands are already first-order terms whereas the cross-terms ${\beta_1} \cdot {\gamma_1}$ vanish.⁷⁰ Thus, there is no need for a further linearization. The remaining expression is

$$ \begin{array} {c} \Delta p \approx 2{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}\int\limits_{t = - \infty }^0 {t \cdot \Phi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}}} \right)dt} \\ - 2{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}\frac{{{f_Y}\left( {q_\alpha^{(\infty )}} \right)}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}\int\limits_{t = - \infty }^0 {{t^2} \cdot \varphi \left( {\frac{t}{{\sigma \left( {q_\alpha^{(\infty )}} \right)}}} \right)dt} . \\ \end{array} $$

(4.123)

In order to solve the integrals, the substitution $y: = t/\sigma (q_\alpha^{(\infty )})$ is performed, with $dy/dt = 1/\sigma (q_\alpha^{(\infty )})$, $y(t = - \infty ) = - \infty$ and $y(t = 0) = 0$:

$$ \begin{array} {c} \Delta p \approx 2{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}\int\limits_{y = - \infty }^0 {y \cdot \sigma \left( {q_\alpha^{(\infty )}} \right) \cdot \Phi (y) \cdot \sigma \left( {q_\alpha^{(\infty )}} \right)dy} \\ - 2{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}\frac{{{f_Y}\left( {q_\alpha^{(\infty )}} \right)}}{{{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)}}\int\limits_{y = - \infty }^0 {{{\left[ {y \cdot \sigma \left( {q_\alpha^{(\infty )}} \right)} \right]}^2} \cdot \varphi (y) \cdot \sigma \left( {q_\alpha^{(\infty )}} \right)dy} \\ = 2{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}{\sigma^2}\left( {q_\alpha^{(\infty )}} \right)\underbrace {\int\limits_{y = - \infty }^0 {y \cdot \Phi (y)dy} }_* \\ - 2{\left. {\frac{{d\sigma (y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}{f_Y}\left( {q_\alpha^{(\infty )}} \right) \cdot \sigma \left( {q_\alpha^{(\infty )}} \right)\underbrace {\int\limits_{y = - \infty }^0 {{y^2} \cdot \varphi (y)dy} }_{**}. \\ \end{array} $$

(4.124)

For the second integral (**), it is used that the integrand is axially symmetric to the y-axis. Furthermore, the definition of the variance is utilized, considering that the standard normal distribution has mean ${\mu_Y} = 0$ and variance $\sigma_Y^2 = 1$:

$$ \int\limits_{y = - \infty }^0 {{y^2} \cdot \varphi (y)dy} = \frac{1}{2}\int\limits_{y = - \infty }^\infty {{y^2} \cdot \varphi (y)dy} = \frac{1}{2}\int\limits_{y = - \infty }^\infty {{{\left( {y - {\mu_Y}} \right)}^2} \cdot \varphi (y)dy} = \frac{1}{2}{\sigma_Y}^2 = \frac{1}{2}. $$

(4.125)

The first integral (*) can be calculated with integration by parts:

$$ \int\limits_{y = - \infty }^0 {y \cdot \Phi (y)dy} = \left[ {\frac{1}{2}{y^2} \cdot \Phi (y)} \right]_{y = - \infty }^0 - \int\limits_{y = - \infty }^0 {\frac{1}{2}{y^2} \cdot \varphi (y)} \,dy. $$

(4.126)

For $y = 0$, the first term is zero but for $y = - \infty$, the result is not obvious. Using l’Hôpital’s rule several times leads to⁷¹

$$ \begin{array} {c} \mathop {{\lim }}\limits_{y \to - \infty } \frac{1}{2}{y^2} \cdot \Phi (y) = \mathop {{\lim }}\limits_{y \to \infty } \frac{1}{2}\frac{{\Phi \left( { - y} \right)}}{{{y^{ - 2}}}}\mathop { = }\limits^{{\text{l'H\^o pital}}} \mathop {{\lim }}\limits_{y \to \infty } \frac{1}{2}\frac{{ - \varphi \left( { - y} \right)}}{{ - 2{y^{ - 3}}}} = \mathop {{\lim }}\limits_{y \to \infty } \frac{1}{4}\frac{{{y^3}}}{{{e^{{{{{y^2}}} \left/ {2} \right.}}}}}\mathop { = }\limits^{{\text{l'H\^o pital}}} \mathop {{\lim }}\limits_{y \to \infty } \frac{1}{4}\frac{{3{y^2}}}{{y \cdot {e^{{{{{y^2}}} \left/ {2} \right.}}}}} \\ = \mathop {{\lim }}\limits_{y \to \infty } \frac{3}{4}\frac{y}{{{e^{{{{{y^2}}} \left/ {2} \right.}}}}}\mathop { = }\limits^{{\text{l'H\^o pital}}} \mathop {{\lim }}\limits_{y \to \infty } \frac{3}{4}\frac{1}{{y \cdot {e^{{{{{y^2}}} \left/ {2} \right.}}}}} = 0, \\ \end{array} $$

(4.127)

so that the first term of (4.126) vanishes. Using the result of the previous integration, (4.126) equals $- {1/4}$. Hence, $\Delta p$ from (4.124) is given as

$$ \Delta p \approx - \frac{1}{2}{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}{\sigma^2}\left( {q_\alpha^{(\infty )}} \right) - {\left. {\frac{{d\sigma (y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}{f_Y}\left( {q_\alpha^{(\infty )}} \right) \cdot \sigma \left( {q_\alpha^{(\infty )}} \right). $$

(4.128)

Because of $\sigma \frac{{d\sigma }}{{dy}} = \frac{1}{2}\frac{{d{\sigma^2}}}{{d\sigma }}\frac{{d\sigma }}{{dy}} = \frac{1}{2}\frac{{d{\sigma^2}}}{{dy}}$, (4.128) is equivalent to

$$ \begin{array} {c} \Delta p \approx - \left[ {\frac{1}{2}{{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}{\sigma^2}\left( {q_\alpha^{(\infty )}} \right) + {{\left. {\frac{1}{2}\frac{{d{\sigma^2}(y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}{f_Y}\left( {q_\alpha^{(\infty )}} \right)} \right] \\ = - \frac{1}{2}{\left[ {\frac{{d{f_Y}(y)}}{{dy}}{\sigma^2}(y) + \frac{{d{\sigma^2}(y)}}{{dy}}{f_Y}(y)} \right]_{y = q_\alpha^{(\infty )}}} \\ = - \frac{1}{2}\frac{d}{{dy}}{\left. {\left( {{f_Y}(y) \cdot {\sigma^2}(y)} \right)} \right|_{y = q_\alpha^{(\infty )}}}. \\ \end{array} $$

(4.129)

This expression is the linearized deviation of the specified probability $p = 1 - \alpha$ if only the systematic loss is considered for calculation of the loss quantile.

As initially noticed, the determined shift of the probability has to be transformed into a shift of the loss quantile (cf. Fig. 4.12). If the probability density function of the portfolio loss is assumed to be almost linear in a region around the quantile, the required transformation is

$$ \Delta p \approx \frac{1}{2}\left[ {{f_Y}\left( {q_\alpha^{(\infty )}} \right) + {f_Y}\left( {q_\alpha^{(\infty )} + \Delta q} \right)} \right]\Delta q. $$

(4.130)

Fig. 4.12

Relation between the shift of the probability and the loss quantile

Two last first-order approximations lead to

$$ \begin{array} {c} \Delta p \approx \frac{1}{2}\left[ {{f_Y}\left( {q_\alpha^{(\infty )}} \right) + \left( {{f_Y}\left( {q_\alpha^{(\infty )}} \right) + \Delta q{{\left. {\frac{{d{f_Y}(y)}}{{dy}}} \right|}_{y = q_\alpha^{(\infty )}}}} \right)} \right]\Delta q \\ = {f_Y}\left( {q_\alpha^{(\infty )}} \right) \cdot \Delta q + {\left. {\frac{1}{2}\frac{{d{f_Y}(y)}}{{dy}}} \right|_{y = q_\alpha^{(\infty )}}}{\left( {\Delta q} \right)^2} \\ \approx {f_Y}\left( {q_\alpha^{(\infty )}} \right) \cdot \Delta q. \\ \end{array} $$

(4.131)

Inserting (4.129) into (4.131) finally leads to

$$ \begin{array} {c} \Delta q \approx \frac{{\Delta p}}{{{f_Y}\left( {q_\alpha^{(\infty )}} \right)}} \approx - \frac{1}{2}\frac{1}{{{f_Y}(y)}}\frac{d}{{dy}}{\left. {\left( {{f_Y}(y) \cdot {\sigma^2}(y)} \right)} \right|_{y = q_\alpha^{(\infty )}}} \\ = - \frac{1}{2}\frac{1}{{{f_Y}(y)}}\frac{d}{{dy}}{\left. {\left( {{f_Y}(y) \cdot b{V}\left( {\tilde{L}|\tilde{Y} = y} \right)} \right)} \right|_{y = q_\alpha^{(\infty )}}}. \\ \end{array} $$

(4.132)

Using (4.8), this can be written as

$$ \Delta q \approx - \frac{1}{{2{f_x}(x)}}\frac{d}{{dx}}{\left. {\left( {\frac{{{f_x}(x)b{V}\left[ {\tilde{L}|\tilde{x} = x} \right]}}{{\frac{d}{{dx}}b{E}\left[ {\tilde{L}|\tilde{x} = x} \right]}}} \right)} \right|_{x = {q_{1 - \alpha }}\left( {\tilde{x}} \right)}}, $$

(4.133)

which is identical to the first-order granularity adjustment of Sect. 4.2.1.1.⁷²

4.1.2 First and Second Derivative of VaR

The derivatives of VaR will be determined on the basis of Rau-Bredow (2002, 2004) in the following. Consider two continuous random variables $\tilde{Y}$ and $\tilde{Z}$ with joint probability density function $f(y,z)$ and a variable $\lambda \in b{R}$. The VaR (the quantile) $q: = {q_\alpha }\left( {\tilde{L}} \right)$ of $\tilde{L} = \tilde{Y} + \lambda \tilde{Z}$ can implicitly be defined as⁷³

$$ b{P}\left( {\tilde{L} \leq q} \right) = \alpha . $$

(4.134)

Furthermore, the formula of the conditional density function will be used:⁷⁴

$$ {f_{Z|Y = y}}(z) = \frac{{{f_{Y,Z}}(y,z)}}{{{f_Y}(y)}} $$

(4.135)

leading to⁷⁵

$$ {f_{Z|Y + \lambda Z = q}}(z) = \frac{{{f_{Y + \lambda Z,Z}}\left( {q,z} \right)}}{{{f_{Y + \lambda Z}}(q)}} = \frac{{{f_{Y,Z}}\left( {q - \lambda z,z} \right)}}{{{f_{Y + \lambda Z}}(q)}}. $$

(4.136)

4.1.2.1 First Derivative

As the derivative of the constant α is zero, the derivative of (4.134) is

$$ \begin{array} {c} 0 = \frac{\partial }{{\partial \lambda }}b{P}\left( {\tilde{Y} + \lambda \tilde{Z} \leq q} \right) \\ = \frac{\partial }{{\partial \lambda }}\int\limits_{z = - \infty }^\infty {\int\limits_{y = - \infty }^{q - \lambda z} {{f_{Y,Z}}(y,z)\,dy\,dz} } \\ = \int\limits_{z = - \infty }^\infty {\frac{\partial }{{\partial \lambda }}\int\limits_{y = - \infty }^{q - \lambda z} {{f_{Y,Z}}(y,z)\,dy\,dz} } . \\ \end{array} $$

(4.137)

Performing the inner integration and the differentiation leads to

$$ 0 = \int\limits_{z = - \infty }^\infty {\left( {\frac{{dq}}{{d\lambda }} - z} \right)\,{f_{Y,Z}}\left( {q - \lambda z,z} \right)\,dz} . $$

(4.138)

Using the formula for the conditional density function (4.135) and the integral representation of the conditional expectation, we get

$$ \begin{array} {c} 0 = \int\limits_{z = - \infty }^\infty {\left( {\frac{{dq}}{{d\lambda }} - z} \right)\,{f_{Y + \lambda Z}}(q)\,{f_{Z|Y + \lambda Z = q}}(z)\,dz} \\ = {f_{Y + \lambda Z}}(q)\left( {\frac{{dq}}{{d\lambda }}\int\limits_{z = - \infty }^\infty {{f_{Z|Y + \lambda Z = q}}(z)\,dz} - \int\limits_{z = - \infty }^\infty {z\,{f_{Z|Y + \lambda Z = q}}(z)\,dz} } \right) \\ = {f_{Y + \lambda Z}}(q)\left( {\frac{{dq}}{{d\lambda }} \cdot 1 - b{E}\left[ {\tilde{Z}|\tilde{Y} + \lambda \tilde{Z} = q} \right]} \right). \\ \end{array} $$

(4.139)

This leads to the first derivative of VaR:

$$ \frac{{dVa{R_\alpha }\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}{{d\lambda }} = b{E}\left[ {\tilde{Z}|\tilde{Y} + \lambda \tilde{Z} = {q_\alpha }\left( {\tilde{Y} + \lambda \tilde{Z}} \right)} \right]. $$

(4.140)

The first derivative at $\lambda = 0$ is

$$ {\left. {\frac{{dVa{R_\alpha }\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}{{d\lambda }}} \right|_{\lambda = 0}} = b{E}\left[ {\tilde{Z}|\tilde{Y} = {q_\alpha }\left( {\tilde{Y}} \right)} \right]. $$

(4.141)

4.1.2.2 Second Derivative

Similar to (4.137), the second derivative of (4.134) is

$$ 0 = \frac{{{\partial^2}}}{{\partial {\lambda^2}}}b{P}\left( {\tilde{Y} + \lambda \tilde{Z} \leq q} \right) = \frac{{{\partial^2}}}{{\partial {\lambda^2}}}\int\limits_{z = - \infty }^\infty {\int\limits_{y = - \infty }^{q - \lambda z} {{f_{Y,Z}}(y,z)\,dy\,dz} } . $$

(4.142)

With the first derivative of (4.138) and applying the product rule, this leads to

$$ \begin{array} {c} 0 = \frac{\partial }{{\partial \lambda }}\int\limits_{z = - \infty }^\infty {\left( {\frac{{dq}}{{d\lambda }} - z} \right){f_{Y,Z}}\left( {q - \lambda z,z} \right)\,dz} \\ = \int\limits_{z = - \infty }^\infty {\left( {\frac{{{d^2}q}}{{{d^2}\lambda }}} \right){f_{Y,Z}}\left( {q - \lambda z,z} \right) + \left( {\frac{{dq}}{{d\lambda }} - z} \right)\underbrace {\frac{{\partial {f_{Y,Z}}\left( {q - \lambda z,z} \right)}}{{\partial \lambda }}}_*dz} . \\ \end{array} $$

(4.143)

The derivative (*) can be determined with the chain rule:

$$ \begin{array} {c} \frac{{\partial {f_{Y,Z}}\left( {q - \lambda z,z} \right)}}{{\partial \lambda }} = \frac{{\partial \left( {q - \lambda z} \right)}}{{\partial \lambda }}\frac{{\partial {f_{Y,Z}}\left( {q - \lambda z,z} \right)}}{{\partial \left( {q - \lambda z} \right)}}\frac{{\partial q}}{{\partial q}} \\ = \left( {\frac{{dq}}{{d\lambda }} - z} \right)\frac{{\partial {f_{Y,Z}}\left( {q - \lambda z,z} \right)}}{{\partial q}}\frac{1}{{{{{\partial \left( {q - \lambda z} \right)}} \left/ {{\partial q}} \right.}}} \\ = \left( {\frac{{dq}}{{d\lambda }} - z} \right)\frac{{\partial {f_{Y,Z}}(q - \lambda z,z)}}{{\partial q}}. \\ \end{array} $$

(4.144)

Inserting (4.144) and the conditional density (4.136) into (4.143) results in

$$ \begin{array} {c} 0 = \int\limits_{z = - \infty }^\infty {\left( {\frac{{{d^2}q}}{{{d^2}\lambda }}} \right){f_{Y,Z}}\left( {q - \lambda z,z} \right) + {{\left( {\frac{{dq}}{{d\lambda }} - z} \right)}^2}\frac{{\partial {f_{Y,Z}}\left( {q - \lambda z,z} \right)}}{{\partial q}}dz} \\ = \left( {\frac{{{d^2}q}}{{{d^2}\lambda }}} \right)\int\limits_{z = - \infty }^\infty {{f_{Y + \lambda Z}}(q){f_{Z|Y + \lambda Z = q}}(z)dz} \\ + \int\limits_{z = - \infty }^\infty {{{\left( {\frac{{dq}}{{d\lambda }} - z} \right)}^2}\frac{{\partial \left( {{f_{Y + \lambda Z}}(q)\,{f_{Z|Y + \lambda Z = q}}(z)} \right)}}{{\partial q}}dz} . \\ \end{array} $$

(4.145)

The first summand of (4.145) equals

$$ \left( {\frac{{{d^2}q}}{{{d^2}\lambda }}} \right){f_{Y + \lambda Z}}(q)\int\limits_{z = - \infty }^\infty {{f_{Z|Y + \lambda Z = q}}(z)dz} = \left( {\frac{{{d^2}q}}{{{d^2}\lambda }}} \right){f_{Y + \lambda Z}}(q). $$

(4.146)

In order to calculate the second summand of (4.145), the first derivative from (4.140) as well as the integral representation of the conditional variance is used:

$$ \begin{array} {c} \int\limits_{z = - \infty }^\infty {{{\left( {\frac{{dq}}{{d\lambda }} - z} \right)}^2}\frac{{\partial \left( {{f_{Y + \lambda Z}}(q)\,{f_{Z|Y + \lambda Z = q}}(z)} \right)}}{{\partial q}}dz} \\ = \int\limits_{z = - \infty }^\infty {{{\left( {z - b{E}\left[ {\tilde{Z}|\tilde{Y} + \lambda \tilde{Z} = q} \right]} \right)}^2}\frac{{\partial \left( {{f_{Y + \lambda Z}}(q){f_{Z|Y + \lambda Z = q}}(z)} \right)}}{{\partial q}}dz} \\ = {\left. {\frac{d}{{dy}}\left( {{f_{Y + \lambda Z}}(y)\int\limits_{z = - \infty }^\infty {{{\left( {z - b{E}\left[ {\tilde{Z}|\tilde{Y} + \lambda \tilde{Z} = q} \right]} \right)}^2}{f_{Z|Y + \lambda Z = y}}(z)dz} } \right)} \right|_{y = q}} \\ = {\left. {\frac{d}{{dy}}\left( {{f_{Y + \lambda Z}}(y)b{V}\left[ {\tilde{Z}|\tilde{Y} + \lambda \tilde{Z} = y} \right]} \right)} \right|_{y = q}}. \\ \end{array} $$

