Let $N$ be a finite subset of $\mathbb{N}$, where $|N|>1$. For $i\in\mathbb{N}$, let $a\_i$ and $b\_i$ be chosen uniformly at random from $N$. Is it true that $\mathbb{P}[\sup\_{n\in\mathbb{N}}\{\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}\}=\infty]=1$? Note that given the symmetry involved, I believe this should be the same as asking if $\mathbb{P}[\inf\_{n\in\mathbb{N}}\{\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}\}=0]=1$.


This question came up while looking at the coarse geometry of trees, and if it is true then it has a nice corollary in that area. Unfortunately I do not have the background in probability to tell if this is an easy question or not, let alone answer it. I suspect it is true based only on the intuition of the people I have already asked, and doing some tests in python.



Probability that the ratio of products of randomly chosen natural numbers is unbounded
Yes, this is true. Indeed, let
$$S\_n:=\sum\_{i=1}^n Y\_i=\frac1s\,\ln\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n},$$
where $Y\_i:=X\_i/s$, $X\_i:=\ln\frac{a\_i}{b\_i}=\ln a\_i-\ln b\_i$, and $s:=\sqrt{Var\, X\_i}\in(0,\infty)$. Note that the $Y\_i$'s are i.i.d. zero-mean unit-variance random variables. So,
$$P\Big(\sup\_{n\ge1}\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}=\infty\Big)
=P(\sup\_{n\ge1}S\_n=\infty)\ge P\Big(\limsup\_n \frac{S\_n}{\sqrt{2n\ln\ln n}}>0\Big)=1,$$
by the [law of the iterated logarithm](https://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm).
So,
$$P\Big(\sup\_{n\ge1}\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}=\infty\Big)=1.$$




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Similarly,
$$P\Big(\inf\_{n\ge1}\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}=0\Big)=1;$$
alternatively, write
$$P\Big(\inf\_{n\ge1}\frac{a\_1\cdots a\_n}{b\_1\cdots b\_n}=0\Big)=
P\Big(\sup\_{n\ge1}\frac{b\_1\cdots b\_n}{a\_1\cdots a\_n}=\infty\Big)=1.$$




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Also, in view of Strassen's [converse to the law of the iterated logarithm](https://link.springer.com/article/10.1007/BF00539114) (and [Kolmogorov's zero–one law](https://en.wikipedia.org/wiki/Kolmogorov%27s_zero%E2%80%93one_law)), we do not need to assume that the $a\_i$'s and $b\_i$'s are chosen uniformly from a finite subset of $\mathbb N$ -- they can be sampled, independently, from any non-degenerate distribution whatsoever on $(0,\infty)$. Another proof of Strassen's result was given by [Heyde](https://www.cambridge.org/core/journals/journal-of-applied-probability/article/abs/on-the-converse-to-the-iterated-logarithm-law/B475B8821B2A214C43A6BA16FFC4E086).


