3
$\begingroup$

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.)

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @ George: I realized Ito would get me out once I posted the questions. And I have edited the question to $\alpha \in \mathbb{Z} \cap \{0,1\}^c$. Thanks $\endgroup$ – user17762 Nov 12 '10 at 21:34
2
$\begingroup$

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 $\alpha=1$.

$\endgroup$
  • $\begingroup$ @George and Sivaram: I don't have enough rep to comment on your solution, but $x\mapsto x^3$ is not convex and $E(W_t^3)=0$ for all $t \geq 0$. $\endgroup$ – user940 Nov 12 '10 at 21:41
  • $\begingroup$ it is convex on the positive reals though $\endgroup$ – George Lowther Nov 12 '10 at 21:44
  • $\begingroup$ But that is also a neat alternative method. Essentially, it is the same technique used to show that a one dimensional diffusion has a unique scale function (up to rescaling and addition of a constant). $\endgroup$ – George Lowther Nov 12 '10 at 21:47
  • $\begingroup$ And what happened to your rep!! (I hope you don't mind me asking). Did you give it all away? $\endgroup$ – George Lowther Nov 12 '10 at 21:50
  • 1
    $\begingroup$ I attached a large bounty to an interesting problem about random walks on graphs. My secret strategy is to get George Lowther interested, then sit back and wait for the solution. :) $\endgroup$ – user940 Nov 12 '10 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy