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Suppose you have the following function:

$u(t,x)=\mathbb{E}[f(xe^{W_t+\frac{1}{2}t})]$, where $W_t$ is a Wiener process.

Let us first differentiate:

$du=\mathbb{E}[f'(xe^{W_t+\frac{1}{2}t})(e^{W_t-\frac{1}{2}t}dx+xd(e^{W_t-\frac{1}{2}t}))]$ and using some properties of the Brownian motion

$d(e^{W_t-\frac{1}{2}t})=e^{W_t-\frac{1}{2}t}(dW_t-\frac{1}{2}dt+\frac{1}{2}(dW_t)^2)=e^{W_t-\frac{1}{2}t}dW_t$

Clearly, $\mathbb{E}[f'(xe^{W_t+\frac{1}{2}t})xd(e^{W_t-\frac{1}{2}t})dW_t]=0$

and, therefore, $du=\mathbb{E}[e^{W_t-\frac{1}{2}t}f'(xe^{W_t+\frac{1}{2}t})]dx+0 dt$

Could someone tell me from this how to find the PDE that the function u satisfies?

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Since we have $\frac{\partial\mathbb{E}[u(t,x)]}{\partial t}=0$, the infinitesimal generator is:

$Af(x)=\frac{1}{2}x^2\frac{\partial}{\partial x^2}$

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