- 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 whosecumulative distribution function is theCantor function .This distribution has neither a
probability density function nor aprobability mass function , as it is not absolutely continuous with respect toLebesgue 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 asingular 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, itsexpected 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 thevariance 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 evencumulants [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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