Infinite products of scaled indicator variables: almost sure convergence vs. uniform convergence of the sample mean

Let $\frac{X_i}{2}\sim Ber(0.5) \implies E[X_i]=1$, and let $Y_n=\prod\limits_{i=1}^n X_i$.

Since the $X_i$ are iid, $E[Y_n]=1,\;\forall n<\infty$. However, something weird appears to be happening as $n\to \infty$, since it seems that $P(\lim\limits_{n\to \infty} Y_n > 0) = 0$ yet it is equal to 1 for all finite $n$. The SLLN would also suggest that $Y_n$ will converge to $0$ almost surely.

How is it possible that a product of bounded random variables can have an expected value of 1 for every finite sequence, yet converge to $0$ almost surely? I guess its a question of why doesn't the SLLN imply uniform convergence of the mean?

• Isn't it $\lim_{n\to\infty} \mathbb{P}\{Y_n > 0\}$ that goes to $0$? Dec 17 '14 at 7:33
• The products are not uniformly bounded.
– Did
Dec 17 '14 at 9:13

The expectation $\mathbb{E}[Y_n]$ is equal to 1 for all $n$; the probability $\mathbb{P}\{Y_n > 0\} = \frac{1}{2^n}$ goes to $0$. However, note that with probability $1/2^n$ (i.e., when $Y_n$ is not $0$), $Y_n = 2^n$, which itself goes to infinity.
The expectation is fixed and equal to $1$, the probability $p_n$ to be non-zero becomes smaller and smaller: but this is possible, as to "make up for that" the value when $Y_n$ is non-zero becomes larger and larger, proportionally to the inverse of $p_n$.