Example 2.31 Suppose that X is a random variable with the Poisson distribution, parameter , and we

Example 2.31 Suppose that X is a random variable with the Poisson distribution, parameter

λ, and we wish to find the expected value of Y = eX . Without Theorem 2.29 we would have to find the mass function of Y. Actually this is not difficult, but it is even easier to apply the theorem to find that E(Y ) = E(eX )

=

∞X k=0 ekP(X = k) =

∞X k=0 ek 1 k!

λke−λ

= e−λ ∞X k=0 1

k!

(λe)k = eλ(e−1). △