I know that for the diffusion process $$ X_t = \mu(t, X_t) dt + \sigma(t, X_t) dB_t, $$ the function $$ u(t, x) = \mathbb{E}_{x, t}[e^{\int_t^T r(s, X_s) ds} g(X_T)] $$ with the boundary condition $u(T, x) = \phi(x)$ satisfies (from the Feynman Kac formula) $$ u_t(t, x) + \mu(t, x)u_x(t, x) + \frac{\sigma^2(t, x)}{2}u_{xx}(t, x) - r(t, x)u(t, x) = 0 $$ with $u(T, x) = \phi(x)$.

My question is what is the form of the Feynman Kac formula (PDE) if $$ X_t = \mu(t, X_t) dt + \sigma(t, X_t) dB_t + (Y_t − 1)dN_t, $$ where $N_t$ is a Poisson process with parameter $\lambda>0$ and $Y_t$ is a sequence of i.i.d. random variables with log-normal distribution.

I'm just starting to learn about jump processes so I do not know how to obtain the PDE for above mentioned $u(t, x)$ function in case of the jump process $X_t$.


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