Extreme at-the-money skew in a local volatility model

Abstract

We consider a local volatility model, with volatility taking two possible values, depending on the value of the underlying with respect to a fixed threshold. When the threshold is taken at the money, we establish exact pricing formulas for European call options and compute short-time asymptotics of the implied volatility surface. We derive an exact formula for the at-the-money implied volatility skew which explodes as $T^{-1/2}$, reproducing the empirical steep short end of the smile. This behaviour is a consequence of the singularity of the local volatility at the money. Finally, we look at continuous, non-differentiable versions of such a model. We still find, in simulations, exploding implied skews.

Appendix: Auxiliary computations

Lemma A.1

Recall$G$in (2.5) and$Z$in (3.8). For all$b,c \in \mathbb{R}$, $t>0$, we have the limit

$$ \lim _{a\downarrow 0}\partial _{a} G(t,a,b,c)=-\frac{c+b}{2\sqrt{bc}}-t \sqrt{bc}+Z\big(t,\max (b,c),\min (b,c)\big). $$

Proof

From the definition of $G$, we get

$$\begin{aligned} &\partial _{a} G(t,a,b,c) \\ &=- \frac{1}{\pi }\bigg( \int _{0}^{\infty }\frac{ \sin (a\sqrt{ z}) \sqrt{z} }{\sqrt{z+b-c}} \frac{(b+z)tc+(\exp (-(b+z)t)-1)(c-b-z)}{(b+z)^{2}} dz \\ & \phantom{=:-\frac{1}{\pi }\bigg(}{}+\int _{0}^{b-c} \frac{\exp (-a \sqrt{b-c-z}) \sqrt{b-c-z}}{\sqrt{z}} \\ & \phantom{=:-\frac{1}{\pi }\bigg(+\int _{0}^{b-c}}{} \times \frac{(c+z)tc-z(\exp (-(c+z)t)-1)}{(c+z)^{2}} dz \bigg). \end{aligned}$$

(A.1)

Since $\lim _{a\downarrow 0}\sin (a\sqrt{z})=0$, the integrable factors multiplying the $\sin (\cdot )$ function give a vanishing contribution in the limit due to dominated convergence. So we get

$$\begin{aligned} &\lim _{a\downarrow 0}\int _{0}^{\infty }\sin (a\sqrt{ z})\frac{ \sqrt{z} }{\sqrt{z+b-c}} \frac{(b+z)tc+(\exp (-(b+z)t)-1)(c-b-z)}{(b+z)^{2}} dz \\ &=\lim _{a\downarrow 0}\int _{0}^{\infty }\sin (a\sqrt{ z})\frac{ \sqrt{z} }{\sqrt{z+b-c}} \frac{(b+z)tc+b+z}{(b+z)^{2}} dz \\ &= \lim _{a\downarrow 0} \int _{0}^{\infty }\sin (a\sqrt{ z}) \frac{tc+1}{b+z} dz \\ &=\pi (tc+1). \end{aligned}$$

(A.2)

In case $b=c$, the second integral in (A.1) vanishes and

$$ \lim _{a\downarrow 0} \partial _{a} G(t,a,b,c)=- tb-1=-tc-1. $$

(A.3)

We suppose now that $b> c$ and compute the limit for $a\downarrow 0$ of the second integral. We have

$$ \begin{aligned} &\lim _{a\downarrow 0} \int _{0}^{b-c}\exp (-a\sqrt{b-c-z}) \frac{ \sqrt{b-c-z}}{\sqrt{z}} \frac{(c+z)tc-z(\exp (-(c+z)t)-1)}{(c+z)^{2}} dz \\ &=t\int _{0}^{b-c} \frac{\sqrt{b-c-z}}{\sqrt{z}}+ \int _{0}^{b-c} \sqrt{(b-c-z)z} \frac{1-\exp (-(c+z)t)-(c+z)t}{(c+z)^{2}} dz. \end{aligned} $$

