# Infinitesimal Generator of Ito Diffusion Process

Suppose one has the an Ito process of the form:

$$dX_t = b(X_t)dt + \sigma(X_t)dW_t$$

The following is an excerpt from wikipedia My question is on how to derive this operator? It looks very similar to what you get when using Ito's Lemma. So I start with applying Ito's Lemma with f to get:

$$df = \frac{\partial f}{\partial t}dt + \sum_i\frac{\partial f}{\partial x_i}dx_i + \frac{1}{2}\sum_{i,j}\frac{\partial^2 f}{\partial x_ix_j}[dx_i,dx_j]$$

Which then becomes:

$$df = \left[ \frac{\partial f}{\partial t}dt + \sum_i b_i(X_t)\frac{\partial f}{\partial x_i}dt + \frac{1}{2}\sum_{i,j}(\sigma\sigma^T)_{i,j}\frac{\partial^2 f}{\partial x_i \partial x_j} \right]dt + \sum_i \sigma_i(X_t) \frac{\partial f}{\partial x_i}dW_t$$

Hopefully I got that correct. What I'm unsure about is how to proceed in order to compute $Af(x)$. I would think the next step is to integrate $df$, but it's not clear to me what happens after (I know these infinitesimal generators are rooted in semigroup theory, but I have very little experience in that).

1. Note that $f$ does not depend on the time $t$, therefore the term $\frac{\partial}{\partial t} f$ is superfluous.
2. Take expectation on both sides, then the stochastic integral $\dots dW_t$ vanishes, because it is a martingale.
3. Use Fubini's theorem and the fundamental theorem of calculus, $$\frac{1}{t} \int_0^t \mathbb{E}^xg(X_s) \, ds \stackrel{t \to 0}{\to} \mathbb{E}^xg(X_0)= g(x).$$
• Thanks for the help! Doing steps 1 and 2 gives me this: $E[f] = E\left[ \int_0^t \left[ \sum_i b_i(X_t)\frac{\partial f}{\partial x_i}dt + \frac{1}{2}\sum_{i,j}(\sigma\sigma^T)_{i,j}\frac{\partial^2 f}{\partial x_i \partial x_j} \right] ds \right]$. I'm not sure I'm understanding how to use Step 3 though. – Brenton Jul 3 '15 at 15:39
• @Brenton ... because you didn't write up Itô's formula properly. You can use these short-notations if you know what you do, but at the beginning it's much better to write it in full length. by Itô's formula $$f(X_t)-f(X_0) = \int_0^t \dots \, dW_s + \int_0^t \dots \, ds.$$ Fill in the "$\dots$" and take then the expectation on both sides. Then you'll see that e.g. the left-hand side in your comments reads $\mathbb{E}^x f(X_t)-\mathbb{E}^x f(X_0)$. – saz Jul 3 '15 at 16:27
• So if I do it this way (to my understanding since I've never taken an SDE course): $$f(X_t) = f(X_0) + \int_0^t \sum_i \frac{\partial f}{\partial x_i}b_i + \frac{1}{2}\sum_{i,j}\frac{\partial^2 f}{\partial x_ix_j}(\sigma \sigma^T)_{ij} ds + \int_0^t \sum_i \frac{\partial f}{\partial x_i}\sigma_i dW_s$$ So taking expectations should give: $$E^x \left[f(X_t)\right] = E^x \left[f(X_0)\right] + E\left[\int_0^t \sum_i \frac{\partial f}{\partial x_i}b_i + \frac{1}{2}\sum_{i,j}\frac{\partial^2f}{\partial x_ix_j}(\sigma \sigma^T)_{ij} ds\right]$$ I feel like I made a mistake – Brenton Jul 3 '15 at 17:38
• @Brenton Actually, it should read $$f(X_t) = f(X_0) + \int_0^t \left( \sum_i \frac{\partial f(X_s)}{\partial x_i}b_i(X_s) + \frac{1}{2}\sum_{i,j}\frac{\partial^2 f(X_s)}{\partial x_ix_j}(\sigma(X_s) \sigma(X_s)^T)_{ij} \right) ds + \int_0^t \sum_i \frac{\partial f(X_s)}{\partial x_i}\sigma_i(X_s) dW_s.$$ – saz Jul 3 '15 at 17:57
• @user46944 Note that $(X_t)_{t \geq 0}$ has continuous sample paths and $y \mapsto Af(y)$ is continuous (for any "nice" $f$). Therefore the continuity of $s \mapsto \mathbb{E}^x (Af(X_s))$ is a direct consequence of the dominated convergence theorem. – saz Dec 2 '16 at 6:45