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Let $(X_n)$ be a sequence of I.I.D. random variables uniformly distributed on $(0,1)$. Let $Y_n=\prod_{k=1}^nX_k$. My task is to find the almost sure limit of $Y_n$. Here is my work:

It is somewhat obvious that the almost sure limit will be $0$, so I prove that this is the almost sure limit. Next, note the following: $$0\leq \prod_{k=1}^nX_k \leq [\max_{k=1,...,n}X_n]^n, $$

and because $s^n$ converges to zero for all $|s|<1$, it follows that the the product must converge to zero.

I'm pretty sure that this proof is wrong because it seems for too simple. What's the issue if there is one?

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    $\begingroup$ You are assuming $s$ is a uniform bound, whereas in actuality you have $\prod_{k=1}^n X_k \leq S_n^n$, where $S_i = \max_{k=1,...,i} X_k.$ It is not even true for scalars $0 \leq s_i < 1$ that $\lim_{n\rightarrow\infty} s_n^n = 0$ -- for instance, $s_i = 1 - 1/i$. $\endgroup$ – snar Oct 21 at 11:16
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As pointed out by snar, the problem in your approach is that the obtained bound for the maximum depends on $n$, that is $Y_n\leqslant M_n^n$ where $M_n=\max_{1\leqslant k\leqslant n}X_k$. And it is a priori possible that $M_n$ converges to $1$ hence taking the limit may lead to an undetermined form. For example if $X_i=1-1/i$, then $M_n=1-1/n$ and $M_n^n\to 1/e$.

However, the previous configuration $X_i=1-1/i$ is actually almost surely impossible. The point is that for almost every $\omega$, an infinite amount of $X_n(\omega)$ will be smaller than $1/2$. This follows from the second Borel-Cantelli lemma applied to the sequence of independent events $A_n=\left\{X_n\leqslant 1/2\right\}$.

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    $\begingroup$ "However, the previous configuration is actually almost surely impossible." : isn't $M_n$ typically of size $1-1/n$ ? I wouldn't be surprised if $M_n^n$ has $[0,1]$ as limit set. $\endgroup$ – D. Thomine Oct 21 at 11:46
  • $\begingroup$ @D.Thomine I meant the configuration that $X_i=1-1/i$ and did not realize the ambiguity. $\endgroup$ – Davide Giraudo Oct 21 at 11:57
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$\ln Y_n= \sum_{i=1}^n \ln X_i$, where $-\ln X_i$'s are i.i.d. $\exp(1)$ r.v.s. By the SLLN, $\ln Y_n^{1/n}\to -1$ a.s. and $Y_n^{1/n}\to e^{-1}$ a.s. Now use the fact that for each $k\ge 1$, $\limsup_{n\to\infty}Y_n\le e^{-k}$ a.s.

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As D. Thomine's comment points out, not only does this proof not work (because $\max_{k\leq n}X_k$ depends on $n$ and can be close to $1$), but in fact this approach can't possibly work because almost surely $(\max_{k\leq n}X_k)^n\not\to 0$. To see this, note that $\Pr(X_n>1-1/(n+1))=1/(n+1)$, and so Borel-Cantelli implies there are infinitely many $n$ for which $X_n>1-1/(n+1)$, and hence $(\max_{k\leq n}X_k)^n>1/e$ infinitely often.

This means that your upper bound is simply too weak to give you what you need. An alternative bound which is good enough is that $Y_n\leq \left(\frac{X_1+\cdots+X_n}{n}\right)^n$ by AM-GM. Then the weak law of large numbers implies that almost surely $\frac{X_1+\cdots+X_n}{n}<0.99$ for sufficiently large $n$.

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