[Math] Feynman-Kac for jump-diffusion

Question

I'm looking for a more general Feynman-Kac formula that works in the case of jump-diffusion processes.
I know that, given a pure diffusion process like
$$dS_t=\mu_tdt+\sigma_tdW_t,$$ if $u(t,s)$ satisfies the PDE
$$f_t(t,s)+\mu_tf_s(t,s)+\frac{\sigma_t^2}{2}f_{ss}(t,s)-V(s)f(t,s)=0$$ with terminal condition $f(T,s)=H(S_T)$
then $u$ is of the form $$u(t,s)=\mathbb{E}\big[e^{-\int_t^T{V(S_x)dx}}H(S_T)|S_t=s\big].$$
Is it possible to extend such a result when the process dynamics are given by
$$dS_t=\mu_tdt+\sigma_tdW_t+\gamma_tdN_t$$ with $N$ a Poisson process independent fom $W$?

Answer 1

Hi it is possible to get some Feynman-Kac formula in this case. The proof only use the martingale property and Itô's formula for jump-diffusion processes.

So let's have $X$ s.t. (I took the compensated version of your sde):

$dX_t=[\mu(t,X_t)+\lambda(t)\gamma(t,X_t)]dt + \sigma(t,X_t)dW_t+ \gamma(t,X_{t-})d\tilde{N}_t$ where $\tilde{N}_t$ is a compensated Poisson process of intensity $\lambda(t)$.

Please notice the explicit dependence in $t$ and $X_t$ of the above equation that is necessary to have Markov property for the solution which is necessary for the Feynman-Kac theorem to apply.

Now let's us be given $F(t,X_t)=e^{-\int_s^t V(X_r)dr}u(t,X_t)=e^{-IV(s,t)}.u(t,X_t)$ and apply Itô to this formula. You get :

$$dY_t=dF(t,X_t)=e^{-IV(s,t)}\Big(\big(\partial_t u(t,X_t)+\lambda(t)[u(t,X_t+\gamma(t,X_t))-u(t,X_t)]+\mu(t,X_t)\partial_x u(t,X_t)+\frac{\sigma^2(t,X_t)\partial_{xx}u(t,X_t)}{2}-V(t,X_t).u(t,X_t)\big)dt+ (u(t,X_t+\gamma(t,X_t))-u(t,X_t))d\tilde{N}_t+(\sigma(t,X_t)\partial_{x}u(t,X_t))dW_t\Big)$$

Now if the $dt$ term is null then $Y_t$ is martingale and for $t=T$ : $$Y_s=F(s,X_s=x)=E[Y_T|X_s=x]=E[e^{-\int_s^T V(X_r)dr}H(X_T)|X_s=x]$$

So in this case the PIDE that solves the Feynman-Kac formula is : $$\partial_t u(t,X_t)+\mu(t,x)\partial_x u(t,x)+\frac{\sigma^2(t,x)}{2}\partial_{xx}u(t,x)+\lambda(t)[u(t,x+\gamma(t,x))-u(t,x)]=V(t,x)u(t,x) $$

With final condition $u(T,x)=H(x)$

Best regards