A random variable doesn\'t have an expected value if its calculation give infini

Question

A random variable doesn't have an expected value if its calculation give infinity. (i.e., if the random variable X has expected value E[X]=infinity; X doesn't have an expected value.) Which of the following random variables do not have an expected value?

a) Let Z be a random variable with the PMF give by: P(Z=z) = 6 / (pi^2 * z^2) for z=1,2,3...,infinity

b) Let R be a random variable with the PMF give by: P(R=r) = 1 / [(e)(r!)] for r=0,1,2,3....,inifinty

c) Let Y be a random variable with the PMF give by: P(Y=y) = 2y / [n(n+1)] for y=1,2,3,....n, where n is a fixed integer greater than 0

Explanation / Answer

Which of the following random variables do not have an expected value?

a) Let Z be a random variable with the PMF give by: P(Z=z) = 6 / (pi^2 * z^2) for z=1,2,3...,infinity

b) Let R be a random variable with the PMF give by: P(R=r) = 1 / [(e)(r!)] for r=0,1,2,3....,inifinty

because the expected value depends of the values that take the random variable

if Z and R take the value of infinity that means that the expecetd value is infinity that means that does not have an expected vlue

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