Consider the random process where you keep drawing samples from [0,1] uniformly at random as long as the current sample is larger than the previous sample. What are the Expected value and the Variance of the number of samples you draw?

I have tried to think of a discrete case with Wald's Equality with step of 0.1. Summing up p * (1 - p) where p is 0.1, 0.2 ..., equals 0.327, matches my simulation mean=array([0.359285]), variance=array([0.05881588]).

import numpy as np
import matplotlib.pyplot as plt
from scipy import stats

cur_num = None
M = 1000000
li =[]
for i in range(M):

    if cur_num is None:
        cur_num = np.random.uniform(0,1,1)
        tmp = 0

    while tmp < cur_num:
        tmp = cur_num
        cur_num = np.random.uniform(0,1,1)
        # print(f" while loop: {cur_num}")

    # print(f" for loop: {cur_num}")

    cur_num =None


## DescribeResult(nobs=10000000, minmax=(array([1.29312049e-08]), array([0.9996544])), 
## mean=array([0.359285]), variance=array([0.05881588]), skewness=array([0.44984084]), kurtosis=array([-0.76854895]))

This gives me the idea of integrate p * (1 - p) dp from 0 to 1. But it gives 1/2 - 1/3, 0.16667, not 0.35


1 Answer 1


Although Math SE might be a bit more suited for this one, I wanted to give it a try.

The answer relies on the law of total expectation, the law of total variance, and the relationship between Euler's number $e$ and series involving the factorial function $n!$.

In the first half of the answer, I will present the derivation of $E(n)$ and $Var(n)$, i.e. mean and variance of the stopping time, in the second half I will show how to derive $E(x_n)$ and $Var(x_n)$, i.e. the expected value and variance of the stopped process. Finally, in the addendum, I provide the full distribution of the stopped process, $f(x_n)$ for $x_k$ drawn from a uniform and from a generic distribution.

Part I: The distribution of the stopping time $n$

Given the $n$th draw $x_{n}$ from the unit interval $(0,1)$, we accept the ($n+1$)th draw $x_{n+1}$ if $x_n<x_{n+1}$; else we stop at the $n$th draw and the process is said to be stopped at $n$ with value $x_n$. We say that $n$ is the stopping time.

Let’s say we are about to draw $k$ numbers $x_1, x_2, \ldots x_k$ uniformly from $(0,1)$ and arrange the results in increasing order, e.g.

$$ \begin{align} x_1&<x_2<\ldots<x_k \end{align} $$

Then there are $k!=k\times(k-1)\times(k-2)\ldots\times2\times1$ possible permutations of the ordering of the draws, each permutation with the same probability $1/k!$ - but only one permutation represents the necessary ordering $x_1<x_2<\ldots<x_k$ under which the process is not yet stopped at draw $k$ (of course, it will be stopped if $x_k>x_{k+1}$). Thus, the unconditional probability of $n\geq k$, i.e. of drawing at least $k$ times, is

$$ \begin{align} P(n\geq k)&=\frac{1}{k!} \\ \Rightarrow P(n=k)&=P(n\geq k)-P(n\geq k+1)\\ &=\frac{1}{k!}-\frac{1}{(k+1)!} \end{align} $$

Calculating the expected stopping time

We calculate $E(n)=\sum_k k\times P(n=k)$, rearranging the factorials in the meantime:

$$ \begin{align} E(n)&=\sum\limits_{k=1}^\infty \frac{k}{k!}-\frac{k}{(k+1)!}\\ &=\left(\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\ldots\right)-\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots\right)\\ &=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots\\ &=\sum\limits_{k=1}^{\infty}\frac{1}{k!}\\ &=\sum\limits_{k=0}^{\infty}\frac{1}{k!}-1 \end{align} $$