(4.147)

With these summands, (4.145) can be written as

$$ 0 = \left( {\frac{{{d^2}q}}{{{d^2}\lambda }}} \right){f_{Y + \lambda Z}}(y) + {\left. {\frac{d}{{dy}}\left( {{f_{Y + \lambda Z}}(y)b{V}\left[ {\tilde{Z}|\tilde{Y} + \lambda \tilde{Z} = y} \right]} \right)} \right|_{y = q}}. $$

(4.148)

Thus, the second derivative of VaR is equal to

$$ \frac{{{d^2}Va{R_\alpha }\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}{{{d^2}\lambda }} = - \frac{1}{{{f_{Y + \lambda Z}}(y)}} \cdot {\left. {\frac{d}{{dy}}\left( {{f_{Y + \lambda Z}}(y)b{V}\left[ {\tilde{Z}|\tilde{Y} + \lambda \tilde{Z} = y} \right]} \right)} \right|_{y = {q_\alpha }\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}. $$

(4.149)

The second derivative at $\lambda = 0$ is

$$ {\left. {\frac{{{d^2}Va{R_\alpha }\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}{{{d^2}\lambda }}} \right|_{\lambda = 0}} = - \frac{1}{{{f_Y}(y)}}{\left. {\frac{d}{{dy}}\left( {{f_Y}(y)b{V}\left[ {\tilde{Z}|\tilde{Y} = y} \right]} \right)} \right|_{y = {q_\alpha }\left( {\tilde{Y}} \right)}}. $$

(4.150)

4.1.3 Probability Density Function of Transformed Random Variables

Let $\tilde{X}$ be a random variable with density ${f_X}(x)$ and let $\tilde{Y}$ be a random variable with $\tilde{Y} = g(\tilde{X})$. If g is strictly monotonous and differentiable, the probability density function (PDF) of $\tilde{Y}$ can be transformed using the inverse function theorem⁷⁶:

$$ {f_Y}(y) = {f_X}\left( {{g^{ - 1}}(y)} \right) \cdot \left| {\frac{{d{g^{ - 1}}(y)}}{{dy}}} \right| $$

(4.151)

With ${g^{ - 1}}(y) = x$, we obtain

$$ \left| {\frac{{d{g^{ - 1}}(y)}}{{dy}}} \right| = \left| {\frac{{dx}}{{dy}}} \right| = \left| {\frac{1}{{{{{dy}} \left/ {{dx}} \right.}}}} \right| $$

(4.152)

which leads to

$$ {f_Y}(y) = \frac{{{f_X}(x)}}{{\left| {{{{dy}} \left/ {{dx}} \right.}} \right|}}. $$

(4.153)

4.1.4 VaR-Based First-Order Granularity Adjustment for a Normally Distributed Systematic Factor

The granularity adjustment (4.10) can be expressed as

$$ \begin{array} {c} \Delta {l_1} = {\left. { - \frac{1}{{2\varphi }}\frac{d}{{dx}}\left( {\frac{{\varphi \,{\eta_{2,c}}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right|_{x = {\Phi^{ - 1}}\left( {1 - \alpha } \right)}} \\ = {\left. { - \frac{1}{{2\varphi }}\left[ {\frac{d}{{dx}}\left( {\varphi \,{\eta_{2,c}}} \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + \varphi \,{\eta_{2,c}}\frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right]} \right|_{x = {\Phi^{ - 1}}\left( {1 - \alpha } \right)}} \\ = {\left. { - \frac{1}{2}\left[ {\frac{1}{\varphi }\frac{d}{{dx}}\left( {\varphi \,{\eta_{2,c}}} \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + {\eta_{2,c}}\frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right]} \right|_{x = {\Phi^{ - 1}}\left( {1 - \alpha } \right)}} \\ = {\left. { - \frac{1}{2}\left[ {\left( {\frac{{{\eta_{2,c}}}}{\varphi }\frac{{d\varphi }}{{dx}} + \frac{{d{\eta_{2,c}}}}{{dx}}} \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - {\eta_{2,c}}\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right]} \right|_{x = {\Phi^{ - 1}}\left( {1 - \alpha } \right)}}. \\ \end{array} $$

(4.154)

Because of

$$ \frac{1}{\varphi }\frac{{d\varphi }}{{dx}} = \frac{{d(\ln \varphi )}}{{dx}} = \frac{d}{{dx}}\left( {\ln \left[ {\frac{1}{{\sqrt {{2\pi }} }}\exp \left( { - \frac{{{x^2}}}{2}} \right)} \right]} \right) = \frac{d}{{dx}}\left( {\ln \frac{1}{{\sqrt {{2\pi }} }} - \frac{{{x^2}}}{2}} \right) = - x, $$

(4.155)

the granularity adjustment (4.154) can be written as

$$ \Delta {l_1} = {\left. {\frac{1}{2}\left[ {\frac{{x \cdot {\eta_{2,c}}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - \frac{{{{{d{\eta_{2,c}}}} \left/ {{dx}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + \frac{{{{{{\eta_{2,c}} \cdot {d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right]} \right|_{x = {\Phi^{ - 1}}\left( {1 - \alpha } \right)}}. $$

(4.156)

For the calculation of (4.156), the conditional expectation and variance have to be determined. Assuming stochastically independent LGDs and with ELGD and VLGD for the expectation and the variance of the LGD, respectively, the required moments are given as⁷⁷

$$ \begin{array} {c} {\mu_{1,c}} = b{E}\left( {\sum\limits_{i = 1}^n {{w_i} \cdot \widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}} |\tilde{x} = x} \right) \\ = \sum\limits_{i = 1}^n {{w_i} \cdot ELG{D_i} \cdot b{E}\left( {{1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x} = x} \right)} \\ = \sum\limits_{i = 1}^n {{w_i} \cdot ELG{D_i}} \cdot {p_i}(x), \\ \end{array} $$

(4.157)

$$ \begin{array} {c} {\eta_{2,c}} = b{V}\left( {\sum\limits_{i = 1}^n {{w_i} \cdot \widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}} |\tilde{x} = x} \right) \\ = \sum\limits_{i = 1}^n {w_i^2 \cdot b{V}\left( {\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x} = x} \right)} \\ = \sum\limits_{i = 1}^n {w_i^2 \cdot \left[ {b{E}\left( {{{\left[ {\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x} = x} \right]}^2}} \right) - {b{E}^2}\left( {\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x} = x} \right)} \right]} \\ = \sum\limits_{i = 1}^n {w_i^2 \cdot \left[ {b{E}\left( {{{\widetilde{{LG{D_i}}}}^2}} \right) \cdot b{E}\left( {{{\left[ {{1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x} = x} \right]}^2}} \right) - {{\left( {ELG{D_i} \cdot {p_i}(x)} \right)}^2}} \right]} \\ = \sum\limits_{i = 1}^n {w_i^2 \cdot \left[ {\left( {ELGD_i^2 + VLG{D_i}} \right) \cdot {p_i}(x) - ELGD_i^2 \cdot p_i^2(x)} \right]} . \\ \end{array} $$

(4.158)

4.1.5 VaR-Based First-Order Granularity Adjustment for Homogeneous Portfolios

For homogeneous portfolios, the granularity adjustment formula (4.28) can be simplified to

$$ \begin{array} {c} \Delta {l_1} = \frac{1}{{2n}}\left[ {{\Phi^{ - 1}}\left( \alpha \right)\frac{{\left( {ELG{D^2} + VLGD} \right)\Phi (z) - ELG{D^2}\,{\Phi^2}(z)}}{{ELGD\left( {{{{\sqrt {\rho } }} \left/ {{\sqrt {{1 - \rho }} }} \right.}} \right)\varphi (z)}}} \right. \\ - \frac{{\left( {ELG{D^2} + VLGD} \right) - 2ELG{D^2}\,\Phi (z)}}{{ELGD}} \\ {\left. { - \frac{{\left( {ELG{D^2} + VLGD} \right)\Phi (z)z - ELG{D^2}\,{\Phi^2}(z)z}}{{ELGD \cdot \varphi (z)}}} \right]_{z = \frac{{{\Phi^{ - 1}}(PD) + \sqrt {\rho } {\Phi^{ - 1}}\left( \alpha \right)}}{{\sqrt {{1 - \rho }} }}}} \\ = \frac{1}{{2n}}\left( {\frac{{ELG{D^2} + VLGD}}{{ELGD}}\left[ {\frac{{\sqrt {{1 - \rho }} \,{\Phi^{ - 1}}\left( \alpha \right)\Phi (z)}}{{\sqrt {\rho } \,\varphi (z)}}} \right. - 1 - \left. {\frac{{\Phi (z)z}}{{\varphi (z)}}} \right]} \right. \\ {\left. { - ELGD\,\Phi (z)\left[ {\frac{{\sqrt {{1 - \rho }} \,{\Phi^{ - 1}}\left( \alpha \right)\Phi (z)}}{{\sqrt {\rho } \,\varphi (z)}}} \right. - 2 - \left. {\frac{{\Phi (z)z}}{{\varphi (z)}}} \right]} \right)_{z = \frac{{{\Phi^{ - 1}}(PD) + \sqrt {\rho } {\Phi^{ - 1}}\left( \alpha \right)}}{{\sqrt {{1 - \rho }} }}}} \\ = \frac{1}{{2n}}\left( {\frac{{ELG{D^2} + VLGD}}{{ELGD}}\left[ {\frac{{\Phi (z)}}{{\varphi (z)}}\frac{{{\Phi^{ - 1}}\left( \alpha \right)\left( {1 - 2\rho } \right) + {\Phi^{ - 1}}(PD)\sqrt {\rho } }}{{\sqrt {\rho } \sqrt {{1 - \rho }} }} - 1} \right]} \right. \\ {\left. { - ELGD \cdot \Phi (z)\left[ {\frac{{\Phi (z)}}{{\varphi (z)}}\frac{{{\Phi^{ - 1}}\left( \alpha \right)\left( {1 - 2\rho } \right) + {\Phi^{ - 1}}(PD)\sqrt {\rho } }}{{\sqrt {\rho } \sqrt {{1 - \rho }} }} - 2} \right]} \right)_{z = \frac{{{\Phi^{ - 1}}(PD) + \sqrt {\rho } {\Phi^{ - 1}}\left( \alpha \right)}}{{\sqrt {{1 - \rho }} }}}}. \\ \end{array} $$

(4.159)

4.1.6 Arbitrary Derivatives of VaR

The following determination of all derivatives of VaR is based on Wilde (2003). The quantile q _α of $\tilde{L} = \tilde{Y} + \lambda \tilde{Z}$ can be written as $q(\lambda )$ to denote that the quantile depends on the parameter λ. Using this notation, the quantile can be defined implicitly as an argument of the distribution function F by $F(q(\lambda ),\lambda ): = b{P}\left( {\tilde{Y} + \lambda \tilde{Z} \leq {q_\alpha }(\tilde{Y} + \lambda \tilde{Z})} \right) = \alpha$. In order to calculate the derivatives of q _α , at first all derivatives of F are determined in Sect. 4.5.6.2.1. As the quantile is defined implicitly, the implicit derivatives of $F(q(\lambda ),\lambda ) - \alpha = 0$ have to be determined. This is done by application of the residue theorem in Sect. 4.5.6.2.2. As a next step, the result will be expressed in combinatorial form in Sect. 4.5.6.2.3. Using the results of the derivatives of the distribution function and the implicit derivatives, it is possible to determine all derivatives of VaR. This is performed in Sect. 4.5.6.2.4. As the resulting formula is quite complex, an expression for the first five derivatives of VaR is determined in Sect. 4.5.7. The mathematical basics to the Laplace transform, complex residues, and partitions, which are needed within the derivation, are presented in the following Sect. 4.5.6.1.

4.1.6.1 Mathematical Basics

4.1.6.1.1 Laplace Transform and Dirac’s Delta Function

The Laplace transform $L$ of a function $f(t)$ with $t \in {b{R}^{+} }$ is given as⁷⁸

$$ \left[ {L\left\{ {f(t)} \right\}} \right](s): = \int\limits_{t = - 0}^\infty {f(t){e^{ - st}}dt} = :\Theta (s) $$

(4.160)

with $s = c + i\omega \,\, \in \,\,b{C}$, where $b{C}$ denotes the set of all complex numbers. The inverse Laplace transform ${L^{ - 1}}$ can be represented as⁷⁹

$$ \left[ {{L^{ - 1}}\left\{ {\Theta (s)} \right\}} \right](t): = \frac{1}{{2\pi i}}\int\limits_{s = c - i\infty }^{c + i\infty } {\Theta (s){{\text{e}}^{st}}ds} = {L^{ - 1}}\left\{ {L\left\{ {f(t)} \right\}} \right\} = f(t). $$

(4.161)

Dirac’s delta function $\delta (x)$ can be defined as⁸⁰

$$ \int\limits_{ - \infty }^\infty {\delta (x)f\left( {x - {x_0}} \right)dx} = f\left( {{x_0}} \right) $$

(4.162)

A more illustrative, heuristic definition of $\delta (x)$ is given by

$$ \delta (x) = \left\{ {\begin{array} {c}{*{20}{c}} 0 \\ {{\text{if }}x \ne 0,} \\ \infty \\ {{\text{if }}x = 0,} \\ \end{array} } \right.\quad {\text{and}}\quad \int\limits_{ - \infty }^\infty {\delta (x)dx} = 1. $$

(4.163)

Using the definition of the Laplace transform and the inverse Laplace transform, Dirac’s delta function can be written as

$$ \begin{array} {c} \delta (t) = {L^{ - 1}}\left\{ {L\left\{ {\delta (t)} \right\}} \right\} = {L^{ - 1}}\left\{ {\int\limits_{t = - 0}^\infty {\delta (t){{\text{e}}^{ - st}}dt} } \right\} \\ = {L^{ - 1}}\left\{ {{{\text{e}}^{ - s \cdot 0}}} \right\} = {L^{ - 1}}\left\{ 1 \right\} = \frac{1}{{2\pi i}}\int\limits_{s = c - i\infty }^{c + i\infty } {1 \cdot {{\text{e}}^{st}}} ds. \\ \end{array} $$

(4.164)

4.1.6.1.2 Laurent Series, Singularities, and Complex Residues

If $f(z)$ is differentiable in all points of an open subset of the complex plane $H \subset b{C}$, then we call $f(z)$ holomorphic on H.⁸¹ For a function $f(z)$, which is holomorphic in a simply connected region H, according to the Cauchy integral theorem we have⁸²

$$ \oint\limits_C {f(z)dz = 0} $$

(4.165)

with C being a closed path in H. If a function $f(z)$ is holomorphic in z ₀ and in a circular region around z ₀, we can perform a Taylor series expansion, which is analogous to the real plane:⁸³

$$ f(z) = \sum\limits_{n = 0}^\infty {\frac{{{f^{(n)}}({z_0})}}{{n!}}} {(z - {z_0})^n} $$

(4.166)

However, if a function $f(z)$ is only holomorphic inside the annulus between two concentric circles with center z ₀ and radii r ₁ and r ₂, which is the region $H = \left\{ {z|0 \leq {r_1} < \left| {z - {z_0}} \right| < {r_2}} \right\}$, the function $f(z)$ can be expressed as a generalized power series, the so-called Laurent series:⁸⁴

$$ f(z) = \sum\limits_{n = - \infty }^\infty {{a_n}} {(z - {z_0})^n} = \underbrace {\sum\limits_{n = - \infty }^{ - 1} {{a_n}} {{(z - {z_0})}^n}}_{\text{principal part}} + \underbrace {\sum\limits_{n = 0}^\infty {{a_n}} {{(z - {z_0})}^n}}_{\text{analytic part}} $$

(4.167)

Thus, the function has to be holomorphic only inside the annulus and not inside the inner circle or outside the outer circle.

If a function $f(z)$ is holomorphic in a neighborhood of z ₀ but not in the point z ₀, then z ₀ is called an isolated singularity of the function $f(z)$. The concrete type of a singularity can be classified according to the analytic part of the Laurent series:⁸⁵

• The point z ₀ is a removable singularity if ${a_n} = 0\,\,\forall n < 0$. In this case, the Laurent series is identical to the Taylor series above.

• The point z ₀ is a pole of order m if the principal part consists of a finite number of terms with ${a_m} \ne 0$ and ${a_n} = 0$ for $n < m < 0$.

• The point z ₀ is an essential singularity if the principal part consists of an infinite number of terms.

