Let $(W_t)$ be standard Brownian motion. Compute:
$$ E[W_1W_2W_3 | W_4 = 5]$$
I've tried transforming $W_1W_2W_3$ into independent random variables (since Brownian motion has independent increments) but I've been stuck on this problem for a very long time.
If anyone has any idea, please tell me – thank you in advance.
Best Answer
I suppose there is some clever way to compute this, but currently I don't see how. The approaches which came to my mind require quite some computations. I'm going to use the following result:
From $(1)$, we see that $$\mathbb{E}(W_t \mid W_s = x, W_u=y) = m \stackrel{\text{def}}{=} \frac{u-t}{u-s} x+ \frac{t-s}{u-s} y \tag{2}$$ and $$\mathbb{E}(W_t^2 \mid W_s = x, W_u = y) = \sigma^2+m^2 \stackrel{\text{def}}{=} \frac{(u-t)(t-s)}{u-s} + \left( \frac{u-t}{u-s} x+ \frac{t-s}{u-s} y \right)^2. \tag{3}$$
Moreover, we will need the following
Since the proofs of the three statements are similar, I only prove a). By the tower property, we have
\begin{align*} \mathbb{E}(W_1 W_2 \mid W_4) &= \mathbb{E} \bigg[ \mathbb{E}(W_1 W_2 \mid W_1, W_4) \mid W_1 \bigg] \\ &= \mathbb{E} \bigg[ W_1 \underbrace{\mathbb{E}(W_2 \mid W_1,W_4)}_{\stackrel{(2)}{=} \frac{2}{3} W_1 + \frac{1}{3} W_4} \mid W_4 \bigg] \\ &= \frac{2}{3} \mathbb{E}(W_1^2 \mid W_4) + \frac{W_4}{3} \mathbb{E}(W_1 \mid W_4) \\ &\stackrel{(2),(3)}{=} \frac{2}{3} \left( \frac{3}{4} + \left[ \frac{W_4}{4} \right]^2 \right) + \frac{W_4}{3} \frac{W_4}{4} \\ &= \frac{1}{2} + \frac{1}{8} W_4^2. \end{align*}
Now, finally, we can compute the conditional expectation which we are interested in. To this end, we note that
$$B_t := W_{4-t}-W_4, \qquad t \in [0,4],$$
is also a Brownian motion and
$$\mathbb{E}(W_1 W_2 W_3 \mid W_4 = x) = \mathbb{E}((B_1+x)(B_2+x)(B_3+x) \mid B_4 = -x).$$
Expanding the brackets on the right-hand side and applying our lemma, we get
\begin{align*} \mathbb{E}(W_1 W_2 W_3 \mid W_4=x) &= \mathbb{E}(B_1 B_2 B_3 \mid B_4 = -x) + x \mathbb{E}(B_2 B_3 \mid B_4 = -x) + x \mathbb{E}(B_1 B_3 \mid B_4=-x) \\ &+ x \mathbb{E}(B_1 B_2 \mid B_4=-x) + x^2 \mathbb{E}(B_2 \mid B_4=-x) + x^2 \mathbb{E}(B_1 \mid B_4=-x) \\ & +x^2 \mathbb{E}(B_3 \mid B_4=-x)+x^3 \\ &= \mathbb{E}(B_1 B_2 B_3 \mid B_4 = -x) + \frac{x}{2} + \frac{3}{8} x^3 + \frac{x}{2} + \frac{1}{8} x^3 + \frac{x}{4} \\ &+ \frac{9}{48} x^3 - \frac{x}{2} x^2 - \frac{x}{4} x^2 - \frac{3x}{4} x^2 + x^3 \\ &=\mathbb{E}(B_1 B_2 B_3 \mid B_4 = -x) + \frac{5}{4} x+ \frac{9}{48} x^3 \end{align*}
Finally, we note that by the symmetry of Brownian motion
$$\mathbb{E}(B_1 B_2 B_3 \mid B_4 = -x) = - \mathbb{E}(B_1 B_2 B_3 \mid B_4 = x)= - \mathbb{E}(W_1 W_2 W_3 \mid W_4=x)$$
and so
$$\mathbb{E}(W_1 W_2 W_3 \mid W_4 = x) = \frac{1}{2} \left( \frac{5}{4} x + \frac{9}{48} x^3 \right).$$
In particular,
$$\mathbb{E}(W_1 W_2 W_3 \mid W_4 = 5) = \frac{25}{8} + 125 \frac{9}{48} = 26,5625.$$
Remark: An alternative approach would be to use the conditional density $p_{(W_1,W_2,W_3) \mid W_4}$ (it's not the difficult to calculate it explicitly) and then to use that
$$\mathbb{E}(f(W_1,W_2,W_3) \mid W_4 = x) = \int \int \int f(u,v,w) p_{(W_1,W_2,W_3) \mid W_4}(u,v,w \mid x) \, du \, dv \, dw.$$
However, as far as I can see, the computations are somewhat lengthy.