# Estimating the value of $e$ using a random function

I encountered the question which asks you to estimate the value of $$\pi$$ by using a function which generates a random real number in $$[0,1]$$ (and it's uniformly distributed). The way is to use the function twice to get a real pair which we can view it as a point on the Euclidean plane, then use the idea of the Monte Carlo method. I wonder if we can also use this function to estimate $$e$$, however, I cannot find any geometrical interpretation of the number, or can this problem be solved in other ways?

• Related Sep 8, 2021 at 12:20
• Sep 8, 2021 at 13:21

Generate $$N$$ random permutations (possibly with Knuth's shuffle).

Then count how many of them are derangements: $$N_D$$.

Then $$\frac N{N_D}\simeq e$$

• A beautiful solution! Sep 8, 2021 at 14:07
• But it only holds when N is sufficiently large, computing N! would be painful… Sep 8, 2021 at 15:33
• @JWang No, the true ratio converges quickly to $1/e$ as $k$ grows ($k$ being the permutation length). However, convergence of the monte carlo approximation is much slower. For permutations on $k=10$ elements, with $N=10^8$ samples, I get $N/N_D\simeq2.718006253262627$. Not too bad. And whatever the permutation length, you need many samples anyway: with only one, even though the probability is close to $1/e$, there are only two possible values for $N_D/N$: $0$ or $1$. Sep 8, 2021 at 15:58
• A quick and dirty coding exercise provides e = 2.718 +-0.036 with about a 1000 samples of about 1000 elements each. pastebin.com/gAEXhdWf Sep 8, 2021 at 22:51
• If you're taking fewer than $k!$ samples, the sample size is the majority of your error: with $N$ samples, your error is on the order of $\pm \frac1{\sqrt N}$, whereas the true fraction of derangements is within $\frac1{k!}$ of $\frac1e$. If, on the other hand, you're taking $k!$ or more samples, you should just go ahead and count all the derangements, instead of sampling them randomly :) Sep 9, 2021 at 1:21

Draw $$X_i \sim \text{Unif}(1,3)$$ and $$Y_i \sim \text{Unif}(0,1)$$ for $$i=1,\ldots,N$$. Reject the samples where $$Y_i > 1/X_i$$. Sort the accepted $$X_i$$ values and take the $$(N/2)$$th.

This relies on the observation that $$\int_1^e dx/x = 1$$. So we do rejection sampling in the rectangle $$[1,3] \times [0,1]$$ to get points under the $$1/x$$ curve. By taking the $$(N/2)$$th smallest value we are finding the point where we have $$1/2$$ of the original points, that is, area $$1$$ (observing that the original rectangle has area $$2$$).

Here is the simulation with $$N=10000$$ points. The rejected points are in cyan. From the accepted points, the leftmost $$N/2=5000$$ are in black and the rest are in blue. • Nice! Here is a way to do this with way fewer rejections. Just generate $n$ iid points uniformly between $1$ and $3$, sort them so $1<x_1<x_2<\dots<x_n<3$, then use them as a mesh to approximate $\int_1^{x_i} dx/x$ with the midpoint rule. Keep increasing $i$ until the first time the approximate integral exceeds $1$. Then $e\approx x_i$. Sep 8, 2021 at 21:19
• Mike, your solution is infinitely better than mine in terms of efficiency! I was striving to emulate the original "points in the disk for $\pi$" spirit as closely as possible. Sep 9, 2021 at 6:26

Flip a biased coin $$K$$ times which has probability of heads $$\frac{1}{K}$$; let $$X_1$$ denotes the number of heads that show up. Repeating this process $$N$$ times gives you $$N$$ counts of heads $$X_1,...,X_N$$. Then $$\frac{N}{\Big|\big\{1 \leq i \leq N:X_i=1\big\}\Big|}\approx e$$ when $$N$$ and $$K$$ are large.

Pick $$N$$ random integers $$X_1,X_2,\ldots,X_N$$ uniformly and independently in $$\{1,2,\ldots,N\}$$, and then count the number $$N_d$$ of distinct integers in this list, or equivalently, let

$$N_d = \#\{X_1, X_2, \ldots, X_N\}.$$

Then for large $$N$$,

$$\frac{N}{N-N_d} \approx e.$$

(That is, the size of the random set $$\{X_1,\ldots,X_N\}$$ is approximately $$(1-e^{-1})N$$.)

The minimum number of Uniform(0,1) random variables required in order for their sum to exceed $$1$$ is on average (exactly) $$e$$. More generally, if $$0 < a \le 1$$ then the minimum number of Uniform(0,1) variables required in order to exceed $$a$$ is $$e^a$$.

