Cantor distribution

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Cantor
Probability function
Cumulative distribution function of the Cantor distribution
Enlarge
Cumulative distribution function of the Cantor distribution
Cumulative distribution function
Parameters
Support Cantor set
Template:Probability distribution/link
Cumulative distribution function (cdf)
Mean 1 \over 2
Median between 1 \over 3 and 2 \over 3
Mode n/a
Variance 1 \over 8
Skewness 0
Excess Kurtosis
Entropy
mgf
Char. func.

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 a discrete probability distribution with one that has a density function. Instead it is called a singular distribution.

[edit] Characterization

The Cantor set is the intersection of all of the sets

C_{0} = [0,1]\,
C_{1} = [0,1/3]\bigcup[2/3,1]\,
C_{2} = [0,1/9]\bigcup[2/9,1/3]\bigcup[2/3,7/9]\bigcup[8/9,1]\,
C_{3} = [0,1/27]\bigcup[2/27,1/9]\bigcup[2/9,7/27]\bigcup[8/27,1/3]
\bigcup[2/3,19/27]\bigcup[20/27,7/9]\bigcup[8/9,25/27]\bigcup[26/27,1]\,

etc...

On each set Ct, the probability of the interval containing the Cantor-distributed random variable X is the discrete uniform distribution on the set of all of these intervals. In other words, the probability that X is in a interval of the set Ct is 1/2t, which is 1 divided by the number of intervals that compose Ct.

The Cantor distribution is the unique probability distribution for which the above remains true for all values of t ∈ { 1, 2, 3, ... }.

The support of the Cantor distribution is the Cantor set.

[edit] Moments

It is easy to see by symmetry that the expected value of X is E(X) = 1/2.

The law of total variance can be used to find the variance var(X), as follows. Let Y = 1 or 0 according as "heads" or "tails" appears on the first coin-toss. Then:

\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}\ \mbox{probability}\ 1/2 \\ 5/6 & \mbox{with}\ \mbox{probability}\ 1/2\end{matrix}\right\}=\frac{1}{9}\operatorname{var}(X)+\frac{1}{9}.

From this we get:

\operatorname{var}(X)=\frac{1}{8}.
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