Editing Cantor distribution (section)

== 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&nbsp;0.

The [[law of total variance]] can be used to find the [[variance]] var(''X''), as follows.  For the above set ''C''<sub>1</sub>, let ''Y'' = 0 if ''X''&nbsp;∈&nbsp;[0,1/3], and 1 if ''X''&nbsp;∈&nbsp;[2/3,1].  Then:

: <math>
\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}
</math>

From this we get:

:<math>\operatorname{var}(X)=\frac{1}{8}.</math>

A closed-form expression for any even [[central moment]] can be found by first obtaining the even [[cumulants]]<ref>{{cite web |last=Morrison |first=Kent |url=http://www.calpoly.edu/~kmorriso/Research/RandomWalks.pdf |title=Random Walks with Decreasing Steps |publisher=Department of Mathematics, California Polytechnic State University |date=1998-07-23 |access-date=2007-02-16 |archive-date=2015-12-02 |archive-url=https://web.archive.org/web/20151202055102/http://www.calpoly.edu/~kmorriso/Research/RandomWalks.pdf |url-status=dead }}</ref>

:<math>
 \kappa_{2n} = \frac{2^{2n-1} (2^{2n}-1) B_{2n}}
                    {n\, (3^{2n}-1)}, \,\!
</math>

where ''B''<sub>2''n''</sub> is the 2''n''th [[Bernoulli number]], and then [[Cumulant#Cumulants and moments|expressing the moments as functions of the cumulants]].