Expectation of Brownian Motion Question: $E[w(t) e^{w(s)}]$

Question

I am trying to solve problems to revise for my exam. This question has two parts, the first asks to find:

$$ E[W(t)e^{W(t)}] $$

where $W$ is a Brownian motion. I got the result to be $te^{\frac{t}{2}}$ (which is what someone got on another stack post). The next part asks to find:

$$ E[W(s)e^{W(t)}] $$

where $0 \leq s \leq t$.

I assume I'm meant to use the previous part to solve this, but I have no idea how. Could someone please explain?

Thanks

Answer 1

You can proceed as follow:

\begin{align*} \mathbb{E}\left(W_s e^{W_t} \right) &= \mathbb{E}\left( \mathbb{E} \left(W_s e^{W_t - W_s + W_s} \rvert W_s\right) \right)\\ &= \mathbb{E}\left(W_s e^{W_s} \mathbb{E} \left(e^{W_t - W_s} \rvert W_s\right) \right)\\ &= \mathbb{E}\left(W_s e^{W_s} \mathbb{E} \left(e^{W_t - W_s}\right) \right)\\ &= \mathbb{E}\left(W_s e^{W_s}\right)\mathbb{E} \left(e^{W_t - W_s}\right) \end{align*}

Knowing that $\mathbb{E} \left(e^{W_t - W_s}\right) = e^{\frac{1}{2}(t-s)}$ you can use your first result