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?
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 $B_t$...