# Expecation of a Brownian Motion

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?

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

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