Suppose we are given a martingale process $(X_t)$, which is strictly positive (almost surely), with its natural filtration $\mathcal{F}_t$ so $\mathbb{E}[ X_{t+j} | \mathcal{F}_t ] = X_t$, for all $j \geq 1$.
We assume that $(X_t)$ is not the trivial identity martingale (always taking the value one). $\star$
Under what regularity conditions can we say that $\lim_{K \rightarrow \infty} \mathbb{E}[ ( \prod_{k=1}^K X_{t+k} )^n | \mathcal{F}_t ] = \infty$, where $n$ is an integer taking the values $\lbrace -2, -1, 2 \rbrace$?
I emphasize that no assumptions are made about $(X_t)$ such as it has independent or bounded increments.
By Jensen's inequality, we know $\mathbb{E}[ ( \frac{X_{t+1}}{X_t} )^n | \mathcal{F}_t ] \geq 1$ for all those values of $n$ and, in fact, the inequality is strict (by $\star$). However, I do not know how to show anything about the expectation of the infinite product. (PS. This is interesting because of the connection to the issue of whether martingale processes definitely are, may be, or definitely are not weakly stationary processes - it seems ``obvious", they cannot be weakly stationary but I am looking for a proof (or, perhaps, counterexample)).
My attempt at a proof:
With little loss of generality, assume $X_t$ is normalised to take the value unity at time $t$. Since the process $(X_t)$ is not the trivial (constant) martingale, we know by Jensen's inequality that, for $n=2$, the process $(Y_t)$ defined by $Y_t \equiv (X_t)^n$ is a submartingale with $\mathbb{E}[ Y_{t+j} | \mathcal{F}_t ] = \mathbb{E}[ (X_{t+j})^n | \mathcal{F}_t ] > (X_t)^n$ almost surely, for any $j \geq 1$. Now use the Doob decomposition and write, for all $t$, $Y_t = Z_t + A_t$ where $(Z_t)$ is a martingale process and $(A_t)$ is a strictly increasing predictable process (normalise $Z_t$ and $A_t$ to take the values unity and zero, respectively, at time $t$). Now consider $\mathbb{E}[ Y_{t+1} Y_{t+2} | \mathcal{F}_t ]$ and use the Tower Law so $\mathbb{E}[ Y_{t+1} Y_{t+2} | \mathcal{F}_t ] = \mathbb{E}[ Y_{t+1} ( Z_{t+1} + A_{t+2} ) | \mathcal{F}_t ]$ $= \mathbb{E}[ ( Z_{t+1} + A_{t+1} ) ( Z_{t+1} + A_{t+2} ) | \mathcal{F}_t ] > Z_t^2 + ( A_{t+1} + A_{t+2} ) Z_t + ( A_{t+1} A_{t+2} )$ $ > 1 + ( A_{t+1} + A_{t+2} )$. Having spotted the pattern, one can see that for $n=2$ and any $K > 1$, $\mathbb{E}[ \prod_{k=1}^K Y_{t+k} | \mathcal{F}_t ] = \mathbb{E}[ \prod_{k=1}^K (X_{t+k})^n | \mathcal{F}_t ] > 1 + \sum_{k=1}^K A_{t+k}$, for all $K>1$. Now let $K \rightarrow \infty$. Then for $n=2$, $\lim_{K \rightarrow \infty} \mathbb{E}[ \prod_{k=1}^K (X_{t+k})^n | \mathcal{F}_t ] = \infty$ because $\lim_{K \rightarrow \infty} \sum_{k=1}^K A_{t+k} = \infty$.
I think with slight cost in notation and with tidying up of loose ends, the proof would work for all $n \geq 2$. I am not sure about for $n=-2,-1$?