An analytical approximation for single barrier options under stochastic volatility models

Abstract

The aim of this paper is to derive an approximation formula for a single barrier option under local volatility models, stochastic volatility models, and their hybrids, which are widely used in practice. The basic idea of our approximation is to mimic a target underlying asset process by a polynomial of the Wiener process. We then translate the problem of solving first hit probability of the asset process into that of a Wiener process whose distribution of passage time is known. Finally, utilizing the Girsanov’s theorem and the reflection principle, we show that single barrier option prices can be approximated in a closed-form. Furthermore, ample numerical examples will show the accuracy of our approximation is high enough for practical applications.

Appendices

Appendix A: Proof of Proposition 3.1

In order to calculate the probability distribution of $X_t$, we derive an approximated characteristic function of $X_t$ and invert it back to derive an approximation of the probability distribution of $X_t$.

Let the characteristic function of $X_t$ be $\Psi (\xi ) = \mathbb {E}[ \mathrm{e}^{i \xi X_t} ]$. We approximate it as

$$\begin{aligned} \Psi (\xi ) \approx \left\{ \begin{array}{lll} \displaystyle 1 + i \xi \mathbb {E}[ a_{2}(t) ],&{} \quad \ if \ a_1(t) = 0, \ a.s., \\ \mathbb {E}[ \mathrm{e}^{ i \xi a_{1}(t)} ] + i \xi \mathbb {E}\left[ \mathrm{e}^{ i \xi a_{1}(t)} \mathbb {E}[ a_{2}(t)\, |\, a_{1}(t) ] \right] ,&{} \quad otherwise. \end{array} \right. \end{aligned}$$

where $R_3$ consists of the forth or higher-order multiple stochastic integrals. Note that, since

$$\begin{aligned} \left| \mathbb {E}\left[ \mathrm{e}^{ i \xi a_{1}(t) } R_3 \right] \right|\le & {} \mathbb {E}\left[ |\mathrm{e}^{ i \xi a_{1}(t) } R_3 |\right] \\\le & {} \left( \mathbb {E}\left[ |\mathrm{e}^{ i \xi a_{1}(t) }|^2\right] \right) ^{\frac{1}{2}} \left( \mathbb {E}\left[ |R_3 |^2\right] \right) ^{\frac{1}{2}} \\= & {} \left( \mathbb {E}\left[ |R_3 |^2\right] \right) ^{\frac{1}{2}} \ \approx \ 0 , \end{aligned}$$

we regard $\mathbb {E}\left[ \mathrm{e}^{ i \xi a_{1}(t) } R_3 \right] \approx 0$ as for the previous case.

Taking the conditional expectation on $a_1(t)$, which follows normal distribution with 0 mean and variance $\Sigma _t$, we then have

$$\begin{aligned} \Psi (\xi ) \approx \left\{ \begin{array}{ll} \displaystyle 1, &{}\quad \Sigma _t = 0, \\ \mathbb {E}[ \mathrm{e}^{ i \xi a_{1}(t)} ] + i \xi \mathbb {E}\left[ \mathrm{e}^{ i \xi a_{1}(t)} \mathbb {E}[ a_{2}(t)\, |\, a_{1}(t) ] \right] , &{}\quad otherwise . \end{array} \right. \end{aligned}$$

Further, if X follows a normal distribution with zero mean and variance $\Sigma$, then by differentiating both sides of

$$\begin{aligned} \frac{1}{2 \pi } \int _{\mathcal {R}} \mathrm{e}^{-i k y} \mathbb {E}\left[ h(X) \mathrm{e}^{ikX} \right] \mathrm{d}k = h(y) n(y;0,\Sigma ) \end{aligned}$$

w.r.t. y, we have for any polynomial functions f(x) and g(x)

$$\begin{aligned} \frac{1}{2 \pi } \int _{\mathcal {R}} \mathrm{e}^{-iky} g(-ik) \mathbb {E}\left[ f(X) \mathrm{e}^{ikX} \right] \mathrm{d}k = g\left( \frac{\partial }{\partial y}\right) f(y) n(y;0,\Sigma ) , \end{aligned}$$

(A.1)

where n(x; a, b) denotes the normal density function with mean a and variance b.

