For what value of $\alpha$ is $W_t^{\alpha}$ a martingale? For what value of $\alpha$ is $W_t^{\alpha}$ a martingale where $\alpha \in \mathbb{Z}$? ($W_t$ standard Brownian motion)
I can prove for $\alpha = 0$ and $\alpha = 1$, it is a martingale. (rather it is obvious). But my question is:
How to prove that $W_t^{\alpha}$ is not a martingale for $\alpha \in \mathbb{Z}\cap\{0,1\}^{c}$?
(This question was asked in an phone interview. They were only interested in the answer which I told correctly.)
 A: First, the question doesn't even make sense if $\alpha$ is not an integer, because $W^\alpha$ is undefined as a real valued process when W goes negative. So, in that case, I'll assume that W is started from a positive value and stopped when it hits zero.
There's two methods: Use Ito's formula for $X=W^\alpha$,
$$
dX = \alpha W^{\alpha-1}\,dW + \frac12\alpha(\alpha-1) W^{\alpha-2}\,dt.
$$
The first term on the right is a local martingale. So the only way that X can be a martingale is for $\alpha(\alpha-1)\int W_t^{\alpha-2}\,dt$ to be a local martingale. It is a standard result that continuous finite variation processes cannot be a local martingale unless they are constant, giving $\alpha(\alpha-1)=0$.
Or, use Jensen's inequality. As $f(x)=x^\alpha$ is strictly convex for $\alpha\not\in[0,1]$ and strictly concave for $\alpha\in(0,1)$ (restricting to the positive reals), the function
$$t\mapsto\mathbb{E}[W_t^\alpha]$$
is strictly increasing for $\alpha\not\in[0,1]$ and strictly decreasing for $\alpha\in(0,1)$. So, $W^\alpha$ is not a martingale in these cases. In fact, it is a submartingale for $\alpha \not\in[0,1]$ and a supermartingale for $\alpha\in(0,1)$.
In this case, Jensen's inequality is the simplest method. However, Ito's formula applies much more generally, to any twice differentiable function of a semimartingale.
A: Assume that $\alpha>0$. 
Apply the optional stopping theorem to the process $W^\alpha_t$ using the hitting time $T$ of $\{-1 , 2\}$. If $W^\alpha_t$ were a martingale, we'd have 
$$0=E(W^\alpha_0)=E(W^\alpha_T)=(2/3)(-1)^\alpha+(1/3)2^\alpha.$$ 
It follows that  $2^{\alpha-1}=1$, so $\alpha=1$.   