The coefficient ${a_{ - 1}}$ of the Laurent series (4.167) around an isolated singularity z ₀ is the residue of $f(z)$ in z ₀. This will subsequently be denoted by ${\text{Re}}{{\text{s}}_{{z_0}}}(f)$. The residue can also be defined as

$$ {a_{ - 1}} = {\text{Re}}{{\text{s}}_{{z_0}}}(f) = \frac{1}{{2\pi i}} \cdot \oint\limits_C {f(z)} \,dz $$

(4.168)

where C is a contour with winding number 1 in a holomorphic region H around an isolated singularity in z ₀. If the contour C encloses a finite number of isolated singularities ${z_1},{z_2},...,{z_m}$ with corresponding residues ${a_{ - 1}}({z_\mu })$ $(\mu = 1,...,m)$, we have

$$ \oint\limits_C {f(z)dz} = 2\pi i\sum\limits_{\mu = 1}^m {{a_{ - 1}}({z_\mu })} $$

(4.169)

which is the residue theorem.⁸⁶

The residue ${\text{Re}}{{\text{s}}_{{z_0}}}(f)$ with z ₀ being a pole of order m can be calculated as⁸⁷

$$ {\text{Re}}{{\text{s}}_{{z_0}}}(f) = \mathop {{\lim }}\limits_{z \to {z_0}} \frac{1}{{(m - 1)!}}\frac{{{d^{m - 1}}}}{{d{z^{m - 1}}}}\left[ {{{\left( {z - {z_0}} \right)}^m} \cdot f(z)} \right]. $$

(4.170)

For a function $f = {{{g(z)}} \left/ {{h(z)}} \right.}$, where h has a simple zero in z ₀, the residue can be determined with

$$ {\text{Re}}{{\text{s}}_{{z_0}}}(f) = \frac{{g({z_0})}}{{h'({z_0})}}. $$

(4.171)

4.1.6.1.3 Partitions

A partition p of a positive integer m is a way to express m as a sum of positive integers in non-decreasing order. A partition p of m will be denoted by $p \prec m$. A partition p can be indicated by $p = {1^{{e_1}}},{2^{{e_2}}},...,{m^{{e_m}}}$, where ${e_i}$ is the frequency of the number i in the partition. The number of summands of p is expresses by $\left| p \right|$, which is the sum $\left| p \right| = {e_1} + {e_2} + ... + {e_m}$. The notation $\hat{p}$ indicates the partition which results if each summand of a partition p is increased by 1. This means that for $p \prec m$ the partition $\hat{p}$ refers to a specific partition of $m + \left| p \right|$.⁸⁸

Example

• For $m = 5$, there exist seven partitions $p \prec m$: $p \prec m = \left\{ {1 + 1 + 1 + 1 + 1,\;1 + 1 + 1 + 2,\;1 + 2 + 2,\;1 + 1 + 3,\;2 + 3,\;1 + 4,\;5} \right\}.$ Thus, a concrete partition for $m = 5$ is $p = 3 + 1 + 1$.

• This partition can also be denoted by $p = {1^{{e_1}}}\,{2^{{e_2}}}\,...\,{m^{{e_m}}} = {1^2}{3^1}$, leading to ${e_1} = 2,$ ${e_2} = 0,$ ${e_3} = 1,$ ${e_4} = 0$, and ${e_5} = 0$. Thus, the number m results from: $m = 1 \cdot {e_1} + 2 \cdot {e_2} + ... + m \cdot {e_m} = 1 \cdot 2 + 3 \cdot 1 = 5$.

• The number of summands of this partition is $\left| {p = {1^2}{3^1}} \right| = {e_1} + {e_2} + ... + {e_m} = 2 + 1 = 3$.

• The partition $\hat{p}$ appendant to the partition $p = 3 + 1 + 1$ is $\hat{p} = 4 + 2 + 2$, which is a specific partition of $m + \left| p \right| = 5 + 3 = 8$.

4.1.6.2 Determination of the Derivatives

4.1.6.2.1 Derivatives of the Distribution Function

Proposition

The derivatives of the distribution function of losses ${F_{Y + \lambda Z}}(y) = b{P}(\tilde{Y} + \lambda \tilde{Z} \leq y)$ at $\lambda = 0$ are given as⁸⁹

$$ {\left. {\frac{{{\partial^m}}}{{\partial {\lambda^m}}}{F_{Y + \lambda Z}}(y)} \right|_{\lambda = 0}} = {( - 1)^m}\frac{{{d^{m - 1}}}}{{d{y^{m - 1}}}}\left( {b{E}\left( {{{\tilde{Z}}^m}|\tilde{Y} = y} \right){f_Y}(y)} \right). $$

(4.172)

Proof

Using the definition of the Laplace transform (4.160) and recognizing that the loss $\tilde{L} = \tilde{Y} + \lambda \tilde{Z}$ cannot go below zero so that the probability density function is ${f_{Y + \lambda Z}}(y) = 0$ for all $y < 0$, we get for the Laplace transform of ${f_{Y + \lambda Z}}(y)$

$$ L\left\{ {{f_{Y + \lambda Z}}(y)} \right\} = \int\limits_{y = - 0}^\infty {{{\text{e}}^{ - sy}}{f_{Y + \lambda Z}}(y)dy} = \int\limits_{y = - \infty }^\infty {{{\text{e}}^{ - sy}}{f_{Y + \lambda Z}}(y)dy} . $$

(4.173)

With the definition of the expectation operator

$$ b{E}\left( {g\left( {\tilde{X}} \right)} \right) = \int\limits_{x = - \infty }^\infty {g(x){f_X}(x)dx} . $$

(4.174)

(4.173) is equivalent to

$$ L\left\{ {{f_{Y + \lambda Z}}(y)} \right\} = \int\limits_{y = - \infty }^\infty {{{\text{e}}^{ - sy}}{f_{Y + \lambda Z}}(y)dy} = b{E}\left( {{{\text{e}}^{ - s\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}} \right) $$

(4.175)

Applying the definition of the inverse Laplace transform (4.161) and using the moment generating function M of $\tilde{Y} + \lambda \tilde{Z}$, which is defined as⁹⁰

$$ {M_{Y + \lambda Z}}(s) = b{E}\left( {{{\text{e}}^{s\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}} \right) $$

(4.176)

the probability density function equals⁹¹

$$ {f_{Y + \lambda Z}}(y) = {L^{ - 1}}\left\{ {L\left\{ {{f_{Y + \lambda Z}}(y)} \right\}} \right\} = {L^{ - 1}}\left\{ {{M_{Y + \lambda Z}}\left( { - s} \right)} \right\} = \frac{1}{{2\pi i}}\int\limits_{s = c - i\infty }^{c + i\infty } {{M_{Y + \lambda Z}}(s){{\text{e}}^{ - sy}}ds} . $$

(4.177)

Thus, the derivatives of the probability density function at $\lambda = 0$ can be determined using the approach

$$ {\left. {\frac{{{\partial^m}}}{{\partial {\lambda^m}}}{f_{Y + \lambda Z}}(y)} \right|_{\lambda = 0}} = \frac{1}{{2\pi i}}{\left. {\int\limits_{s = c - i\infty }^{c + i\infty } {\frac{{{\partial^m}}}{{\partial {\lambda^m}}}{M_{Y + \lambda Z}}(s){{\text{e}}^{ - sy}}} ds} \right|_{\lambda = 0}}. $$

(4.178)

Applying definition (4.176), we obtain for the derivatives of M

$$ \begin{array} {c} {\left. {\frac{{{\partial^m}{M_{Y + \lambda Z}}(s)}}{{\partial {\lambda^m}}}} \right|_{\lambda = 0}} = {\left. {\frac{{{\partial^m}}}{{\partial {\lambda^m}}}b{E}\left( {{{\text{e}}^{s\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}} \right)} \right|_{\lambda = 0}} \\ = {\left. {b{E}\left( {\frac{{{\partial^m}}}{{\partial {\lambda^m}}}{{\text{e}}^{s\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}} \right)} \right|_{\lambda = 0}} \\ = {\left. {b{E}\left( {{s^m}{{\tilde{Z}}^m}{{\text{e}}^{s\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}} \right)} \right|_{\lambda = 0}} \\ = b{E}\left( {{s^m}{{\tilde{Z}}^m}{{\text{e}}^{s\tilde{Y}}}} \right). \\ \end{array} $$

(4.179)

With (4.179) and ${s^m}{{\text{e}}^{s\left( {\tilde{Y} - y} \right)}} = {( - 1)^m}\frac{{{\partial^m}}}{{\partial {y^m}}}{{\text{e}}^{s\left( {\tilde{Y} - y} \right)}}$, (4.178) is equivalent to

$$ \begin{array} {c} {\left. {\frac{{{\partial^m}}}{{\partial {\lambda^m}}}{f_{Y + \lambda Z}}(y)} \right|_{\lambda = 0}} = \frac{1}{{2\pi i}}\int\limits_{s = c - i\infty }^{c + i\infty } {b{E}\left( {{s^m}{{\tilde{Z}}^m}{{\text{e}}^{s\tilde{Y}}}} \right){e^{ - sy}}} ds \\ = b{E}\left( {\frac{1}{{2\pi i}}{{\tilde{Z}}^m}\int\limits_{s = c - i\infty }^{c + i\infty } {{s^m}{{\text{e}}^{s\left( {\tilde{Y} - y} \right)}}} ds} \right) \\ = {( - 1)^m}\frac{{{d^m}}}{{d{y^m}}}b{E}\left( {{{\tilde{Z}}^m}\frac{1}{{2\pi i}}\int\limits_{s = c - i\infty }^{c + i\infty } {{{\text{e}}^{s\left( {\tilde{Y} - y} \right)}}} ds} \right). \\ \end{array} $$

(4.180)

According to (4.164), Dirac’s delta function can be written as

$$ \delta (t) = \frac{1}{{2\pi i}}\int\limits_{s = c - i\infty }^{c + i\infty } {1 \cdot {{\text{e}}^{st}}} ds $$

(4.181)

which leads to

$$ \delta \left( {\tilde{Y} - y} \right) = \frac{1}{{2\pi i}}\int\limits_{s = c - i\infty }^{c + i\infty } {1 \cdot {{\text{e}}^{s\left( {\tilde{Y} - y} \right)}}} ds $$

(4.182)

for $t = \tilde{Y} - y$. Hence, (4.180) is equivalent to

$$ {\left. {\frac{{{\partial^m}}}{{\partial {\lambda^m}}}{f_{Y + \lambda Z}}(y)} \right|_{\lambda = 0}} = {( - 1)^m}\frac{{{d^m}}}{{d{y^m}}}b{E}\left( {{{\tilde{Z}}^m}\delta \left( {\tilde{Y} - y} \right)} \right). $$

(4.183)

With $b{E}[{\tilde{Z}^m}\delta (\tilde{Y} - y)] = b{E}[{\tilde{Z}^m}|\tilde{Y} = y] \cdot {f_Y}(y)$, the derivatives of the distribution function result after integration of (4.183):

$$ {\left. {\frac{{{\partial^m}}}{{\partial {\lambda^m}}}{F_{Y + \lambda Z}}(y)} \right|_{\lambda = 0}} = {( - 1)^m}\frac{{{d^{m - 1}}}}{{d{y^{m - 1}}}}\left( {b{E}\left( {{{\tilde{Z}}^m}|\tilde{Y} = y} \right){f_Y}(y)} \right), $$

(4.184)

which is proposition (4.172). In order to determine the derivatives of the quantile ${{{{d^m}q}} \left/ {{d{\lambda^m}}} \right.}$, the implicit derivatives of $F(q(\lambda ),\lambda ) - \alpha = 0$ with $F(q(\lambda ),\lambda ): = {F_{\tilde{Y} + \lambda \tilde{Z}}}({q_\alpha }(\tilde{Y} + \lambda \tilde{Z}))$ $= b{P}\left( {\tilde{Y} + \lambda \tilde{Z} \leq {q_\alpha }(\tilde{Y} + \lambda \tilde{Z})} \right)$ will be calculated in the following.

4.1.6.2.2 Implicit Derivatives: Complex Residue Form

Consider a function $G(z,w)$ of two variables $z,w \in b{C}$. Suppose there exists an analytic function $w = w(z)$ in a region around a pole $z = {z_0}$, such that $G(z,w(z)) = 0$. The first derivative ${{{dw}} \left/ {{dz}} \right.}$ can be determined as follows:⁹²

$$ \begin{array} {c} \\ 0 = \frac{{\partial G}}{{\partial z}} + \frac{{\partial G}}{{\partial w}} \cdot \frac{{dw}}{{dz}} \\ \Leftrightarrow \frac{{dw}}{{dz}} = - \frac{{\partial G/\partial z}}{{\partial G/\partial w}} = : - \frac{{{G_z}}}{{{G_w}}}. \\ \end{array} $$

(4.185)

Proposition

For ${G_w}({z_0},{w_0}) \ne 0$, the derivatives ${{{{d^m}w}} \left/ {{d{z^m}}} \right.}$ are given as

$$ \frac{{{d^m}w}}{{d{z^m}}} = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {{{\left. {\frac{{{\partial^{m - 1}}}}{{\partial {z^{m - 1}}}}\left( {\frac{{{G_z}(z,w)}}{{G(z,w)}}} \right)} \right|}_{z = {z_0}}}} \right]. $$

(4.186)

Proof

According to (4.186), the first derivative is

$$ \frac{{dw}}{{dz}} = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {{{\left. {\left( {\frac{{{G_z}(z,w)}}{{G(z,w)}}} \right)} \right|}_{z = {z_0}}}} \right] = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {\frac{{{G_z}({z_0},w)}}{{G({z_0},w)}}} \right]. $$

(4.187)

As z ₀ is a pole of G and $G({z_0},w) = 0$, an application of (4.171) leads to

$$ \frac{{dw}}{{dz}} = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {\frac{{{G_z}({z_0},w)}}{{G({z_0},w)}}} \right] = - \frac{{{G_z}}}{{{G_w}}}. $$

(4.188)

which is equal to (4.185). This shows that the formula is correct for $m = 1$.

Applying the residue theorem (4.169)

$$ \sum\limits_{\mu = 1}^m {{a_{ - 1}}({z_\mu })} = \frac{1}{{2\pi i}}\oint\limits_C {f(z)dz} $$

(4.189)

and recognizing that there is only a singularity at $z = {z_0}$ leads to

$$ \frac{{dw}}{{dz}} = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {{{\left. {\frac{{{G_z}(z,w)}}{{G(z,w)}}} \right|}_{z = {z_0}}}} \right] = - \frac{1}{{2\pi i}}\oint\limits_C {{{\left. {\frac{{{G_z}(z,w)}}{{G(z,w)}}} \right|}_{z = {z_0}}}} dw. $$

(4.190)

Differentiating and applying the residue theorem again results in

$$ \begin{array} {c} \frac{{{d^m}w}}{{d{z^m}}} = \frac{{{\partial^{m - 1}}}}{{\partial {z^{m - 1}}}}\left( { - \frac{1}{{2\pi i}}\oint\limits_C {{{\left. {\frac{{{G_z}(z,w)}}{{G(z,w)}}} \right|}_{z = {z_0}}}} dw} \right) \\ = - \frac{1}{{2\pi i}}\oint\limits_C {\frac{{{\partial^{m - 1}}}}{{\partial {z^m}}}{{\left. {\frac{{{G_z}(z,w)}}{{G(z,w)}}} \right|}_{z = {z_0}}}} dw \\ = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {\frac{{{\partial^{m - 1}}}}{{\partial {z^m}}}{{\left. {\frac{{{G_z}(z,w)}}{{G(z,w)}}} \right|}_{z = {z_0}}}} \right], \\ \end{array} $$

(4.191)

which is the proposition presented in (4.186). This result is a generalization of the Lagrange inversion theorem.⁹³

4.1.6.2.3 Implicit Derivatives: Combinatorial Form

In order to express the implicit derivatives (4.191) in combinatorial form, Faà di Bruno’s formula will be used. According to this formula, the following equation holds for a function $g = g(y)$ with $y = y(x)$:⁹⁴

$$ \frac{{{d^m}g}}{{d{x^m}}} = \sum\limits_{p \prec m} {{\alpha_p}\frac{{{d^{\left| p \right|}}g}}{{d{y^{\left| p \right|}}}}} \frac{{{d^p}y}}{{d{x^p}}} $$

(4.192)

with ${\alpha_p} = \frac{{m!}}{{{{(1!)}^{{e_1}}} \cdot {e_1}!\, \cdot ... \cdot {{(m!)}^{{e_m}}} \cdot {e_m}!}}$, $\frac{{{d^{\left| p \right|}}g}}{{d{y^{\left| p \right|}}}}$ as ordinary ${\left| p \right|^{\text{th}}}$ derivative, and

$$ \frac{{{d^p}y}}{{d{x^p}}}: = {\left( {\frac{{dy}}{{dx}}} \right)^{{e_{p1}}}} \cdot {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^{{e_{p2}}}} \cdot ... \cdot {\left( {\frac{{{d^m}y}}{{d{x^m}}}} \right)^{{e_{pm}}}} = {\prod\limits_{i = 1}^m {\left( {\frac{{{d^i}y}}{{d{x^i}}}} \right)}^{{e_{pi}}}}. $$

(4.193)

Proposition

Equation (4.191) is equivalent to

$$ \frac{{{d^m}w}}{{d{z^m}}} = \sum\limits_{ p \prec m, \\ u \prec s \leq |p| - 1 } {{\alpha_p}{\alpha_{\hat{u}}}\frac{{{{( - 1)}^{\left| p \right| + \left| u \right|}}\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!\left( {\left| p \right| - 1 - s} \right)!}}{{\left. {{G_w}^{ - \left| p \right| - \left| u \right|}\frac{{{\partial^{\hat{u}}}G}}{{\partial {w^{\hat{u}}}}}\frac{{{\partial^{\left| p \right| - 1 - s}}}}{{\partial {w^{\left| p \right| - 1 - s}}}}\frac{{{\partial^p}G}}{{\partial {z^p}}}} \right|}_{z,w = 0}}} . $$

(4.194)

Proof

For ease of notation, it will be assumed that ${z_0} = {w_0} = 0$, so that $G(0,0) = 0$. With ${{{\partial \ln G}} \left/ {{\partial z}} \right.} = {{{{G_z}}} \left/ {G} \right.}$, (4.191) is equivalent to

$$ \frac{{{d^m}w}}{{d{z^m}}} = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {{{\left. {\frac{{{\partial^{m - 1}}}}{{\partial {z^{m - 1}}}}\left( {\frac{{{G_z}}}{G}} \right)} \right|}_{z = 0}}} \right] = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {{{\left. {\frac{{{\partial^m}}}{{\partial {z^m}}}\ln G} \right|}_{z = 0}}} \right]. $$

(4.195)

The mth derivative of $\ln G$ can be calculated using Faà di Bruno’s formula:

$$ \begin{array} {c} \frac{{{\partial^m}}}{{\partial {z^m}}}\ln G = \sum\limits_{p \prec m} {{\alpha_p}\frac{{{d^{\left| p \right|}}\ln G}}{{d{G^{\left| p \right|}}}}} \frac{{{\partial^p}G}}{{\partial {z^p}}} = \sum\limits_{p \prec m} {{\alpha_p}\frac{{{d^{\left| p \right| - 1}}}}{{d{G^{\left| p \right| - 1}}}}\left( {\frac{1}{G}} \right)} \frac{{{\partial^p}G}}{{\partial {z^p}}} \\ = \sum\limits_{p \prec m} {{\alpha_p} \cdot {{( - 1)}^{\left| p \right| - 1}} \cdot \left( {\left| p \right| - 1} \right)! \cdot {G^{ - \left| p \right|}} \cdot {G_{z,p}}}, \\ \end{array} $$

