Expectation of the product of Brownian motions

Question

I'm new to Stack Exchange. I'd like to find the expectation of the product of three Brownian motions:

$E(B(t_1)B(t_2)B(t_3))$

I know from a previous post here that

$E(B(t_1)B(t_2)B(t_3)B(t_4))=2t_1t_2+t_1t_3$

I know that $E(B_tB_s) = min${$s,t$} and $E(B_t) = 0$.

Answer 1

Solution I: (Similar to @Did's suggestion)

Set $x := \mathbb{E}(B_{t_1} B_{t_2} B_{t_3})$. Then, since $(-B_t)_{t \geq 0}$ is also a Brownian motion, we get

$$x = \mathbb{E}((-B_{t_1}) \cdot (-B_{t_2}) \cdot (-B_{t_3})) = - \mathbb{E}(B_{t_1} B_{t_2} B_{t_3}) = -x;$$

hence, $x=0$.

Solution 2:

Suppose (without loss of generality) that $t_1 \leq t_2 \leq t_3$. Recall that $(B_t)_{t \geq 0}$ and $(B_t^2-t)_{t \geq 0}$ are martingales. It follows from the tower property that

$$\begin{align*} \mathbb{E}(B_{t_1} B_{t_2} B_{t_3}) &= \mathbb{E}[ \mathbb{E}(B_{t_1} B_{t_2} B_{t_3} \mid \mathcal{F}_{t_2})] \\ &= \mathbb{E}(B_{t_1} B_{t_2} \underbrace{\mathbb{E}(B_{t_3} \mid \mathcal{F}_{t_2})}_{B_{t_2}}) \\ &= \mathbb{E}(B_{t_1} B_{t_2}^2). \end{align*}$$

(Here we have used that $(B_t)_{t \geq 0}$ is a martingale.) Using the fact that $(B_t^2-t)_{t \geq 0}$ is a martingale, we get

$$\begin{align*} \mathbb{E}(B_{t_1} B_{t_2} B_{t_3}) &= \mathbb{E}[ \mathbb{E}(B_{t_1} B_{t_2}^2 \mid \mathcal{F}_{t_1})] \\ &= \mathbb{E}(B_{t_1} (B_{t_1}^2-(t_1-t_2)) \\ &= 0 \end{align*}$$

as $\mathbb{E}(B_{t_1}) = \mathbb{E}(B_{t_1}^3)=0$.