Conditional Brownian Motion Cubic Moments Using Independent Increments

Summary

The document derives the conditional third moment of Brownian motion at a later time given its value at an earlier time. It writes the later value as the known earlier value plus an independent increment, expands the cube, and takes the conditional expectation. The increment has zero mean, variance equal to the elapsed time, and zero third moment, so only the variance term and the known earlier value contribute. The resulting conditional third moment is the earlier value cubed plus three times elapsed time times that value.

The explanation highlights a useful distinction: Brownian motion has zero unconditional odd moments because it is centered and symmetric, but conditioning on its current value generally makes an odd moment nonzero. The question also asks for the conditional fourth moment, but the provided answer derives only the third moment; it gives no fourth-moment formula or numerical example. The method, however, extends through expansion and the independent-increments property.

Key ideas

• Express the later Brownian value as the earlier value plus an independent increment.
• Conditioning on the earlier value lets it be treated as fixed when taking expectations over the increment.
• The conditional third moment equals the earlier value cubed plus three times elapsed time times that value.
• Unconditional odd moments vanish for centered Brownian motion, while conditional odd moments need not vanish.
• The response does not derive the fourth conditional moment requested in the question.

Tags

# Expectation of Bt^4 given BS


# Expectation of Bt^4 given BS












What is the expectation of Bt^4 and Bt^3 given Bs? Given t>s.

I understand that the expectation of Bt given Bs is Bs and that the expectation of Bt^2 given Bs is something like Bs - s + t.

## Answer by Bennnn (score 2)

https://quant.stackexchange.com/a/74035

In general for any question like this the trick is to use the independent increments property:

$$B_t^3 = (B_t-B_s + B_s)^3$$ Then from $(a+b)^3 = a^3 + 3a^2b + 3 ab^2 + b^3$ you have

$$E[B_t^3|B_s] =E\left[ (B_t-B_s)^3 +3(B_t-B_s)^2B_s + 3 (B_t-B_s)B_s^2 + B_s^3 | B_s \right]\\ =E[(B_t-B_s)^3] + 3 B_s E[(B_t-B_s)^2] + 3 B_s^2E[B_t-B_s] + B_s^3\\ =0+3B_s(t-s) + 0 + B_s^3 \\ = B_s^3 + 3(t-s) B_s$$

Ie. using the property that $B_t-B_s$ is independent of $B_s$.

Brownian motion processes centred at zero have odd moments of zero expectation while conditional ones don't in general.

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.