Let $N(dt,dy)$ be Poisson Random measure with a intensity probability measure $\nu$ on $[0,1]$, $b(x,y)$ be continuous function on $[0,\infty)\times [0,1]$ and a cad lag solution $X$ to the following SDE:
$X_t-X_0= \int_{[0,t]\times[0,1]}b(X_{s-},y) N(ds,dy)$
I want to calculate just formally the infinitesimal generator $A$ associated with $X$ for $f\in C^2(\mathbb{R})$. Usually we can apply Ito-formula or Taylor expansion for:
$Af(x)= \lim_{t\rightarrow 0} \frac{\mathbb{E}[f(X_t)-f(x)]}{t}$
But I have very little experience with SDE with jumps.
The solution is $Af(x)=\int_{[0,1]}f(x+b(x,y))-f(x)\nu(dy)$.
I dont see how the term $f(x+b(x,y))$ comes up in the derivation. As far I understood we have to handle possible jumps of the cad lag process X and jumps happen according to the intensity measure $\nu$. Does any one know how to calculate the generator in this case?