Demystifying Singular Functions: A Mathematical Adventure

July 23, 2023

Singular Function

Question: “If a function like F(x) is nondecreasing and hence has almost everywhere a derivative, does integrating its derivative lead back to F(x)?

Answer: “No, even if the function F(x) is assumed continuous.”

A function is singular if its derivative is almost everywhere 0. A singular function is continuous but not differentiable.

The success probability of a game under the bold play strategy is singular.

A Class of Singular Continuous Functions

Hsu (1992) shows that if $X_n,n=1,2,\dots$ be a sequence of independent, identically distributed random variables with the common distribution:

P[X_1=i]=p_i ,i = 0,\dots, N-1 ,

Define a random variable X with values in the interval 0, 1:

X=\sum\nolimits_{n=1}^{\infty }\dfrac{X_n}{N^n}

X in base N expansion

$X=0.X_1X_2 \dots$ , since X is a number in 0, 1. The digits $X_i$ are chosen randomly and independently according to the probability $p_i$ .

The distribution function of X:

F(x)=P[X\le x], x=0.x_1x_2\dots

F(x)=\sum\nolimits_{n=0}^{\infty } P[x^n\lt X\le x^{(n+1)}]=\sum\nolimits_{n=0}^{\infty } p_{x_1} \dots p_{x_n}(p_0+p_1+\dots +p_{x_{n+1}-1})

Bold Play Singular Distribution

F(x)=pF(2x)\nobreakspace if\nobreakspace 0\le x\le 0.5 ,\nobreakspace p+(1-p)F(2x-1)\nobreakspace if\nobreakspace 0.5\le x\le 1,\nobreakspace where\nobreakspace F(0)=0, F(1)=1.

It is the distribution function of $X=\sum\nolimits_{i=1}^{\infty } {X_i}{2^{-i}}$ X in base 2 expansion: $0.X_1X_2 \dots$ , where $P[X_i=0]=1-p$ and $P[X_i=1]=p.$
F(x) is a singular function, also a singular continuous distribution function. This is the success probability of a game under the bold play strategy.

What is Bold play strategy?

Suppose that an initial capital of a gambler is $x \in \{0,1,\dots ,a\}$ where a is the target. The fact that the large bets boost the chance of success persuade the gambler to stake all his initial capital, which leads to the bold play strategy.

If the gambler fortune is less than or equal to half of the target: $0\le x\le 0.5$ , the first wager is x (all he has):

Win $x$ with probability $p$ , and continue to succeed with the new fortune $x+x=2x$ .
Lose $x$ with probability $1-p$ , it ruins him.

If the gambler’s fortune is more than or equal to half of the target: $0.5\le x\le 1$ , the first wager is $1-x$ :

Win with probability $p$ , gain $x+(1-x)=1$ , the target.
Lose with probability $1-p$ , continue with a new fortune of $x-(1-x)=2x-1$ .

What is Timid play strategy?

Suppose that he bets unit stakes on each turn of the roulette wheel until he stops. A single turn has two outcomes:

Winning one unit with probability $p$ .
Losing one unit with probability $1-p$ .