Find $\beta \in \mathbb{R}$ for which $$2W_t^3+\beta tW_t$$ is a martingale, where $W_t$ is standard Wiener process.

My attempt:

$$E(2W_t^3+\beta tW_t|F_s)=2E(W_t^3|F_s)+\beta tE(W_t|F_s)=2E(W_t-W_s)^3+6W_sE(W_t-W_s)^2+6W_s^2E(W_t-W_s)+W_s^3+\beta tW_s=2E(W_t-W_s)^3+6W_s(t-s)+W_s^3+\beta tW_s=\\6W_s(t-s)+W_s^3+\beta tW_s$$ Now I have to solve: $$6W_s(t-s)+W_s^3+\beta tW_s=2W_s^3+\beta tW_s$$ so $$6(t-s)+\beta t =W_s^2+\beta t$$ I must have done sth wrong. Can someone check my attempt?


Using Ito's Lemma

$d(2 W_t^3 + \beta t W_t) = 6 W_t^2 dW_t + \beta t dW_t + (6 W_t dt + \beta W_t dt)$

The first two terms above don't matter because they will be martingales if you take the integral of both sides. The term in parenthesis is what you want to make $0$. Therefore of course $\beta = -6$

This is actually also consistent with the method you were applying; except you made an algebra mistake. You need to have $2 W_s^3$ instead of $W_s^3$ and then you will see that your expressions also imply $\beta = -6$. Oh I guess you also need to have $\beta s W_s$ on the right hand side of your equation instead of $\beta t W_s$. I think my way is much cleaner. It is important to remember that in the ito interpretation any integral with respect to $dW_t$ is a martingale.


$W_t - W_s$ is distributed like a Gaussian with mean $0$ and variance $\sigma^2 = t-s$. The pdf of such a Gaussian is symmetric about the origin. Since cubing is odd about the origin, what can you say about $E(W_t - W_s)^3$?

  • $\begingroup$ It is equal to zero? $\endgroup$ – luka5z Feb 2 '14 at 19:33
  • $\begingroup$ Am I right? Thanks for any help $\endgroup$ – luka5z Feb 3 '14 at 11:55
  • 1
    $\begingroup$ @luka5z Yes, you are right. $\endgroup$ – saz Feb 3 '14 at 12:05
  • $\begingroup$ edited, but I must have done sth wrong $\endgroup$ – luka5z Feb 3 '14 at 12:31

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