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I was doing some software engineering and wanted to have a thread do something in the background to basically just waste CPU time for a certain test.

While I could have done something really boring like for(i < 10000000) { j = 2 * i }, I ended up having the program start with $1$, and then for a million steps choose a random real number $r$ in the interval $[0,R]$ (uniformly distributed) and multiply the result by $r$ at each step.

  • When $R = 2$, it converged to $0$.
  • When $R = 3$, it exploded to infinity.

So of course, the question anyone with a modicum of curiosity would ask: for what $R$ do we have the transition. And then, I tried the first number between $2$ and $3$ that we would all think of, Euler's number $e$, and sure enough, this conjecture was right. Would love to see a proof of this.

Now when I should be working, I'm instead wondering about the behavior of this script.

Ironically, rather than wasting my CPUs time, I'm wasting my own time. But it's a beautiful phenomenon. I don't regret it. $\ddot\smile$

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    $\begingroup$ If the threshold really is $e$, I'm ready for my mind to be blown. $\endgroup$ – littleO Sep 13 '19 at 17:22
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    $\begingroup$ Same question popped up more recently here: math.stackexchange.com/questions/3355832/… $\endgroup$ – Gerry Myerson Sep 14 '19 at 3:55
  • $\begingroup$ @littleO. Now, what ? $\endgroup$ – Tom-Tom Sep 18 '19 at 6:27
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EDIT: I saw that you solved it yourself. Congrats! I'm posting this anyway because I was most of the way through typing it when your answer hit.

Infinite products are hard, in general; infinite sums are better, because we have lots of tools at our disposal for handling them. Fortunately, we can always turn a product into a sum via a logarithm.

Let $X_i \sim \operatorname{Uniform}(0, r)$, and let $Y_n = \prod_{i=1}^{n} X_i$. Note that $\log(Y_n) = \sum_{i=1}^n \log(X_i)$. The eventual emergence of $e$ as important is already somewhat clear, even though we haven't really done anything yet.

The more useful formulation here is that $\frac{\log(Y_n)}{n} = \frac 1 n \sum \log(X_i)$, because we know from the Strong Law of Large Numbers that the right side converges almost surely to $\mathbb E[\log(X_i)]$. We have $$\mathbb E \log(X_i) = \int_0^r \log(x) \cdot \frac 1 r \, \textrm d x = \frac 1 r [x \log(x) - x] \bigg|_0^r = \log(r) - 1.$$

If $r < e$, then $\log(Y_n) / n \to c < 0$, which implies that $\log(Y_n) \to -\infty$, hence $Y_n \to 0$. Similarly, if $r > e$, then $\log(Y_n) / n \to c > 0$, whence $Y_n \to \infty$. The fun case is: what happens when $r = e$?

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    $\begingroup$ I accepted your answer, as it is an excellent explanation! Thank you for taking the time! $\endgroup$ – Jake Mirra Sep 13 '19 at 17:33
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    $\begingroup$ Was thinking a bit about your question of "what happens when r = e", and all I can say is that, once you look at it on a logarithmic scale, it's a weird, sort of lopsided random walk through the reals where you sometimes take giant steps backwards and then lots of small steps forward. $\endgroup$ – Jake Mirra Sep 13 '19 at 18:01
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    $\begingroup$ Yep! And you can convince yourself that even though those increments are unbounded (on the negative side), they still have a finite variance... $\endgroup$ – Aaron Montgomery Sep 13 '19 at 18:13
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    $\begingroup$ @AaronMontgomery - what happens when $r=e$? I am not good with the details of probability theory. Does $\{Y_n\}$ converge (to $1$) or does it not converge? And what has the finite variance (of $\log X_i$) got to do with it? Intuitively I would guess the sequence does not converge, but your mention of finite variance seems to hint that it would... $\endgroup$ – antkam Sep 13 '19 at 19:53
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    $\begingroup$ When $r = e$, the fact that we are taking an average of the $\log(X_i)$ variables (which have finite variance) means that we can use the Central Limit Theorem to proceed. This implies that $\sqrt n \overline X$ converges (in distribution only, NOT almost surely) to a normal variable with mean $0$ and variance $\sigma^2$ (i.e. the variance of $\log(X_i)$), so $\log(Y_n)/\sqrt n$ does the same. Consequently, $Y_n$ just becomes diffuse, and on individual realizations it will wander, much like an ordinary random walk will do. $\endgroup$ – Aaron Montgomery Sep 13 '19 at 20:03
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I found the answer! One starts with the uniform distribution on $ [0,R] $. The natural logarithm pushes this distribution forward to a distribution on $ (-\infty, \ln(R) ] $ with density function given by $ p(y) = e^y / R, y \in (-\infty, \ln(R)] $. The expected value of this distribution is $$ \int_{-\infty}^{\ln(R)}\frac{y e^y}{R} \,\mathrm dy = \ln(R) - 1 .$$ Solving for zero gives the answer to the riddle! Love it!

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