Brownian Motion $E[B_t \cdot B_s \cdot B_v]$ with $0 < t < s <v$

Question

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

Answer 1

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$

Answer 2

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.

Answer 3

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.