(4.196)

with ${{{{\partial^p}G}} \left/ {{\partial {z^p}}} \right.} = :{G_{z,p}}$. This leads to

$$ \frac{{{d^m}w}}{{d{z^m}}} = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {{{\left. {\frac{{{\partial^m}}}{{\partial {z^m}}}\ln G} \right|}_{z = 0}}} \right] = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {{{\left. {\sum\limits_{p \prec m} {{\alpha_p} \cdot {{( - 1)}^{\left| p \right| - 1}} \cdot \left( {\left| p \right| - 1} \right)! \cdot {G^{ - \left| p \right|}} \cdot {G_{z,p}}} } \right|}_{z = 0}}} \right]. $$

(4.197)

According to (4.170), the residue of a function h(w) in w ₀, with w ₀ being a pole of order r, can be calculated as

$$ {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {h(w)} \right] = \mathop {{\lim }}\limits_{w \to {w_0}} \frac{1}{{\left( {r - 1} \right)!}}\frac{{{d^{r - 1}}}}{{d{w^{r - 1}}}}\left( {{{\left( {w - {w_0}} \right)}^r} \cdot h(w)} \right). $$

(4.198)

With $r = \left| p \right|$, we obtain for the derivative (4.197)

$$ \begin{array} {c} \frac{{{d^m}w}}{{d{z^m}}} = - {\text{Re}}{{\text{s}}_{{w_0}}}\left[ {{{\left. {\sum\limits_{p \prec m} {{\alpha_p} \cdot {{( - 1)}^{\left| p \right| - 1}} \cdot \left( {\left| p \right| - 1} \right)! \cdot {G^{ - \left| p \right|}} \cdot {G_{z,p}}} } \right|}_{z = 0}}} \right] \\ = - \frac{1}{{\left( {\left| p \right| - 1} \right)!}}{\left. {\frac{{{\partial^{\left| p \right| - 1}}}}{{\partial {w^{\left| p \right| - 1}}}}\left[ {{{\left. {{w^{\left| p \right|}} \cdot \sum\limits_{p \prec m} {{\alpha_p} \cdot {{( - 1)}^{\left| p \right| - 1}} \cdot \left( {\left| p \right| - 1} \right)! \cdot {G^{ - \left| p \right|}} \cdot {G_{z,p}}} } \right|}_{z = 0}}} \right]} \right|_{w = 0}} \\ = - {\left. {\sum\limits_{p \prec m} {{\alpha_p} \cdot {{( - 1)}^{\left| p \right| - 1}} \cdot \frac{{{\partial^{\left| p \right| - 1}}}}{{\partial {w^{\left| p \right| - 1}}}}\left( {{{\left. {{{\left( {\frac{G}{w}} \right)}^{ - \left| p \right|}} \cdot {G_{z,p}}} \right|}_{z = 0}}} \right)} } \right|_{w = 0}}. \\ \end{array} $$

(4.199)

Using the Leibniz identity for arbitrary-order derivatives of products of functions, we get:⁹⁵

$$ \begin{array} {c} \frac{{{d^m}w}}{{d{z^m}}} = {\left. { - \sum\limits_{p \prec m} {{\alpha_p} \cdot {{( - 1)}^{\left| p \right| - 1}} \cdot \frac{{{\partial^{\left| p \right| - 1}}}}{{\partial {w^{\left| p \right| - 1}}}}\left( {{{\left. {{{\left( {\frac{G}{w}} \right)}^{ - \left| p \right|}} \cdot {G_{z,p}}} \right|}_{z = 0}}} \right)} } \right|_{w = 0}} \\ = {\left. { - \sum\limits_{p \prec m} {{\alpha_p} \cdot {{( - 1)}^{\left| p \right| - 1}} \cdot \sum\limits_{s = 0}^{\left| p \right| - 1} {\left( {\begin{array} {c}{*{20}{c}} {\left| p \right| - 1} \\ s \\ \end{array} } \right)\frac{{{\partial^s}}}{{\partial {w^s}}}{{\left( {\frac{{G(0,w)}}{w}} \right)}^{ - \left| p \right|}} \cdot \frac{{{\partial^{\left| p \right| - 1 - s}}}}{{\partial {w^{\left| p \right| - 1 - s}}}}\left( {{G_{z,p}}(0,w)} \right)} } } \right|_{w = 0}}. \\ \end{array} $$

(4.200)

As a next step, the derivative $\frac{{{\partial^s}}}{{\partial {w^s}}}{\left( {\frac{{G(0,w)}}{w}} \right)^{ - \left| p \right|}}$ contained in (4.200) will be calculated. Performing a Taylor series expansion of $G(0,w)$ at $w = 0$, we have

$$ \begin{array} {c} G(0,w) = G(0,0) + \frac{w}{{1!}} \cdot \frac{\partial }{{\partial w}}G(0,0) + \frac{{{w^2}}}{{2!}} \cdot \frac{{{\partial^2}}}{{\partial {w^2}}}G(0,0) + \frac{{{w^3}}}{{3!}} \cdot \frac{{{\partial^3}}}{{\partial {w^3}}}G(0,0) + ... \\ = 0 + w \cdot {G_w}(0,0) + \sum\limits_{r \geq 2} {\frac{{{w^r}}}{{r!}} \cdot \frac{{{\partial^r}}}{{\partial {w^r}}}G(0,0)} \\ = w \cdot {G_w}(0,0) + \sum\limits_{r \geq 1} {\frac{{{w^{r + 1}}}}{{(r + 1)!}} \cdot \frac{{{\partial^{r + 1}}}}{{\partial {w^{r + 1}}}}G(0,0)} \\ = w \cdot {G_w}(0,0) + w \cdot {G_w}(0,0) \cdot \sum\limits_{r \geq 1} {\frac{{{w^r}}}{{(r + 1)!}} \cdot \frac{{{\partial^{r + 1}}}}{{\partial {w^{r + 1}}}}G(0,0) \cdot \frac{1}{{{G_w}(0,0)}}} \\ = w \cdot {G_w}(0,0) \cdot \left( {1 + \sum\limits_{r \geq 1} {\frac{{{w^r}}}{{r!}} \cdot \frac{1}{{r + 1}} \cdot \frac{{\frac{{{\partial^{r + 1}}}}{{\partial {w^{r + 1}}}}G(0,0)}}{{\frac{\partial }{{\partial w}}G(0,0)}}} } \right). \\ \end{array} $$

(4.201)

Thus, for ${{{G(0,w)}} \left/ {w} \right.}$, we obtain

$$ \begin{array} {c} \frac{{G(0,w)}}{w} = {G_w}(0,0) \cdot \left( {1 + \sum\limits_{r \geq 1} {\frac{{{w^r}}}{{r!}} \cdot \frac{1}{{r + 1}} \cdot \frac{{\frac{{{\partial^{r + 1}}}}{{\partial {w^{r + 1}}}}G(0,0)}}{{\frac{\partial }{{\partial w}}G(0,0)}}} } \right) \\ = {G_w}(0,0) \cdot \left( {1 + \sum\limits_{r \geq 1} {\frac{{{w^r}}}{{r!}} \cdot {\varphi_r}} } \right), \\ \end{array} $$

(4.202)

with ${\varphi_r} = \frac{1}{{r + 1}} \cdot \frac{{{{{{\partial^{r + 1}}}} \left/ {{\partial {w^{r + 1}}G(0,0)}} \right.}}}{{{{\partial } \left/ {{\partial w}} \right.}G(0,0)}}.$ Another application of Faà di Bruno’s formula results in:⁹⁶

$$ \begin{array} {c} \frac{{{\partial^s}}}{{\partial {w^s}}}{\left( {\frac{{G(0,w)}}{w}} \right)^{ - \left| p \right|}} = G_w^{ - \left| p \right|}(0,0) \cdot \frac{{{\partial^s}}}{{\partial {w^s}}}{\left( {1 + \sum\limits_{r \geq 1} {{\varphi_r} \cdot \frac{{{w^r}}}{{r!}}} } \right)^{ - \left| p \right|}} \\ = G_w^{ - \left| p \right|}(0,0) \cdot \sum\limits_{u \prec s} {{\alpha_u} \cdot {\varphi_u} \cdot {{( - 1)}^{\left| u \right|}} \cdot \frac{{\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {\left| p \right| - 1} \right)!}}}, \\ \end{array} $$

(4.203)

with⁹⁷

$$ {\alpha_u} \cdot {\varphi_u} = \frac{{s!}}{{\left( {s + \left| u \right|} \right)!}} \cdot {\alpha_{\hat{u}}} \cdot \frac{{{\partial^{\hat{u}}}}}{{\partial {w^{\hat{u}}}}}G\left( {0,0} \right) \cdot {G_w}^{ - \left| u \right|}\left( {0,0} \right). $$

(4.204)

Applying (4.203) and (4.204) to (4.200) leads to

$$ \begin{array} {c} \frac{{{d^m}w}}{{d{z^m}}} = {\left. { - \sum\limits_{p \prec m} {{\alpha_p} \cdot {{( - 1)}^{\left| p \right| - 1}} \cdot \sum\limits_{s = 0}^{\left| p \right| - 1} {\left( {\begin{array} {c}{*{20}{c}} {\left| p \right| - 1} \\ s \\ \end{array} } \right)\frac{{{\partial^s}}}{{\partial {w^s}}}{{\left( {\frac{{G(0,w)}}{w}} \right)}^{ - \left| p \right|}} \cdot \frac{{{\partial^{\left| p \right| - 1 - s}}}}{{\partial {w^{\left| p \right| - 1 - s}}}}\left( {{G_{z,p}}(0,w)} \right)} } } \right|_{w = 0}} \\ = - \sum\limits_{p \prec m} {{\alpha_p} \cdot {{( - 1)}^{\left| p \right| - 1}} \cdot \sum\limits_{s = 0}^{\left| p \right| - 1} {\left( {\begin{array} {c}{*{20}{c}} {\left| p \right| - 1} \\ s \\ \end{array} } \right) \cdot G_w^{ - \left| p \right|}(0,0) \cdot \sum\limits_{u \prec s} {\frac{{s!}}{{\left( {s + \left| u \right|} \right)!}} \cdot {\alpha_{\hat{u}}} \cdot \frac{{{\partial^{\hat{u}}}}}{{\partial {w^{\hat{u}}}}}G\left( {0,0} \right)} } } \\ \cdot {G_w}^{ - \left| u \right|}\left( {0,0} \right) \cdot {( - 1)^{\left| u \right|}} \cdot \frac{{\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {\left| p \right| - 1} \right)!}} \cdot {\left. {\frac{{{\partial^{\left| p \right| - 1 - s}}}}{{\partial {w^{\left| p \right| - 1 - s}}}}\left( {{G_{z,p}}(0,w)} \right)} \right|_{w = 0}} \\ = - \sum\limits_{p \prec m} {{\alpha_p} \cdot {{( - 1)}^{\left| p \right| - 1}} \cdot \sum\limits_{s = 0}^{\left| p \right| - 1} {\left( {\begin{array} {c}{*{20}{c}} {\left| p \right| - 1} \\ s \\ \end{array} } \right) \cdot \sum\limits_{u \prec s} {{\alpha_{\hat{u}}} \cdot {{( - 1)}^{\left| u \right|}} \cdot G_w^{ - \left| p \right| - \left| u \right|}(0,0)} } } \\ \cdot \frac{{s!\, \cdot \left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!\, \cdot \left( {\left| p \right| - 1} \right)!}} \cdot \frac{{{\partial^{\hat{u}}}}}{{\partial {w^{\hat{u}}}}}G\left( {0,0} \right) \cdot {\left. {\frac{{{\partial^{\left| p \right| - 1 - s}}}}{{\partial {w^{\left| p \right| - 1 - s}}}}\left( {{G_{z,p}}(0,w)} \right)} \right|_{w = 0}}. \\ \end{array} $$

(4.205)

Summarizing the sums, using $( - 1) \cdot {( - 1)^{\left| p \right| - 1}} \cdot {( - 1)^{\left| u \right|}} = {( - 1)^{\left| p \right| + \left| u \right|}}$, and

$$ \begin{array} {c} \left( {\begin{array} {c}{*{20}{c}} {\left| p \right| - 1} \\ s \\ \end{array} } \right) \cdot \frac{{s!}}{{\left( {\left| p \right| - 1} \right)!}} \cdot \frac{{\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!}} = \frac{{\left( {\left| p \right| - 1} \right)!}}{{s! \cdot \left( {\left| p \right| - 1 - s} \right)!}} \cdot \frac{{s!}}{{\left( {\left| p \right| - 1} \right)!}} \cdot \frac{{\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!}} \\ = \frac{{\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {\left| p \right| - 1 - s} \right)!\, \cdot \left( {s + \left| u \right|} \right)!}}, \\ \end{array} $$

(4.206)

(4.205) can be simplified to

$$\begin{array}{lll} \frac{{{d^m}w}}{{d{z^m}}} = & \sum\limits_{ p \prec m, u \prec s \leq |p| - 1 } {{\alpha_p} \cdot {\alpha_{\hat{u}}} \cdot {{( - 1)}^{\left| p \right| + \left| u \right|}} \cdot \frac{{\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {\left| p \right| - 1 - s)! \cdot (s + \left| u \right|} \right)!}}}\\ & \cdot G_w^{ - \left| p \right| - \left| u \right|}(0,0) \cdot \frac{{{\partial^{\hat{u}}}}}{{\partial {w^{\hat{u}}}}}G\left( {0,0} \right)\\ & \cdot {\left. {\frac{{{\partial^{\left| p \right| - 1 - s}}}}{{\partial {w^{\left| p \right| - 1 - s}}}}\left( {{G_{z,p}}(0,w)} \right)} \right|_{w = 0}},\end{array}$$

(4.207)

which concludes the proof.

4.1.6.2.4 Completion of the Derivation

Application of (4.207) can be used to determine the derivatives of a quantile, which will be calculated subsequently. With $F(q(\lambda ),\lambda ) - \alpha = 0 = G(w(z),z)$, the derivatives are given as

$$ {\left. {\frac{{{d^m}q}}{{d{\lambda^m}}}} \right|_{\lambda = 0}} = {\left. {\frac{{{d^m}w}}{{d{z^m}}}} \right|_{z = 0}}, $$

(4.208)

where the right-hand side can be determined with (4.207). The derivatives of G contained in (4.207) can be calculated with (4.172):

$$ \begin{array} {c} {\left. {\frac{{{\partial^{r + s}}G}}{{\partial {w^r}\partial {z^s}}}} \right|_{z = 0}} = {\left. {\frac{{{\partial^{r + s}}F}}{{\partial {y^r}\partial {\lambda^s}}}} \right|_{\lambda = 0}} = \frac{{{\partial^r}}}{{\partial {y^r}}}{\left. {\left( {\frac{{{\partial^s}F}}{{\partial {\lambda^s}}}} \right)} \right|_{\lambda = 0}} \\ = \frac{{{d^r}}}{{d{y^r}}}\left( {{{( - 1)}^s}\frac{{{d^{s - 1}}}}{{d{y^{s - 1}}}}\left( {b{E}\left( {{{\tilde{Z}}^s}|\tilde{Y} = y} \right){f_Y}(y)} \right)} \right) \\ = {( - 1)^s}\frac{{{d^{r + s - 1}}}}{{d{y^{r + s - 1}}}}\left( {b{E}\left( {{{\tilde{Z}}^s}|\tilde{Y} = y} \right){f_Y}(y)} \right) \\ = {( - 1)^s}\frac{{{d^{r + s - 1}}}}{{d{y^{r + s - 1}}}}\left( {{\mu_{s,c}}f} \right), \\ \end{array} $$

(4.209)

where we define ${\mu_{s,c}}: = b{E}({\tilde{Z}^s}|\tilde{Y} = y)$ and $f: = {f_Y}(y)$ for convenience. Using definition (4.193) for the pth derivative with $p \prec m$, this leads to

$$ {\left. {\frac{{{\partial^p}G}}{{\partial {z^p}}}} \right|_{z = 0}} = {\left. {{{\prod\limits_{i = 1}^m {\left( {\frac{{{\partial^i}G}}{{\partial {z^i}}}} \right)} }^{{e_{pi}}}}} \right|_{z = 0}} = {\prod\limits_{i = 1}^m {\left( {{{( - 1)}^i}\frac{{{d^{i - 1}}({\mu_{i,c}}f)}}{{d{y^{i - 1}}}}} \right)}^{{e_{pi}}}} = {( - 1)^m}{\prod\limits_{i = 1}^m {\left( {\frac{{{d^{i - 1}}({\mu_{i,c}}f)}}{{d{y^{i - 1}}}}} \right)}^{{e_{pi}}}}. $$

(4.210)

Similarly the ${\hat{u}^{\text{th}}}$ derivative can be determined with $u \prec s$. It has to be considered that for each partition u the elements of the corresponding partition $\hat{u}$ are increased by 1. Thus, the smallest number is 2 and the largest is $s + 1$. Hence, we obtain

$$ \frac{{{\partial^{\hat{u}}}G}}{{\partial {w^{\hat{u}}}}} = {\prod\limits_{i = 2}^{s + 1} {\left( {\frac{{{\partial^i}G}}{{\partial {w^i}}}} \right)}^{{e_{\hat{u}i}}}} = {\prod\limits_{i = 2}^{s + 1} {\left( {\frac{{{\partial^i}G}}{{\partial {w^i}}}} \right)}^{{e_{u(i - 1)}}}} = {\prod\limits_{i = 1}^s {\left( {\frac{{{\partial^{i + 1}}G}}{{\partial {w^{i + 1}}}}} \right)}^{{e_{ui}}}} = {\prod\limits_{i = 1}^s {\left( {\frac{{{\partial^{i + 1}}F}}{{\partial {y^{i + 1}}}}} \right)}^{{e_{ui}}}} = {\prod\limits_{i = 1}^s {\left( {\frac{{{d^i}f}}{{d{y^i}}}} \right)}^{{e_{ui}}}}. $$

(4.211)

Furthermore, we have ${G_w} = {{{dF}} \left/ {{dy}} \right.} = f$ and ${( - 1)^{\left| p \right| + \left| u \right|}} \cdot {f^{\left| p \right| + \left| u \right|}} = {( - f)^{\left| p \right| + \left| u \right|}}$. Using these formulas, we finally get for (4.207) or (4.208):

$$ \begin{array} {lll} {\left. {\frac{{{d^m}q}}{{d{\lambda^m}}}} \right|_{\lambda = 0}} &= {( - 1)^m}\left[ {\sum\limits_{ p \prec m, u \prec s \leq |p| - 1 } {\frac{{{\alpha_p}{\alpha_{\hat{u}}}\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!\left( {\left| p \right| - 1 - s} \right)!}}} \cdot {{\left( { - f} \right)}^{ - \left| p \right| - \left| u \right|}}} \right. \cdot \left( {\prod\limits_{i = 1}^s {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^{{e_{ui}}}}} } \right) \\ &{\left. { \cdot \frac{{{d^{\left| p \right| - 1 - s}}}}{{d{y^{\left| p \right| - 1 - s}}}}\left( {\prod\limits_{i = 1}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right)} \right]_{y = {q_\alpha }\left( {\tilde{Y}} \right)}}, \\ \end{array} $$

(4.212)

which is the formula for arbitrary derivatives of VaR. Written without abbreviations this is

$$ \begin{array} {c} {\left. {\frac{{{d^m}Va{R_\alpha }\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}{{d{\lambda^m}}}} \right|_{\lambda = 0}} = {( - 1)^m}\left[ {\sum \limits_{-} { p \prec m, u \prec s \leq |p| - 1 } {\frac{{{\alpha_p}{\alpha_{\hat{u}}}\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!\left( {\left| p \right| - 1 - s} \right)!}}} \cdot {{\left( { - {f_Y}(y)} \right)}^{ - \left| p \right| - \left| u \right|}}} \right. \\ \left[ \cdot \left( {\prod \limits_{i = 1}^s {{{\left[ {\frac{{{d^i}{f_Y}(y)}}{{d{y^i}}}} \right]}^{{e_{ui}}}}} } \right) \cdot \frac{{{d^{\left| p \right| - 1 - s}}}}{{d{y^{\left| p \right| - 1 - s}}}}\left( {\prod\limits_{i = 1}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {b{E}\left( {{{\tilde{Z}}^m}|\tilde{Y} = y} \right){f_Y}(y)} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \right] _{y = {q_\alpha }\left( {\tilde{Y}} \right)}, \\ \end{array} $$

(4.213)

with ${\alpha_p} = \frac{{m!}}{{{{(1!)}^{{e_p}_1}}{e_p}_{,1}!\, \cdot ... \cdot {{(m!)}^{{e_{p,m}}}}{e_{p,m}}!}}$.

