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Suppose $t\geq s$ and $W_i$ is a Brownian motion

$\mathbb{E}_{0}\left[W_{s}^{2} W_{t}\right]$

Now I found online that it can be written as

$\mathbb{E}_{0}\left[W_{s}^{2} W_{t}\right]=\mathbb{E}_{0}\left[W_{s}^{3}\right]+\mathbb{E}_{0}\left[W_{s}^{2}\left(W_{t}-W_{s}\right)\right]$

However, I am unable to follow this line of reasoning

Could someone explain me which steps are taken to rewrite $\mathbb{E}_{0}\left[W_{s}^{2} W_{t}\right]$ in this form?

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    $\begingroup$ What have you tried? Where did you get stuck? What do you know about the expected value? $\endgroup$ Commented Jun 9, 2022 at 10:39

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This follows from the linearity of expectation: $$\mathbb{E}_{0}\left[W_{s}^{3}\right]+\mathbb{E}_{0}\left[W_{s}^{2}\left(W_{t}-W_{s}\right)\right] = \mathbb{E}_{0}\left[W_{s}^{3}\right]+\mathbb{E}_{0}\left[W_{s}^{2}W_{t}-W_{s}^{2}W_{s}\right] = \mathbb{E}_{0}\left[W_{s}^{3}\right]+\mathbb{E}_{0}\left[W_{s}^{2}W_{t}\right]-\mathbb{E}_0\left[W_{s}^{3}\right] = \mathbb{E}_0\left[W_{s}^{2}W_{t}\right]$$

Edit: The reasoning behind writing it this way is you reduce computing of your expectation of the product of two correlated normal random variables to expectations of products of uncorrelated normal random variables, which is fairly easy to compute.

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  • $\begingroup$ Thank you, I was looking for a way too difficult solution. This is extremely clear $\endgroup$
    – WHN
    Commented Jun 9, 2022 at 12:30

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