Let X have probability mass function fX(x) = exp[] x/x! for x {0, 1, 2, . . .}, where is a positive real number. This is a Poisson distribution with mean parameter . In what follows, you may also assume without proof the well-known result that Var[X] = .

a. Use E[X] and V ar[X] to find E[X2 ].

b. Find E[X(X 1)(X 2)] by evaluating P x=0 x(x 1)(x 2) exp[] x/x!.

c. Use parts a and b to find E[X3 ].