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Let $(X_n)$ be a sequence of independent uniform random variables in $[-1,1]$. Prove of disprove: $\lim_{n\to\infty} (1+X_1)(1+\frac{X_2}{2})...(1+\frac{X_n}{n})$ exists and belongs to $(0,\infty)$ almost surely.

Attempt: If we denote $Y_n=(1+X_1)(1+\frac{X_2}{2})...(1+\frac{X_n}{n})$ then it is well known that $(Y_n)$ is a martingale. Since it is non negative we conclude from the martingale convergence theorem that it must converge almost surely to a finite limit. So the only question is: is the limit almost surely in $(0,\infty)$ or it might be $0$ with a positive probability?

The limit is zero if and only if (we can forget about the case when $X_1=-1$ because it happens with probability $0$) for each $n$ we have $\prod_{k=n}^\infty (1+\frac{X_k}{k})=0$. So the event $\{Y_n\to 0\}$ belongs to the tail sigma algebra of $(X_n)$ and by Kolmogorov's $0-1$ law its probability is either $0$ or $1$. So now I just need to check if the probability of the limit of $Y_n$ being zero is positive or not. I had an idea to show that if at a point $\omega$ we have $(1+X_1(\omega))...(1+\frac{X_n(\omega)}{n})\to 0$ then at this same point we have $(1-X_1(\omega))...(1-\frac{X_n(\omega)}{n})\to\infty$. Since $Z_n:=(1-X_1(\omega))...(1-\frac{X_n(\omega)}{n})$ is a non negative martingale it must converge to a finite limit almost surely as well. So if my idea was correct then we would get that the set of such points $\omega$ must have probability zero, hence that would imply that almost surely the limit of $Y_n$ is in $(0,\infty)$. However, I failed to prove that my idea was correct. Any ideas?

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  • $\begingroup$ Kolmogorov 3-series theorem tells that $$\sum_{n=1}^{\infty}\frac{X_n}{n}$$ converges almost surely. Now notice that the infinite product in question converges to a positive value if and only if this series converges. $\endgroup$
    – Sangchul Lee
    May 28, 2019 at 19:21
  • $\begingroup$ I can see that the series converges almost surely. I suppose you mean that $\prod_{n=1}^\infty (1+a_n)$ converges to a positive number if and only if $\sum_{n=1}^\infty a_n$ converges. This is a known theorem, but don't we need the terms of the sequence $a_n$ to be non negative to use it? $\endgroup$
    – Mark
    May 28, 2019 at 19:58
  • $\begingroup$ Good point. I was lazy enough to skip explaining that I am using the fact that $\log(1 + X_k/k) = (X_k/k) + \mathcal{O}(1/k^2)$. $\endgroup$
    – Sangchul Lee
    May 28, 2019 at 20:00
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    $\begingroup$ Oh, so here it's just Taylor expansion of logarithm. I guess this is the way to work with infinite products in general. Thanks a lot for your help. $\endgroup$
    – Mark
    May 28, 2019 at 20:05

1 Answer 1

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Here is a problem-specific solution. Notice that

$$ \prod_{k=2}^{n} \left( 1 - \frac{X_k}{k} \right) = \prod_{k=2}^{n} \left( 1 + \frac{X_k}{k} \right)^{-1} e^{\mathcal{O}(1/k^2)}, $$

which follows from $1 - x = \frac{1-x^2}{1+x}$. From this, we find that $\left[ \prod_{k=1}^{n} \left( 1 + \frac{X_k}{k} \right) \right]^{-1}$ converges almost surely in $(0, \infty)$, and so, its reciprocal cannot be zero.

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