# Poisson distribution moment generating function

• Mar 13th 2010, 08:08 PM
redwings6
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.
• Mar 13th 2010, 11:31 PM
mr fantastic
Quote:

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?
• Mar 14th 2010, 05:17 PM
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.
• Mar 14th 2010, 05:43 PM
mr fantastic
Quote:

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.
• Mar 14th 2010, 05:57 PM
redwings6
Quote:

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!
• Mar 14th 2010, 06:04 PM
mr fantastic
Quote:

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.
• Mar 14th 2010, 06:20 PM
redwings6
Quote:

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!