In the last row, we recognize the definition of Euler's constant $e=\sum_k 1/k!=2.718282\ldots$; hence:

$$ E(n)=e-1 $$

Calculating the variance of the stopping time

As $Var(x)=E(x^2)-E(x)^2$, we now try to estimate $E(n^2)$, again applying some rearranging, and observing relationships with the definition of $e$:

$$ \begin{align} E(n^2)&=\sum\limits_{k=1}^\infty k^2\times P(n=k)\\ &=\sum\limits_{k=1}^\infty\frac{k^2}{k!}-\frac{k^2}{(k+1)!}\\ &=\left(\frac{1^2}{1!}+\frac{2^2}{2!}+\frac{3^2}{3!}+\ldots\right)-\left(\frac{1^2}{2!}+\frac{2^2}{3!}+\frac{3^2}{4!}+\ldots\right)\\ &=\sum\limits_{k=1}^{\infty}\frac{k^2-(k-1)^2}{k!}\\ &=\sum\limits_{k=1}^{\infty}\frac{2k-1}{k!}\\ &=2\sum\limits_{k=1}^{\infty}\frac{k}{k!}-\left(e-1\right)\\ &=2\sum\limits_{k=0}^{\infty}\frac{1}{k!}-\left(e-1\right)\\ &=e+1 \end{align} $$


$$ \begin{align} Var(n)=&E(n^2)-E(n)^2\\ &=e+1-(e-1)^2\\ &=3e-e^2 \end{align} $$

A simulation of the stopping time $n$

Let's simulate this using R. I choose a cutoff level n=14, the corresponding probability is below 1E-10.

nSim <- 1E6
k    <- 14
mu <- exp(1)-1
v  <- 3*exp(1)-exp(2)
  x <- rle( diff( runif(k) ) > 0 )  
  1 + x$values[1] * x$lengths[1]

print( mean(u)- mu )
print( var(u) - v )

with output

[1] -0.0001078285
[1] -0.0003605153

Part II: Expectation and variance of $x_n$

Here, we will make use of the law of total expectation and the law of total variance.

Expectation of $x_n$

We start with the law of total expectation, applying it to $x_n$ as

$$ \begin{align} E(x_n)&=E\left(E\left(x_n|n=k\right)\right)\\ &=\sum\limits_{k=1}^{\infty}E\left(x_n|n=k\right)P(n=k) \end{align} $$

We already know the distribution of $n$, all that remains is calculating $E(x_n|n=k)$.

For a (fixed) stopping time $k$, it must hold that $x_1\leq x_2\leq\ldots \leq x_{k-1}\leq x_k > x_{k+1}$, i.e. the $k$th component from $k+1$ draws represents the maximum over $(k+1)$ draws. The distribution of the maximum is:

$$ \begin{align} F^{max}_{k+1}(x)&\equiv P(\max(x_1,x_2,\ldots,x_k,x_{k+1})\leq x)\\ &=P(x_1\leq x, x_2\leq x, \ldots x_{k-1}\leq x,x_{k+1}\leq x)\\ &=F(x)^{k+1}\\ &=x^{k+1} \end{align} $$

The corresponding density, $f^{max}_{k+1}(x)=\partial F^{max}_{k+1}(x)/\partial x$, is $$ f^{max}_{k+1}(x)=(k+1)x^k $$

and the expectation of $x_n$ given stopping at $k$ is

$$ \begin{align} E(x_n|n=k)&=\int\limits_0^1 xf^{max}_{k+1}(x)\mathrm{d}x\\ &=(k+1)\int\limits_0^1 x^{k+1}\mathrm{d}x\\ &=\frac{k+1}{k+2} \end{align} $$

We now have both ingredients for the expectation:

$$ \begin{align} E(x_n)&=\sum\limits_{k=1}^{\infty}E\left(x_n|n=k\right)P(n=k)\\ &=\sum\limits_{k=1}^{\infty}\frac{k+1}{k+2}\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)\\ &=\left(\frac{2/3}{1!}+\frac{3/4}{2!}+\frac{4/5}{3!}+\frac{5/6}{4!}+\ldots\right)-\left(\frac{2/3}{2!}+\frac{3/4}{3!}+\frac{4/5}{4!}+\frac{5/6}{5!}+\ldots\right)\\ &=0.5+\sum\limits_{k=1}^{\infty}\frac{1}{(k+2)!}\\ &=0.5+\sum\limits_{k=1}^{\infty}\frac{1}{k!}-1.5\\ &=\sum\limits_{k=1}^{\infty}\frac{1}{(k)!}-1\\ &=e-2 \end{align} $$ where we have reused the relationships between the inverse factorial sum and Euler's number, above.