With the change of variables $c+z=u$, we have

$$ \int _{0}^{b-c} \frac{\sqrt{b-c-z}}{\sqrt{z}} dz = \int _{c}^{b}\sqrt{ \frac{b-u}{u-c}} du = \frac{t\pi (b-c)}{2} $$

and

$$\begin{aligned} &\int _{0}^{b-c} \sqrt{(b-c-z)z} \frac{1-\exp (-(c+z)t)-(c+z)t}{(c+z)^{2}} dz \\ &= \int _{c}^{b} \sqrt{\Big(\frac{b}{u}-1\Big)\Big(1-\frac{c}{u} \Big)} \frac{1-\exp (-ut)-ut}{u} du. \end{aligned}$$

Now,

$$ t\int _{c}^{b}\sqrt{\Big(\frac{b}{u}-1\Big)\Big(1-\frac{c}{u}\Big)}du =\pi \bigg(\frac{c+b}{2}-\sqrt{bc}\bigg)t $$

and

$$ \int _{c}^{b}\sqrt{\Big(\frac{b}{u}-1\Big)\Big(1-\frac{c}{u}\Big)} \frac{du}{u} =\pi \bigg(\frac{c+b}{2\sqrt{bc}}-1\bigg). $$

(A.4)

Therefore,

$$\begin{aligned} &\int _{c}^{b} \sqrt{\Big(\frac{b}{u}-1\Big)\Big(1-\frac{c}{u}\Big)} \frac{1-\exp (-ut)-ut}{u} du \\ &= \pi \bigg(\frac{c+b}{2\sqrt{bc}}-1-\Big(\frac{c+b}{2}- \sqrt{bc}\Big)t-Z(t,b,c)\bigg). \end{aligned}$$

Recalling also (A.3), we have

$$ \lim _{a\downarrow 0}\partial _{a} G(t,a,b,c)=-\frac{c+b}{2\sqrt{bc}}-t \sqrt{bc}+Z(t,b,c) . $$

Suppose now $c>b$. In this case, it is not clear from (2.5) that $G$ is a real function. For this choice of parameters, $G$ can be rewritten with standard manipulations as

$$\begin{aligned} &G(t,a,b,c) \\ &= \frac{1}{\pi }\bigg( \int _{c-b}^{\infty }\frac{\cos (a\sqrt{z}) }{\sqrt{z+b-c}} \frac{(b+z)tc+(\exp (-(b+z)t)-1)(c-b-z)}{(b+z)^{2}} dz \\ & \phantom{=:\frac{1}{\pi }\bigg(}{}+\int _{0}^{c-b} \frac{\sin (a \sqrt{z}) }{\sqrt{-z-b+c}} \frac{(b+z)tc+(\exp (-(b+z)t)-1)(c-b-z)}{(b+z)^{2}} dz \bigg). \end{aligned}$$

(A.5)

In this form, one can see that $G$ is a real function. Completely analogous computations can be made starting from (A.5). One can show that

$$\begin{aligned} &\partial _{a} G(t,a,b,c) \\ &= \frac{1}{\pi }\bigg( -\int _{c-b}^{\infty }\frac{ \sin (a \sqrt{ z})\sqrt{z} }{\sqrt{z+b-c}} \frac{(b+z)tc+(\exp (-(b+z)t)-1)(c-b-z)}{(b+z)^{2}} dz \\ & \phantom{=:\frac{1}{\pi }\bigg(}{}+\int _{0}^{c-b} \frac{\cos (a \sqrt{z})\sqrt{z}}{\sqrt{c-b-z}} \frac{(b+z)tc+(c-b-z)(\exp (-(b+z)t)-1)}{(b+z)^{2}} dz \bigg) \end{aligned}$$

(A.6)

and

$$\begin{aligned} &\lim _{a\downarrow 0} \int _{0}^{c-b}\cos (a\sqrt{z}) \frac{ \sqrt{z}}{\sqrt{c-b-z}} \frac{(b+z)tc+(c-b-z)(\exp (-(b+z)t)-1)}{(b+z)^{2}} dz \\ & =\pi \left (tc+1-\sqrt{bc}-\frac{b+c}{2\sqrt{cb}} +Z(t,c,b)\right ). \end{aligned}$$

(A.7)

From (A.2), (A.6) and (A.7) (it clearly does not matter if the lower integration bound in the first integral is 0 or $c-b$), we get

$$ \lim _{a\downarrow 0}\partial _{a} G(t,a,b,c)=-\frac{c+b}{2\sqrt{bc}}-t \sqrt{bc}+Z(t,c,b). $$

Therefore, the statement holds for all $b,c,t$. □