The random variable X has a truncated Poisson distribution with mean mu: the probability
mass function is pX(x) =(e^(-mu)*mu^(x))/(1-e^(-mu)x!)
; x = 1; 2; 3; : : : :
(i) Given that
sum(i=0 to infinity) k^(i)/i!= e^k
, show that the probability generating function for this distribution is (e^(mu*t) -1)/(e^(mu)-1)

(ii) Hence find the moment generating function for this distribution and use it to derive
the first two moments, E(X) and E(X^2).

Question

The random variable X has a truncated Poisson distribution with mean mu: the probability
mass function is pX(x) =(e^(-mu)*mu^(x))/(1-e^(-mu)x!)
; x = 1; 2; 3; : : : :
(i) Given that
sum(i=0 to infinity) k^(i)/i!= e^k
, show that the probability generating function for this distribution is (e^(mu*t) -1)/(e^(mu)-1)

(ii) Hence find the moment generating function for this distribution and use it to derive
the first two moments, E(X) and E(X^2).