I am trying to prove: Let $g \in C^{2}_{0}(\mathbb{R})$ and let $B_{t}$ be a Brownian Motion. Then, the following process is a martingale: $$X_{t}=g(B_{t})-\frac{1}{2}\int_{0}^{t}g''(B_{s})ds$$
My attempt: Let $(F_{t})$ be the canonical filtration. I need to show that $$E[X_{t}-X_{s}|F_{s}] = E[g(B_{t})|F_{s}]-g(B_{s})-\frac{1}{2}E\bigg[\int_{s}^{t}g''(B_{r})dr\bigg|F_{s}\bigg]=0$$ My questions:
- Is it true that $E[g(B_{t})|F_{s}]=E[g(B_{s})]$? How to prove it?
- I can pull the conditional expectation into the integral using Riemann Sums. Using the same argument as above, it is then a normal expectation. Is that correct?
- Then, I can use Fubini and get the integral $\int_{s}^{t}\int_{-\infty}^{\infty}g''(x)p(x,r)dxdr$ where $p$ is the normal density in $x$ with variance $r$ and mean $0$. I have integrated by parts twice to somehow get $g$ instead of $g''$, but this didn't help.
You are better off using Ito's formula: $$ g(B_t) = g(0)+\int_0^tg'(B_s) dB_s + \frac12 \int_0^t g''(B_s)ds. $$ Since $ \int_0^t E(g'(B_s))^2 ds< Kt <\infty$ and due to the definition of the Ito-integral $\int_0^t g'(B_s) dB_s$ is progressively measurable, $ \int_0^t g'(B_s)dB_s$ is a martingale. Now $$ X_t = g(B_t) - \frac12 \int_0^t g''(B_s) ds = g(0) + \int_0^t g'(B_s)dB_s,$$ hence it is a martingale.
Without Ito's formula things are a bit more complicated. $$ E[ g(B_t)| F_s] = E[ g(B_t-B_s + B_s) | F_s] = \big. E [ g(B_t-B_s +x)]\big|_{x=B_s} $$ due to the fact that $B_t-B_s$ is independent of $F_s$. Furthermore $$ E[\int_s^t g''(B_r) dr| F_s] =\big. E[ \int_s^t g''( B_r- B_s +x)dr] \big|_{x= B_s} $$ Can you proceed from here?
The lemma. Let $X,Y$ be two random variables and let $F$ be a sigma-algebra, such that $X$ is $F$ measurable and $Y$ is independent of $F$. Let $g:\mathbb R^2\to \mathbb R$ be a measurable function satisfying $E[|g(X,Y)|]<\infty$. Then $$ E[g(X,Y)| F] = \big. E [ g(x,Y)] \big |_{x=X} \quad a.s.$$