# Conditional expectation of Brownian motion given its absolute value.

Assume that $W_t$ is Brownian motion (1-D) and that $t<T$.

How can I compute $$E(W_t||W_T|),$$

the conditional expectation of $W_t$ given $|W_T|$, i.e. with respect to the $\sigma$-algebra $F$ induced by $|W_T|$?

The $\sigma$-algebra induced by $|W_T|$ would be formed by sets of the form $S\cup S'$, where $S=X_T^{-1}(A)$, $A\subset \mathbb{R}_+$ and $S'=X_T^{-1}(-A)$. Since $X_T$ is $N(0,1)$, $P(S)=P(S')$. We need $\int _{S\cup S'}E(W_t||W_T|)dP=\int_{S\cup S'}W_tdP=0$, for all $S\cup S'$ in $F$.But $E(W_t||W_T|)$ should be $F$-measurable. Therefore ... I guess I should use that last condition now.

I am not clear what to deduce next.

• Got something from the answers below? – Did Nov 8 '14 at 14:20

The distributions of $(W_t,W_T)$ and $(-W_t,-W_T)$ coincide hence the distributions of $(W_t,|W_T|)$ and $(-W_t,|W_T|)$ coincide. Conditional expectations depend only on distributions$^{(\ast)}$ hence $E(W_t\mid|W_T|)=E(-W_t\mid|W_T|)=-E(W_t\mid|W_T|)$, which implies $E(W_t\mid|W_T|)=0$.
The argument shows the much stronger result that the conditional distribution of $W_t$ conditionally on $|W_T|$ is symmetric.
$^{(\ast)}$ ...In the following sense: if the random variables $(U,V)$ and $(U',V')$ have the same joint distribution and $E(U\mid V)=g(V)$, then $E(U'\mid V')=g(V')$.
Another way of looking at it is to write $W_t = W_t \mathbf{1}_{ \left( W_T > 0 \right)} + W_t \mathbf{1}_{ \left( W_T < 0 \right)}$, then using the conditional expectation formula $$\mathbf{E}\left[ X \vert A\right] = \frac{ \mathbf{E} \left[ X \mathbf{1}_{A} \right ]}{\mathbf{P}(A)},$$ for events $A = \left\{ W_T > 0 \right\}$, and $A = \left \{ W_T < 0 \right\}$ we get that $$\mathbf{E} \left[ W_t \right. \left| |W_T| \right] = \frac{1}{2} \mathbf{E} \left[ W_t \right. \left| W_T>0, |W_T| \right] + \frac{1}{2} \mathbf{E} \left[ W_t \right. \left| W_T<0, |W_T| \right] \\ =\frac{1}{2} \mathbf{E} \left[ W_t \right. \left| W_T =|W_T| \right] + \frac{1}{2} \mathbf{E} \left[ W_t \right. \left| W_T = -|W_T| \right].$$ Each of the expectations above are the expectations of a brownian bridge that starts at $0$ at time $0$ and ends at $\pm |W_T|$ at time $T$. That is, $$\mathbf{E} \left[ W_t \right. \left| W_T = \pm|W_T| \right] = \pm\frac{t}{T} |W_T|,$$ and hence $\mathbf{E} \left[ W_t \right. \left| |W_T| \right] = 0$.