Conditional Product Expectation for Brownian Motion

Summary

The note derives the conditional expectation of the product of Brownian motion values at two times, given the value at one of those times. It treats separately the cases where the conditioned time comes first and where it comes second.

When s is earlier than t, the later Brownian value is decomposed into the known value at s plus an independent, mean-zero increment, giving a conditional product equal to the square of the known value. When t is earlier than s, joint normality and zero covariance show that the earlier value minus its linear projection on the later value is independent of that later value. The result is the time ratio multiplied by the square of the conditioned value. The explanation is a short derivation rather than an empirical result, and assumes standard Brownian motion with positive times.

Key ideas

• For s earlier than t, the conditional expectation is W_s squared.
• For t earlier than s, the conditional expectation is (t/s) times W_s squared.
• The later Brownian increment is independent of the value at the earlier time.
• For the reverse ordering, joint normality makes the residual after linear projection independent of W_s.

Tags

# How do we calcualte $E[W_sW_t|W_s]$


# How do we calcualte $E[W_sW_t|W_s]$












$W_t$ is a Brownian motion. How do we calculate this expectation? there are two cases:

- $s < t$

- $t < s$

Do we have to distinguish the two cases or there is a unified way of calculating it

## Answer by Gordon (score 7, accepted)

https://quant.stackexchange.com/a/49487

For $s<t$, then \begin{align*} E\big(W_sW_t \,|\, W_s\big) &= W_sE\big((W_t-W_s + W_s)\,|\,W_s\big) \\ &=W_s^2. \end{align*}

For $0 < t < s$, then $$E\left(W_s \Big(W_t-\frac{t}{s}W_s\Big) \right)= 0,$$ and, given their joint normality, $W_s$ and $W_t-\frac{t}{s}W_s$ are independent. Therefore, \begin{align*} E\big(W_sW_t \,|\, W_s\big) &= W_sE\left(\Big(W_t-\frac{t}{s}W_s +\frac{t}{s} W_s\Big)\,|\,W_s\right) \\ &= \frac{t}{s} W_s^2. \end{align*}

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.