# Cantor distribution

(Redirected from Cantor measure)
Parameters none Cantor set none Cantor function 1/2 anywhere in [1/3, 2/3] n/a 1/8 0 −8/5 $e^{t/2} \prod_{i= 1}^{\infty} \cosh{\left(\frac{t}{3^{i}} \right)}$ $e^{\mathrm{i}\,t/2} \prod_{i= 1}^{\infty} \cos{\left(\frac{t}{3^{i}} \right)}$

The Cantor distribution is the probability distribution whose cumulative distribution function is the Cantor function.

This distribution has neither a probability density function nor a probability mass function, as it is not absolutely continuous with respect to Lebesgue measure, nor has it any point-masses. It is thus neither a discrete nor an absolutely continuous probability distribution, nor is it a mixture of these. Rather it is an example of a singular distribution.

Its cumulative distribution function is sometimes referred to as the Devil's staircase, although that term has a more general meaning.

## Characterization

The support of the Cantor distribution is the Cantor set, itself the intersection of the (countably infinitely many) sets

\begin{align} C_{0} = & [0,1] \\ C_{1} = & [0,1/3]\cup[2/3,1] \\ C_{2} = & [0,1/9]\cup[2/9,1/3]\cup[2/3,7/9]\cup[8/9,1] \\ C_{3} = & [0,1/27]\cup[2/27,1/9]\cup[2/9,7/27]\cup[8/27,1/3]\cup \\ & [2/3,19/27]\cup[20/27,7/9]\cup[8/9,25/27]\cup[26/27,1] \\ C_{4} = & \cdots . \end{align}

The Cantor distribution is the unique probability distribution for which for any Ct (t ∈ { 0, 1, 2, 3, ... }), the probability of a particular interval in Ct containing the Cantor-distributed random variable is identically 2-t on each one of the 2t intervals.

## Moments

It is easy to see by symmetry that for a random variable X having this distribution, its expected value E(X) = 1/2, and that all odd central moments of X are 0.

The law of total variance can be used to find the variance var(X), as follows. For the above set C1, let Y = 0 if X ∈ [0,1/3], and 1 if X ∈ [2/3,1]. Then:

\begin{align} \operatorname{var}(X) & = \operatorname{E}(\operatorname{var}(X\mid Y)) + \operatorname{var}(\operatorname{E}(X\mid Y)) \\ & = \frac{1}{9}\operatorname{var}(X) + \operatorname{var} \left\{ \begin{matrix} 1/6 & \mbox{with probability}\ 1/2 \\ 5/6 & \mbox{with probability}\ 1/2 \end{matrix} \right\} \\ & = \frac{1}{9}\operatorname{var}(X) + \frac{1}{9} \end{align}

From this we get:

$\operatorname{var}(X)=\frac{1}{8}.$

A closed-form expression for any even central moment can be found by first obtaining the even cumulants[1]

$\kappa_{2n} = \frac{2^{2n-1} (2^{2n}-1) B_{2n}} {n\, (3^{2n}-1)}, \,\!$

where B2n is the 2nth Bernoulli number, and then expressing the moments as functions of the cumulants.