Approximate option pricing and hedging in the CEV model via path-wise comparison of stochastic processes

Abstract

This paper presents a methodology of finding explicit boundaries for some financial quantities via comparison of stochastic processes. The path-wise comparison theorem is used to establish domination of the stock price process by a process with a known distribution that is relatively simple. We demonstrate how the comparison theorem can be applied in the constant elasticity of variance model to derive closed-form expressions for option price bounds, an approximate hedging strategy and a conditional value-at-risk estimate. We also provide numerical examples and compare precision of our method with the distribution-free approach.

Appendix

In order to prove Proposition 1, we shall use the fact that the conditional density of a CEV stock price may be expressed in terms of power series (see e.g. Randal 1998).

Denote by $f_{\tau }(s)$ the conditional density of $S_{t+\tau }$ given $S_{t}$:

$$\begin{aligned} f_{\tau }(s~|~S_{t})=\kappa {\mathrm {e}}^{-x-z}\sum \limits _{n=0}^{\infty } \frac{x^{n+1}z^{n}}{n!(n+1)!}, \end{aligned}$$

where

$$\begin{aligned} x=\kappa S_{t}{\mathrm {e}}^{r\tau },\quad z=\kappa s,\quad \kappa =\frac{2r}{ \sigma ^{2}\left( {\mathrm {e}}^{r\tau }-1\right) }. \end{aligned}$$

Note that

$$\begin{aligned} f_{\tau }(s~|~S_{t})ds={\mathrm {e}}^{-x-z}\sum \limits _{n=0}^{\infty }\frac{ x^{n+1}z^{n}}{n!(n+1)!}dz, \end{aligned}$$

therefore

$$\begin{aligned} {\mathbb {E}}\left( S_{t+\tau }^{m}~|~S_{t}\right)= & {} \int \limits _{0}^{\infty }\left( \frac{z}{\kappa }\right) ^{m}{\mathrm {e}}^{-x-z}\sum \limits _{n=0}^{\infty } \frac{x^{n+1}z^{n}}{n!(n+1)!}dz \\= & {} \left( \frac{x}{\kappa }\right) \kappa ^{1-m}\sum \limits _{n=0}^{\infty } \frac{{\mathrm {e}}^{-x}x^{n}}{n!}\int \limits _{0}^{\infty }\frac{{\mathrm {e}} ^{-z}z^{n+m}}{(n+1)!}dz \\= & {} S_{t}{\mathrm {e}}^{r\tau }\kappa ^{1-m}\sum \limits _{n=0}^{\infty }\frac{ {\mathrm {e}}^{-x}x^{n}}{n!}\frac{(n+m)!}{(n+1)!} \\= & {} S_{t}{\mathrm {e}}^{r\tau }\kappa ^{1-m}\sum \limits _{n=0}^{\infty }g_{x}(n)p_{m-1}(n), \end{aligned}$$

where $g_{x}(n)$ is a probability mass function of a Poisson random variable with mean x.

If we choose such $\alpha _{0},\alpha _{1},\dots \alpha _{m-1}$ that

$$\begin{aligned} p_{m-1}(n)=\sum \limits _{i=0}^{m-1}\alpha _{i}\prod \limits _{j=1}^{i}(n-j+1) \end{aligned}$$

and notice that

$$\begin{aligned} \prod \limits _{j=1}^{i}(n-j+1)=\left\{ \begin{array}{ll} 0 &{}\quad \hbox { if }n=0,1,\dots i-1,\\ \dfrac{n!}{(n-i)!} &{}\quad \hbox { if }n=i,i+1,\dots , \end{array} \right. \end{aligned}$$

we can rewrite the expectation as

$$\begin{aligned} {\mathbb {E}}\left( S_{t+\tau }^{m}~|~S_{t}\right)= & {} S_{t}{\mathrm {e}}^{r\tau }\kappa ^{1-m}\sum \limits _{i=0}^{m-1}\sum \limits _{n=i}^{\infty }\alpha _{i}\frac{ {\mathrm {e}}^{-x}x^{n-i}}{(n-i)!}x^{i} \\= & {} S_{t}{\mathrm {e}}^{r\tau }\kappa ^{1-m}\sum \limits _{i=0}^{m-1}\alpha _{i}x^{i}\sum \limits _{n=i}^{\infty }g_{x}(n-i) \\= & {} S_{t}{\mathrm {e}}^{r\tau }\kappa ^{1-m}\sum \limits _{i=0}^{m-1}\alpha _{i}\left( \kappa S_{t}{\mathrm {e}}^{r\tau }\right) ^{i} \\= & {} \sum \limits _{i=0}^{m-1}\alpha _{i}S_{t}^{i+1}{\mathrm {e}}^{(i+1)r\tau }\kappa ^{i+1-m}, \end{aligned}$$

which concludes the proof. $\square$