4.1.7 Determination of the First Five Derivatives of VaR

The general form of the mth derivative of VaR is given by (4.213). Subsequently, the first five derivatives will be determined with this formula. For each derivative, we have summands for all partitions $p \prec m$ and $u \prec s \leq \left| p \right| - 1$. For the considered cases $1 \leq m \leq 5$, the following partitions $p \prec m$ exist:

$$ \begin{array} {c} p \prec 1 = \left\{ {{1^1}} \right\}; \\ p \prec 2 = \left\{ {{1^2},\;{2^1}} \right\}; \\ p \prec 3 = \left\{ {{1^3},\;{1^1}{2^1},\;{3^1}} \right\}; \\ p \prec 4 = \left\{ {{1^4},\;{1^2}{2^1},\;{2^2},\;{1^1}{3^1},\;{4^1}} \right\}; \\ p \prec 5 = \left\{ {{1^5},\;{1^3}{2^1},\;{1^1}{2^2},\;{1^2}{3^1},\;{2^1}{3^1},\;{1^1}{4^1},\;{5^1}} \right\}. \\ \end{array} $$

(4.214)

By construction, the expectation of the unsystematic loss is zero:

$$ {\mu_{1,c}}(y) = b{E}\left( {{{\tilde{Z}}^1}|\tilde{Y} = y} \right) = 0, $$

(4.215)

which is called the “granularity adjustment condition”. Consequently, for all partitions with ${e_{p1}} \ne 0$, the summands of (4.213) are zero, too:

$$ \prod\limits_{i = 1}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} = {0^{{e_{p1}}}} \cdot \prod\limits_{i = 2}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} = \left\{ \begin{array} {c} \prod\limits_{i = 2}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} \\ {\text{if }}{e_{p1}} = 0, \\ 0 \\ {\text{if }}{e_{p1}} \ne 0. \\ \end{array} \right. $$

(4.216)

Hence, the only relevant partitions $p \prec m$ of (4.214) with non-zero terms and the corresponding numbers $\left| p \right|$ are given as⁹⁸

$$ \begin{array} {c} p \prec 1 = \left\{ {{1^1}} \right\}\quad {\text{with }}\left| {p = {1^1}} \right| = 1, \\ p \prec 2 = \left\{ {{2^1}} \right\}\quad {\text{with }}\left| {p = {2^1}} \right| = 1, \\ p \prec 3 = \left\{ {{3^1}} \right\}\quad {\text{with }}\left| {p = {3^1}} \right| = 1, \\ p \prec 4 = \left\{ {{4^1},\;{2^2}} \right\}\quad {\text{with }}\left| {p = {4^1}} \right| = 1,\;\left| {p = {2^2}} \right| = 2, \\ p \prec 5 = \left\{ {{5^1},\;{2^1}{3^1}} \right\}\quad {\text{with }}\left| {p = {5^1}} \right| = 1,\;\left| {p = {2^1}{3^1}} \right| = 2. \\ \end{array} $$

(4.217)

For the associated terms

$$ {\alpha_p} = \frac{{m!}}{{{{(1!)}^{{e_p}_1}}{e_p}_{,1}!\, \cdot ... \cdot {{(m!)}^{{e_{p,m}}}}{e_{p,m}}!}}, $$

(4.218)

we obtain

$$ \begin{array} {c} {\alpha_{{1^1}}} = \frac{{1!}}{{{{(1!)}^1} \cdot 1!}} = 1, \\ {\alpha_{{2^1}}} = \frac{{2!}}{{{{(2!)}^1} \cdot 1!}} = 1, \\ {\alpha_{{3^1}}} = \frac{{3!}}{{{{(3!)}^1} \cdot 1!}} = 1, \\ {\alpha_{{4^1}}} = \frac{{4!}}{{{{(4!)}^1} \cdot 1!}} = 1,\;\;\;\;{\alpha_{{2^2}}} = \frac{{4!}}{{{{(2!)}^2} \cdot 2!}} = \frac{{24}}{8} = 3, \\ {\alpha_{{5^1}}} = \frac{{5!}}{{{{(5!)}^1} \cdot 1!}} = 1,\;\;\;\;{\alpha_{{2^1}{3^1}}} = \frac{{5!}}{{{{(2!)}^1} \cdot 1! \cdot {{(3!)}^1} \cdot 1!}} = \frac{{120}}{{12}} = 10. \\ \end{array} $$

(4.219)

According to (4.217), we only have $\left| p \right| = 1$ and $\left| p \right| = 2$, leading to the following partitions $u \prec s \leq \left| p \right| - 1$:

$$ \begin{array} {c} \left| p \right| = 1: \\ u \prec \left( {s = 0} \right) = \left\{ 0 \right\} \\ \left| p \right| = 2: \\ u \prec \left\{ {s = 0,\;s = 1} \right\} = \left\{ {0,\;{1^1}} \right\}. \\ \end{array} $$

(4.220)

As we have one summand for each $p \prec m$ and $u \prec s \leq (|p| - 1)$, we obtain one summand for $m = 1,\;2,\;3$ and three summands for $m = 4,\;5$:

$$ {\left. {\frac{{{d^m}q}}{{d{\lambda^m}}}} \right|_{\lambda = 0}} = \left\{ {\begin{array} {c}{*{20}{c}} {(I),} \\ {(I) + (II) + (III),} \\ \end{array} } \right.\begin{array} {c}{*{20}{c}} {\quad {\text{if }}m = 1,\;2,\;3,} \\ {\quad {\text{if }}m = 4,\;5,} \\ \end{array} $$

(4.221)

where the summands are determined with the following variables:

$$ \begin{array} {c} (I)\quad m = 1,\;...,\;5:p = {m^1},\quad \left| p \right| = 1,\;u \prec \left( {s = 0} \right) = \left\{ 0 \right\}, \\ (II)\quad \begin{array} {c}{*{20}{c}} {m = 4:} \\ {m = 5:} \\ \end{array} \left. {\begin{array} {c}{*{20}{c}} {p = {2^2},} \\ {p = {2^1}{3^1},} \\ \end{array} } \right\}\quad \left| p \right| = 2,\;u \prec \left( {s = 0} \right) = \left\{ 0 \right\}, \\ (III)\quad \begin{array} {c}{*{20}{c}} {m = 4:} \\ {m = 5:} \\ \end{array} \left. {\begin{array} {c}{*{20}{c}} {p = {2^2},} \\ {p = {2^1}{3^1},} \\ \end{array} } \right\}\quad \left| p \right| = 2,\;u \prec \left( {s = 1} \right) = \left\{ {{1^1}} \right\}. \\ \end{array} $$

(4.222)

The first summand (I), with $p = {m^1},\;\left| p \right| = 1,\;s = 0,\;u = 0,\left| u \right| = 0,\;\hat{u} = {1^1}$, ${e_{pm}} = 1$, and ${e_{pi}} = 0$ for all $i \ne m$, equals:⁹⁹

$$ \begin{array} {c} (I) = \frac{{{\alpha_p}{\alpha_{\hat{u}}}\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!\left( {\left| p \right| - 1 - s} \right)!}}{\left( { - f} \right)^{ - \left| p \right| - \left| u \right|}}\left( {\prod\limits_{i = 1}^s {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^{{e_{ui}}}}} } \right) \cdot \frac{{{d^{\left| p \right| - 1 - s}}}}{{d{y^{\left| p \right| - 1 - s}}}}\left( {\prod\limits_{i = 1}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \\ = \frac{{1 \cdot 1 \cdot \left( {1 + 0 - 1} \right)!}}{{\left( {0 + 0} \right)!\left( {1 - 1 - 0} \right)!}}{\left( { - f} \right)^{ - 1 - 0}}\left( {\prod\limits_{i = 1}^0 {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^{{e_{ui}}}}} } \right) \cdot \frac{{{d^{1 - 1 - 0}}}}{{d{y^{1 - 1 - 0}}}}\left( {\prod\limits_{i = 1}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \\ = - \frac{1}{f} \cdot \frac{{{d^{m - 1}}\left( {{\mu_{m,c}}f} \right)}}{{d{y^{m - 1}}}}. \\ \end{array} $$

(4.223)

For $m = 4$, the second summand (II.[4]), with values $p = {2^2},\;\left| p \right| = 2,\;s = 0,\;u = 0,\left| u \right| = 0,\;\hat{u} = {1^1}$, ${e_{p2}} = 2$, and ${e_{pi}} = 0$ for all $i \ne 2$, is equivalent to

$$ \begin{array} {c} II.[4] = \frac{{{\alpha_p}{\alpha_{\hat{u}}}\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!\left( {\left| p \right| - 1 - s} \right)!}}{\left( { - f} \right)^{ - \left| p \right| - \left| u \right|}}\left( {\prod\limits_{i = 1}^s {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^{{e_{ui}}}}} } \right) \\ \cdot \frac{{{d^{\left| p \right| - 1 - s}}}}{{d{y^{\left| p \right| - 1 - s}}}}\left( {\prod\limits_{i = 1}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \\ = \frac{{3 \cdot 1 \cdot \left( {2 + 0 - 1} \right)!}}{{\left( {0 + 0} \right)!\left( {2 - 1 - 0} \right)!}}{\left( { - f} \right)^{ - 2 - 0}}\left( {\prod\limits_{i = 1}^0 {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^{{e_{ui}}}}} } \right) \cdot \frac{{{d^{2 - 1 - 0}}}}{{d{y^{2 - 1 - 0}}}}\left( {\prod\limits_{i = 1}^4 {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \\ = 3 \cdot \frac{1}{{{f^2}}} \cdot \frac{d}{{dy}}{\left[ {\frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}} \right]^2}. \\ \end{array} $$

(4.224)

For $m = 5$, we have $p = {2^1}{3^1},\;\left| p \right| = 2,\;s = 0,\;u = 0,\left| u \right| = 0,\;\hat{u} = {1^1},\;{e_{p2}} = 1,\;{e_{p3}} = 1,$ and ${e_{pi}} = 0$ for all $i \ne 2,3$, leading to

$$ \begin{array} {c} II.[5] = \frac{{{\alpha_p}{\alpha_{\hat{u}}}\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!\left( {\left| p \right| - 1 - s} \right)!}}{\left( { - f} \right)^{ - \left| p \right| - \left| u \right|}}\left( {\prod\limits_{i = 1}^s {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^{{e_{ui}}}}} } \right) \\ \cdot \frac{{{d^{\left| p \right| - 1 - s}}}}{{d{y^{\left| p \right| - 1 - s}}}}\left( {\prod\limits_{i = 1}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \\ = \frac{{10 \cdot 1 \cdot \left( {2 + 0 - 1} \right)!}}{{\left( {0 + 0} \right)!\left( {2 - 1 - 0} \right)!}}{\left( { - f} \right)^{ - 2 - 0}}\left( {\prod\limits_{i = 1}^0 {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^{{e_{uj}}}}} } \right) \cdot \frac{{{d^{2 - 1 - 0}}}}{{d{y^{2 - 1 - 0}}}}\left( {\prod\limits_{i = 1}^5 {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \\ = 10 \cdot \frac{1}{{{f^2}}} \cdot \frac{d}{{dy}}\left( {\left[ {\frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}} \right]\left[ {\frac{{{d^2}\left( {{\mu_{3,c}}f} \right)}}{{d{y^2}}}} \right]} \right). \\ \end{array} $$

(4.225)

The third summand for $m = 4$ (III.[4]), with $p = {2^2},\;\left| p \right| = 2,\;s = 1,\;u = {1^1},\;\left| u \right| = 1,\;\hat{u} = {2^1},\;{e_{p2}} = 2,$ ${e_{pi}} = 0$ for all $i \ne 2$, and ${e_{u1}} = 1$ equals

$$ \begin{array} {c} III.[4] = \frac{{{\alpha_p}{\alpha_{\hat{u}}}\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!\left( {\left| p \right| - 1 - s} \right)!}}{\left( { - f} \right)^{ - \left| p \right| - \left| u \right|}}\left( {\prod\limits_{i = 1}^s {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^{{e_{ui}}}}} } \right) \\ \cdot \frac{{{d^{\left| p \right| - 1 - s}}}}{{d{y^{\left| p \right| - 1 - s}}}}\left( {\prod\limits_{i = 1}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \\ = \frac{{3 \cdot 1 \cdot \left( {2 + 1 - 1} \right)!}}{{\left( {1 + 1} \right)!\left( {2 - 1 - 1} \right)!}}{\left( { - f} \right)^{ - 2 - 1}}\left( {\prod\limits_{i = 1}^1 {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^1}} } \right) \cdot \frac{{{d^{2 - 1 - 1}}}}{{d{y^{2 - 1 - 1}}}}\left( {\prod\limits_{i = 1}^4 {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \\ = - 3 \cdot \frac{1}{{{f^3}}} \cdot \frac{{df}}{{dy}} \cdot {\left[ {\frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}} \right]^2}. \\ \end{array} $$

(4.226)

For $m = 5$, we have $p = {2^1}{3^1},\;\left| p \right| = 2,\;s = 1,\;u = {1^1},\;\left| u \right| = 1,\;\hat{u} = {2^1},\;{e_{p2}} = 1,\;{e_{p3}} = 1,$ ${e_{pi}} = 0$ for all $i \ne 2,3$, and ${e_{u1}} = 1$. Hence, we get

$$ \begin{array} {c} III.[5] = \frac{{{\alpha_p}{\alpha_{\hat{u}}}\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!\left( {\left| p \right| - 1 - s} \right)!}}{\left( { - f} \right)^{ - \left| p \right| - \left| u \right|}}\left( {\prod\limits_{i = 1}^s {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^{{e_{ui}}}}} } \right) \\ \cdot \frac{{{d^{\left| p \right| - 1 - s}}}}{{d{y^{\left| p \right| - 1 - s}}}}\left( {\prod\limits_{i = 1}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \\ = \frac{{10 \cdot 1 \cdot \left( {2 + 1 - 1} \right)!}}{{\left( {1 + 1} \right)!\left( {2 - 1 - 1} \right)!}}{\left( { - f} \right)^{ - 2 - 1}}\left( {\prod\limits_{i = 1}^1 {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^1}} } \right) \cdot \frac{{{d^{2 - 1 - 1}}}}{{d{y^{2 - 1 - 1}}}}\left( {\prod\limits_{i = 1}^5 {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right) \\ = - 10 \cdot \frac{1}{{{f^3}}} \cdot \frac{{df}}{{dy}} \cdot \left[ {\frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}} \right] \cdot \left[ {\frac{{{d^2}\left( {{\mu_{3,c}}f} \right)}}{{d{y^2}}}} \right]. \\ \end{array} $$

(4.227)

Summing up the relevant elements from (4.223) to (4.227) and multiplying by ${( - 1)^m}$ leads to

$$ {\left. {\frac{{dq}}{{d\lambda }}} \right|_{\lambda = 0}} = {( - 1)^1} \cdot \left( { - \frac{1}{f}} \right) \cdot \frac{{{d^{1 - 1}}\left( {{\mu_{1,c}}f} \right)}}{{d{y^{1 - 1}}}} = {\mu_{1,c}} = 0 $$

(4.228)

$$ {\left. {\frac{{{d^2}q}}{{d{\lambda^2}}}} \right|_{\lambda = 0}} = {( - 1)^2} \cdot \left( { - \frac{1}{f}} \right) \cdot \frac{{{d^{2 - 1}}\left( {{\mu_{2,c}}f} \right)}}{{d{y^{2 - 1}}}} = - \frac{1}{f} \cdot \frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}} $$

(4.229)

$$ {\left. {\frac{{{d^3}q}}{{d{\lambda^3}}}} \right|_{\lambda = 0}} = {\left( { - 1} \right)^3} \cdot \left( { - \frac{1}{f}} \right) \cdot \frac{{{d^{3 - 1}}\left( {{\mu_{3,c}}f} \right)}}{{d{y^{3 - 1}}}} = \frac{1}{f} \cdot \frac{{{d^2}\left( {{\mu_{3,c}}f} \right)}}{{d{y^2}}} $$

