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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?

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    $\begingroup$ 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. $\endgroup$ – saz Mar 6 at 7:22
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A fundamental property of Brownian motion is that $W(t)-W(s)$ is independent of $\sigma(W(u),u\leqslant s)$ for all $s<t$. 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.
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