Conditional Expectation of Brownian Motion Given Its Absolute Value
Let $W_t$ be a standard Brownian motion. For $0 \leq s \leq t$, compute: $$E[W_s \mid |W_t|]$$
What is the numerical answer when $s = 1$, $t = 3$, and $|W_t| = 6$?
Hints
- Knowing $|W_t|$ tells you the magnitude but not the direction. Try conditioning on $W_t > 0$ and $W_t \leq 0$ separately, then recombine using the Law of Total Expectation.
- Given $W_t = c$ (known sign and magnitude), the conditional mean $E[W_s \mid W_t = c]$ is the Brownian bridge formula: $\frac{s}{t} c$.
- By symmetry of the normal distribution, $P(W_t > 0) = P(W_t \leq 0) = 1/2$. The positive-direction term is $+\frac{s}{t}|W_t| \cdot \frac{1}{2}$ and the negative-direction term is $-\frac{s}{t}|W_t| \cdot \frac{1}{2}$. They cancel.
Worked Solution
How to Think About It: We condition on the *magnitude* $|W_t|$ but not the *sign* of $W_t$ — strictly weaker than conditioning on $W_t$ itself. If we knew $W_t = +6$, the Brownian bridge gives $E[W_s\mid W_t] = \frac{s}{t}W_t$; if $W_t=-6$ we'd get the negative. Knowing only $|W_t|=6$ leaves the sign a fair coin, and the two signed answers are exact negatives. The heuristic is symmetry / odd-function cancellation: $W_s$ is an odd functional of the path, we've thrown away the only information that could break the $\pm$ symmetry, so the conditional mean must be $0$ — independent of $s$, $t$, and the magnitude.
Quick Estimate (sanity check). No number to estimate — the answer is exactly $0$ — so probe the structure instead. Under $W_t = c$, the bridge mean of $W_s$ is $\frac{s}{t}c$, *linear and odd in $c$*. Given $|W_t|=6$, the sign is $\pm$ with probability $\tfrac12$ each (Brownian motion is symmetric about $0$). Average: $\tfrac12\cdot\frac{s}{t}(+6) + \tfrac12\cdot\frac{s}{t}(-6) = 0$. Plug the specific numbers ($s=1$, $t=3$): $\tfrac12\cdot\tfrac13(6) + \tfrac12\cdot\tfrac13(-6) = 1 - 1 = 0$. The two signed drifts ($+2$ and $-2$) cancel exactly.
Formal Solution:
Condition on the sign of $W_t$ (Law of Total Expectation):
$$E[W_s\mid |W_t|] = E[W_s\mid |W_t|, W_t>0]\,P(W_t>0) + E[W_s\mid |W_t|, W_t\le 0]\,P(W_t\le 0).$$
Since $W_t\sim N(0,t)$, symmetry gives $P(W_t>0)=P(W_t\le 0)=\tfrac12$. The Brownian bridge mean $E[W_s\mid W_t=c]=\frac{s}{t}c$ gives, with $v=|W_t|\ge 0$:
$$E[W_s\mid |W_t|=v, W_t>0] = \frac{s}{t}v, \qquad E[W_s\mid |W_t|=v, W_t\le 0] = -\frac{s}{t}v.$$
Therefore
$$E[W_s\mid |W_t|=v] = \tfrac12\cdot\frac{s}{t}v + \tfrac12\cdot\left(-\frac{s}{t}v\right) = \boxed{0},$$
for every $s\le t$ and every $v\ge 0$. At $s=1$, $t=3$, $|W_t|=6$: the answer is $0$.
Answer: $E[W_1 \mid |W_3|=6] = \mathbf{0}$ (and $E[W_s\mid|W_t|]=0$ generally).
Intuition
The answer being zero (regardless of the magnitude $|W_t|$ and regardless of $s$) is a clean illustration of how partial information can leave you with no predictive power. Conditioning on $|W_t|$ tells you the process strayed far from zero by time $t$, but gives you no information about which direction. Since Brownian motion is symmetric around zero, the positive and negative scenarios are equally likely and their effects on $E[W_s]$ perfectly cancel.
The Brownian bridge is the key building block here: conditional on a future endpoint, the Brownian motion interpolates linearly in expectation. This result shows up in practice when you observe the magnitude of a price move but not its direction -- for example, when you know a stock moved a lot overnight but have not yet seen the direction. Your best estimate of where it was at any intermediate time remains the unconditional expectation (zero, or whatever the drift implies), precisely because the two directions are equally likely given only the magnitude.