Cantor distribution

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Parameters none
Support Cantor set
pmf none
CDF Cantor function
Mean 1/2
Median anywhere in [1/3, 2/3]
Mode n/a
Variance 1/8
Skewness 0
Ex. kurtosis −8/5
MGF e^{t/2} 
                   \prod_{i= 1}^{\infty} \cosh{\left(\frac{t}{3^{i}}
CF e^{\mathrm{i}\,t/2} 
                   \prod_{i= 1}^{\infty} \cos{\left(\frac{t}{3^{i}}

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.


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

 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 .

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 2t on each one of the 2t intervals.


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:

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

From this we get:


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.


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