Therefore, we can apply (A.1) to obtain the approximation of the density function as

$$\begin{aligned} f_{X_t}(x) \approx \left\{ \begin{array}{lll} \displaystyle \delta (x),&{} \quad \Sigma _t = 0, \\ n\left( x; 0, \Sigma _{t} \right) - \frac{\partial }{\partial {x}} \left\{ \mathbb {E}[ a_{2}(t) | a_{1}(t) = x ] n\left( x; 0, \Sigma _{t} \right) \right\} + \cdots ,&{} \quad otherwise. \end{array} \right. \end{aligned}$$

Appendix B: Formulas for conditional expectation

Let $W^p_t$, $W^q_t$, and $W^r_t$ be standard Brownian motions with correlation $\rho _{i,j} \mathrm{d}t = \mathrm{d}W^i_t \mathrm{d}W^j_t$, and let $y_{i}(x)$ for $i=1,2,3$ be some deterministic functions. Moreover, let $\Sigma := \int _{0}^{T} y^2_{1}(t) \mathrm{d}t$, and denote $J_T(y_1)= \int _{0}^{T} y_{1}(t) \mathrm{d}W^p_t$.

Then, the following formulas are derived:

$$\begin{aligned} E\left[ \int _{0}^{T} y_{3}(t) \left( \int _{0}^{t} y_{2}(s) \mathrm{d}W^q_s \right) \mathrm{d}W^r_t \bigg | J_T(y_1) = x \right] = v_{1} \left( \frac{x^{2}}{\Sigma ^{2}}- \frac{1}{\Sigma } \right) , \end{aligned}$$

(B.1)

where

$$\begin{aligned} v_{1} = \int _{0}^{T} \rho _{p,r} y_{3}(t) y_{1}(t) \left( \int _{0}^{t} \rho _{p,q} y_{2}(s) y_{1}(s) \mathrm{d}s \right) \mathrm{d}t . \end{aligned}$$

The interested reader can find more details in Funahashi and Kijima (2015).

Appendix C: Proof of Theorem 5.1

If r, $\kappa$, and $\theta$ are zero, we have $F(0,t) = S_0$ and $V(0,t) = v_0$. Therefore, ${\bar{\sigma }} := \sigma (S_0,v_0)$, ${\bar{\sigma }}_S := \sigma _S(S_0,v_0)$, ${\bar{\sigma }}_v := \sigma _v(S_0,v_0)$, and ${\bar{\gamma }} := \gamma _0$ become constants.

The variance of the 1st-order Wiener–Ito chaos expansion $V_1(t) := \mathbb {E}[\left( a_1(t) - \widetilde{a}_1(t) \right) ^2]$ is

$$\begin{aligned} V_1(t) = \int _0^t \left( \sigma ^{(0)}(s) - \sqrt{\Sigma _t / t} \right) ^2 \mathrm{d}s. \end{aligned}$$

But, since $\sigma ^{(0)}(s) - \sqrt{\Sigma _t / t} = 0$, we get $V_1(t) = 0$.

Similarly, let ${\bar{p}}_1 := {\bar{\sigma }} + S_0 {\bar{\sigma }}_S$, ${\bar{p}}_2 := {\bar{\sigma }} _v$, ${\bar{p}}_3 := {\bar{\gamma }}$, $\Sigma _t = {\bar{p}}_1^2 t$, and $q(t) = \left( {\bar{\sigma }}^3 {\bar{p}}_1 + \rho {\bar{p}}_2 {\bar{p}}_3 {\bar{\sigma }}^2 \right) t^2 / 2$ the variance of the 2st-order Wiener–Ito chaos expansion $V_2(t):= \mathbb {E}[\left( a_1(t) - \widetilde{a}_1(t) \right) ^2]$ can be computed as

$$\begin{aligned}&V_2(t) = \left( {\bar{p}}^2_1 {\bar{\sigma }}^2 + {\bar{p}}^2_2 {\bar{p}}^2_3 + 4 \left( \frac{q(t)}{t \Sigma _t} \right) ^2 + 2 \rho \bar{p}_1 \bar{p}_2 \bar{p}_3 \bar{\sigma } - 4 \rho \frac{q(t)}{t \Sigma _t} {\bar{p}}_2 {\bar{p}}_3 - 4 \frac{q(t)}{t \Sigma _t} {\bar{p}}_1 {\bar{\sigma }} \right) \int _0^t \nonumber \\&\quad \left( \int _0^s \mathrm{d}u \right) \mathrm{d}s . \end{aligned}$$

Inserting $q(t)/(t \Sigma _t) = \left[ {\bar{p}}_1 {\bar{\sigma }} + \rho {\bar{p}}_2 {\bar{p}}_3\right] /2$ and $\rho ^2 = 1$ into the left-hand side of the last equation, we obtain $V_2(t) = 0$.

Hence, from the definition of $V_Y(t)$, we obtain the desired result.