Let $W_t$ be the standard Brownian motion. Is the random process a martingale? - $Y_t = exp(\int_0^t sdW_s)$ (Find $dY_t$ using Ito formula in its differential form)
Base on what I have learned we should calculate the first and second derivate of that function, however I don't know what to do with the $dW_s$ or how to proceed tackling this problem.
Please help.
Define $Z_t := \int_0^t s dW_s$. Then $Z_t$ is an Ito process and to compute $dY_t$ we can apply Ito's lemma to $f(Z_t)=e^{Z_t}$. The Ito's formula is given by
$$df(Z_t)=f'(Z_t)dZ_t+\frac{1}{2}f''(Z_t)d[Z,Z]_t,$$
where $[Z,Z]_t$ is the quadratic variation of $Z_t$. Now, $Z_t$ is an Ito integral. Thus, we use the formula for the quadratic variation of Ito integral, which states that
$$ [Z,Z]_t=\left[\int_0^{\cdot}sdW_s,\int_0^{\cdot}sdW_s\right]=\int_0^ts^2ds$$
So that $d[Z,Z]_t=t^2dt$. To establish whether $Y_t$ is a martingale, use now Ito's fomula above, substituting for $d[Z,Z]_t=t^2dt$.
In general: let $X_t=\int_0^tf_sdW_s$. Then the quadratic variation of $X_t$ is given by $$ [X,X]_t=\int_0^tf_s^2ds $$