Hunting for Black Swans in the European Banking Sector Using Extreme Value Analysis

Abstract

In financial risk management, a Black Swan refers to an event that is deemed improbable yet has massive consequences. In this communication we propose a way to investigate if the recent financial crisis was a Black Swan event for a given bank based on weekly closing prices and derived log-returns. More specifically, using techniques from extreme value methodology we estimate the tail behavior of the negative log-returns over two specific horizons:

• Pre-crisis: from January 1, 1994 until August 7, 2007 (often referred to as the official starting date of the credit crunch crisis);

• Post-crisis: from August 8, 2007 until September 23, 2014 (the cut-off date of our study).

We illustrate this approach with Barclays and Credit Suisse data, and argue that Barclays can be considered as having experienced a Black Swan and Credit Suisse not. We then link the differences in tail risk behavior between these banks with capitalization and leverage indicators. We emphasize the statistical methods for modeling univariate extremes linked with graphical support.

Appendices

Appendix 1: Derivation of the Scale Estimators $\hat{A}_{k,n}$ and $\hat{A}^{EP}_{k,n}$

Starting from the Hall model (7) and ignoring the second order terms yields the approximation

$$\begin{aligned} \bar{F} (x) \sim Ax^{-1/\xi }, \text{ as } x \rightarrow \infty . \end{aligned}$$

(15)

Alternatively, for intermediate order statistics $X_{n-k,n}$, the tail probability $\bar{F} (X_{n-k,n})$ can be estimated by the empirical probability $k/n\approx (k+1)/(n+1)$, leading to the defining equation

$$\begin{aligned} \hat{A}_{k,n} X_{n-k,n}^{-1/H_{k,n}} = \frac{k+1}{n+1}, \end{aligned}$$

where $\xi$ in (15) is estimated by the Hill estimator $H_{k,n}$. This immediately gives (12).

In order to reduce the bias in estimating the scale parameter, $\xi$ first needs to be estimated by the EPD estimator $\hat{\xi }_{k,n}$ to lift up the bias caused by the estimation of $\xi$. The other source of bias originates from ignoring the second order terms when approximating A. Following a similar reasoning as before, now taking the second order terms into account, the defining equation is

$$\begin{aligned} \hat{A}X_{n-k,n}^{-1/\hat{\xi }_{k,n}}\left( 1+ b X_{n-k,n}^{-\beta }(1+o(1)) \right) = \frac{k+1}{n+1}. \end{aligned}$$

Since $\kappa = \kappa _t = \xi b t^{-\beta }(1+o(1))$, we can estimate $b X_{n-k,n}^{-\beta }(1+o(1))$ by $\hat{\kappa }_{k,n}/\hat{\xi }_{k,n}$ with $\hat{\kappa }_{k,n}$ the EPD estimator for $\kappa$ at the threshold $t=X_{n-k,n}$. In order to obtain numerically stable results, we can use that $(1+\kappa _t/\xi )^{-1}\sim 1-\kappa _t/\xi$ since $\kappa _t \rightarrow 0$ as $t \rightarrow \infty$, which leads to the bias reduced scale estimator in (13).

Appendix 2: Proofs for Sect. 3

Proof

(Proof of Theorem 1)

Remark that

$$ \sqrt{k} \left( \log \hat{A}_{k,n} -\log A \right) = T_{k,n}^{(1)}+ T_{k,n}^{(2)}, $$

with

$$\begin{aligned} T_{k,n}^{(1)}= & {} \sqrt{k} \left( \frac{ \log X_{n-k,n}}{H_{k,n} }- \frac{\log U(n/k)}{\xi }\right) \\ T_{k,n}^{(2)}= & {} \sqrt{k} \left( \frac{\log U(n/k)}{\xi } + \log \left( \frac{k+1}{n+1}\right) - \log A \right) . \end{aligned}$$

