Given an SDE $$dX_t = b(t,X_t)dt + dZ_t$$ where $Z_t$ is a Levy process. I am curious about the infinitesimal generator of this process.
If the SDE was say $$dX_t = b(X_t)dt + dW_t$$ where $W_t$ was a Wiener process then we would have the generator being
$$Af(x) = \sum_i b_i(x)\frac{\partial f}{\partial x_i} + \frac{1}{2}\sum_{i,j}(\sigma\sigma^T)_{i,j}(x)\frac{\partial^2 f}{\partial x_i\partial x_j}$$
where $f \in C_{0}^{2}(\mathbb{R}^n)$.
My question is does time-dependent drift make a change to what the generator would be? I know there is a difference for the Lévy process vs just Wiener process but for the example only wanted to put the Wiener process version. I am only considered in the case where the drift is the only coefficient containing time-dependency.
I looked for a reference covering this case but could not find one.
Too long for a comment:
Writing $Y_t=(t,X_t)$ and $y=(t,x)$ the SDE followed by $Y$ is \begin{align} dY_t=\beta(Y_t)\,dt+\alpha(Y_t)\,dW_t \end{align} where $\beta(y)=\begin{pmatrix}1\\b(t,x)\end{pmatrix}$ ($b(t,x)$ being your old drift vector of $X_t$) and $$ \alpha(y)=\begin{pmatrix}0&0\\0&\sigma \end{pmatrix} $$ Here, $\sigma$ being your old diffusion matrix of $X$ which was simply a constant identity matrix.
Can you finish?