# What's wrong with this proof of almost sure limits?

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?

• 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, $s_i = 1 - 1/i$. – snar Oct 21 at 11:16

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 $$A_n=\left\{X_n\leqslant 1/2\right\}$$.
• "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. – D. Thomine Oct 21 at 11:46
• @D.Thomine I meant the configuration that $X_i=1-1/i$ and did not realize the ambiguity. – Davide Giraudo Oct 21 at 11:57
$$\ln Y_n= \sum_{i=1}^n \ln X_i$$, 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.
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 $$n$$.