Cantor Distribution - Moments

Moments

It is easy to see by symmetry 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 ∈, and 1 if X ∈ . 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:

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 B2n is the 2nth Bernoulli number, and then expressing the moments as functions of the cumulants.

Read more about this topic:  Cantor Distribution

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