Efficiently pricing double barrier derivatives in stochastic volatility models

Abstract

Imposing a symmetry condition on returns, Carr and Lee (Math Financ 19(4):523–560, 2009) show that (double) barrier derivatives can be replicated by a portfolio of European options and can thus be priced using fast Fourier techniques (FFT). We show that prices of barrier derivatives in stochastic volatility models can alternatively be represented by rapidly converging series, putting forward an idea by Hieber and Scherer (Stat Probab Lett 82(1):165–172, 2012). This representation turns out to be faster and more accurate than FFT. Numerical examples and a toolbox of a large variety of stochastic volatility models illustrate the practical relevance of the results.

Appendices

Appendix 1: Parameters of the Stein–Stein model

The functions $L(u)$, $M(u)$, and $N(u)$ in the characteristic function are defined as

$$\begin{aligned} A&:= -\frac{\xi }{k^2},\quad B := \frac{\varkappa \xi }{k^2},\quad C_u := -\frac{u}{k^2T},\quad a_u :=\sqrt{A^2-2C_u},\quad b_u = -\frac{A}{a_u}, \\ L(u)&:= -A - a_u\left( \frac{\sin \,\mathrm{h}(a_uk^2T) + b_u\cos \,\mathrm{h}(a_uk^2T)}{\cos \,\mathrm{h}(a_uk^2T)+b_u\sin \,\mathrm{h}(a_uk^2T)}\right) ,\\ M(u)&:= B\left( \frac{b_u\sin \,\mathrm{h}(a_uk^2T) + b_u^2\,\cos \,\mathrm{h}(a_uk^2T) + 1 - b_u^2}{\cos \,\mathrm{h}(a_uk^2T)+b_u\,\sin \,\mathrm{h}(a_uk^2T)}-1\right) ,\\ N(u)&:= \frac{a_u-A}{2a_u^2} \left( a_u^2-AB^2-B^2a_u\right) k^2T\\&\quad + \frac{B^2(A^2-a_u^2)}{2a_u^3}\left( \frac{(2A+a_u)+(2A-a_u)e^{2a_uk^2T}}{A+a_u+(a_u-A)e^{2a_uk^2T}}\right) \\&\quad +\frac{2AB^2(a_u^2-A^2)e^{a_uk^2T}}{a_u^3\big ( A+a_u+(a_u-A)e^{2a_uk^2T}\big )}\\&\quad - \frac{1}{2}\ln \left( \frac{1}{2}\left( \frac{A}{a_u}+1\right) + \frac{1}{2}\left( 1-\frac{A}{a_u}\right) e^{2a_uk^2T}\right) . \end{aligned}$$

Appendix 2: Single barrier limit

In our series representation the limit ${a:=\ln (P)\rightarrow \infty }$ cannot be exchanged with the infinite summation over $n$ as the series representation for $X_{D,\infty }^{g(S_T)}(S_0)$ is not absolutely convergent. To derive the limiting expression, one has to change the series representation. Then, the limiting option price $X_{D,\infty }^{g(S_T)}(S_0)$ is given by Theorem 1. For $a:=\ln (P)$, $b:=\ln (D)$, and $x:=\ln (S_0)$, we obtain

$$\begin{aligned} X_{D,\infty }^{g(S_T)}(S_0)&= \lim _{a\rightarrow \infty } \, \frac{1}{B_T}\frac{2e^{\frac{x}{2}}}{a-b} \sum \limits _{n=1}^\infty \vartheta _T^c\left( \frac{1}{8}+\frac{n^2\pi ^2}{2(a-b)^2}\right) \,\sin \left( \frac{n\pi (x-b)}{a-b}\right) Z_n^{g(S_T)}\\&= \lim _{a\rightarrow \infty } \,\mathbb {E}\left[ \frac{1}{B_T}\frac{2e^{\frac{x}{2}}}{a-b} \sum \limits _{n=1}^\infty \exp \left( -\frac{\varLambda _T}{8}\!-\!\frac{n^2\pi ^2\varLambda _T}{2(a\!-\!b)^2}\right) \,\sin \left( \frac{n\pi (x\!-\!b)}{a\!-\!b}\right) Z_n^{g(S_T)} \right] \\&= \lim _{a\rightarrow \infty } \,\mathbb {E}\Bigg [ \frac{1}{B_T}\frac{2}{a-b}\int \limits _b^a \exp \left( -\frac{x-y}{2}-\frac{\varLambda _T}{8}\right) \,g(e^y) \\&\quad \times \sum \limits _{n=1}^\infty \exp \left( -\frac{n^2\pi ^2\varLambda _T}{2(a-b)^2}\right) \,\sin \left( \frac{n\pi (x-b)}{a-b}\right) \,\sin \left( \frac{n\pi (y-b)}{a-b}\right) \,dy \Bigg ]\,. \end{aligned}$$

If we change the parameterization [see He et al. 1998, Equations (2.3) and (2.4)], we get

$$\begin{aligned}&= \lim _{a\rightarrow \infty } \,\mathbb {E}\Bigg [ \frac{1}{B_T}\frac{2}{a-b}\int \limits _b^a \frac{a-b}{2} \exp \left( -\frac{x-y}{2}-\frac{\varLambda _T}{8}\right) \,g(e^y) \\&\quad \times \sum \limits _{n=-\infty }^\infty \Bigg ( \varphi \left( \frac{y-x-2n(a-b)}{\sqrt{\varLambda _T}}\right) - \varphi \left( \frac{y+x-2na + (2n-2)b}{\sqrt{\varLambda _T}}\right) \Bigg )\,dy \Bigg ]\,. \end{aligned}$$

This series is absolutely convergent, thus we can change limit and summation. In the limit $a\rightarrow \infty$ only the “$n=0$” term remains, i.e.

$$\begin{aligned}&= \mathbb {E}\Bigg [ \frac{1}{B_T}\int \limits _b^\infty \exp \left( -\frac{x-y}{2}-\frac{\varLambda _T}{8}\right) \,g(e^y)\Bigg ( \varphi \left( \frac{y-x}{\sqrt{\varLambda _T}}\right) - \varphi \left( \frac{y+x -2b}{\sqrt{\varLambda _T}}\right) \Bigg )\,dy\Bigg ]\\&= \mathbb {E}\Bigg [ \frac{1}{B_T}\int \limits _b^\infty g(e^y)\Bigg ( \varphi \left( \frac{y-x+\varLambda _T/2}{\sqrt{\varLambda _T}}\right) \\&\quad - \exp \big (-(b-x)\big )\varphi \left( \frac{y+x -2b+\varLambda _T/2}{\sqrt{\varLambda _T}}\right) \Bigg )\,dy\Bigg ]\\&= \frac{1}{B_T}\Bigg (\mathbb {E}_{\mathbb {Q},S_0}\Big [1\!\!1_{\{S_T>D_T\}}\, g(S_T)\Big ]- \frac{S_0}{D}\,\mathbb {E}_{\mathbb {Q},D^2/S_0}\Big [1\!\!1_{\{S_T>D_T\}}\,g\left( S_T\right) \Big ]\Bigg )\,. \end{aligned}$$