I have the following PDE

\begin{aligned} \frac{\partial}{\partial t} f(t, x)+\mu x \frac{\partial}{\partial x} f(t, x)+\frac{1}{2} \sigma^{2} x^{2} \frac{\partial^{2}}{\partial x^{2}} f(t, x) &=0 \\ f(T, x) &=\log \left(x^{2}\right) \end{aligned}

with $\mu$ and $\sigma$ some fixed parameters.

Now, I know through Feynman-Kac that I can write $$ f(t,X_t) = E_t [ log(X_T^2) ] $$

where $X_t$ is geometric brownian motion ($dX_t = \mu dt + \sigma dB_t$).

Now the question is, how do I find the function $f(t,x)$?

I tried applying Itto's rule on $X_t^2$, and I get (not sure if it is correct):

$$ d(X_t^2) = (2 \mu X_t + \sigma^2 )dt + 2 \sigma X_t dB_t $$

I'm not sure how to proceed from here...


As $X$ is a geometric Brownian motion, we know the SDE solved by $X$ has a unique solution: \begin{equation} X_t = X_0 e^{(\mu - \frac12\sigma^2 )t + \sigma B_t} \end{equation} where $\left\{B_t\right\}_{t \geq 0}$ is $\mathcal{F}_t$-Brownian motion. We can also rewrite the above quantity at time $T$: \begin{equation} X_T = X_te^{(\mu - \frac12\sigma^2 )(T-t) + \sigma (B_T - B_t)} \end{equation} Using Feynam-Kac theorem (verifying we fill in the conditions), the solution of the PDE can be written as : \begin{align} f(t,X_t) &= \mathbb{E}_t [ \log(X_T^2) ]\\ &=2\mathbb{E}_t [ \log(X_T) ]\\ &=2\mathbb{E}_t [ \log(X_t) + (\mu - \frac12\sigma^2)(T-t) + \sigma (B_T - B_t) ]\\ &=2\log(X_t) + (2\mu - \sigma^2)(T-t) + 2\sigma \mathbb{E}_t [B_T - B_t]\\ &=2\log(X_t) + (2\mu - \sigma^2)(T-t) \end{align}
Where in the last equation, we used that $B_T - B_t \sim\mathcal{N}(0, T-t)$ and is independent of $\mathcal{F}_t$

You can remark that the function $f$ verifies the PDE.


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