I'm just having a bit of trouble figuring out what the what the expectation of Brownian motion cubed is. I know

$E(W_t)$ = 0

$E(W_{t}^{2})$ = $t$

but what would $E(W_{t}^{3})$ (The expectation of Brownian motion cubed) be

Any help would be appreciated.


2 Answers 2


$W_t$ is just a Normal random variable with mean $0$ and variance $t$. This means that its easy to even just look up its moments. All of its odd moments are $0$ and its even moments are given by $$\mathbb{E}[W_t^{2k}] = t^{k} (2k-1)!!$$ where $(2k-1)!!$ is the product of all odd integers between $1$ and $2k-1$.


$EW_t^{3}=0$ because distribution of $W_t$ is symmetric about the origin.

  • $\begingroup$ Hi thank you for you help. I just wanted to confirm that Wt is the Wiener process $\endgroup$ Mar 13, 2020 at 23:29

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