SECTION 6.1 Discrete Random Variables 339
We expect that the mean of the Poisson random variable isμ; how-
ever, a direct proof is possible as soon as we remember the Maclaurin
series expansion forex(see Exercise 1 on page 302). We have that
E(X) =
∑∞
k=0kP(X=k)=
∑∞
k=0ke−μμk
k!=∑∞
k=0e−μμk+1
k!= μe−μ∑∞
k=0μk
k!=μe−μeμ=μ,as expected.
Similarly,Var(X) = E(X^2 )−μ^2
=∑∞
k=0k^2 P(X=k)−μ^2=
∑∞
k=0k^2e−μμk
k!−μ^2= e−μ∑∞
k=0k
μk+1
k!−μ^2= μe−μ∑∞
k=0(k+ 1)μk
k!−μ^2= μe−μ∑∞
k=0kμk
k!+μe−μ∑∞
k=0μk
k!−μ^2= μ^2 e−μ∑∞
k=0μk
k!
+μ−μ^2= μ^2 +μ−μ^2 =μ.That is to say, Var(X) =μ=E(X).