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 $B_{t}^{3}$ ?

commented Mar 23, 2016 by (100 points)
edited Mar 23, 2016

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

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

commented Mar 24, 2016 by (330 points)
edited Mar 24, 2016

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)

commented Mar 24, 2016 by (1,390 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.