# Thread: Bernoulli distribution and PGF

1. ## Bernoulli distribution and PGF

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?

2. ## Re: Bernoulli distribution and PGF

What have you tried so far?

3. ## Re: Bernoulli distribution and PGF

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

4. ## Re: Bernoulli distribution and PGF

The arithmetic looks good to me but it seems that you are working on the mgf instead of the pgf.

5. ## The Following User Says Thank You to BGM For This Useful Post:

aytajalli (02-12-2015)

6. ## Re: Bernoulli distribution and PGF

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?

8. ## Re: Bernoulli distribution and PGF

You are right. I uede the pgf and the result was the same.

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