Brownian motion conditional on the future

brownian motionconditional-expectation

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:

Proposition. Let $(W_t)_{t \geq 0}$ be a one-dimensional Brownian motion and $s<t<u$. Then $$\mathbb{P}(W_t \in A \mid W_s = x, W_u=y) = \frac{1}{\sqrt{2\pi \sigma^2}} \int_A \exp \left(- \frac{(z-m)^2}{2\sigma^2} \right) \, dz \tag{1}$$ with $$m := \frac{u-t}{u-s} x + \frac{t-s}{u-s} y \qquad \sigma^2 = \frac{(u-t)(t-s)}{u-s}.$$ (Note that the right-hand side of $(1)$ is the Gaussian distribution with mean $m$ and variance $\sigma^2$.)

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

Lemma: Let $(W_t)_{t \geq 0}$ be a one-dimensional Brownian motion. Then

a) $\mathbb{E}(W_1 W_2 \mid W_4 = x) = \frac{1}{2} + \frac{1}{8} x^2$.

b) $\mathbb{E}(W_1 W_3 \mid W_4 = x) = \frac{1}{4} + \frac{9}{48} x^2$.

c) $\mathbb{E}(W_2 W_3 \mid W_4 = x) = \frac{1}{2} + \frac{3}{8} x^2$.

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.

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