Supermartingale of product of random variables

Question

Let $Y_1, Y_2, \ldots$ be independent and identically distributed non-negative random variables. Define $X_n := \prod_{i=1}^n Y_i$ for $n\geq 1$.

I have to show that if $(X_n)_n$ is a supermartingale and $\mathrm{P}(Y_1=1)<1$, then $X_n \overset{n\rightarrow\infty}{\longrightarrow} 0$ almost surely.

I'm really stuck with this exercise.

If $\Bbb{E}[Y_1] = 1$, which means $\Bbb{E}[Y_s] = 1$ for any $s\in\Bbb{N}$, and $\mathcal{F}_t := \sigma(Y_1, \ldots, Y_t)$, then I think that the following idea might be helpful: $$\Bbb{E}[X_{s+1}\mid \mathcal{F}_s] = \Bbb{E}[X_{s}Y_{s+1}\mid \mathcal{F}_s] = X_{s}\Bbb{E}[Y_{s+1}\mid \mathcal{F}_s] = X_s \, .$$

In the last step I used that $Y_{s+1}$ and $\mathcal{F}_s$ are independent.

So it seems that if $\Bbb{E}[Y_1] = 1$ then $(X_n)_n$ is a martingale (and so a supermartingale). Probably $\Bbb{E}[Y_1] \leq 1$ must hold for $(X_n)_n$ to be a supermartingale?

Does anybody have any idea? Why should random variables like $\mathrm{P}(Y_s = 0.8) = \mathrm{P}(Y_s = 1.2) = 0.5$ not work? Thank you.

Answer 1

Here are some hints:

1. By a calculation similar to your idea, you can show that the supermartingale assumption implies that $E[Y_i]\leq 1$.
2. You can then use Jensen's inequality to conclude that $E[\log Y_i]<0$. (In the case $E[Y_i]=1$, you'll need to use the strict form of Jensen's inequality, using the condition $P(Y_i=1)<1$ to ensure the nondegeneracy of $Y_i$).
3. Use the strong law of large numbers.
4. If you run into any trouble due to $Y_i$ possibly being zero, you can take care of this case separately using Borel-Cantelli.