Conditional Brownian Motion

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?

• why did the conditioning on $W_u$ suddenly disappear in the second equality? Also why are you multiplying by $t/u$? – Alex R. Mar 28 '15 at 4:11
• @AlexR. I am using the fact that $(W_t - \frac{t}{u} W_u) \bot W_u$, so I remove it from the conditioning in the second equality – dimebucker Mar 28 '15 at 4:14

@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 $$\mathbb{E}(f(X) \mid Y,Z) = \mathbb{E}(f(X) \mid Z).$$

• How can I prove this, if for example I want to see what the covariance here, how does one go about solving: $$Cov ( W_t - \frac{t}{u} W_u , (W_s, W_u) )$$ – dimebucker Mar 28 '15 at 8:44
• @dimebucker91 What do you mean by an expression of the form $\text{cov}\, (X, (Y,Z))$ for random variables $X,Y,Z$...? – saz Mar 28 '15 at 8:57
• I guess it doesn't make sense. Im just trying to solve this expectation by using the 'trick' of adding and subtracting an increment. For example if I instead started off with: \begin{align*} E(W_t | W_u,W_s) & = E(W_t - W_s + W_s | W_u,W_s)\\ &=E( W_t-W_s| W_u,W_s) +W_s\\ &=E( (W_t-W_s) -\frac{t-s}{u-s}(W_u-W_s) + \frac{t-s}{u-s}(W_u-W_s)| W_u,W_s) +W_s\\ &=E( (W_t-W_s) -\frac{t-s}{u-s}(W_u-W_s) | W_u,W_s)+ \frac{t-s}{u-s}(W_u-W_s) +W_s\\ \end{align*} The part outside of the expectation is the answer given in the book. That means the expectation is zero, but im not sure how to show this.... – dimebucker Mar 28 '15 at 9:12
• @dimebucker91 I would calculate the conditional expectation using the joint density of $(W_s,W_t,W_u)$. However, right now, I'm not sure how to do this using this increment-trick. Might be better to ask a new question. – saz Mar 28 '15 at 10:04