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?
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.