Efficient Simulation of Value-at-Risk Under a Jump Diffusion Model: A New Method for Moderate Deviation Events

Abstract

Importance sampling is a powerful variance reduction technique for rare event simulation, and can be applied to evaluate a portfolio’s Value-at-Risk (VaR). By adding a jump term in the geometric Brownian motion, the jump diffusion model can be used to describe abnormal changes in asset prices when there is a serious event in the market. In this paper, we propose an importance sampling algorithm to compute the portfolio’s VaR under a multi-variate jump diffusion model. To be more precise, an efficient computational procedure is developed for estimating the portfolio loss probability for those assets with jump risks. And the tilting measure can be separated for the diffusion and the jump part under the assumption of independence. The simulation results show that the efficiency of importance sampling improves over the naive Monte Carlo simulation from 9 to 277 times under various situations.

Appendix: Proofs

Proof

(Proof of Proposition 1) We first note that $\psi (\theta )$ is the cumulant generating function of X, and therefore its second derivative $\psi ''(\theta )>0$ for $\theta \in \mathbf{I}$. This implies that $\psi '(\theta )$ is strictly increasing. Since $\Psi (\theta )$ is convex and steep by assumption, we have $\psi '(\theta ) \rightarrow \infty$ as $\theta \rightarrow \theta _{\max }$. Furthermore, consider the conditional measure of $\bar{Q}$ on the set A, denoted by $\bar{Q}_A$, which is defined as

$$\begin{aligned} d{\bar{Q}}_A = \frac{\mathbf{1 }_{\{X\in A \}} d{\bar{Q}}}{ \int {\mathbf{1}}_{\{X\in A\}} d{\bar{Q}}}. \end{aligned}$$

(28)

Then we have

$$\begin{aligned} \frac{d E_{\bar{Q}_\theta }[X|X\in A]}{d\theta }= & {} \frac{d}{d\theta }\left( \frac{E[\mathbf 1 _{\{X\in A\}} X \text {e}^{-\theta X}]}{E[\mathbf 1 _{\{X\in A\}} \text {e}^{-\theta X}]}\right) \nonumber \\= & {} -\frac{E[\mathbf 1 _{\{X\in A\}} X^2\text {e}^{-\theta X}]}{E[ \mathbf 1 _{\{X\in A\}} \text {e}^{-\theta X}]} + \frac{E^2 [\mathbf 1 _{\{X\in A\}} X \text {e}^{-\theta X}]}{E^2[ \mathbf 1 _{\{X\in A\}} \text {e}^{-\theta X}]} =-\mathrm{var}_{\bar{Q}_A}(X) < 0.\nonumber \\ \end{aligned}$$

(29)

Since under the assumption in Proposition 1, the second and third terms in (29) are bounded for all $\theta \in \mathbf{I}$, by Fubini’s theorem, the second equality in (29) holds. This implies that $E_{\bar{Q}_\theta }[X|X\in A]$ is strictly decreasing. $\square$

Proof

(Proof of Theorem 1 ) The existence of the optimization problem (3) requires the following fact,

$$\begin{aligned} \psi '(\theta ) ~\mathrm{is~ strictly ~increasing~and}~E_{\bar{Q}}[X|X\in A]~\mathrm{is~strictly~decreasing}, \end{aligned}$$

(30)

which has been proved in Proposition 1.

The existence of the optimization problem (3) follows from $E_{\bar{Q}_0}[X|X\in A] = E[X|X\in A] > \mu = \psi '(0)$.

To prove the uniqueness, we note that the second derivative of G equals

$$\begin{aligned} \displaystyle \frac{d^2 G(\theta )}{d \theta ^2}= & {} \frac{d^2}{d \theta ^2} E\left[ \mathbf 1 _{\{X\in A\}} \frac{d P}{d Q}\right] =\frac{d^2}{d \theta ^2}E\left[ \mathbf 1 _{\{X\in A\}}e^{-\theta {X}+\psi (\theta )}\right] \nonumber \\= & {} \frac{d}{d \theta } E{\left[ \mathbf 1 _{\{X\in A\}} (+\psi '(\theta ))e^{-\theta {X}+\psi (\theta )}\right] } \nonumber \\= & {} E{\left[ \mathbf 1 _{\{X\in A\}}((-X+\psi '(\theta ))^2+\psi ''(\theta ))e^{-\theta {X}+\psi (\theta )}\right] }. \end{aligned}$$

(31)

Since $\psi (\theta )$ is the cumulant generating function of X, its second derivative $\psi ''(\theta )>0$. It then follows from (31) that $\frac{d^2 G(\theta )}{d \theta ^2}>0$, which implies that there exists a unique minimum of $G(\theta )$.

To prove (8), we need to simplify the RHS of (5) under $\bar{Q}$. Standard algebra gives

$$\begin{aligned} \frac{E[\mathbf 1 _{\{X\in A\}}X\text{ e }^{-\theta X}]}{E[\mathbf 1 _{\{X\in A\}}\text{ e }^{-\theta X}]}= \frac{\int \mathbf 1 _{\{x\in A\}}x\text{ e }^{-\theta x} dP / \tilde{\Psi }(\theta )}{\int \mathbf 1 _{\{x\in A\}}\text{ e }^{-\theta x} dP/ \tilde{\Psi }(\theta )} = \frac{\int \mathbf 1 _{\{x\in A\}}x d\bar{Q}}{\int \mathbf 1 _{\{x\in A\}}d\bar{Q}}. \end{aligned}$$

(32)

As a result, (32) equals

$$\begin{aligned} \int \mathbf 1 _{\{x\in A\}} x d\bar{Q}_A = E_{\bar{Q}_A}[X]= E_{\bar{Q}}[X|X\in A], \end{aligned}$$

which implies the desired result. $\square$