Extension of Dynkin's formula, conclude that process is a martingale

Question

This question was asked here, but it did not get enough attention, so I'm crossposting it to MO.

Let $u: \mathbb{R}_+ \times \mathbb{R}^d$ be a bounded $C^2$ function whose first and second partial derivatives are uniformly bounded (or, more generally, have at most polynomial growth as $|x| \to \infty)$ on $[0, T] \times \mathbb{R}^d$, for any $0 \le T < \infty$. For any $t \ge 0$ and any $x \in \mathbb{R}^d$ it is not hard to see that,$$E^x u(t, W_t) = u(0, x) + E^x \int_0^y \left( {\partial\over{\partial s}} + {1\over2}\Delta_s\right) u(s, W_s)\,ds.$$$($$W_t$ is a Brownian motion process which takes the value $x$ at $t=0$ so the dependence of $x$ is implicit here, most of the references will consider $W_t$ a standard Brownian motion which means it takes the value $0$ at $t=0$ and in the above formula we will have $x+W_t$ instead. $\Delta_s$ here means the sum of all the mixed derivatives.$)$

My question is, how do we conclude that under $P^x$ the process$$u(t, W_t) - u(0, x) - \left({\partial\over{\partial t}} + {1\over2}\Delta_x\right)u(t, W_t)$$is a martingale?

Answer 1

What you wrote is incorrect - try the function $u(t,x)=x^2$ - it is of course not true that $W_t^2-1/2$ is a martingale when $W_t$ is standard Brownian motion..... You are missing an integral.

You conclude the correct form from Ito's formula, applied to the function $u(t,x)$. Namely, $W_t$ be a Brownian motion started at $x$, and write $X_t=u(t,W_t)$. From Ito's formula you get $$dX_t=(\partial_t u +\frac12 \partial_{xx} u)(t,W_t) dt+ (\partial_x u)(t,W_t)dW_t.$$ Rearranging you get (remember that $X_t=u(t,X_t), X_0=u(0,x)$) $$u(t,W_t)-u(0,x)=X_t-X_0=\int_0^t (\partial_t u+\frac12 \partial_{xx} u)(s,W_s)ds+M_t$$ where $M_t$ is a stochatic integral, hence a martingale under your growth assumptions on $\partial_x u$.

Of course, once you know it is a martingale, by taking expectations you get the first formula that you wrote.