Calculating $E(2W(s)+W(u)|W(u)=2)$

Let $$W(t)$$ be standard Brownian motion and let $$u.

I know that $$W(s)\sim\mathcal{N}(0,\sqrt{s}), W(u)\sim\mathcal{N}(0,\sqrt{u})$$ and $$2W(s)+W(u)\sim\mathcal{N}(0,\sqrt{8s+u})$$.

How should I calculate $$E(2W(s)+W(u)|W(u)=2)$$?

If I'm not missing something: conditioned on $$W_u$$, the random variable $$W_s - W_u$$ is independent with distribution $$\mathcal{N}(0, s - u)$$. Therefore you can compute the expectation by rewriting $$2W_s = 2(W_s - W_u) + 2W_u$$:

$$\mathbf{E}[2 W_s + W_u | W_u=2] = \mathbf{E}[2(W_s - W_u) + 3W_u |W_u=2] = \mathbf{E}[2(W_s - W_u) | W_u] + 6 =6.$$

Edit: if $$s < u$$, you can write

$$W_s = \underbrace{\left(W_s - \frac{s}{u}W_u\right)}_{g_1} + \frac{s}{u} W_u$$

and verify that $$g_1$$ is independent of $$W_u$$, since both are Gaussian random variables and their covariance function is

$$\mathbf{E}((W_s - (s/u)W_u) W_u) = \mathbf{E}(W_s W_u) - \frac{s}{u}\mathbf{E}(W_u^2) = \underbrace{\min(s, u)}_{(*)} - s = 0,$$ where $$(*)$$ is a standard property of Brownian motion (see e.g. [Le Gall, '16]). Then you can replace $$W_s$$ in your expectation appropriately.

• But how would this now change if s<u? – user570271 May 16 at 18:55
• @user570271: See edit. – VHarisop May 16 at 19:02
• $\operatorname E = 6$ is correct if we ignore the condition on the distribution of $2 W_s + W_u$, which is not compatible with $s > u$. – Maxim May 26 at 18:28

I believe the $$\mathcal N(0, s)$$ notation is more common ($$s$$ is the variance). A Wiener process is Gaussian, therefore $$(W_s, W_u)$$ is jointly normal. The condition $$\operatorname{Var}(2 W_s + W_u) = \begin{pmatrix} 2 & 1 \end{pmatrix} \begin{pmatrix} s & \sigma \\ \sigma & u \end{pmatrix} \begin{pmatrix} 2 & 1 \end{pmatrix}^t = 8 s + u$$ gives $$\sigma = s$$. Since $$\sigma = \operatorname{cov}(W_s, W_u) = \min(s, u)$$, the problem statement is contradictory. If $$s \leq u$$, then the problem reduces to finding the conditional distribution of $$W_s \mid W_u$$. The matrix product in the general formula becomes simply $$s/u$$, therefore $$\operatorname{E}(W_s \mid W_u = w_u) = \frac s u w_u.$$