Implied Volatility of Basket Options at Extreme Strikes

Abstract

In the paper, we characterize the asymptotic behavior of the implied volatility of a basket call option at large and small strikes in a variety of settings with increasing generality. First, we obtain an asymptotic formula with an error bound for the left wing of the implied volatility, under the assumption that the dynamics of asset prices are described by the multidimensional Black-Scholes model. Next, we find the leading term of asymptotics of the implied volatility in the case where the asset prices follow the multidimensional Black-Scholes model with time change by an independent increasing stochastic process. Finally, we deal with a general situation in which the dependence between the assets is described by a given copula function. In this setting, we obtain a model-free tail-wing formula that links the implied volatility to a special characteristic of the copula called the weak lower tail dependence function.

We thank the anonymous reviewer for the careful reading of our manuscript and many constructive comments.

Proof of Lemma 1

The function F satisfies

$$ F(t,w) = \max _{\lambda >0} \{\theta t + \lambda w^\perp (\mathbf 1+\mu t) - \frac{\lambda ^2 w^\perp \mathfrak {B}w t}{2}\}, $$

where $\mathbf 1$ stands for the n-dimensional vector with all elements equal to 1. Therefore,

$$ \max _{w\in \Delta _n} F(t,w) = \max _{u \in \mathbb R_+^n } \widetilde{F}(t,u), $$

with

$$ \widetilde{F}(t,u) = \{\theta t + u^\perp (\mathbf 1+\mu t) - \frac{u^\perp \mathfrak {B}u t}{2}\}. $$

Since for every $t>0$, $\widetilde{F}(t,u)$ is strictly concave in u, there exists a unique $\bar{u}(t)\in \mathbb R^n_+$ with $\bar{u}(t)\ne 0$ such that $\widetilde{F}(t,\bar{u}) = \max _{u \in \mathbb R_+^n } \widetilde{F}(t,u)$. This in turn implies that there exists a unique $\bar{w}(t)$ such that $F(t,\bar{w}) = \max _{w\in \Delta _n } F(t,w)$. It is also easy to see that $\bar{u}(t)$ depends continuously on t.

Let $\bar{f}(t) =\widetilde{F}(t,\bar{u}(t))$. We would like to show that $\bar{f}$ is differentiable in t and compute its derivative. $\bar{u}(t)$ may be characterized as follows: for $i=1,\ldots ,n$

$$\begin{aligned}&[\mathbf 1+ \mu t - t\mathfrak {B}\bar{u}(t) ]_i = 0\quad \text {if}\quad \bar{u}(t)_i >0\end{aligned}$$

(64)

$$\begin{aligned}&[\mathbf 1+ \mu t - t\mathfrak {B}\bar{u} (t)]_i \le 0\quad \text {if}\quad \bar{u}(t)_i =0. \end{aligned}$$

(65)

Let I(t) denote the set of indices $i\in \{1,\ldots ,n\}$ such that $\bar{u}(t)_i>0$, and, for a vector $x\in \mathbb R^n$, let $x_{I(t)}$ denote the subset of components of x with indices in I(t): $x_{I(t)} = \{x_i: i\in I(t)\}$. Furthermore, let $\mathfrak {B}_{I(t),I(t)}$ denote the submatrix of the covariance matrix, containing the elements $b_{{ ij}}$ with $i\in I(t)$ and $j\in I(t)$. Then, the vector $\bar{u}(t)$ satisfies

$$ \bar{u}(t)_{I(t)} = \frac{1}{t}\mathfrak {B}_{I(t),I(t)}^{-1} (\mathbf 1+\mu t)_{I(t)},\quad \bar{u}(t)_{\tilde{I}(t)} = 0, $$

where the set $\tilde{I}(t)$ contains the indices $i\in \{1,\ldots ,n\}$ which are not in I(t).

Now, fix $t\in (0,\infty )$ and for $t'\in (0,\infty )$, define

$$ v(t')_{I(t)} = \frac{1}{t'}\mathfrak {B}_{I(t),I(t)}^{-1} (\mathbf 1+\mu t')_{I(t)},\quad v(t)_{\tilde{I}(t)} = 0 $$

First, assume that for all i such that $\bar{u}(t)_i =0$, either $[\mathbf 1+ \mu t - t\mathfrak {B}\bar{u} (t)]_i <0$ (with strict inequality) or

$$ [\mathbf 1+\mu t' - t' \mathfrak {B}v(t')]_i = 0 $$

for all $t'\in (0,\infty )$. We shall call this Assumption 1. Then we can find $\delta >0$, such that for every $t' \in (0,\infty )$ with $|t'-t|<\delta$, $v(t')$ satisfies the characterization (64) and (65). Therefore, $v(t') = \bar{u}(t')$. This means that

$$ \bar{f}(t') = \theta t' + \frac{1}{2t'} (\mathbf 1+\mu t')_{I(t)}^{\perp }\mathfrak {B}_{I(t),I(t)}^{-1} (\mathbf 1+\mu t')_{I(t)}. $$

Therefore, $\bar{f}$ is differentiable at t with first derivative given by

$$\begin{aligned} \bar{f}'(t) = \theta - \frac{1}{2t^2} \mathbf 1_{I(t)}^{\perp }\mathfrak {B}_{I(t),I(t)}^{-1} \mathbf 1_{I(t)} + \frac{1}{2} \mu _{I(t)}^{\perp }\mathfrak {B}_{I(t),I(t)}^{-1} \mu _{I(t)} = \theta - \frac{1}{2t} \bar{u}(t)^\perp (\mathbf 1-\mu t) \end{aligned}$$

(66)

and second derivative

$$ \bar{f}^{\prime \prime }(t) = \frac{1}{t^3} \mathbf 1_{I(t)}^{\perp }\mathfrak {B}_{I(t),I(t)}^{-1} \mathbf 1_{I(t)} . $$

Now assume that there exists at least one i such that $\bar{u}(t)_i = 0$ and $[\mathbf 1+ \mu t - t\mathfrak {B}\bar{u} (t)]_i =0$, or, equivalently,

$$ [\mathbf 1+ \mu t' - t'\mathfrak {B}v (t')]_i =0 $$

with $t'=t$. The case when the above equality holds for all $t'$ is covered by Assumption 1. Since the left-hand side is linear in $t'$, this means that for a given index set I(t) and for a given i, there exists only one $t'\in (0,\infty )$ which satisfies the above equality. Since the number of possible index sets is finite, we conclude that there is at most a finite number of elements $t\in (0,\infty )$ which do not satisfy Assumption 1. But then, we can conclude by continuity that $\bar{f}$ is strictly convex (which entails uniqueness of $\bar{t}$) and differentiable for all $t\in (0,\infty )$, with the derivative given by (66) or alternatively by

$$ \bar{f}'(t) = \theta -\frac{1}{2 t^2 \bar{w}(t)^\perp \mathfrak {B}\bar{w}(t)} + \frac{(\bar{w}(t)^\perp \mu )^2}{2 \bar{w}(t)^\perp \mathfrak {B}\bar{w}(t)}. $$

Comparing this with the derivative of f, which is easily computed, we see that at the point $\bar{t}$, these derivatives coincide. Since this point is characterized by the first order condition $\bar{f}'(\bar{t})=0$, and the function f is strictly convex, f also attains its unique minumum at $\bar{t}$.