Determine $h$ process so that the following stochastic representation result holds

Question

I want to find a process $h$ such that $$m(T) = Em(T) + \int_0^T h(t) dW(t), $$

where $m(T) = e^{ \int_0^T t dW(t)}$. Here, $T$ is some positive constant, and $W(t)$ is Brownian motion.

I get $Em(T) = \frac{1}{6}T^3$. But then I am unsure how to continue.

Is the exponent in the expresion for $m(T)$ to be $\int_0^T T,dW(t)$ or $\int_0^T t,dW(t)$? — John Dawkins, Jan 07 '17 at 18:05
See my edit. I've also got an idea which I am adding shortly, but I am unsure if it'll work out. — Imean H, Jan 07 '17 at 18:07

John Dawkins · Accepted Answer · 2017-01-08T17:20:49.433

1

Hint: Use Ito's formula to expand the martingale $e^{-t^3/6}m(t)$, $0\le t\le T$. Having done this set $t=T$ and multiply through by $e^{T^3/6}$.

edited Jan 08 '17 at 17:20

answered Jan 08 '17 at 16:40

John Dawkins

Thanks. Any comment on why you know that this function is a martingale beforehand? – Imean H Jan 11 '17 at 09:37
It has the "exponential martingale" form $\exp(Y_t-\langle Y\rangle_t/2)$, where $Y_t=\int_0^t s,dW(s)$ is a martingale with quadratic variation $\langle Y\rangle_t=\int_0^t s^2,ds$. – John Dawkins Jan 11 '17 at 16:00

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