# Numerical phenomenon. Who can explain?

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. $$\ddot\smile$$

• If the threshold really is $e$, I'm ready for my mind to be blown. – littleO Sep 13 '19 at 17:22
• Same question popped up more recently here: math.stackexchange.com/questions/3355832/… – Gerry Myerson Sep 14 '19 at 3:55
• @littleO. Now, what ? – Tom-Tom Sep 18 '19 at 6:27

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 $$r = e$$?

• I accepted your answer, as it is an excellent explanation! Thank you for taking the time! – Jake Mirra Sep 13 '19 at 17:33
• 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. – Jake Mirra Sep 13 '19 at 18:01
• Yep! And you can convince yourself that even though those increments are unbounded (on the negative side), they still have a finite variance... – Aaron Montgomery Sep 13 '19 at 18:13
• @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... – antkam Sep 13 '19 at 19:53
• 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. – Aaron Montgomery Sep 13 '19 at 20:03

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!