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I am studying a course on Stochastic Calculus for Finance and am struggling with the following question:

Given $dY_t = b(t,Y_t) \, dt + \sigma(t, Y_t) \, dW_t$ where $\gamma \neq 0$, and $$f(t,Y_t) = \exp\left(- \frac{\gamma^2}{2} t + \gamma W_t\right) \mathbb{E}\left[\exp\left(\frac{\gamma^2}{2} T - \gamma W_T\right) F(Y_T) \mid \mathcal{F}_t \right]$$ Find a PDE for $f$.

This clearly relates to Feynman-Kac Theorem but as there are Brownian Motion terms in the expression for $f$, this is not in the form that I am familiar with. Here is the version of the theorem that I am familiar with:

Feynman Kac: If a process $Y$ satisfies the SDE $$dY_t = b(t,Y_t)dt + \sigma (t, Y_t) dW_t$$ Then suppose $f(t,y) \in C^{1,2}$ and solves the PDE: $$f_t + \frac{1}{2} \sigma ^2 (t,y) f_{yy} + b(t,y) f_y - \delta(t,y) f = 0$$ with the terminal condition: $$f(T,y) = F(y)$$ Then under suitable assumptions, the function $f$ has the expression: $$ f(t,Y_t) = \mathbb{E}[ \exp ( - \int _t^T \delta(s,Y_s) ds) F(Y_T) | \mathscr{F}_t]$$

As mentioned, the Brownian Motion terms in the exponential (in the question) make it unclear whether or not I am able to apply the theorem without some sort of manipulation.

A possible avenue to explore: I have also noticed that the exponentials both look like changes of measure. I know that we can define $\exp\left(- \frac{\gamma^2}{2} t + \gamma W_t\right)$ as $\frac{d \mathbb{Q}}{d \mathbb{P}}$ given $\mathscr{F}_t$ and so $\exp\left(\frac{\gamma^2}{2} T - \gamma W_T\right)$ would be $\frac{d \mathbb{P}}{d \mathbb{Q}}$ given $\mathscr{F}_T$. But this seems to complicate things when I try to substitute these expressions in.

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Note that we can rewrite the random variable in the following way: $$\begin{aligned} &e^{-\gamma^2t/2+\gamma W_t}E^P[e^{\gamma^2T/2-\gamma W_T}F(Y_T)|\mathscr{F}_t]\\ &=e^{-\gamma^2t/2+\gamma W_t}E^P[e^{\gamma^2T/2-\gamma W_T}F(Y_T)e^{\gamma^2(T-t)}e^{-\gamma^2(T-t)}|\mathscr{F}_t]\\ &=e^{\gamma^2t/2+\gamma W_t}E^P[e^{-\gamma^2T/2-\gamma W_T}F(Y_T)e^{\gamma^2(T-t)}|\mathscr{F}_t] \end{aligned}$$ Now note that the martingale $L_s:=e^{-\gamma^2s/2-\gamma W_s},\,s\leq T$ defines the change of probability measure $dQ=L_TdP$ s.t., in particular, for any $Z$ $\mathscr{F}_T$-measurable we have $L_t^{-1}E^P[L_TZ|\mathscr{F}_t]=E^Q[Z|\mathscr{F}_t]$. Therefore we can write: $$e^{-\gamma^2t/2+\gamma W_t}E^P[e^{\gamma^2T/2-\gamma W_T}F(Y_T)|\mathscr{F}_t]=E^Q[F(Y_T)e^{\gamma^2(T-t)}|\mathscr{F}_t]$$ Recall the theorem of Girsanov which tells us that $W_t^Q:=W_t+\gamma t$ is a $Q$-Brownian motion for $t\leq T$. We get the $Q$-dynamics of $(Y_t)_{t\leq T}$ as following: $$\begin{aligned} dY_t&=b(t,Y_t)dt+\sigma(t,Y_t)dW_t\\ &=(b(t,Y_t)-\gamma \sigma(t,Y_t))dt+\sigma(t,Y_t)dW_t^Q \end{aligned}$$ What is left is to apply Feynman-Kac: the function $f(t,Y_t)=E^Q[F(Y_T)e^{\gamma^2(T-t)}|\mathscr{F}_t]$ solves the terminal value problem on $[0,T]\times \mathbb{R}$ $$\begin{cases} \partial_tf(t,y)+(b(t,y)-\gamma \sigma(t,y))\partial_y f(t,y)+\frac{1}{2}\sigma^2(t,y)\partial_{yy}f(t,y)+\gamma^2f(t,y)=0\\ f(T,y)=F(y) \end{cases}$$

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