# Conditional expectation of a function of brownian motion.

Assuming that $$\{W(t) | t \geq 0\}$$ is a Brownian motion, I am trying to find following conditional expectation

$$\mathbb{E}\left[W^{2}(4) | W(1), W(2)\right]$$

My try:

What I think is that I should introduce the conditional terms $$W(1)$$ and $$W(2)$$ into $$W^{2}(4)$$ to solve the problem. I tried

$$\mathbb{E}\left[([W(4)-W(2)]+[W(2)-W(1)+W(1)])^2\right]$$

but this introduces too many terms that don't get any simpler, like $$(W(4)W(2)^2)$$.

Is the approach incorrect?

• Write $$W^2(4) = (W(4)-W(2)+W(2))^2 = W(4)^2 +2 (W(4)-W(2))W(2) + W(2)^2$$ and then pull out the $W(2)$-terms from the conditional expectation; use that $W(4)-W(2)$ is independent from $W(2)$ to compute the remaining conditional expectation. – saz Mar 6 '19 at 7:22

A fundamental property of Brownian motion is that $$W(t)-W(s)$$ is independent of $$\sigma(W(u),u\leqslant s)$$ for all $$s. Here, in this context, we know that $$W(4)-W(2)$$ is independent of $$\mathcal F:=\sigma(W(1),W(2))$$. But unfortunately, we have to compute the conditional expectation of $$W(4)^2$$ with respect to $$\mathcal F$$. So in the last expression, we make the term $$W(4)-W(2)$$ appear like this: $$W(4)^2=(W(4)-W(2)+W(2))^2= (W(4)-W(2) )^2+2(W(4)-W(2) )W(2)+W(2)^2.$$ Now we take the conditional expectation with respect to $$\mathcal F$$: $$\mathbb E\left[W(4)^2\mid\mathcal F\right]= \mathbb E\left[(W(4)-W(2) )^2\mid\mathcal F\right]+\mathbb E\left[2(W(4)-W(2) )W(2)\mid\mathcal F\right]+\mathbb E\left[W(2)^2\mid\mathcal F\right].$$
• first term: $$(W(4)-W(2) )^2$$ is independent of $$\mathcal F$$ hence $$\mathbb E\left[(W(4)-W(2) )^2\mid\mathcal F\right]=\mathbb E\left[(W(4)-W(2) )^2 \right]=2$$.
• second term: $$W(2)$$ is $$\mathcal F$$-measurable and $$W(4)-W(2)$$ is independent of $$\mathcal F$$: we get zero.
• third term: $$W(2)^2$$ is $$\mathcal F$$-measurable.