# Thread: Poisson distribution moment generating function

1. ## Poisson distribution moment generating function

Let Y1, Y2,...,Yn be independent Poisson random variables with means ƛ1, ƛ2, ...,ƛn, respectively.
Find the: Probability function of Yi using the method of moment generating functions.

I am unsure of how to do this question so any help would be great.

2. Originally Posted by redwings6
Let Y1, Y2,...,Yn be independent Poisson random variables with means ƛ1, ƛ2, ...,ƛn, respectively.
Find the: Probability function of Yi using the method of moment generating functions.

I am unsure of how to do this question so any help would be great.
Do you mean find the pdf of U = Y1 + Y2 + .... + Yn using the method of moment generating functions? What have you tried and where do you get stuck?

3. Yeah that is exactly what i mean. The sum of Yi (ie. from Y=1 to Y=n). I am unsure of how to set it up and then do the first line of calculations.

4. Originally Posted by redwings6
Yeah that is exactly what i mean. The sum of Yi (ie. from Y=1 to Y=n). I am unsure of how to set it up and then do the first line of calculations.
Well, do you know what the mgf is for a Poisson distribution? And do you realise that for a sum of independent rv's you multiply the mgf's? Please show what you've done.

5. Originally Posted by mr fantastic
Well, do you know what the mgf is for a Poisson distribution? And do you realise that for a sum of independent rv's you multiply the mgf's? Please show what you've done.
Well I know that the mgf of Poisson is: exp[ƛ(e^t - 1)]
I also realize that since they are indepedent you can multiply them together. ie. My1(t) = exp[ƛ1(e^t - 1)], My2(t) = exp[ƛ2(e^t - 1)],....,Myn(t) = exp[ƛn(e^t - 1)] ------> My1(t)*My2(t)*...*Myn(t)

I just don't know how that when multiplying these together you obtain the Poisson probability function: p(n) = (ƛ^n*e^-ƛ)/n!

6. Originally Posted by redwings6
Well I know that the mgf of Poisson is: exp[ƛ(e^t - 1)]
I also realize that since they are indepedent you can multiply them together. ie. My1(t) = exp[ƛ1(e^t - 1)], My2(t) = exp[ƛ2(e^t - 1)],....,Myn(t) = exp[ƛn(e^t - 1)] ------> My1(t)*My2(t)*...*Myn(t)

I just don't know how that when multiplying these together you obtain the Poisson probability function: p(n) = (ƛ^n*e^-ƛ)/n!
Use a basic rule for multiplying exponentials to get $M(t) = \exp[(e^t - 1) (\lambda_1 + \lambda_2 + .... + \lambda_n)]$. The conclusion is crystal clear.

7. Originally Posted by mr fantastic
Use a basic rule for multiplying exponentials to get $M(t) = \exp[(e^t - 1) (\lambda_1 + \lambda_2 + .... + \lambda_n)]$. The conclusion is crystal clear.
Thank you very much!