Consider the following PDE on $[0,T]\times \mathbb{R}$:
$$
\begin{cases}
\dfrac{\partial F}{\partial t}+\mu(t,x) \dfrac{\partial F}{\partial x}+ \frac12 \sigma^2(t,x)\dfrac{\partial^2 F}{\partial x^2} = 0 \\
F(T,x)=g(x)
\end{cases}
$$
with $\mu(t,x)=-\frac{1}{1-t}, \ \sigma(t,x)=1, \ g(x)=x^2$.
Question: What is the solution of the problem?
I know that if $X=\{X_t: t \geq 0\}$ satisfies
$$dX_t=\mu(t,X_t)dt+\sigma(t,X_t)dW_t$$
then (by Feynman-Kac's theorem)
$$F(t,x)=\mathbb{E}^Q[g(X_T)|X_t=x]$$
I know that
$$X_T=X_t-\int_{t}^T\frac{1}{1-s}ds+\int_{t}^TdW_s$$
but then I don't know how to deal with $(X_T)^2$.
How can I get the solution?
Best Answer
Strictly speaking the Feynman-Kac formula should be written as $F(t,x)=\mathbb{E}_{t}[g(X_T)|X_t=x]$, where the $t$ subscript in the expectation tells you that you're conditioning on a $\sigma$-algebra (namely $\mathscr{F}_t = \sigma(B_{u}: u \leq t)$) and the result is a random variable, not a constant. Only when you condition on $X_t = x$ does this become a constant.
I'm assuming that $T<1$, otherwise there will be a blow-up when solving this. From what you wrote: \begin{equation} X_T = X_t + \log\left(\dfrac{1-T}{1-t} \right)+ W_T-W_t. \end{equation} By the Markov property, $W_T - W_t = B_{T-t}$ is also normally distributed with mean zero and variance $T-t$. Now: \begin{equation} \mathbb{E}_{t}[X_{T}^{2}|X_t=x] = x^2 + \log ^2\left(\dfrac{1-T}{1-t} \right) + 2x \log\left(\dfrac{1-T}{1-t} \right)+ \\ +2\mathbb{E}_{t}\left[B_{T-t}\left(X_t + \log\left(\dfrac{1-T}{1-t} \right) \right)|X_t=x\right] + \mathbb{E}_t [B^{2}_{T-t}] = \\ = x^2 + \log ^2\left(\dfrac{1-T}{1-t} \right) + 2x \log\left(\dfrac{1-T}{1-t} \right) + T-t, \end{equation} since the penultimate expectation vanishes as the thing multiplying the Brownian motion is measurable with respect to $\mathscr{F}_t$ and can be pulled out in front of the expectation.