Is $E[X_sX_t^2]=0 \,\,s<t$ for a MDS?

Question

Let $(X_t,\mathcal F_t)$ be a stationary martingale difference sequence (MDS). Can we say that $$E[X_sX_t^2]=0 \quad s<t \quad ?$$

For $s>t$ we can use the the law of iterated expectations and the pull-out property.

Thanks a lot for your help.

Answer 1

This is not true in general.

One example would be to take $Y_n \sim \mathcal{N}(0, 2)$ and $Z_n \sim \mathcal{N}(0, 1)$ independent with $$ X_0 = \begin{cases} Y_0 & \text{w.p. }1/2,\\ Z_0 & \text{w.p. }1/2, \end{cases} \quad X_{n+1} = \begin{cases} Y_{n+1} & \text{if } X_n \geq 0, \\ Z_{n+1} & \text{if } X_n < 0, \end{cases} $$ and the natural filtration $\mathcal{F}_n = \sigma(X_0, \dotsc, X_n)$.

But now, we have that, since $Y_{n+1}$ and $Z_{n+1}$ are independent of $X_n$, \begin{align*} \mathbf{E} X_n X_{n+1}^2 &= \mathbf{E}[Y_{n+1}^2 X_n; X_n \geq 0] + \mathbf{E}[Z_{n+1}^2 X_n; X_n < 0] \\ &= \mathbf{E} Y_{n+1}^2 \mathbf{E}[X_n; X_n \geq 0] + \mathbf{E} Z_{n+1}^2 \mathbf{E}[X_n; X_n < 0] \\ &= (\mathbf{E} Y_{n+1}^2 - \mathbf{E} Z_{n+1}^2) \mathbf{E}[X_n; X_n \geq 0] + \mathbf{E} Z_{n+1}^2 \mathbf{E} X_n \\ &= \mathbf{E}[X_n; X_n \geq 0] \\ &> 0. \end{align*}