Cantor distribution


Cantor distribution

Probability distribution
name =Cantor
type =mass
pdf_

cdf_

parameters =none
support =Cantor set
pdf =none
cdf =Cantor function
mean =1/2
median =anywhere in [1/3, 2/3]
mode =n/a
variance =1/8
skewness =0
kurtosis =-8/5
entropy =
mgf =e^{t/2} prod_{i=1}^{infty} cosh{left(frac{t}{3^{i
ight)}
char =e^{mathrm{i},t/2} prod_{i=1}^{infty} cos{left(frac{t}{3^{i ight)}

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 a 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 (countably infinite) intersection of the sets

:egin{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 "C""t" ("t" ∈ { 0, 1, 2, 3, ... }), the probability of a particular interval in "C""t" containing the Cantor-distributed random variable is identically 2-"t" on each one of the 2"t" 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 "C"1, let "Y" = 0 if "X" ∈ [0,1/3] , and 1 if "X" ∈ [2/3,1] . Then:

: egin{align}operatorname{var}(X) & = operatorname{E}(operatorname{var}(Xmid Y)) + operatorname{var}(operatorname{E}(Xmid Y)) \ & = frac{1}{9}operatorname{var}(X) + operatorname{var} left{ egin{matrix} 1/6 & mbox{with probability} 1/2 \ 5/6 & mbox{with probability} 1/2 end{matrix} ight} \ & = 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 [http://www.calpoly.edu/~kmorriso/Research/RandomWalks.pdf]

: kappa_{2n} = frac{2^{2n-1} (2^{2n}-1) B_{2n {n (3^{2n}-1)},

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

External links

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