the stochastic process$\{X_t\}_{t\ge 0}$ is a martingale

Question

Let $f$ be twice continuously differentiable, and $B$ be a Brownian motion. Using the transition density $p(t,x,y)=\frac{1}{\sqrt(2\pi t)}e^{-\frac{(x-y)^2)}{2t}}$ how can we prove that the stochastic process $$X_t = f(B_t)-\frac{1}{2}\int_{0}^{t}f''(B_s)ds 􀀀$$ is a martingale.

I am stuck at this problem and don't know where to start. Could anyone give a help ?

Answer 1

I think this works.

Let $g(t,x)$ be the function defined by $g(t,x) : = f(x) - \frac{1}{2} \int_0^t f''(s) ds$. We observe that $\partial_t g(t,x) = - \frac{1}{2} f''(t,x)$ and $\partial_x^2 g(t,x) = f''(t,x)$. Hence we see that $\partial_t g(t,x) = -\frac{1}{2} \partial_x^2 g(t,x)$. This shows that the process $(g(t,B_t))_{t \geq 0}$ is a local martingale. To show that we actually have a full martingale, we need to show that for all $T >0$, $$\mathbb{E} \left( \int_0^T \left| \partial_x g(t,x) \right|^2 dt \right) < \infty.$$ This is clear however, since $g(t,x)$ is differentiable, and therefore bounded.