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, since although its cumulative distribution function is a continuous function, the distribution is not absolutely continuous with respect to Lebesgue measure, nor does it have 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 continuous everywhere but horizontal almost everywhere, so 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:
\begin{align} C0={}&[0,1]\\[8pt] C1={}&[0,1/3]\cup[2/3,1]\\[8pt] C2={}&[0,1/9]\cup[2/9,1/3]\cup[2/3,7/9]\cup[8/9,1]\\[8pt] C3={}&[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] C4={}&[0,1/81]\cup[2/81,1/27]\cup[2/27,7/81]\cup[8/81,1/9]\cup[2/9,19/81]\cup[20/81,7/27]\cup\\[4pt] &[8/27,25/81]\cup[26/81,1/3]\cup[2/3,55/81]\cup[56/81,19/27]\cup[20/27,61/81]\cup\\[4pt] &[62/81,21/27]\cup[8/9,73/81]\cup[74/81,25/27]\cup[26/27,79/81]\cup[80/81,1]\\[8pt] C5={}& … \end{align}
The Cantor distribution is the unique probability distribution for which for any Ct (t ∈ ), 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.
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\midY))+ \operatorname{var}(\operatorname{E}(X\midY))\\ &=
| 1 | |
| 9 |
\operatorname{var}(X)+ \operatorname{var} \left\{ \begin{matrix}1/6&withprobability 1/2\\ 5/6&withprobability 1/2 \end{matrix} \right\}\\ &=
| 1 | |
| 9 |
\operatorname{var}(X)+
| 1 | |
| 9 |
\end{align}
From this we get:
| \operatorname{var}(X)= | 1 |
| 8 |
.
A closed-form expression for any even central moment can be found by first obtaining the even cumulants[1]
\kappa2n=
| 22n-1(22n-1)B2n | |
| n(32n-1) |
,
where B2n is the 2nth Bernoulli number, and then expressing the moments as functions of the cumulants.