What have you tried so far?
let
Yi,j be independent Bernoulli random variables with E[Yi,j ] = 1/j. Let Zj = sumYi,j for i=1 to j . Let MZj(s) be the probability generating function of Zj. Calculate
M∞(s) = limj→∞ MZj(s). Do you recognize this function?
What have you tried so far?
Yij~Bernoulli
My(s)= ((1-P)+pe^s)
Mzj(s)= ((1-P)+pe^s)^j
Limit Mzj(s)= limit((1-P)+pe^s)^j, when j goes infinity
= (1-(1/j)+(e^s)/(i/j))^j j goes infinity
=(1+(e^s-1)/(1/j))^j j goes infinity
=e^((e^s)-1) ~ Poisson (mean=1, Var=1)
This is my solution but not sure if it is correct
The arithmetic looks good to me but it seems that you are working on the mgf instead of the pgf.
aytajalli (02-12-2015)
As I underestand the difference between mgf and pgf is the fisrt one is for continouse distribution and the second is for discrete. Is there any other diferences?
You are right. I uede the pgf and the result was the same.
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