(4.230)

$$ \begin{array} {c} {\left. {\frac{{{d^4}q}}{{d{\lambda^4}}}} \right|_{\lambda = 0}} = {\left( { - 1} \right)^4} \cdot \left[ {\left( { - \frac{1}{f}} \right) \cdot \frac{{{d^{4 - 1}}\left( {{\mu_{4,c}}f} \right)}}{{d{y^{4 - 1}}}} + 3 \cdot \frac{1}{{{f^2}}} \cdot \frac{d}{{dy}}{{\left( {\frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}} \right)}^2}} \right. \\ \left. { - 3 \cdot \frac{1}{{{f^3}}} \cdot \frac{{df}}{{dy}} \cdot {{\left( {\frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}} \right)}^2}} \right] \\ = \left( { - \frac{1}{f}} \right) \cdot \left( {\frac{{{d^3}\left( {{\mu_{4,c}}f} \right)}}{{d{y^3}}} - 3 \cdot \frac{d}{{dy}}\left[ {\frac{1}{f}{{\left( {\frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}} \right)}^2}} \right]} \right), \\ \end{array} $$

(4.231)

and

$$ \begin{array} {c} {\left. {\frac{{{d^5}q}}{{d{\lambda^5}}}} \right|_{\lambda = 0}} = {\left( { - 1} \right)^5} \cdot \left[ {\left( { - \frac{1}{f}} \right) \cdot \frac{{{d^{5 - 1}}\left( {{\mu_{5,c}}f} \right)}}{{d{y^{5 - 1}}}} + 10 \cdot \frac{1}{{{f^2}}} \cdot \frac{d}{{dy}}\left( {\left[ {\frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}} \right]\left[ {\frac{{{d^2}\left( {{\mu_{3,c}}f} \right)}}{{d{y^2}}}} \right]} \right)} \right. \\ \left. { - 10 \cdot \frac{1}{{{f^3}}} \cdot \frac{{df}}{{dy}} \cdot \left[ {\frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}} \right] \cdot \left[ {\frac{{{d^2}\left( {{\mu_{3,c}}f} \right)}}{{d{y^2}}}} \right]} \right] \\ = \frac{1}{f} \cdot \left[ {\frac{{{d^4}\left( {{\mu_{5,c}}f} \right)}}{{d{y^4}}} - 10 \cdot \frac{d}{{dy}}\left( {\frac{1}{f} \cdot \frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}\frac{{{d^2}\left( {{\mu_{3,c}}f} \right)}}{{d{y^2}}}} \right)} \right]. \\ \end{array} $$

(4.232)

Comparing these terms, we find that the derivatives for $m = 1,\;...,\;5$ can be written as

$$ {\left. {\frac{{{d^m}q}}{{d{\lambda^m}}}} \right|_{\lambda = 0}} = {\left( { - 1} \right)^m}\left( { - \frac{1}{f}} \right)\left[ {\frac{{{d^{m - 1}}\left( {{\mu_{m,c}}f} \right)}}{{d{y^{m - 1}}}}} \right. - \kappa (m) \cdot \left. {\frac{d}{{dy}}\left( {\frac{1}{f} \cdot \frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}\frac{{{d^{m - 3}}\left( {{\mu_{m - 2,c}}f} \right)}}{{d{y^{m - 3}}}}} \right)} \right] $$

(4.233)

or without abbreviations as

$$ \begin{array} {c} {\left. {\frac{{{d^m}Va{R_\alpha }\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}{{d{\lambda^m}}}} \right|_{\lambda = 0}} = {\left( { - 1} \right)^m}\left( { - \frac{1}{{{f_Y}(y)}}} \right)\left[ {\frac{{{d^{m - 1}}\left( {{\mu_m}\left( {\tilde{Z}|\tilde{Y} = y} \right){f_Y}(y)} \right)}}{{d{y^{m - 1}}}}} \right. \\ - \kappa (m) \cdot {\left. {\frac{d}{{dy}}\left( {\frac{1}{{{f_Y}(y)}} \cdot \frac{{d\left( {{\mu_2}\left( {\tilde{Z}|\tilde{Y} = y} \right){f_Y}(y)} \right)}}{{dy}}\frac{{{d^{m - 3}}\left( {{\mu_{m - 2}}\left( {\tilde{Z}|\tilde{Y} = y} \right){f_Y}(y)} \right)}}{{d{y^{m - 3}}}}} \right)} \right]_{y = {q_\alpha }\left( {\tilde{Y}} \right)}}, \\ \end{array} $$

(4.234)

with $\kappa (1) = \kappa (2) = 0,\,\,\kappa (3) = 1,\,\,\kappa (4) = 3$, and $\kappa (5) = 10$, which is the result of Wilde (2003).

4.1.8 Order of the Derivatives of VaR

For any $m \in b{N}$, the (m+1)th element of the Taylor series can be written as¹⁰⁰

$$ \frac{{{\lambda^m}}}{{m!}}{\left[ {\frac{{{\partial^m}Va{R_\alpha }(\tilde{Y} + \lambda \tilde{Z})}}{{\partial {\lambda^m}}}} \right]_{\lambda = 0}} = g \circ {\left. {\left( {\frac{{{\lambda^m}}}{{m!}}\sum\limits_{p \prec m} {\prod\limits_{i = 1}^m {{{\left( {{\mu_i}\left[ {\tilde{Z}|\tilde{Y} = y} \right]} \right)}^{{e_{pi}}}}} } } \right)} \right|_{y = {q_\alpha }(\tilde{Y})}}, $$

(4.235)

with g being a function that is independent of the number of credits n. With ${\mu_i}$ as the ith moment about the origin and ${\eta_i}$ as the ith moment about the mean, it is possible to write¹⁰¹

$$ \begin{array} {c} \lambda \cdot \sum\limits_{p \prec 5} {\prod\limits_{i = 1}^5 {{{\left( {{\mu_i}\left( {\tilde{Z}} \right)} \right)}^{{e_{pi}}}}} } = \lambda \cdot \left( {{\mu_5}\left( {\tilde{Z}} \right) + {\mu_4}\left( {\tilde{Z}} \right) \cdot {\mu_1}\left( {\tilde{Z}} \right) + {\mu_3}\left( {\tilde{Z}} \right) \cdot {{\left( {{\mu_1}\left( {\tilde{Z}} \right)} \right)}^2}} \right. \\ + {\mu_3}\left( {\tilde{Z}} \right) \cdot {\mu_2}\left( {\tilde{Z}} \right) + {\mu_2}\left( {\tilde{Z}} \right) \cdot {\left( {{\mu_1}\left( {\tilde{Z}} \right)} \right)^3} + {\mu_2}{\left( {\tilde{Z}} \right)^2} \cdot {\mu_1}\left( {\tilde{Z}} \right)\left. { + {{\left( {{\mu_1}\left( {\tilde{Z}} \right)} \right)}^5}} \right) \\ = {\mu_5}\left( {\lambda \tilde{Z}} \right) + {\mu_4}\left( {\lambda \tilde{Z}} \right) \cdot {\mu_1}\left( {\lambda \tilde{Z}} \right) + {\mu_3}\left( {\lambda \tilde{Z}} \right) \cdot {\left( {{\mu_1}\left( {\lambda \tilde{Z}} \right)} \right)^2} \\ + {\mu_3}\left( {\lambda \tilde{Z}} \right) \cdot {\mu_2}\left( {\lambda \tilde{Z}} \right) + {\mu_2}\left( {\lambda \tilde{Z}} \right) \cdot {\left( {{\mu_1}\left( {\lambda \tilde{Z}} \right)} \right)^3} + {\mu_2}{\left( {\lambda \tilde{Z}} \right)^2} \cdot {\mu_1}\left( {\lambda \tilde{Z}} \right) \\ + {\left( {{\mu_1}\left( {\lambda \tilde{Z}} \right)} \right)^5}. \\ \end{array} $$

(4.236)

for each m. Thus, the derivatives are given as

$$ \begin{array} {c} {\lambda^m}\sum\limits_{p \prec m} {\prod\limits_{i = 1}^m {{{\left. {{{\left( {{\mu_i}\left[ {\tilde{Z}|\tilde{Y} = y} \right]} \right)}^{{e_{pi}}}}} \right|}_{y = {q_\alpha }\left( {\tilde{Y}} \right)}}} } = \sum\limits_{p \prec m} {\prod\limits_{i = 1}^m {{{\left. {{{\left( {{\mu_i}\left[ {\lambda \tilde{Z}|\tilde{Y} = y} \right]} \right)}^{{e_{pi}}}}} \right|}_{y = {q_\alpha }\left( {\tilde{Y}} \right)}}} } \\ = \sum\limits_{p \prec m} {{{\prod\limits_{i = 1}^m {\left. {{{\left( {{\mu_i}\left[ {\tilde{L} - b{E}\left( {\tilde{L}|\tilde{x}} \right)|\tilde{x} = x} \right]} \right)}^{{e_{pi}}}}} \right|} }_{x = {q_{1 - \alpha }}\left( {\tilde{x}} \right)}}} \\ = \sum\limits_{p \prec m} {\prod\limits_{i = 1}^m {{{\left. {{{\left( {{\mu_i}\left[ {\left( {\tilde{L}|\tilde{x} = x} \right) - b{E}\left( {\tilde{L}|\tilde{x} = x} \right)} \right]} \right)}^{{e_{pi}}}}} \right|}_{x = {q_{1 - \alpha }}\left( {\tilde{x}} \right)}}} } \\ = \sum\limits_{p \prec m} {{{\prod\limits_{i = 1}^m {\left. {{{\left( {{\eta_i}\left[ {\tilde{L}|\tilde{x} = x} \right]} \right)}^{{e_{pi}}}}} \right|} }_{x = {q_{1 - \alpha }}\left( {\tilde{x}} \right)}}} \\ = \sum\limits_{p \prec m} {\prod\limits_{i = 1}^m {{{\left. {{{\left( {{\eta_i}\left[ {\tilde{L}|\tilde{Y} = y} \right]} \right)}^{{e_{pi}}}}} \right|}_{y = {q_\alpha }\left( {\tilde{Y}} \right)}}} } \\ \end{array} $$

(4.237)

Due to¹⁰²

$$ \frac{{{\lambda^m}}}{{m!}}{\left[ {\frac{{{\partial^m}Va{R_\alpha }(\tilde{Y} + \lambda \tilde{Z})}}{{\partial {\lambda^m}}}} \right]_{\lambda = 0}} = g \circ {\left. {\left( {\frac{1}{{m!}}\sum\limits_{p \prec m} {\prod\limits_{i = 1}^m {{{\left( {{\eta_i}\left[ {\tilde{L}|\tilde{Y} = y} \right]} \right)}^{{e_{pi}}}}} } } \right)} \right|_{y = {q_\alpha }(\tilde{Y})}}. $$

with ${\eta_i}\left( {\tilde{L}|\tilde{x} = x} \right) = {\eta_i}^*(x) \cdot \sum\limits_{j = 1}^n {{w_j}^i} \leq {\eta_i}^*(x) \cdot {\left( {\frac{b}{a}} \right)^i} \cdot \frac{1}{{{n^{i - 1}}}} = O\left( {\frac{1}{{{n^{i - 1}}}}} \right),$ for all i, and revisiting (4.235) and (4.236), it is straightforward to see that only for $0 < a \leq EA{D_i} \leq b$ and $m = 3$ there exist terms which are at maximum of order O(1/n ²):

$$ m = 4 $$

(4.238)

All terms with higher derivatives of VaR are at least of Order O(1/n ³).

4.1.9 VaR-Based Second-Order Granularity Adjustment for a Normally Distributed Systematic Factor

For convenience, the summands of the second-order granularity add-on $\begin{array} {c} \sum\limits_{p \prec 3} {\prod\limits_{i = 1}^3 {{{\left( {{\eta_i}\left[ {\tilde{L}|\tilde{Y} = y} \right]} \right)}^{{e_{pi}}}}} } = {\eta_3}\left[ {\tilde{L}|\tilde{Y} = y} \right] = O\left( {\frac{1}{{{n^2}}}} \right), \\ \sum\limits_{p \prec 4} {\prod\limits_{i = 1}^4 {{{\left( {{\eta_i}\left[ {\tilde{L}|\tilde{Y} = y} \right]} \right)}^{{e_{pi}}}}} } = {\eta_4}\left[ {\tilde{L}|\tilde{Y} = y} \right] + {\left( {{\eta_2}\left[ {\tilde{L}|\tilde{Y} = y} \right]} \right)^2} = O\left( {\frac{1}{{{n^3}}}} \right) + O\left( {\frac{1}{{{n^2}}}} \right). \\ \end{array}$ will be calculated separately:

$$ \Delta {l_2} $$

(4.239)

The term $\begin{array} {c} \Delta {l_2} = \frac{1}{{6\varphi }}\frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}\frac{d}{{dx}}\left[ {\frac{{{\eta_{3,c}}\varphi }}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right]} \right) \\ + \frac{1}{{8\varphi }}\frac{d}{{dx}}{\left. {\left[ {\frac{1}{\varphi }\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}{{\left( {\frac{d}{{dx}}\left[ {\frac{{{\eta_{2,c}}\varphi }}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right]} \right)}^2}} \right]} \right|_{x = {\Phi^{ - 1}}(1 - \alpha )}} \\ = :{\left. {\Delta {l_{2,1}} + \Delta {l_{2,2}}} \right|_{x = {\Phi^{ - 1}}(1 - \alpha )}}. \\ \end{array}$ equals

$$ \Delta {l_{2,1}} $$

(4.240)

For the calculation, we need the first and second derivative of the density function $\begin{array} {c} \Delta {l_{2,1}} = \frac{1}{6}\left[ {\frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)\frac{1}{\varphi }\frac{d}{{dx}}\left( {\frac{{{\eta_{3,c}}\varphi }}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right) + \frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}\frac{1}{\varphi }\frac{{{d^2}}}{{d{x^2}}}\left( {\frac{{{\eta_{3,c}}\varphi }}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right] \\ = \frac{1}{6}\left[ {\frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)\left( {\underbrace {\frac{1}{\varphi }\frac{d}{{dx}}\left( {{\eta_{3,c}}\varphi } \right)}_{ = :A}\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + {\eta_{3,c}}\frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right)} \right. \\ + \frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}\frac{1}{\varphi }\frac{d}{{dx}}\left[ {\underbrace {\frac{d}{{dx}}\left( {{\eta_{3,c}}\varphi } \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}}_{ = :B} + \underbrace {{\eta_{3,c}}\varphi \frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)}_{ = :C}} \right]. \\ \end{array}$. As the systematic factor is assumed to be normally distributed, we have

$$ \varphi $$

(4.241)

$$ \varphi = \frac{1}{{\sqrt {{2\pi }} }}{e^{ - \frac{{{x^2}}}{2}}} $$

(4.242)

$$ \frac{{d\varphi }}{{dx}} = \left( { - x} \right)\frac{1}{{\sqrt {{2\pi }} }}{e^{ - \frac{{{x^2}}}{2}}} = - x\,\varphi $$

(4.243)

Furthermore, we need the derivative

$$ \frac{{{d^2}\varphi }}{{d{x^2}}} = \left( { - 1} \right)\frac{1}{{\sqrt {{2\pi }} }}{e^{ - \frac{{{x^2}}}{2}}} - x\left( { - x} \right)\frac{1}{{\sqrt {{2\pi }} }}{e^{ - \frac{{{x^2}}}{2}}} = ({x^2} - 1)\varphi . $$

(4.244)

Herewith, the term A form (4.240) can easily be calculated:

$$ \frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right) = - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}. $$

(4.245)

Furthermore, $A = \frac{1}{\varphi }\frac{d}{{dx}}\left( {{\eta_{3,c}}\varphi } \right) = \frac{{d{\eta_{3,c}}}}{{dx}} + \frac{{{\eta_{3,c}}}}{\varphi }\frac{{d\varphi }}{{dx}} = \frac{{d{\eta_{3,c}}}}{{dx}} - {\eta_{3,c}}x.$ is equal to

$$ {{{dB}} \left/ {{dx}} \right.} $$

(4.246)

Similarly, $\begin{array} {c} \frac{{dB}}{{dx}} = \frac{d}{{dx}}\left( {\frac{d}{{dx}}\left( {{\eta_{3,c}}\varphi } \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right) \\ = \frac{{{d^2}}}{{d{x^2}}}\left( {{\eta_{3,c}}\varphi } \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + \frac{d}{{dx}}\left( {{\eta_{3,c}}\varphi } \right)\frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right) \\ = \frac{d}{{dx}}\left( {\frac{{d{\eta_{3,c}}}}{{dx}}\varphi + {\eta_{3,c}}\frac{{d\varphi }}{{dx}}} \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + \left( {\frac{{d{\eta_{3,c}}}}{{dx}}\varphi + {\eta_{3,c}}\frac{{d\varphi }}{{dx}}} \right)\left( { - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right) \\ = \left( {\frac{{{d^2}{\eta_{3,c}}}}{{d{x^2}}}\varphi + 2\frac{{d{\eta_{3,c}}}}{{dx}}\frac{{d\varphi }}{{dx}} + {\eta_{3,c}}\frac{{{d^2}\varphi }}{{d{x^2}}}} \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} \\ - \left( {\frac{{d{\eta_{3,c}}}}{{dx}}\varphi + {\eta_{3,c}}\frac{{d\varphi }}{{dx}}} \right)\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}. \\ \end{array}$ is equivalent to

$$ {{{dC}} \left/ {{dx}} \right.} $$

(4.247)

Using these terms, $\begin{array} {c} \frac{{dC}}{{dx}} = \frac{d}{{dx}}\left( {{\eta_{3,c}}\varphi \left( { - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right)} \right) \\ = - \frac{d}{{dx}}\left( {{\eta_{3,c}}\varphi } \right)\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}} - {\eta_{3,c}}\varphi \frac{d}{{dx}}\left( {\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right) \\ = \left( { - \frac{{d{\eta_{3,c}}}}{{dx}}\varphi - {\eta_{3,c}}\frac{{d\varphi }}{{dx}}} \right)\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}} \\ - {\eta_{3,c}}\varphi \left( {\frac{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}\left( {{{{{d^3}{\mu_{1,c}}}} \left/ {{d{x^3}}} \right.}} \right) - 2\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right){{\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}^2}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^4}}}} \right). \\ \end{array}$ results in

$$ \Delta {l_{2,1}} $$

(4.248)

