If $0\leq X_k < 1$, then $\lim_{k\to\infty} \prod_{i}^{k}X_i =0$ almost surely?

Question

For $k=0,1,\dots,$ consider a sequence of random variables $0\leq X_k<1$ with arbitrary distribution. I wonder if $$ \lim_{k\to \infty} \prod_{i=0}^{k-1}X_i =?=0 $$almost surely.

Updated: In deterministic case, it is true (FALSE Statement, see comments below). However, when $X_i$ are stochastic, I think I am only allowed to say that $ \limsup_{k\to \infty} \prod_{i=0}^{k-1}X_i =0 $ because, while $X_i$ are bounded, we lose the monotonicity of $X_i$. Am I correct? If so, I wonder under which condition we allow to say the limit exist. Any comment/suggestion is appreciated.

Answer 1

It is not true in general. For instance, consider $X_k$ to be a degenerate random variable taking value $1 - \frac{1}{k+2}$ with probability 1, for $k = 0, 1, 2,\dots.$ Then the product of $X_0, X_1,\dots, X_{k-1}$ equals $$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots \left(1-\frac{1}{(k+1)^2}\right)$$ with probability 1. Note that the above product telescopes and simplifies to $\frac{k+2}{2(k+1)}$ which converges to $\frac{1}{2}.$ Thus, for this example, $X_0X_1\cdots X_{k-1}$ converges to $\frac{1}{2}$ with probability 1.

Answer 2

Just for completeness, I would add that you can generalise the idea in the comments to show that any random variable $Y$ with $0 \le Y < 1$ can be written as $\prod_{i=1}^\infty X_i$ where the $X_i$ are random variables with $0\le X_i < 1$.

To do this, put:

$$ X_k = \left\{ \begin{array}{lcl} \exp\left(-\left({\frac{\ln Y}{\ln Y - 1}}\right)^k\right)&\quad&\mbox{where $Y \neq 0$}\\0 &&\mbox{where $Y= 0$} \end{array} \right. $$

Then taking logarithms and using the formula for the limit of a geometric series you find that, where $Y \neq 0$:

$$ \lim_{k \to \infty}\sum_{i=1}^k\ln X_i = \ln Y $$

Hence, whether $ Y \neq 0$ or not: $$ \lim_{k \to \infty}\prod_{i=1}^k X_i = Y $$