First, as $U(x) = A^{\xi }x^{\xi } (1+ \xi b A^{-\xi \beta }x^{-\xi \beta }(1+o(1))$ when $x \rightarrow \infty$,

$$ T_{k,n}^{(2)} = -\sqrt{k}B(n/k)\frac{1}{\xi \beta }(1+o(n/k)), $$

as $n/k\rightarrow \infty$. Next, with $\tilde{H}_{k,n} := \frac{1}{k}\sum _{j=1}^k (\log X_{n-j+1,n}- \log U(n/k))$ and $\mathbb {E}(\tilde{H}_{k,n}) = \xi + B(n/k)/(1+\xi \beta )$, see Hsing [15], we get

$$\begin{aligned} T_{k,n}^{(1)}= & {} -\frac{\log U(n/k)}{H_{k,n} \xi }\sqrt{k} \left( H_{k,n}- \xi \right) + \frac{\sqrt{k}}{H_{k,n}} (\log X_{n-k,n}-\log U(n/k)) \\= & {} -\frac{\log U(n/k)}{H_{k,n} \xi }\sqrt{k} \left( \tilde{H}_{k,n}- \mathbb {E}(\tilde{H}_{k,n}) \right) \\&+ \,\frac{1}{H_{k,n}} \left( \frac{\log U(n/k)}{\xi } +1\right) \sqrt{k} (\log X_{n-k,n}-\log U(n/k)) \\&- \,\frac{\log U(n/k)}{H_{k,n} \xi }\frac{\sqrt{k} B(n/k)}{1+\xi \beta }. \end{aligned}$$

Hence,

$$\begin{aligned} \sqrt{k} \left( \log \hat{A}_{k,n} -\log A \right)= & {} -\frac{\log U(n/k)}{H_{k,n} \xi }\sqrt{k} \left( \tilde{H}_{k,n}- \mathbb {E}(\tilde{H}_{k,n}) \right) \\&+\, \frac{1}{H_{k,n}} \left( \frac{\log U(n/k)}{\xi } +1\right) \sqrt{k} (\log X_{n-k,n}-\log U(n/k)) \\&- \,\frac{1}{\xi }\left( \frac{\log U(n/k)}{H_{k,n} (1+\beta \xi )}+ \frac{1}{\beta } \right) \sqrt{k} B(n/k) . \end{aligned}$$

Using the fact that $\log U(n/k)/ \log (n/k) \rightarrow \xi$ as $n/k \rightarrow \infty$, the result now follows from Lemma 2.1 and Corollary 3.4 in Hsing [15]. $\square$

Proof

(Proof of Theorem 2)

Using the approach from Beirlant et al. [4], we have with $\kappa _n = \kappa (X_{n-k,n})$

$$\begin{aligned} k^{-1/2}Z_{k,n}:= & {} \frac{n}{k}\bar{F}(X_{n-k,n})-1 \\= & {} \frac{ A }{(k/n)X_{n-k,n}^{1/\xi }}\left( 1+ \frac{\kappa _n}{\xi } (1+o_p(1))\right) -1 \\= & {} \frac{A}{\hat{A}_{k,n}^{EP}} X_{n-k,n}^{1/\hat{\xi }_{k,n}-1/\xi } \left( 1+ \frac{\kappa _n}{\xi } (1+o_p(1))\right) \left( 1- \frac{\hat{\kappa }_{k,n}}{\hat{\xi }_{k,n}}\right) -1 \\= & {} \frac{A}{\hat{A}_{k,n}^{EP}}\left( 1-\frac{1}{\xi \hat{\xi }_{k,n}} (\hat{\xi }_{k,n}-\xi )\log X_{n-k,n} (1+o_p(1))\right) \\&\qquad \qquad \quad \times \left( 1+ \{\frac{\kappa _n}{\xi }-\frac{\hat{\kappa }_{k,n}}{\hat{\xi }_{k,n}} \} (1+o_p(1))\right) -1, \end{aligned}$$

from which it follows, using $\hat{\kappa }_{k,n}-\kappa _n = O_p(k^{-1/2})$ and $\hat{\xi }_{k,n}-\xi = O_p(k^{-1/2})$ from Theorem 3.1 in Beirlant et al. [4],

$$\begin{aligned}&\!\!\!\! \frac{A}{\hat{A}_{k,n}^{EP}}-1 \\= & {} \frac{k^{-1/2}Z_{k,n}+1}{\left( 1- \frac{\hat{\xi }_{k,n}-\xi }{\xi \hat{\xi }_{k,n}}\log X_{n-k,n} (1+o_p(1))\right) \left( 1+ \{\frac{\kappa _n}{\xi }-\frac{\hat{\kappa }_{k,n}}{\hat{\xi }_{k,n}} \} (1+o_p(1))\right) }-1 \\= & {} (k^{-1/2}Z_{k,n}+1) \left( 1+\frac{1}{\xi \hat{\xi }_{k,n}} (\hat{\xi }_{k,n}-\xi )\log X_{n-k,n} (1+o_p(1)) \right) \\&\qquad \qquad \quad \times \left( 1- \{\frac{\kappa _n}{\xi }-\frac{\hat{\kappa }_{k,n}}{\hat{\xi }_{k,n}} \} (1+o_p(1)) \right) -1. \end{aligned}$$