Applying the derivatives of $\begin{array} {c} \Delta {l_{2,1}} = \frac{1}{6}\left[ { - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}\left( {\frac{{{{{d{\eta_{3,c}}}} \left/ {{dx}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - \frac{{{\eta_{3,c}}x}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - {\eta_{3,c}}\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right)} \right. \\ + \frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}\frac{1}{\varphi }\left[ {\left( {\frac{{{d^2}{\eta_{3,c}}}}{{d{x^2}}}\varphi + 2\frac{{d{\eta_{3,c}}}}{{dx}}\frac{{d\varphi }}{{dx}} + {\eta_{3,c}}\frac{{{d^2}\varphi }}{{d{x^2}}}} \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right. \\ - 2\left( {\frac{{d{\eta_{3,c}}}}{{dx}}\varphi + {\eta_{3,c}}\frac{{d\varphi }}{{dx}}} \right)\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}} \\ \left. { - {\eta_{3,c}}\varphi \left( {\frac{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}\left( {{{{{d^3}{\mu_{1,c}}}} \left/ {{d{x^3}}} \right.}} \right) - 2\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right){{\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}^2}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^4}}}} \right)} \right]. \\ \end{array}$ from (4.242) and (4.243) leads to

$$ \varphi $$

(4.249)

Henceforward, the summand $\begin{array} {c} \Delta {l_{2,1}} = \frac{1}{6}\left[ { - 3\frac{{\left( {{{{d{\eta_{3,c}}}} \left/ {{dx}} \right.}} \right)\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}} + 3\frac{{{\eta_{3,c}}x\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}} + 3{\eta_{3,c}}\frac{{{{\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}^2}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^4}}}} \right. \\ + \frac{{{{{{d^2}{\eta_{3,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}} - 2x\frac{{{{{d{\eta_{3,c}}}} \left/ {{dx}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}} + \frac{{{\eta_{3,c}}\left( {{x^2} - 1} \right)}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}\left. { - {\eta_{3,c}}\frac{{{{{{d^3}{\mu_{1,c}}}} \left/ {{d{x^3}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}}} \right] \\ = \frac{1}{{6{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}\left[ {{\eta_{3,c}}\left( {{x^2} - 1 - \frac{{{{{{{{d^3}{\mu_{1,c}}}} \left/ {{dx}} \right.}}^3}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + \frac{{3x\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + \frac{{3{{\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}^2}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right)} \right. \\ \left. { + \frac{{d{\eta_{3,c}}}}{{dx}}\left( { - 2x - \frac{{3\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right) + \frac{{{d^2}{\eta_{3,c}}}}{{d{x^2}}}} \right]. \\ \end{array}$ will be simplified:

$$ \Delta {l_{2,2}} $$

(4.250)

The term (*) is the negative twice of the first-order granularity adjustment, so that we can use the resulting equation (4.18). This leads to

$$ \begin{array} {c} \Delta {l_{2,2}} = \frac{1}{{8\varphi }}\frac{d}{{dx}}\left[ {\frac{1}{\varphi }\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}{{\left( {\frac{d}{{dx}}\left[ {\frac{{{\eta_{2,c}}\varphi }}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right]} \right)}^2}} \right] \\ = \frac{1}{{8\varphi }}\frac{d}{{dx}}\left( {\frac{\varphi }{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}{{\underbrace {\left( {\frac{1}{\varphi }\frac{d}{{dx}}\left[ {\frac{{{\eta_{2,c}}\varphi }}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right]} \right)}_*}^2}} \right). \\ \end{array} $$

(4.251)

Using the derivative of a normal distribution $\begin{array} {c} \Delta {l_{2,2}} = \frac{1}{{8\varphi }}\frac{d}{{dx}}\left( {\frac{\varphi }{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}{{\left[ { - \frac{{x\,{\eta_{2,c}}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + \frac{{{{{d{\eta_{2,c}}}} \left/ {{dx}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - \frac{{{\eta_{2,c}}{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right]}^2}} \right) \\ = \frac{1}{8}\left[ {\underbrace {\frac{1}{\varphi }\frac{d}{{dx}}\left( {\frac{\varphi }{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}}} \right)}_{ = :(I)}{{\left( { - x\,{\eta_{2,c}} + \frac{{d{\eta_{2,c}}}}{{dx}} - \frac{{{\eta_{2,c}}{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)}^2}} \right. \\ \left. { + \frac{1}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}}\underbrace {\frac{d}{{dx}}\left( {{{\left[ { - x\,{\eta_{2,c}} + \frac{{d{\eta_{2,c}}}}{{dx}} - \frac{{{\eta_{2,c}}{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right]}^2}} \right)}_{ = :(II)}} \right]. \\ \end{array}$, the term (I) is equivalent to

$$ d\varphi /dx = - x\,\varphi $$

(4.252)

Term (II) can be written as

$$ \begin{array} {c} (I) = \frac{1}{\varphi }\frac{d}{{dx}}\left( {\frac{\varphi }{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}}} \right) \\ = \frac{1}{\varphi }\frac{{d\varphi }}{{dx}}\frac{1}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}} + \frac{d}{{dx}}\left( {\frac{1}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}}} \right) \\ = \frac{{ - x}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}} - 3\frac{{\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^4}}}. \\ \end{array} $$

(4.253)

Using these expressions, $\begin{array} {c} (II) = \frac{d}{{dx}}\left( {{{\left[ { - x\,{\eta_{2,c}} + \frac{{d{\eta_{2,c}}}}{{dx}} - \frac{{{\eta_{2,c}}{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right]}^2}} \right) \\ = 2\left( { - x\,{\eta_{2,c}} + \frac{{d{\eta_{2,c}}}}{{dx}} - \frac{{{\eta_{2,c}}{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)\left( { - {\eta_{2,c}} - x\frac{{d{\eta_{2,c}}}}{{dx}} + \frac{{{d^2}{\eta_{2,c}}}}{{d{x^2}}}} \right. \\ \left. { - \frac{d}{{dx}}\left( {{\eta_{2,c}}\frac{{{d^2}{\mu_{1,c}}}}{{d{x^2}}}} \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - {\eta_{2,c}}\frac{{{d^2}{\mu_{1,c}}}}{{d{x^2}}}\frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right) \\ = 2\left( { - x{\eta_{2,c}} + \frac{{d{\eta_{2,c}}}}{{dx}} - \frac{{{\eta_{2,c}}{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)\left( { - {\eta_{2,c}} - x\frac{{d{\eta_{2,c}}}}{{dx}} + \frac{{{d^2}{\eta_{2,c}}}}{{d{x^2}}}} \right. \\ \left. { - \frac{{d{\eta_{2,c}}}}{{dx}}\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - {\eta_{2,c}}\frac{{{{{{d^3}{\mu_{1,c}}}} \left/ {{d{x^3}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + {\eta_{2,c}}\frac{{{d^2}{\mu_{1,c}}}}{{d{x^2}}}\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right). \\ \end{array}$ from (4.251) is equal to

$$ \Delta {l_{2,2}} $$

(4.254)

which leads to

$$ \begin{array} {c} \Delta {l_{2,2}} = \frac{1}{8}\left[ {\left( {\frac{{ - x}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}} - 3\frac{{\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^4}}}} \right){{\left( { - x\,{\eta_{2,c}} + \frac{{d{\eta_{2,c}}}}{{dx}} - \frac{{{\eta_{2,c}}{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)}^2}} \right. \\ + \frac{2}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}}\left( { - x\,{\eta_{2,c}} + \frac{{d{\eta_{2,c}}}}{{dx}} - \frac{{{\eta_{2,c}}{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)\left( { - {\eta_{2,c}} - x\frac{{d{\eta_{2,c}}}}{{dx}} + \frac{{{d^2}{\eta_{2,c}}}}{{d{x^2}}}} \right. \\ \left. {\left. { - \frac{{d{\eta_{2,c}}}}{{dx}}\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - {\eta_{2,c}}\frac{{{{{{d^3}{\mu_{1,c}}}} \left/ {{d{x^3}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + {\eta_{2,c}}\frac{{{d^2}{\mu_{1,c}}}}{{d{x^2}}}\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right)} \right], \\ \end{array} $$

(4.255)

Adding the terms $\begin{array} {c} \Delta {l_{2,2}} = \frac{1}{{8{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}}{\left[ {\left( { - x - 3\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)\left( {{\eta_{2,c}}\left[ { - x - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right] + \frac{{d{\eta_{2,c}}}}{{dx}}} \right)} \right.^2} \\ + 2\left( {{\eta_{2,c}}\left[ {x + \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right] - \frac{{d{\eta_{2,c}}}}{{dx}}} \right)\left( {{\eta_{2,c}}\left[ {1 + \frac{{{{{{d^3}{\mu_{1,c}}}} \left/ {{d{x^3}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - \frac{{{{\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}^2}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right]} \right. \\ \left. {\left. { + \frac{{d{\eta_{2,c}}}}{{dx}}\left[ {x + \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right] - \frac{{{d^2}{\eta_{2,c}}}}{{d{x^2}}}} \right)} \right]. \\ \end{array}$ and $\Delta {l_{2,1}}$ together results in

$$ \Delta {l_{2,2}} $$

(4.256)

4.1.10 Third Conditional Moment of Losses

Subsequently, the third conditional moment of the portfolios loss about the mean, $\begin{array} {c} \Delta {l_2} = \frac{1}{{6{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}\left[ {{\eta_{3,c}}\left( {{x^2} - 1 - \frac{{{{{{{{d^3}{\mu_{1,c}}}} \left/ {{dx}} \right.}}^3}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + \frac{{3x\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + \frac{{3{{\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}^2}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right)} \right. \\ \left. { + \frac{{d{\eta_{3,c}}}}{{dx}}\left( { - 2x - \frac{{3\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right) + \frac{{{d^2}{\eta_{3,c}}}}{{d{x^2}}}} \right] \\ + \frac{1}{{8{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}}{\left[ {\left( { - x - 3\frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)\left( {{\eta_{2,c}}\left[ { - x - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right] + \frac{{d{\eta_{2,c}}}}{{dx}}} \right)} \right.^2} \\ + 2\left( {{\eta_{2,c}}\left[ {x + \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right] - \frac{{d{\eta_{2,c}}}}{{dx}}} \right)\left( {{\eta_{2,c}}\left[ {1 + \frac{{{{{{d^3}{\mu_{1,c}}}} \left/ {{d{x^3}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - \frac{{{{\left( {{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}} \right)}^2}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right]} \right. \\ {\left. {\left. {\left. { + \frac{{d{\eta_{2,c}}}}{{dx}}\left[ {x + \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right] - \frac{{{d^2}{\eta_{2,c}}}}{{d{x^2}}}} \right)} \right]} \right|_{x = {\Phi^{ - 1}}\left( {1 - \alpha } \right)}}. \\ \end{array}$, shall be expressed in terms of the moments of separated factors ${\eta_{3,c}} = {\eta_3}(\tilde{L}|\tilde{x} = x)$ and $\widetilde{{LG{D_i}}}$. With

$$ {1_{\left\{ {{{\tilde{D}}_i}} \right\}}} $$

(4.257)

which is due to the conditional independence property, we need to determine $\begin{array} {c} {\eta_{3,c}} = {\eta_3}\left( {\tilde{L}|\tilde{x} = x} \right) \\ = {\eta_3}\left( {\sum\limits_{i = 1}^n {{w_i} \cdot \widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}} |\tilde{x} = x} \right) \\ = \sum\limits_{i = 1}^n {{w_i}^3 \cdot {\eta_3}\left( {\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x} = x} \right)}, \\ \end{array}$. In general, the third moment about the mean is equal to

$$ {\eta_3}(\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x}) $$

(4.258)

Thus, the conditional moment $\begin{array} {c} {\eta_3}\left( {\tilde{X}} \right) = b{E}\left( {{{\left[ {\tilde{X} - b{E}\left( {\tilde{X}} \right)} \right]}^3}} \right) \\ = b{E}\left[ {{{\tilde{X}}^3} - 3{{\tilde{X}}^2}b{E}\left( {\tilde{X}} \right) + 3\tilde{X}{b{E}^2}\left( {\tilde{X}} \right) - {b{E}^3}\left( {\tilde{X}} \right)} \right] \\ = b{E}\left( {{{\tilde{X}}^3}} \right) - 3b{E}\left( {{{\tilde{X}}^2}} \right)b{E}\left( {\tilde{X}} \right) + 3b{E}\left( {\tilde{X}} \right){b{E}^2}\left( {\tilde{X}} \right) - {b{E}^3}\left( {\tilde{X}} \right) \\ = b{E}\left( {{{\tilde{X}}^3}} \right) - 3b{E}\left( {{{\tilde{X}}^2}} \right)b{E}\left( {\tilde{X}} \right) + 2{b{E}^3}\left( {\tilde{X}} \right). \\ \end{array}$ can be written as

$$ {\eta_3}(\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x}) $$

(4.259)

Using the conditional independence property again, considering that the LGDs are assumed to be stochastically independent of each other, and with $\begin{array} {c} {\eta_3}\left( {\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x}} \right) = b{E}\left( {{{\left[ {\widetilde{{LGD}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x}} \right]}^3}} \right) - 3\,b{E}\left( {{{\left[ {\widetilde{{LGD}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x}} \right]}^2}} \right) \cdot b{E}\left( {\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x}} \right) \\ + 2\,{b{E}^3}\left( {\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x}} \right). \\ \end{array}$, we have

$$ b{E}[{({1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x})^i}] = b{E}[({1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x})] = p(\tilde{x}) $$

(4.260)

With the abbreviations $\begin{array} {c} {\eta_3}\left( {\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x}} \right) = b{E}\left( {{{\left[ {\widetilde{{LGD}}|\tilde{x}} \right]}^3}} \right)p\left( {\tilde{x}} \right) - 3\,b{E}\left( {{{\left[ {\widetilde{{LGD}}|\tilde{x}} \right]}^2}} \right)b{E}\left( {\widetilde{{LGD}}|\tilde{x}} \right){p^2}\left( {\tilde{x}} \right) \\ + 2\,{b{E}^3}\left( {\widetilde{{LGD}}|\tilde{x}} \right){p^3}\left( {\tilde{x}} \right) \\ = b{E}\left( {{{\widetilde{{LGD}}}^3}} \right)p\left( {\tilde{x}} \right) - 3\,b{E}\left( {{{\widetilde{{LGD}}}^2}} \right)b{E}\left( {\widetilde{{LGD}}} \right){p^2}\left( {\tilde{x}} \right) \\ + 2\,{b{E}^3}\left( {\widetilde{{LGD}}} \right){p^3}\left( {\tilde{x}} \right). \\ \end{array}$, $ELGD = b{E}(\widetilde{{LGD}})$ as well as $VLGD = b{V}(\widetilde{{LGD}})$ and using (4.258) again, we obtain

$$ SLGD = {\eta_3}(\widetilde{{LGD}}) $$

(4.261)

$$ b{E}\left( {{{\widetilde{{LGD}}}^2}} \right) = ELG{D^2} + VLGD $$

(4.262)

Consequently, (4.260) is equivalent to

$$ \begin{array} {c} b{E}\left( {{{\widetilde{{LGD}}}^3}} \right) = SLGD + 3(ELG{D^2} + VLGD)ELGD - 2\,ELG{D^3} \\ = ELG{D^3} + 3\,ELGD \cdot VLGD + SLGD. \\ \end{array} $$

(4.263)

Thus, the conditional moment of the portfolio loss (4.257) can finally be written as

$$ \begin{array} {c} {\eta_3}\left( {\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}|\tilde{x}} \right) = \left( {ELG{D^3} + 3\,ELGD \cdot VLGD + SLGD} \right)p\left( {\tilde{x}} \right) \\ - 3\,\left( {ELG{D^3} + ELGD \cdot VLGD} \right){p^2}\left( {\tilde{x}} \right) + 2\,ELG{D^3}{p^3}\left( {\tilde{x}} \right). \\ \end{array} $$

(4.264)

4.1.11 Difference Between the VaR Definitions

For the case of homogeneous credits and with $\begin{array} {c} {\eta_{3,c}} = \sum\limits_{i = 1}^n {{w_i}^3 \cdot {\eta_3}\left( {\widetilde{{LG{D_i}}} \cdot {1_{\left\{ {{{\tilde{D}}_i}} \right\}}}\left| {\widetilde{x} = x} \right.} \right)} \\ = \sum\limits_{i = 1}^n {{w_i}^3\left[ {\left( {ELG{D_i}^3 + 3 \cdot ELG{D_i} \cdot VLG{D_i} + SLG{D_i}} \right) \cdot {p_i}(x)} \right.} \\ \left. { - 3 \cdot \left( {ELG{D_i}^3 + ELG{D_i} \cdot VLG{D_i}} \right) \cdot {p_i}^2(x) + 2 \cdot ELG{D_i}^3 \cdot {p_i}^3(x)} \right]. \\ \end{array}$, the possible realizations of losses are

$$ LGD = 1 $$

(4.265)

which implies

$$ l \in \left\{ {0,\;\frac{1}{n},\;\frac{2}{n},\;...,\;\frac{{n - 1}}{n},\;1} \right\} $$

(4.266)

If we define $b{P}\left[ {\tilde{L} \leq l} \right] = b{P}\left[ {\tilde{L} < \left( {l + {{1} \left/ {n} \right.}} \right)} \right]$, we get

$$ {l_2}: = {l_1} + {{1} \left/ {n} \right.} $$

(4.267)

4.1.12 Identity of ES Within the Basel Framework

Using the result of the ASRF framework (2.93), the definition of the ES (2.19), the integral representation of the conditional expectation, and the identity of the condition as in (4.9), the ES of the portfolio loss equals

$$ \begin{array} {c} VaR_\alpha^{( - )}\left( {\tilde{L}} \right) = \sup \left\{ {{l_1} \in b{R}|b{P}\left[ {\tilde{L} \leq {l_1}} \right] < \alpha } \right\} \\ = \sup \left\{ {{l_1} \in b{R}|b{P}\left[ {\tilde{L} < \left( {{l_1} + \frac{1}{n}} \right)} \right] < \alpha } \right\} \\ = \sup \left\{ {\left( {{l_2} - \frac{1}{n}} \right) \in b{R}|b{P}\left[ {\tilde{L} < {l_2}} \right] < \alpha } \right\} \\ = \sup \left\{ {{l_2} \in b{R}|b{P}\left[ {\tilde{L} < {l_2}} \right] < \alpha } \right\} - \frac{1}{n} \\ = VaR_\alpha^{( + )}\left( {\tilde{L}} \right) - \frac{1}{n}. \\ \end{array} $$

