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 $0 < t < s < v$.
 A: You lose parentheses and therefore get incorrect equalities.
$$
\mathbb E[B_t \cdot B_s \cdot B_v] = \mathbb E[B_t \cdot \bigl(B_t+(B_s-B_t)\bigr) \cdot \bigl(B_t+(B_s-B_t)+(B_v-B_s)\bigr)]
$$
Denote $X=B_t$, $Y=B_s-B_t$, $Z=B_v-B_s$. Note that this three r.v. are independent and have zero expectation. 
$$
\mathbb E[B_t \cdot B_s \cdot B_v] 
= \mathbb E[X\cdot (X+Y)\cdot (X+Y+Z)]
$$
$$ = 
\mathbb E[X^3]+2\mathbb E[X^2Y]+\mathbb E[XY^2]+\mathbb E[X^2Z]+\mathbb E[XYZ] 
$$
$$ = 
\mathbb E[X^3]+2\mathbb E[X^2]\mathbb E[Y]+\mathbb E[X]\mathbb E[Y^2]+\mathbb E[X^2]\mathbb E[Z]+\mathbb E[X]\mathbb E[Y]\mathbb E[Z]=0 
$$
since all expected values are zero and $\mathbb E[X^3]=0$ too. 
A: This is what I tried: $E [B_s \cdot B_t \cdot B_v]$ 
$ = E[(B_t \cdot \ B_v − B_s + B_s) \cdot B_s]$ Use linearity
$ = E[(B_t \cdot \ B_v − B_s) \cdot B_s] + E[B_s^2]$
independent increments
$=E[B_t \cdot B_v - B_s] \cdot E[B_s] + E[B_s^2]$
with $E[B_s] = 0$
Therefore
$ = VAR[B_s] + (E[B_s])^2$
with $E[B_s] = 0$
Therefore 
$ = VAR[B_s]$
$ = s$
A: 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.