Let $$X_1, X_2, X_3, \dots$$ be a sequence of i.i.d. Uniform(0,1) random variables, and let $$S_n = \sum_{i=1}^n X_i$$.

As a first step, we claim $$P(S_n \le a) =\frac{a^n}{n!} \tag{*}$$ for $$0 < a \le 1$$ and $$n \ge 0$$. Proof by induction on $$n$$: The case $$n=0$$ is trivial. Suppose that $$(*)$$ holds from some $$n$$. Then $$P(S_{n+1} \le a) = \int_0^a \frac{(a-x)^n}{n!} \;dx = \frac{1}{n!} \cdot \frac{a^{n+1}}{n} = \frac{a^{n+1}}{(n+1)!}$$ and the proof is complete.

Now define $$m(a)$$ to be the least value of $$n$$ such that $$S_n > a$$. We have $$m(a) > n$$ exactly when $$S_n \le a$$, so $$E(m(a)) = \sum_{n \ge 0} P(m(a) > n) = \sum_{n \ge 0} P(S_n \le a) = \sum_{n=0}^{\infty} \frac{a^n}{n!} = e^a$$

Choose a sequence of random numbers from $$[0,1]$$, stopping when the $$n$$-th choice exceeds the $$(n-1)$$-th choice. Repeat, averaging the $$(n-1)$$-th values. This average will approach $$3-e$$.

• Nice! However, could it be that this approximates $e - 2$ rather than $3 - e$? Sep 14, 2021 at 20:44
• no, I am pretty sure it is 3-e. How did you get e-2? Sep 15, 2021 at 21:11
• I just tried it out! Note that the average will certainly be more than $1/2$, since that is the expected value of the first draw, and the generated value will always be at least equal to the first draw. Sep 16, 2021 at 6:21
• I'll paste by program (in Python3), so that you can see if I misunderstood your algorithm: Sep 16, 2021 at 6:24
• The average will be less than .5, as that will be the expected value from terminations on second choice; remember we are constructing a decreasing sequence. Sep 16, 2021 at 17:10

Pick $$n$$ numbers uniformly in $$[0,1]$$. Let $$X_n$$ be the value $$x$$ that you picked, for which $$x^x$$ takes the smallest value. Then as $$n\to\infty$$ we have $$\mathbb{E}X_n\to \frac 1e$$.

Proof: First note that $$x^x$$ has a minima at $$x=\frac1e$$. Also note that both branches of the inverse function to $$x^x, x\in[0,1]$$ are continuous. For $$\epsilon>0$$ pick $$\delta>0$$ such that $$x^x<\left(\frac1e\right)^\frac1e+\delta\implies |x-\frac1e|<\epsilon$$. For sufficiently large $$n$$, the probability that none of the numbers sampled satisfies $$x^x<\left(\frac1e\right)^\frac1e+\delta$$ is less than $$\epsilon$$ and if one of the numbers sampled does satisfy this then $$X_n$$ will too, so $$|X_n-\frac1e|<\epsilon$$. Thus $$(\frac1e-\epsilon)(1-\epsilon)<\mathbb{E}X_n\leq (\frac1e+\epsilon)(1-\epsilon)+\epsilon.$$

Ordering the $$x^x$$ reduces to integer arithmetic:

To order the $$x^x, y^y$$ one can sandwich them between rationals: $$\frac{a_x}{b_x}\leq x\leq\frac{c_x}{d_x},\qquad \frac{a_y}{b_y}\leq y\leq\frac{c_y}{d_y},\qquad a_x,b_x,c_x,d_x\in \mathbb{N},$$ so that $$f(\frac{a_x}{b_x}),f(\frac{c_x}{d_x})\leq f(\frac{a_y}{b_y}),f(\frac{c_y}{d_y}) \qquad {\rm or}\quad f(\frac{a_x}{b_x}),f(\frac{c_x}{d_x})\geq f(\frac{a_y}{b_y}),f(\frac{c_y}{d_y}),$$ where $$f(x)=x^x$$.

To compare $$f(\frac{u}{v})$$ and $$f(\frac{z}{w})$$ with $$u,v,w,z\in\mathbb{N}_{>0}$$, note that $$\left(\frac{u}{v}\right)^{\frac{u}{v}}\leq \left(\frac{z}{w}\right)^{\frac{z}{w}}\iff \left(\frac{u}{v}\right)^{{u}{w}}\leq \left(\frac{z}{w}\right)^{{z}{v}}\iff u^{uw}w^{zv}\leq z^{zv}v^{uw}.$$