let $Y_1,Y_2,..$ be a sequence of equally distributed, independent and positive random variables.
Consider $X_n = Y_1…Y_n$. Under which condition is $X_n$ a (super)-martingale? Show that neglecting the case $Y_n=1$ we have $X_n -> 0$.
$E[X_n | Y_1,…,Y_{n-1}] = Y_1 … Y_{n-1} E[Y_n | Y_1 ,…, Y_{n-1}] = Y_1…Y_{n-1} E[Y_n]$
This is a (super)-martingale if $E[Y_n] \le 1$. Since the variables are equally distributed we have $E[Y_i]=E[Y_j]$.
Since $X_n \ge 0$ and $E[X_n] = E[Y_1]…E[Y_n] \le 1$ the martingale is converging $X_n -> X$.
If $E[Y_i] < 1$ we have $E[X_n] = E[Y_1] … E[Y_n] -> 0$ hence with fatuous lemma and since $X_n \ge 0$ we have $X=0$.
But what is with the case $E[Y_i] =1$? Here the result is remarkable, since $E[X_n] =1$ for all n but $E[X] = 0$. So somehow the probability of the events, where $X_n \ge 1$ have to go to 0, while $X_n -> \infty$ there. Any ideas?
In order to identify the limit of $(X_n)$, one can use the strong law of large numbers: we have $$\lim_{n\to \infty}\frac{\log X_n}n=\mathbb E\left[ \log(Y_1)\right].$$ Here we used the not necessarily integrable version of the law of large numbers, since the positive part of $\log (Y_1)$ is integrable.
Using Jensen's inequality, note that $\mathbb E\left[ \log(Y_1)\right]\leqslant 0$, and the inequality is strict if $\log Y_1$ is non-degenerated.