(4.268)

With the conditional independence property as in (2.92), the conditional PD of the Vasicek model (2.66), the integral representation (2.126), and the symmetry of the normal distribution, the ES can be written as

$$ \begin{array} {c} ES_\alpha^{{\text{(Basel)}}}\left( {\tilde{L}} \right) = E{S_\alpha }\left[ {b{E}\left( {\tilde{L}|\tilde{x}} \right)} \right] \\ = E{S_\alpha }\left[ {{\mu_{1,c}}\left( {\tilde{x}} \right)} \right] \\ = \frac{1}{{1 - \alpha }}\left[ {b{E}\left( {{\mu_{1,c}}\left( {\tilde{x}} \right)|{\mu_{1,c}}\left( {\tilde{x}} \right) \geq {q_\alpha }\left( {{\mu_{1,c}}\left( {\tilde{x}} \right)} \right)} \right)} \right] \\ = \frac{1}{{1 - \alpha }}\left[ {b{E}\left( {{\mu_{1,c}}\left( {\tilde{x}} \right)|\tilde{x} \leq {\Phi^{ - 1}}\left( {1 - \alpha } \right)} \right)} \right] \\ = \frac{1}{{1 - \alpha }}\int\limits_{ - \infty }^{{\Phi^{ - 1}}\left( {1 - \alpha } \right)} {{\mu_{1,c}}(x)\varphi (x)dx} . \\ \end{array} $$

(4.269)

4.1.13 Arbitrary Derivatives of ES

According to (2.20), the ES can be written as

$$ \begin{array} {c} ES_\alpha^{{\text{(Basel)}}}\left( {\tilde{L}} \right) = \frac{1}{{1 - \alpha }}\int\limits_{ - \infty }^{{\Phi^{ - 1}}\left( {1 - \alpha } \right)} {\sum\limits_{i = 1}^n {b{E}\left( {{w_i} \cdot {{\widetilde{{LGD}}}_i} \cdot {1_{\left\{ {{D_i}} \right\}}}|x} \right)\varphi (x)dx} } \\ = \frac{1}{{1 - \alpha }}\sum\limits_{i = 1}^n {{w_i} \cdot ELG{D_i} \cdot \int\limits_{ - \infty }^{{\Phi^{ - 1}}\left( {1 - \alpha } \right)} {{p_i}(x)\varphi (x)dx} } \\ = \frac{1}{{1 - \alpha }}\sum\limits_{i = 1}^n {{w_i} \cdot ELG{D_i} \cdot \int\limits_{ - \infty }^{{\Phi^{ - 1}}\left( {1 - \alpha } \right)} {\Phi \left( {\frac{{{\Phi^{ - 1}}(P{D_i}) - \sqrt {{{\rho_i}}} \cdot x}}{{\sqrt {{1 - {\rho_i}}} }}} \right)\varphi (x)dx} } \\ = \frac{1}{{1 - \alpha }}\sum\limits_{i = 1}^n {{w_i} \cdot ELG{D_i} \cdot {\Phi_2}\left( {{\Phi^{ - 1}}(1 - \alpha ),{\Phi^{ - 1}}(P{D_i}),\sqrt {{{\rho_i}}} } \right)} \\ = \frac{1}{{1 - \alpha }}\sum\limits_{i = 1}^n {{w_i} \cdot ELG{D_i} \cdot {\Phi_2}\left( { - {\Phi^{ - 1}}(\alpha ),{\Phi^{ - 1}}(P{D_i}),\sqrt {{{\rho_i}}} } \right)} . \\ \end{array} $$

(4.270)

Thus, for continuous distributions, all derivatives of ES can be expressed as

$$ E{S_\alpha }\left( {\tilde{L}} \right) = \frac{1}{{1 - \alpha }}\int\limits_\alpha^1 {{q^u}\left( {\tilde{L}} \right)du} . $$

(4.271)

The derivative of VaR is a function of $\frac{{{d^m}E{S_\alpha }}}{{d{\lambda^m}}} = \frac{{{d^m}}}{{d{\lambda^m}}}\left( {\frac{1}{{1 - \alpha }}\int\limits_\alpha^1 {{q_u}du} } \right) = \frac{1}{{1 - \alpha }}\int\limits_\alpha^1 {\frac{{{d^m}{q_u}}}{{d{\lambda^m}}}du} .$ and ${f_Y}(y)$ evaluated at ${\mu_{i,c}}(y)$. The substitution ${q_u}(\tilde{Y})$, so that $u = {F_Y}(y)$, ${{{du}} \left/ {{dy}} \right.} = {f_Y}(y)$, and $y(u = \alpha ) = F_Y^{ - 1}(\alpha ) = {q_\alpha }(\tilde{Y})$, leads to:¹⁰³

$$ y(u = 1) = F_Y^{ - 1}(1) = \infty $$

(4.272)

where the expression resulting from the derivative of VaR simply has to be evaluated at y since ${\left. {\frac{{{d^m}E{S_\alpha }}}{{d{\lambda^m}}}} \right|_{\lambda = 0}} = \frac{1}{{1 - \alpha }}\int\limits_{u = \alpha }^1 {{{\left. {\frac{{{d^m}{q_u}}}{{d{\lambda^m}}}} \right|}_{\lambda = 0}}du} = \frac{1}{{1 - \alpha }}\int\limits_{y = {q_\alpha }\left( {\tilde{Y}} \right)}^\infty {{{\left. {\frac{{{d^m}{q_u}}}{{d{\lambda^m}}}} \right|}_{\lambda = 0}}{f_Y}dy},$. Using the derivatives of VaR from (4.212), this leads to

$$ {q_u}(\tilde{Y}) = y $$

(4.273)

with $\begin{array}{ll} \left.\frac{{d^m} E{S_\alpha }}{{d{\lambda^m}}}\right|_{\lambda = 0} & = \frac{1}{{1 - \alpha }} \int\limits_{y = {q_\alpha }\left( {\tilde{Y}} \right)}^\infty{{( - 1)}^m} \left[\sum\limits_{ p \prec m, u \prec s \leq |p| - 1 } \right. {\frac{{{\alpha_p}{\alpha_{\hat{u}}}\left( {\left| p \right| + \left| u \right| - 1} \right)!}}{{\left( {s + \left| u \right|} \right)!\left( {\left| p \right| - 1 - s} \right)!}}} \\ & \cdot{{\left( { - f} \right)}^{ - \left| p \right| - \left| u \right|}} \cdot \left( {\prod\limits_{i = 1}^s {{{\left[ {\frac{{{d^i}f}}{{d{y^i}}}} \right]}^{{e_{ui}}}}} } \right)\left. { \cdot \frac{{{d^{\left| p \right| - 1 - s}}}}{{d{y^{\left| p \right| - 1 - s}}}}\left( {\prod\limits_{i = 1}^m {{{\left[ {\frac{{{d^{i - 1}}\left( {{\mu_{i,c}}f} \right)}}{{d{y^{i - 1}}}}} \right]}^{{e_{pi}}}}} } \right)} \right]f\,dy,\end{array}$.

4.1.14 Determination of the First Five Derivatives of ES

Instead of solving the integral (4.272) for each of the derivatives of VaR (4.228)–(4.232), we will directly evaluate the integral for the first five derivatives. Using the expression for the first five derivatives of VaR (4.233), we obtain

$$ {\alpha_p} = \frac{{m!}}{{{{(1!)}^{{e_p}_1}}{e_p}_{,1}! \cdot ... \cdot {{(m!)}^{{e_{p,m}}}}{e_{p,m}}!}} $$

(4.274)

This term is equal to

$$ \begin{array} {c} {\left. {\frac{{{d^m}ES}}{{d{\lambda^m}}}} \right|_{\lambda = 0}} = \frac{1}{{1 - \alpha }}\int\limits_{y = {q_\alpha }\left( {\tilde{Y}} \right)}^\infty {\frac{{{d^m}q}}{{d{\lambda^m}}}{f_Y}dy} \\ = \frac{1}{{1 - \alpha }}\int\limits_{y = {q_\alpha }\left( {\tilde{Y}} \right)}^\infty {{{\left( { - 1} \right)}^m}\left( { - \frac{1}{f}} \right)\left[ {\frac{{{d^{m - 1}}\left( {{\mu_{m,c}}f} \right)}}{{d{y^{m - 1}}}}} \right.} \\ - \kappa (m) \cdot \left. {\frac{d}{{dy}}\left( {\frac{1}{f} \cdot \frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}}\frac{{{d^{m - 3}}\left( {{\mu_{m - 2,c}}f} \right)}}{{d{y^{m - 3}}}}} \right)} \right]f\,dy. \\ \end{array} $$

(4.275)

or written without abbreviations as

$$ \begin{array} {c} {\left. {\frac{{{d^m}ES}}{{d{\lambda^m}}}} \right|_{\lambda = 0}} = {\left( { - 1} \right)^m} \cdot \frac{1}{{1 - \alpha }} \cdot \left( {\int\limits_{y = {q_\alpha }\left( {\tilde{Y}} \right)}^\infty {\left( { - \frac{{{d^{m - 1}}\left( {{\mu_{m,c}}f} \right)}}{{d{y^{m - 1}}}}} \right)dy} } \right. \\ + \left. {\kappa (m) \cdot \int\limits_{y = {q_\alpha }\left( {\tilde{Y}} \right)}^\infty {\frac{d}{{dy}}\left( {\frac{1}{f} \cdot \frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}} \cdot \frac{{{d^{m - 3}}\left( {{\mu_{m - 2,c}}f} \right)}}{{d{y^{m - 3}}}}} \right)dy} } \right) \\ = {\left( { - 1} \right)^m} \cdot \frac{1}{{1 - \alpha }} \cdot \left( {\left[ { - \frac{{{d^{m - 2}}\left( {{\mu_{m,c}}f} \right)}}{{d{y^{m - 2}}}}} \right]_{{q_\alpha }\left( {\tilde{Y}} \right)}^\infty } \right. \\ \left. { + \kappa (m) \cdot \left[ {\frac{1}{f} \cdot \frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}} \cdot \frac{{{d^{m - 3}}\left( {{\mu_{m - 2,c}}f} \right)}}{{d{y^{m - 3}}}}} \right]_{y = {q_\alpha }\left( {\tilde{Y}} \right)}^\infty } \right) \\ = {\left( { - 1} \right)^m} \cdot \frac{1}{{1 - \alpha }} \cdot \left( {\frac{{{d^{m - 2}}\left( {{\mu_{m,c}}f} \right)}}{{d{y^{m - 2}}}}} \right. \\ {\left. {\left. { - \kappa (m) \cdot \left[ {\frac{1}{f} \cdot \frac{{d\left( {{\mu_{2,c}}f} \right)}}{{dy}} \cdot \frac{{{d^{m - 3}}\left( {{\mu_{m - 2,c}}f} \right)}}{{d{y^{m - 3}}}}} \right]} \right)} \right|_{y = {q_\alpha }\left( {\tilde{Y}} \right)}}, \\ \end{array} $$

(4.276)

with $\begin{array} {c} {\left. {\frac{{{d^m}E{S_\alpha }\left( {\tilde{Y} + \lambda \tilde{Z}} \right)}}{{d{\lambda^m}}}} \right|_{\lambda = 0}} = {\left( { - 1} \right)^m} \cdot \frac{1}{{1 - \alpha }} \cdot \left( {\frac{{{d^{m - 2}}\left( {{\mu_m}\left( {\tilde{Z}|\tilde{Y} = y} \right){f_Y}(y)} \right)}}{{d{y^{m - 2}}}}} \right. \\ - \kappa (m) \cdot \left[ {\frac{1}{{{f_Y}(y)}}} \right.{\left. {\left. {\left. { \cdot \frac{{d\left( {{\mu_2}\left( {\tilde{Z}|\tilde{Y} = y} \right){f_Y}(y)} \right)}}{{dy}} \cdot \frac{{{d^{m - 3}}\left( {{\mu_{m - 2}}\left( {\tilde{Z}|\tilde{Y} = y} \right){f_Y}(y)} \right)}}{{d{y^{m - 3}}}}} \right]} \right)} \right|_{y = {q_\alpha }\left( {\tilde{Y}} \right)}}, \\ \end{array}$, and $\kappa (1) = \kappa (2) = 0,\,\,\kappa (3) = 1,\,\,\kappa (4) = 3$. This is the result of Wilde (2003), except that the algebraic signs of Wilde (2003) seem to be wrong.

4.1.15 ES-Based Second-Order Granularity Adjustment for a Normally Distributed Systematic Factor

The summands of the second-order granularity add-on $\kappa (5) = 10$ can be expressed as

$$ \Delta {l_2} $$

(4.277)

Using the derivative of the normal distribution (4.242), the summand $\begin{array} {c} \Delta {l_2} = \frac{1}{{6\left( {1 - \alpha } \right)}}\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}\frac{d}{{dx}}\left( {\frac{{{\eta_{3,c}}\varphi }}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right) \\ + \frac{1}{{8\left( {1 - \alpha } \right)}}\frac{1}{\varphi }\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}{\left. {{{\left[ {\frac{d}{{dx}}\left( {\frac{{{\eta_{2,c}}\varphi }}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right]}^2}} \right|_{x = {\Phi^{ - 1}}(1 - \alpha )}} \\ = :{\left. {\Delta {l_{2,1}} + \Delta {l_{2,2}}} \right|_{x = {\Phi^{ - 1}}(1 - \alpha )}}. \\ \end{array}$ equals

$$ \Delta {l_{2,1}} $$

(4.278)

Using the same transformations, the summand $\begin{array} {c} \Delta {l_{2,1}} = \frac{1}{{6\left( {1 - \alpha } \right)}}\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}\frac{d}{{dx}}\left( {\frac{{{\eta_{3,c}}\varphi }}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right) \\ = \frac{1}{{6\left( {1 - \alpha } \right)}}\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}\left[ {\frac{d}{{dx}}\left( {{\eta_{3,c}}\varphi } \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} + {\eta_{3,c}}\varphi \frac{d}{{dx}}\left( {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right] \\ = \frac{1}{{6\left( {1 - \alpha } \right)}}\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}\left[ {\left( {\frac{{d{\eta_{3,c}}}}{{dx}}\varphi + {\eta_{3,c}}\frac{{d\varphi }}{{dx}}} \right)\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}} - {\eta_{3,c}}\varphi \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}} \right] \\ = \frac{1}{{6\left( {1 - \alpha } \right)}}\frac{\varphi }{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}\left[ {\frac{{d{\eta_{3,c}}}}{{dx}} - {\eta_{3,c}}\left( {x - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right]. \\ \end{array}$ is equivalent to

$$ \Delta {l_{2,2}} $$

(4.279)

leading to a second-order adjustment of

$$ \begin{array} {c} \Delta {l_{2,2}} = \frac{1}{{8\left( {1 - \alpha } \right)}}\frac{1}{\varphi }\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}{\left[ {\frac{d}{{dx}}\left( {\frac{{{\eta_{2,c}}\varphi }}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right]^2} \\ = \frac{1}{{8\left( {1 - \alpha } \right)}}\frac{1}{\varphi }\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}{\left[ {\frac{1}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}\left[ {\frac{{d{\eta_{2,c}}}}{{dx}}\varphi - {\eta_{2,c}}\varphi \left( {x - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right]} \right]^2} \\ = \frac{1}{{8\left( {1 - \alpha } \right)}}\frac{\varphi }{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}}{\left[ {\frac{{d{\eta_{2,c}}}}{{dx}} - {\eta_{2,c}}\left( {x - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right]^2}, \\ \end{array} $$

(4.280)

4.1.16 Probability Density Function of the Logit-Normal Distribution

The derivation of the density function is based on the inverse function theorem¹⁰⁴

$$ \begin{array} {c} \Delta {l_2} = \frac{1}{{6\left( {1 - \alpha } \right)}}\frac{\varphi }{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^2}}}\left[ {\frac{{d{\eta_{3,c}}}}{{dx}} - {\eta_{3,c}}\left( {x - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right] \\ + \frac{1}{{8\left( {1 - \alpha } \right)}}\frac{\varphi }{{{{\left( {{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}} \right)}^3}}}{\left. {{{\left[ {\frac{{d{\eta_{2,c}}}}{{dx}} - {\eta_{2,c}}\left( {x - \frac{{{{{{d^2}{\mu_{1,c}}}} \left/ {{d{x^2}}} \right.}}}{{{{{d{\mu_{1,c}}}} \left/ {{dx}} \right.}}}} \right)} \right]}^2}} \right|_{x = {\Phi^{ - 1}}(1 - \alpha )}}. \\ \end{array} $$

(4.281)

For the logit function ${f_Y}(y) = {f_X}\left( {{g^{ - 1}}(y)} \right) \cdot \left| {\frac{{d{g^{ - 1}}(y)}}{{dy}}} \right|.$, we have

$$ \tilde{Y} = {{{{e^{\tilde{X}}}}} \left/ {{(1 + {e^{\tilde{X}}})}} \right.} $$

(4.282)

and

$$ \begin{array} {c} \\ g(x) = y = \frac{{{e^x}}}{{1 + {e^x}}} = \frac{1}{{{e^{ - x}} + 1}} \\ \Leftrightarrow \\ {e^{ - x}} = \frac{1}{y} - 1 \\ \Leftrightarrow \\ {g^{ - 1}}(y) = x = - \ln \left( {\frac{1}{y} - 1} \right) \\ \end{array} $$

(4.283)

Using the density of a normal distribution (4.82) for $\frac{{d{g^{ - 1}}(y)}}{{dy}} = \frac{d}{{dy}}\left( { - \ln \left( {\frac{1}{y} - 1} \right)} \right) = - \frac{1}{{\frac{1}{y} - 1}} \cdot \left( { - \frac{1}{{{y^2}}}} \right) = \frac{1}{{y\left( {1 - y} \right)}}.$ and recognizing that y is bounded in the interval [0, 1], we get

$$ {f_X} $$

(4.284)