Using Feynman-Kac, compute the following:

Question

Let $B(t)$ be Brownian Motion and let $\alpha$ be a constant and $T>0$.

Compute $\mathbb{E}_{B_{0} = x}\left[\exp\left(-\alpha \int_0^T B(s)^2 ds\right)\right]$.

I'm just having a hard time with this one, any help?

Answer 1

If you compare your expression with $$ u(x,t) = E^Q\left[ \int_t^T e^{- \int_t^r V(X_\tau,\tau)\, d\tau}f(X_r,r)dr + e^{-\int_t^T V(X_\tau,\tau)\, d\tau}\psi(X_T) \Bigg| X_t=x \right] $$ where $dX = \mu(X,t)\,dt + \sigma(X,t)\,dW^Q$ (this is pasted from the wikipedia article ), we see that in your case the process $X_t$ is simply $B_t$ so $\mu =0$ and $\sigma=1$. Furthermore, $f(x, t)=0$, $V(x)=\alpha x^2$ and terminal condition is $u(x, T)=\psi(x)=1$

Hence $$ \frac{\partial u}{\partial t}(x,t) + \mu(x,t) \frac{\partial u}{\partial x}(x,t) + \tfrac{1}{2} \sigma^2(x,t) \frac{\partial^2 u}{\partial x^2}(x,t) -V(x,t) u(x,t) + f(x,t) = 0 $$ reduces to $$ \frac{\partial u}{\partial t}(x,t) + \tfrac{1}{2} \frac{\partial^2 u}{\partial x^2}(x,t) -\alpha x^2 u(x,t) = 0 $$ so we are still left with solving this PDE, which can possibly be solved following this post