Show that $M_t=e^{X_t-\frac{t^3}{6}}$ is a martingale 
Let $X_t=\int_0^t s\text dB_s$, show that $M_t=e^{X_t-\frac{t^3}{6}}$ is a martingale and give its distribution. 

First encounter with the form of integral and Brownian motion, the tool I came to is only the definition ($E(M_{t+1}|\mathcal{F}_s)=M_t$). What other properties should I know in the question.
 A: It is actually $\mathbb{E}[M_t|\mathcal{F}_s] = M_s$. (definition of martingale) Basically how you do it is to assume that everything up to $M_s$ is known, and compute expressions involving $M_t-M_s$.
$\mathbb{E}[M_{t}|\mathcal{F}_s] = \mathbb{E}[(M_{t}-M_s)+M_s|\mathcal{F}_s]$.
Since we conditioned on $\mathcal{F}_s$, we known that the expectation of $M_s$ is just $M_s$, as it is already known (non random). Then $M_t-M_s$ is independent of any knowledge we know from $\mathcal{F}_s$, so we remove the conditioning.
$ = \mathbb{E}[(M_{t}-M_s)]+M_s$.
So $e^{t^3/6}\mathbb{E}[M_{t}] = \mathbb{E[e^{X_t}]}$. This is the same as finding the MGF of $X_t$ with parameter $1$.
$\mathbb{E}[X_t] = 0$ and $Var(X_t) = \int_0^t s^2 ds = t^3/3$ (use Ito isometry)
Integrating brownian motion gives normal random variates, so $X_t \sim N(0,\frac{t^3}{3})$.
Hence the MGF evaluated at $1$ is $e^{\frac{t^3}{6}}$. Thus $\mathbb{E}[M_{t}] = 1$. 
$\mathbb{E}[M_{t}-M_s] = 0$
Hence our final answer is just $M_s$ as required