Variance of $x_n$

We make use of the law of total variance


and start with the first term:

$$ \begin{align} E(Var(x_n|n=k))&=\sum\limits_{k=1}^{\infty}\left(E(x_n^2|n=k)-E(x_n|n_k)^2\right)P(n=k)\\ &=\sum\limits_{k=1}^{\infty}\left((n+1)\int_{x=0}^1x^2f^{max}_{k+1}(x)\mathrm{d}x-\left(\frac{k+1}{k+2}\right)^2\right)P(n=k)\\ &=\sum\limits_{k=1}^{\infty}\left(\frac{k+1}{k+3}-\left(\frac{k+1}{k+2}\right)^2\right)P(n=k) \end{align} $$

and stop here. For the second term, we get

$$ \begin{align} Var(E(x_n|n=k))&=E(E(x_n|n=k)^2)-E(E(x_n|n=k))^2\\ &=\sum_{k=1}^{\infty}\left(\frac{k+1}{k+2}\right)^2P(n=k)-\left(\sum_{k=1}^{\infty}\frac{k+1}{k+2}P(n=k)\right)^2\end{align} $$

Combining both terms yields

$$ \begin{align} Var(x_n)&=E(Var(x_n|n=k))+Var(E(x_n|n=k))\\ &=\sum\limits_{k=1}^{\infty}\frac{k+1}{k+3}P(n=k)-\left(\sum_{k=1}^{\infty}\frac{k+1}{k+2}P(n=k)\right)^2 \end{align} $$

which, after some more rearranging yields

$$ Var(x_n)=2+2e-e^2 $$

A simulation study of $x_n$

nSim <- 1e6
k    <- 14

u <- sapply(1:nSim,function(i){
  x <- runif(k)
  z <- rle(diff(x)>0)
  x[1 + z$values[1]*z$lengths[1]]
cat("Variable  : "  ,"\t", "simulation", "\t", "theoretical"    , "\n")
cat("mean(x_n) : "  ,"\t",  mean(u)    , "\t", exp(1)-2         , "\n")
cat("var(x_n)  : "  ,"\t",  var(u)     , "\t", 2+2*exp(1)-exp(2), "\n")

resulting in

Variable  :      simulation      theoretical 
mean(x_n) :      0.7182006       0.7182818 
var(x_n)  :      0.04752103      0.04750756


We can even produce the distribution of $x_n$ in closed form by combining the PMF of $n$ and the density $f^{max}_{k+1}(x)$:

$$ \begin{align} f(x)&=\sum\limits_{k=1}^{\infty}f(x_n|n=k)P(n=k)\\ &=\sum\limits_{k=1}^{\infty}(k+1)(x)^k\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)\\ &=\sum\limits_{k=1}^{\infty}(x)^k\frac{1}{(k-1)!}\\ &=x\sum\limits_{k=0}^{\infty}(x)^k\frac{1}{k!}\\ &=xe^{x} \end{align} $$

By the same line of reasoning we can show that the distribution of $x_n$ given draws from a continuous distribution $\Phi$ (and corresponding density $\phi$) equals

$$ f(x)=\phi(x)\Phi(x)e^{\Phi(x)} $$

  • $\begingroup$ Thanks! I only realised now I have to derive E(N), not E(process)! Supposed I have to derive E(process),shouldn't it be (e-1)* E(X_uniform) = (e-1)*(0.5) = 0.859? Why is it not 0.359 as suggested from the simulation? $\endgroup$ Commented Sep 19, 2021 at 8:42
  • $\begingroup$ I think he also wants the expected value of the most recent U(0.1) given that the process has stopped $\endgroup$
    – dm63
    Commented Sep 19, 2021 at 11:19
  • 1
    $\begingroup$ @Kermittfrog Yes! Thank you so much! The proof you provide is very clear. $\endgroup$ Commented Sep 20, 2021 at 10:49
  • 1
    $\begingroup$ The loop is more clear (to me) if written like this: u <- sapply(1:nSim,function(i){ x <- runif(k); x[which(diff(x)<0)[[1]]] }) . It's a bit faster as well. $\endgroup$
    – Bob Jansen
    Commented Sep 20, 2021 at 11:28
  • 1
    $\begingroup$ Hi @BobJansen thanks! I'll give it a try later; I am always looking for ways to improve speed and readability of my code :) $\endgroup$ Commented Sep 20, 2021 at 11:35

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