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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

enter image description here

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).

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Hints:

  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, $$\frac{1}{t} \int_0^t \mathbb{E}^xg(X_s) \, ds \stackrel{t \to 0}{\to} \mathbb{E}^xg(X_0)= g(x).$$
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  • $\begingroup$ 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. $\endgroup$ – Brenton Jul 3 '15 at 15:39
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    $\begingroup$ @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 $\mathbb{E}^x f(X_t)-\mathbb{E}^x f(X_0)$. $\endgroup$ – saz Jul 3 '15 at 16:27
  • $\begingroup$ 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 $\endgroup$ – Brenton Jul 3 '15 at 17:38
  • $\begingroup$ @Brenton Actually, it should read $$f(X_t) = f(X_0) + \int_0^t \left( \sum_i \frac{\partial f(X_s)}{\partial x_i}b_i(X_s) + \frac{1}{2}\sum_{i,j}\frac{\partial^2 f(X_s)}{\partial x_ix_j}(\sigma(X_s) \sigma(X_s)^T)_{ij} \right) ds + \int_0^t \sum_i \frac{\partial f(X_s)}{\partial x_i}\sigma_i(X_s) dW_s.$$ $\endgroup$ – saz Jul 3 '15 at 17:57
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    $\begingroup$ @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. $\endgroup$ – saz Dec 2 '16 at 6:45

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