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

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

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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.

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

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