Almost sure convergence of a product of random variables

Question

Let $X_1, X_2, ...$ be a sequence of random variables such that:

$X_i = \begin{cases} 1 &\mbox{with probability } p = 1-1/n^2 \\ 2 & \mbox{with probability } p = 1/n^2 \end{cases} $

and let $Y_n=\Pi_{i=1}^n X_i $. Show that $Y_n$ converges almost surely to some random variable $Y$.

I already know several methods of proving a.s. convergence to a constant. However, I do not know how to establish convergence to a non-constant random variable except for definition which does not seem to work in this case.

Answer 1

The $Y_n$'s fail to converge exactly when infinitely many of the $X_i$s are $2$. By Kolmogorov's zero-one law, the probability of this happening is either $0$ or $1$.

However, since $\sum_{n=1}^\infty n^{-2}$ is finite (see the Basel problem), from some point the probability that there's even one more $2$ in the sequence of $X_i$s will be less than $1$ -- so the probability that there's infinitely many of them cannot be $1$. Therefore it has to be $0$.

Define the random variable $Y$ to be the limit of the $Y_i$s in outcomes where they converge, and $42$ otherwise. Then the $Y_i\to Y$ almost surely.

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Instead of the slightly handwavy "from some point" argument, one can also show similarly that the probability of the $Y_i$ sequence being precisely $(2,2,2,2,\ldots)$ is at least $e^{-(\pi^2/3-2)}$ (because $\log(1-1/n^2) \ge -2n^2$ for all $n\ge 2$). Since this sequence converges, "the $Y_i$s converge" has positive probability, and again thanks to Kolmogorov's 0-1 law, this positive probability must be $1$.