Let $B_{t}$ denote standard Brownian motion and $F_{t}$ be its natural filtration. What is $E[B_{t}^{3}|F_{s}]$ ?

+1 vote
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My attempt so far:

$$E[B_{t}^{3}|F_{s}]=E[(B_{t}-B_{s}+B_{s})^{3}|F_{s}]=E[(B_{t}-B_{s})^3+B_{s}^{3}|F_{s}]=B_{s}^{3}+E[(B_{t}-B_{s})^3|F_{s}]$$

I'm not too sure what to do from here (assuming what I've done so far is even right). Any suggestions?

The second part of the question then asks how can a martingale be constructed out of $B_{t}^{3}$ ?

by (100 points)
edited

moving from the second to third step in your attempt seems inconsistent Jason.

$$(a+b)^3 = a^3+ 3a^2b+3ab^2+b^3$$

by (330 points)
edited

Once you've created independent intervals, the expectation of all the terms will be zero except for the cubic terms in the expansion so I think my method is correct.

(Never mind the non cubic terms won't all be equal to zero so I will have to expand it fully.Thanks for your help)

by (1.4k points)

Yeah, I also only realised that in the tut this morning. Thanks Sinjun :)

$$B_{t}^{3} - {3}{t}\cdot B_{t}$$
and can show it is a martingale in a similar way to how we showed $$B_{t}^{2} - {t}$$ is a martingale.