Conditional Brownian Motion

Question

What is wrong with the following logic:

let $0\leqslant s \leqslant t \leqslant u$, find $E[W_t | W_s, W_u]$

\begin{align*} E[W_t | W_s, W_u] &= E\left.\left[W_t - \frac{t}{u} W_u + \frac{t}{u}W_u\right|W_s,W_u\right]\\ &=E\left.\left[W_t - \frac{t}{u} W_u \right|W_s\right] + \frac{t}{u}W_u\\ &=E\left[W_t|W_s\right] - \frac{t}{u} E\left[W_u|W_s\right] + \frac{t}{u} W_u\\ &=W_s+\frac{t}{u}(W_u-W_s). \end{align*}

I know this is is the wrong answer, but I am struggling to identify which part of this argument is incorrect?

Answer 1

@AlexR. is correct: Your argumentation fails in the second line. You claim that

$$\mathbb{E} \left( W_t - \frac{t}{u} W_u \mid W_s, W_u \right) = \mathbb{E} \left( W_t - \frac{t}{u} W_u \mid W_s \right)$$

because $(W_t- \frac{t}{u})$ and $W_u$ are independent. This equality does not hold because $W_u$ and $W_s$ are not independent. The following statement is correct

Let $X,Y,Z$ be random variables such that $X,Y$ are jointly independent from $Z$. Then $$\mathbb{E}(f(X) \mid Y,Z) = \mathbb{E}(f(X) \mid Z).$$

whereas this statement is (in general) not correct:

Let $X,Y,Z$ be random variables such that $X$ and $Y$ are independent. Then $$\mathbb{E}(f(X) \mid Y,Z) = \mathbb{E}(f(X) \mid Z).$$