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).

Answer 1

ok part 1 is just plug and chug into the definition of probability generating function.
for part 2, take your answer from part 1 (which they tell u: (e^(mu*t)-1)/(e^mu-1)), and replace t with e^t to get the moment generating function. from there take derivatives and evaluate at t=0 for the first 2 moments.

Answer 2

yh thats what i was trying for part one but i cant seem to get the ' minus 1' in the numerator of (e^(mu*t)-1)/(e^mu-1))

Answer 3

k you're given the formula: sum(i=0 to infinity) k^(i)/i!= e^k
but since the distribution is truncated (x=0 is cut off), your sum actually goes from x=1 to infinity. so you need to modify the formula slightly

Answer 4

oh so am i right in saying that as it is truncated, and the x=0 is cut off this is the same as the probabilty generating function 'sum(i =0 to infinity)Px(x)t^x' minus the term when x=0