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.

  • $\begingroup$ @IosifPinelis Apologies, I'm new to posting here and forgpt about the "accept answer" button! Thank you for the reminder (and the great answer). $\endgroup$
    – Alice Kerr
    Jun 6 at 14:09

1 Answer 1


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. So, $$P\Big(\sup_{n\ge1}\frac{a_1\cdots a_n}{b_1\cdots b_n}=\infty\Big)=1.$$

Similarly, $$P\Big(\inf_{n\ge1}\frac{a_1\cdots a_n}{b_1\cdots b_n}=0\Big)=1;$$ alternatively, write $$P\Big(\inf_{n\ge1}\frac{a_1\cdots a_n}{b_1\cdots b_n}=0\Big)= P\Big(\sup_{n\ge1}\frac{b_1\cdots b_n}{a_1\cdots a_n}=\infty\Big)=1.$$

Also, in view of Strassen's converse to the law of the iterated logarithm (and Kolmogorov's zero–one 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.

  • $\begingroup$ Does your final generalisation cover the case where $E[X_i^2]$ is infinite and perhaps $E[X_i]$ undefined (despite $X_i$ being symmetrically distributed about $0$)? Presumably it does - the probability of $\sum\limits_{i=1}^n X_i$ exceeding any given bound for some $n$ will be $1$. $\endgroup$
    – Henry
    Jun 1 at 10:41
  • $\begingroup$ @Henry : Yes. As I said, the $a_i$'s and $b_i$'s can be sampled, independently, from any non-degenerate distribution whatsoever on $(0,\infty)$. Whatever $X_i$'s one can this way will be just fine. $\endgroup$ Jun 1 at 14:51

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