This implies that $\sqrt{k}(A/\hat{A}^{EP}_{k,n}-1)$ has the same limit distribution as

$$ \frac{\log U(n/k)}{\xi ^2} \sqrt{k}(\hat{\xi }_{k,n}-\xi ) - \xi ^{-1}\sqrt{k}(\hat{\kappa }_{k,n}-\kappa _n)+Z_{k,n}. $$

From Theorem 3.1 in Beirlant et al. [4] it follows that this stochastic sum is asymptotically unbiased when $\sqrt{k} B(n/k) \rightarrow \lambda$, while the asymptotic variance follows from the variance of $\hat{\xi }_{k,n}$ which has the asymptotic dominating coefficient $\log U(n/k)/\xi ^2$ in this asymptotic representation. $\square$

Appendix 3: The Dependence Between Tests on Scale and Shape

We now derive the asymptotic covariance matrix of

$$\begin{aligned} \left( \frac{\xi \sqrt{k}}{\log U\left( \frac{n}{k}\right) }(\log \hat{A}_{k,n}-\log A), \sqrt{k} \left( \frac{H_{k,n}}{\xi }-1\right) \right) . \end{aligned}$$

From

$$\begin{aligned} \frac{\log X_{n-k,n}}{H_{k,n}}-\frac{\log U\left( \frac{n}{k}\right) }{\xi } = \frac{1}{\xi } \left( \log X_{n-k,n}-\log U\left( \frac{n}{k}\right) \right) - \frac{ \log X_{n-k,n}}{\xi H_{k,n}} (H_{k,n}-\xi ), \end{aligned}$$

we have using the notation from the proof of Theorem 1 that

$$\begin{aligned} \frac{\xi T_{k,n}^{(1)} }{\log U\left( \frac{n}{k}\right) }&= \sqrt{k}\left( \frac{\log X_{n-k,n} - \log U\left( \frac{n}{k}\right) }{\log U\left( \frac{n}{k}\right) }\right) -\sqrt{k} \left( \frac{H_{k,n}-\xi }{H_{k,n}}\right) \frac{\log X_{n-k,n}}{\log U(n/k)}\\&\sim _p \sqrt{k}\left( \frac{\log X_{n-k,n} - \log U\left( \frac{n}{k}\right) }{\log U\left( \frac{n}{k}\right) }\right) -\sqrt{k} \frac{H_{k,n}-\xi }{\xi }. \end{aligned}$$

We hence have concerning the asymptotic covariance

$$\begin{aligned} Acov\left( \frac{\xi T_{k,n}^{(1)} }{\log U\left( \frac{n}{k}\right) },\sqrt{k} \frac{H_{k,n}-\xi }{\xi }\right)&= Acov\left( \sqrt{k}\frac{\log X_{n-k,n} - \log U\left( \frac{n}{k}\right) }{\log U\left( \frac{n}{k}\right) } ,\sqrt{k} \frac{H_{k,n}-\xi }{\xi }\right) \\&\quad - Avar\left( \sqrt{k} \frac{H_{k,n}-\xi }{\xi }\right) . \end{aligned}$$

From (11) we know that the asymptotic variance in this expression is asymptotically equal to $1+\chi +\omega -2\psi$.

$$\begin{aligned} Acov\left( \frac{\xi T_{k,n}^{(1)} }{\log U\left( \frac{n}{k}\right) },\sqrt{k} \frac{H_{k,n}-\xi }{\xi }\right)&= \frac{k}{\xi \log U\left( \frac{n}{k}\right) }Acov\left( \log X_{n-k,n} - \log U\left( \frac{n}{k}\right) , H_{k,n}- \xi \right) \\&\quad - (1+\chi +\omega -2\psi ). \end{aligned}$$

