I came across the following problem and below is my approach. I am not sure how to proceed.
Problem: suppose
$$M(t)=\int_0^t W(s)dW(s)$$
where $W(s)$ is a standard Brownian motion, find $f=f(t,x)$ such that $$E(t)=e^{M(t)-\int_0^tf(s,W(s))ds}$$
is a martingale.
What I did was to let
$$g(t,W(t))=E(t)$$
and then by Ito's formula,
$$dg(t,W(t))=\partial_tg(t,W(t))dt+\partial_xg(t,W(t))dW(t)+\frac{1}{2}\partial_x^2g(t,W(t))dt$$
Edit:
My question now is how do we compute those partial derivatives of $g$? I tried the following: $$\partial_tg(t,W(t))=E(t)(-f(t,W(t))$$ $$\partial_xg(t,W(t))=E(t)W(t)$$ $$\partial_x^2g(t,W(t))=E(t)W(t)^2+E(t)$$ Could someone tell me what I did wrong here?
We know that $$ d\bigg(\int_0^t W(s)dW(s)\bigg) = W(t)dW(t) $$ by definition. Now let $N(t) := M(t)- \int_0^t f(s,W(s))ds $, then $$ d(N(t)) = dM(t)-d(\int_0^t f(s,W(s))ds) = W(t)dW(t)-d(\int_0^t f(s,W(s))ds) = W(t)dW(t) -f(t,W(t))dt $$ since $\int_0^t f(s,W(s))ds$ is a "regular" integral, and $$ d[N](t) = W^2(t)dt. $$ By Ito we have $$ d(E(t))= d(e^{N(t)}) = e^{N(t)}dN(t)+\frac12e^{N(t)} d[N](t) = e^{N(t)}(W(t)dW(t)-f(t,W(t)dt+\frac{W^2(t)}{2}dt). $$ In order for this to be a martingale the $dt$ term has to be 0, thus choose $f(t,W(t)) = \frac12 W^2(t) $, i.e. $f(t,x) = \frac{x^2}2$.
Note that $\int_0^t W^2(s)ds = [M](t)$ is actually the quadratic variation of $M(t)$. This means that given any $L^2$ martingale $M(t)$ in the filtration of $(W(t))_{t\geq 0}$, the process $\mathcal E(M)(t) = \exp\bigg( M(t)-\frac12 [M](t) \bigg) $ is a(n exponential) martingale.