Expectation of martingales

Question

If I know that $\{M_t\}_t$ is a martingale, we know that

$$\mathbb{E}({M_tM_s})=\mathbb{E}(M_{\min({t,s})}^2)$$

Is there something I can say about $\mathbb{E}(M_tM_sM_r)$?

Answer 1

Assume without loss of generality that $r \leq s \leq t$. By conditioning with respect to $\mathcal{F}_s$, we find using the martingale property

$$\mathbb{E}(M_r M_s M_t) = \mathbb{E}(M_r M_s \mathbb{E}(M_t \mid \mathcal{F}_s)) = \mathbb{E}(M_r M_s^2).$$

If we denote by $\langle M \rangle_t$ the quadratic variation (i.e. the unique increasing process $(A_t)_{t \geq 0}$ such that $A_0 = 0$ and $M_t^2-A_t$ is a martingale), then we obtain by conditioning with respect to $\mathcal{F}_r$

$$\begin{align*} \mathbb{E}(M_r M_s M_t) &= \mathbb{E}(M_r \mathbb{E}(M_s^2 - \langle M \rangle_s \mid \mathcal{F}_r)) + \mathbb{E}(M_r \langle M \rangle_s) \\ &= \mathbb{E}(M_r (M_r^2-\langle M \rangle_r)) + \mathbb{E}(M_r \langle M \rangle_s) \\ &= \mathbb{E}(M_r^3) + \mathbb{E}(M_r (\langle M \rangle_s- \langle M \rangle_r)). \end{align*}$$

This shows in particular that the identity

$$\mathbb{E}(M_r M_s M_t) = \mathbb{E}(M_{\min\{r,s,t\}}^3) \tag{1}$$

does, in general, not hold true. However, $(1)$ is e.g. correct if $(M_t)_{t \geq 0}$ is a Brownian motion:

Example: Let $(M_t)_{t \geq 0}$ be a Brownian motion. Then it is well-known that $\langle M \rangle_t = t$ and therefore the above computations show

$$\mathbb{E}(M_r M_s M_t) = \mathbb{E}(M_r^3) + (s-r) \underbrace{\mathbb{E}(M_r)}_{0} = \mathbb{E}(M_r^3)$$

for any $r \leq s \leq t$.