Geometric Distribution - Moments and Cumulants

Moments and Cumulants

The expected value of a geometrically distributed random variable X is 1/p and the variance is (1 − p)/p2:

$\mathrm{E}(X) = \frac{1}{p}, \qquad\mathrm{var}(X) = \frac{1-p}{p^2}.$

Similarly, the expected value of the geometrically distributed random variable Y is (1 − p)/p, and its variance is (1 − p)/p2:

$\mathrm{E}(Y) = \frac{1-p}{p}, \qquad\mathrm{var}(Y) = \frac{1-p}{p^2}.$

Let μ = (1 − p)/p be the expected value of Y. Then the cumulants of the probability distribution of Y satisfy the recursion

Outline of proof: That the expected value is (1 − p)/p can be shown in the following way. Let Y be as above. Then

$\begin{align} \mathrm{E}(Y) & {} =\sum_{k=0}^\infty (1-p)^k p\cdot k \\ & {} =p\sum_{k=0}^\infty(1-p)^k k \\ & {} = p\left(1-p) \\ & {} =-p(1-p)\frac{d}{dp}\frac{1}{p}=\frac{1-p}{p}. \end{align}$

(The interchange of summation and differentiation is justified by the fact that convergent power series converge uniformly on compact subsets of the set of points where they converge.)

Read more about this topic: Geometric Distribution

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