Cantor distribution

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Cantor
Probability function
Cumulative distribution function Cumulative distribution function of the Cantor distribution
Parameters	none
Support	Cantor set
Template:Probability distribution/link	none
Cumulative distribution function (cdf)	Cantor function
Mean	1/2
Median	anywhere in [1/3, 2/3]
Mode	n/a
Variance	1/8
Skewness	0
Excess kurtosis	-8/5
Entropy
Moment-generating function (mgf)	$e^{t/2} \prod_{i=1}^{\infty} \cosh{\left(\frac{t}{3^{i}} \right)}$
Characteristic function	$e^{\mathrm{i}\,t/2} \prod_{i=1}^{\infty} \cos{\left(\frac{t}{3^{i}} \right)}$

The Cantor distribution is the probability distribution whose cumulative distribution function is the Cantor function. This distribution is not absolutely continuous with respect to Lebesgue measure, so it has no probability density function; neither is it discrete, since it has no point-masses; nor is it even a mixture of discrete continuous probability distributions. Instead it is an example of a singular distribution.

[edit] Characterization

The support of the Cantor distribution is the Cantor set, itself the (countably infinite) intersection of the 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 C_t (t ∈ { 0, 1, 2, 3, ... }), the probability of a particular interval in C_t containing the Cantor-distributed random variable X is the discrete uniform distribution on the set of all 2^t of these intervals.

[edit] Moments

It is easy to see by symmetry that the expected value of X is E(X) = 1/2, and that all odd central moments 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 B_2n is the 2nth Bernoulli number, and then expressing the moments as functions of the cumulants.[1]

[edit] External links

Morrison, Kent. "Random Walks with Decreasing Steps", Department of Mathematics, California Polytechnic State University, 1998-07-23. Retrieved on February 16, 2007.

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