Prove that serial correlation in error terms makes $X_t e_t$ a non-MDS

Question

Given the simple regression $$Y_t = \beta X_t + e_t$$, with $$e_t = \epsilon_t + \theta_1 \epsilon_{t-1} + \theta_2 \epsilon_{t-2}, \epsilon_t \sim WN(0, \sigma^2)$$ and $$X_t iid, X_s \perp e_t \forall s,t$$ Is ${X_t e_t}$ a Martingale Difference Sequence w.r.t $F_{t-1}=(X_{t-1}, e_{t-1}, X_{t-2}, e_{t-2}, ...)$?

This is my attempt so far:

$E(X_t e_t | F_{t-1}) = E(X_t E(e_t|F_{t-1}, X_t)|F_{t-1})$

In the serially uncorrelated case, $E(e_t|F_{t-1}, X_t)=0$, leading to a MDS. In this serially correlated case, I suspect that it will be different from 0 but cannot calculate the expectation exactly. Ultimate it boils down to finding $E(\epsilon_{t-1}|F_{t-1})$ and $E(\epsilon_{t-2}|F_{t-1})$, which is where I am stuck.

Answer 1

$$E[X_t e_t \mid F_{t-1}] = E[X_t(\epsilon_t + \theta_1 \epsilon_{t-1} + \theta_2 \epsilon_{t-2})\mid \{X_{t-1}, e_{t-1}, X_{t-2}, e_{t-2}, ...\}]$$

$$=E[X_t[\epsilon_t + X_t\theta_1 \epsilon_{t-1} + X_t\theta_2 \epsilon_{t-2})\mid \{X_{t-1}, e_{t-1}, X_{t-2}, e_{t-2}, ...\}]$$

$$=E[X_t]\cdot E[\epsilon_t] + \theta_1E[X_t]\cdot E[\epsilon_{t-1}\mid \{e_{t-1}\}] +\theta_2 E[X_t]\cdot E[\epsilon_{t-2}\mid \{e_{t-1}, e_{t-2}\}]$$

The first term is zero but the other two are not: $\epsilon_{t-1}$ is not independent of $e_{t-1}$ and $\epsilon_{t-2}$ is not independent of $(e_{t-1}, e_{t-2})$. But non-independence implies that

$$E[\epsilon_{t-1}\mid \{e_{t-1}\}] \neq E[\epsilon_{t-1}]=0,\;\; E[\epsilon_{t-2}\mid \{e_{t-1}, e_{t-2}\}] \neq E[\epsilon_{t-2}]=0$$

So $$E[X_t e_t \mid F_{t-1}] \neq 0$$ and therefore it is not a martingale-difference sequence. There is no need to actually calculate the expected values.