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.

  • $\begingroup$ Try to figure out what the generator of the process $(t,X_t)$ should be. $\endgroup$
    – Kurt G.
    Feb 1, 2023 at 6:14
  • $\begingroup$ Extending the example above should be $$Af(t,x) = \sum_i b_i(t,x)\frac{\partial f}{\partial t} + \sum_i b_i(t,x)\frac{\partial f}{\partial x_i} + \frac{1}{2}\sum_{i,j}(\sigma\sigma^T)_{i,j}(t,x)\frac{\partial^2 f}{\partial x_i\partial x_j}$$ However, that include time-dependency in the diffusive portion. $\endgroup$ Feb 1, 2023 at 16:10

1 Answer 1


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?

  • $\begingroup$ Yes I believe that makes sense, I should've clarified more I mean a drift that could be along the lines of say $b(t,x) = \cos(tx)$. $\endgroup$ Feb 1, 2023 at 19:03
  • $\begingroup$ Edited the answer accordingly. What you are trying to do is probably simpler than you think. If not this paper might help further. $\endgroup$
    – Kurt G.
    Feb 1, 2023 at 19:10
  • $\begingroup$ Thank you, yes that clears it up and yes indeed simpler than I had thought $\endgroup$ Feb 1, 2023 at 19:20
  • $\begingroup$ @KurtG. Great answer. I assume a similar trick for turning a time-dependent SDE into one without $t$ works for Lévy-driven SDEs as well, but do you have a reference on how to compute generators for Lévy-driven SDEs? $\endgroup$ Feb 1, 2023 at 20:30
  • $\begingroup$ @SmallDeviation . Unfortunately I don't. An important formula in the Levy process world is the Levy-Khintchine formula. Also: Levy processes other than those that are practically a Brownian motion typically have jumps. The generators of such processes are not differential operators but rather integro differential operators or pseudo differential operators. I guess that are enough buzz words now for you to google. The interplay between operators and Markov processes is a fascinating subject. Unfortunately many nice things did not become very mainstream. $\endgroup$
    – Kurt G.
    Feb 2, 2023 at 16:51

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