Stuck on this question for a very long time: was wondering if any kind soul could help me out:

Suppose $B_t$ is a standard Brownian Motion under measure P.

Question: Create a martingale process that has $B_t^3$ . Justify your process is a martingale under measure P.

My Approach: I have shown how $B_t^3$ is not a martingale under measure P, but I am lost as to how to make it a P-martingale.

This is what I've got: $dB_t^3$ = $3B_t^2$$dB_t$ + $3B_t$$dt$

I know i have to make the dt term zero to make the process a martingale. Any help please?


2 Answers 2


Here are a couple of propositions which could be of help.

Proposition 1 Let $B_t$ be a Brownian motion. Then ${\rm exp}(\theta x - \frac{1}{2} \theta^2 t)$ is a martingale for each $\theta \in \mathbb R$.

Let $H_n(t,x)$ being the $n$-th Hermite polynomial, defined by $$ {\rm exp}(\theta x - \frac{1}{2} \theta^2 t) = \sum_{n=0}^\infty \frac{\theta^n}{n!} H_n(t,x). $$

Proposition 2 Let $B_t$ be a Brownian motion. Then $H_n(t,B_t)$ is a martingale for each $n$.

Since $H_3(t,x) = x^3 - 3 t x$, the proposition above implies that $B^3_t - 3 t B_t$ is a martingale.

Reference: L.C.G. Rogers, D. Williams, Diffusions, Markov Processes, and Martingales: Volume 1, Chapter 1


If this exercise is a step on the discovery of Brownian exponential martingales based on Hermite polynomials, here is a hands-on approach to solve it.

One already knows that $B_t^3-3C_t=\displaystyle\int_0^t3B_s^2\mathrm dB_s$ is a martingale, where $C_t=\displaystyle\int_0^tB_s\mathrm ds$, hence one must deal with $C_t$.

How could $\mathrm dC_t=B_t\mathrm dt$ appear in the RHS of Itô's formula?The simplest way could be to use the identity $$\mathrm d(tB_t)=t\mathrm dB_t+B_t\mathrm dt=t\mathrm dB_t+\mathrm dC_t,$$ thus $C_t=tB_t-D_t$, where $D_t=\displaystyle\int_0^ts\mathrm dB_s$ is a martingale. To sum up, $$ B_t^3-3tB_t=3\int_0^t(B_t^2-t)\mathrm dB_t, $$ hence $B_t^3-3tB_t$ is a martingale.

And naturally, $H_2(x,t)=x^3-3xt$ is the second Hermite polynomial, a fact which suggests how to generalize this result to polynomials of any degree with respect to $B_t$...

  • $\begingroup$ Thanks Did. Really detailed explanation! $\endgroup$
    – Jack Wills
    May 8, 2014 at 9:00
  • $\begingroup$ Welcome. The level of the answer is meant to be adapted to what I perceive of the level of your knowledge. If I am not too far from the target, everything is fine. $\endgroup$
    – Did
    May 8, 2014 at 9:02

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