Following Hsing [15], approximating $H_{k,n}$ by $H^+_{k,n} - \left( \log X_{n-k,n} - \log U\left( \frac{n}{k}\right) \right)$ with $H^+_{k,n}=\frac{1}{k} \sum _{j=1}^k \max \{\log X_{n-j+1,n} - \log U\left( \frac{n}{k}\right) ,0\}$, we find

$$\begin{aligned} k\,Acov\left( \log X_{n-k,n} - \log U\left( \frac{n}{k}\right) , H_{k,n}- \xi \right)&\approx k\, Acov\left( \log X_{n-k,n} - \log U\left( \frac{n}{k}\right) ,H^+_{k,n}-\xi \right) \\&- k\, Avar\left( \log X_{n-k,n} - \log U\left( \frac{n}{k}\right) \right) . \end{aligned}$$

From Corollary 3.4 in Hsing [15] it then follows that

$$\begin{aligned} k\, Acov\left( \log X_{n-k,n} - \log U\left( \frac{n}{k}\right) , H_{k,n}^+- \xi \right)&= \xi ^2 (1+\psi ),\\ k\, Avar\left( \log X_{n-k,n} - \log U\left( \frac{n}{k}\right) \right)&= \xi ^2(1+\omega ), \end{aligned}$$

which results in

$$\begin{aligned} Acov\left( \frac{\xi T_{k,n}^{(1)} }{\log U\left( \frac{n}{k}\right) },\sqrt{k} \frac{H_{k,n}-\xi }{\xi }\right)&= \frac{\xi }{\log U\left( \frac{n}{k}\right) }(\psi -\omega ) -(1+\chi +\omega -2\psi ). \end{aligned}$$

Since $T_{k,n}^{(2)}$ is deterministic it does not play a role in the calculation of the covariance matrix. We then get

$$\begin{aligned} Acov\left( \frac{\xi \sqrt{k}}{\log U\left( \frac{n}{k}\right) }(\log \hat{A}_{k,n}-\log A),\sqrt{k} \frac{H_{k,n}-\xi }{\xi } \right) = -(1+\chi +\omega -2\psi ) + \frac{\xi }{\log U\left( \frac{n}{k}\right) }(\psi -\omega ). \end{aligned}$$

Using the obtained expression for the asymptotic variance of both components (see Theorem 1 and (11)) and the fact that $\log U\left( \frac{n}{k}\right) / \log (n/k) \rightarrow \xi$ as $n/k \rightarrow \infty$ gives the asymptotic covariance matrix of

$\left( \frac{\xi \sqrt{k}}{\log U\left( \frac{n}{k}\right) }(\log \hat{A}_{k,n}-\log A),\sqrt{k} \left( \frac{H_{k,n}}{\xi }-1 \right) \right)$:

$$\begin{aligned}&\begin{pmatrix} 1+\chi +\omega -2\psi &{} -(1+\chi +\omega -2\psi ) + \frac{\psi -\omega }{\log \left( \frac{n}{k}\right) }\\ -(1+\chi +\omega -2\psi ) + \frac{\psi -\omega }{\log \left( \frac{n}{k}\right) } &{} 1+\chi +\omega -2\psi \end{pmatrix} \nonumber \\= & {} (1+\chi + \omega -\psi ) \begin{pmatrix} {-}1 &{} -1 \\ -1 &{} {-}1 \end{pmatrix} + \frac{\psi -\omega }{\log \left( \frac{n}{k}\right) } \begin{pmatrix} 0 &{} 1 \\ 1 &{} 0 \end{pmatrix}. \end{aligned}$$

(16)

Appendix 4: 3D Plots of p-values for Tests

Figs. 11 and 12

Fig. 11

p-values for testing differences in shape using $T^{(\xi )}_{k_1,k_2,n_1,n_2}$ for all possible choices of $k_1$ and $k_2$ for pre- and post-crisis negative log-returns for Barclays and Credit Suisse

Fig. 12

p-values for testing differences in scale using $T^{(A)}_{k_1,k_2,n_1,n_2}$ for all possible choices of $k_1$ and $k_2$ for pre- and post-crisis negative log-returns for Barclays and Credit Suisse