Cantor distribution

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] \[8pt] C_1 = {} & [0,1/3]\cup[2/3,1] \[8pt] C_2 = {} & [0,1/9]\cup[2/9,1/3]\cup[2/3,7/9]\cup[8/9,1] \[8pt] C_3 = {} & [0,1/27]\cup[2/27,1/9]\cup[2/9,7/27]\cup[8/27,1/3]\cup \[4pt] {} & [2/3,19/27]\cup[20/27,7/9]\cup[8/9,25/27]\cup[26/27,1] \[8pt] 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 2t intervals.

Moments

It is easy to see by symmetry and being bounded 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

: \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.

References

References

1. Morrison, Kent. (1998-07-23). "Random Walks with Decreasing Steps". Department of Mathematics, California Polytechnic State University.