Moments and Cumulants
The expected value of a geometrically distributed random variable X is 1/p and the variance is (1 − p)/p2:
Similarly, the expected value of the geometrically distributed random variable Y is (1 − p)/p, and its variance is (1 − p)/p2:
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
(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
Famous quotes containing the word moments:
“The moments of the past do not remain still; they retain in our memory the motion which drew them towards the future, towards a future which has itself become the past, and draw us on in their train.”
—Marcel Proust (18711922)