Conditional Expectation of a Product of Brownian Motion Values

Summary

The document solves a conditional expectation involving two values of standard Brownian motion given an earlier value. Although the later values have different marginal conditional distributions, their product expectation depends on their shared path and cannot be found by multiplying those marginal means.

The derivation writes each Brownian value as the earlier observation plus an increment. Brownian increments after the conditioning time are independent of the past, and the increment from time one to time two has variance one. This yields a conditional product expectation of one plus the square of the conditioned value, and setting that value to zero gives one. The result illustrates how conditional dependence and shared increments matter in moment calculations. It is a probability exercise rather than a trading method, and the discussion does not extend the calculation to more general processes or time points.

Key ideas

• Brownian motion values after a conditioning time share increments, so their conditional product cannot be inferred from marginal distributions alone.
• Decomposing each value into the conditioned value and subsequent increments simplifies the expectation.
• The conditional expectation of the product is one plus the square of the observed value at time one.
• At a conditioned value of zero, the product expectation equals one.

Tags

# The conditional mean of a product of standard Brownian motions


# The conditional mean of a product of standard Brownian motions












Suppose $\{W_t, t>=0\}$ is a Standard Brownian Motion. How to compute $ \mathbb{E} \left[ W_2 W_3 \vert W_1 =0 \right]$? We know $ W_2 \vert W_1 = 0 \sim N(0,1)$ and $ W_3 \vert W_1 = 0 \sim N(0,2)$. Thank you so much.

## Answer by Gordon (score 1)

https://quant.stackexchange.com/a/43927

The conditional expectation with respect to $W_1=0$ can be treated as the conditional expectation with respect to $W_1$ and then set $W_1$ to 0. See more discussions in this question.

Note that $W_3 = W_3-W_2+W_2$, and $W_2 = W_2-W_1+W_1$. Then \begin{align*} E\big(W_2W_3\mid W_1\big) &= E\Big(\big(W_3-W_2\big)\big(W_2-W_1\big)\\ &\qquad+ \big(W_3-W_2\big)W_1 + \big(W_2-W_1\big) W_2 + W_2 W_1\mid W_1\Big)\\ &=E\Big(\big(W_2-W_1\big) W_2 + W_2 W_1\mid W_1\Big)\\ &=E\Big(W_2^2\mid W_1\Big)\\ &=E\Big(\big(W_2-W_1\big)^2 + 2\big(W_2-W_1\big) W_1 + W_1^2\mid W_1\Big)\\ &=1+ W_1^2. \end{align*} That is, \begin{align*} E\big(W_2W_3\mid W_1=0\big)=1. \end{align*}

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