Brownian Motion $E[B_t \cdot B_s \cdot B_v]$ with $0 < t < s Can someone derive analytically the following Brownian Motion question? $E[B_t \cdot B_s \cdot B_v]$ with $0 < t < s < v$.

We can also use the martingale property of the BM. Let us denote $\mathcal{F}_t$ the natural filtration of the Brownian motion $B_t$. 
We now that $\{B_t\}_{t\geq0}$ and $\{B_t^2-t\}_{t\geq0}$ are martingales. I recall that $B_t \sim \mathcal{N}(0,t)$.
We have then
\begin{align*}
E\left[B_sB_tB_v\right] &= E\left[\underbrace{B_sB_t}_{\mathcal{F}_t-\text{mesurable}}E[B_v|\mathcal{F}_t]\right]\\
&= E\left[B_sB_t^2\right] = E\left[B_s(B_t^2-t)+ tB_s\right] \\
&= E\left[B_sE[(B_t^2-t)|\mathcal{F}_s]\right] = E\left[B_s^3-B_s\right]  \\
&= 0
\end{align*}
The last equality comes from the fact the odd moments of gaussian r.v. are 0.
