find the distribution of the sum of 25 ind poisson rv's

• Apr 30th 2009, 09:55 AM
plm2e
find the distribution of the sum of 25 ind poisson rv's
Find the distribution of the sum of 25 independent Poisson random variables with the identical parameter λ = 5
• Apr 30th 2009, 09:59 AM
Moo
Hello,
Quote:

Originally Posted by plm2e
This is confusing me and i would appreciate any help i can get.

Find the distribution of the sum of 25 independent Poisson random variables with the identical parameter λ = 5

Use moment generating function, by remembering that the mgf of the sum of independent rv equals the product of their mgf.

(you should find the mgf of a Poisson distribution with parameter 25x5)
• Apr 30th 2009, 10:06 AM
plm2e
ok so, Mx(t) = e^[λ((e^t)- 1))]

i get e^[5((e^25) - 1)] = my ti 83 says overflow

i think i may be confused on what t is
• Apr 30th 2009, 10:08 AM
plm2e
missed the product part on my last post, but im still concerned about what the t means in the mgf
• Apr 30th 2009, 10:26 AM
Moo
Quote:

Originally Posted by plm2e
ok so, Mx(t) = e^[λ((e^t)- 1))]

i get e^[5((e^25) - 1)] = my ti 83 says overflow

i think i may be confused on what t is

Let $\displaystyle X_1,\dots,X_{25}$ be iid Poisson rv with parameter 5.

Then their mgf is : $\displaystyle M(t)=e^{5(e^t-1)}$

The mgf of $\displaystyle X=X_1+\dots+X_{25}$ is :
$\displaystyle M_X(t)=M_{X_1+\dots+X_{25}}(t)=M_{X_1}(t)\cdots M_{X_{25}}(t)=M(t) \cdots M(t)=[M(t)]^{25}$
$\displaystyle =\left(e^{5(e^t-1)}\right)^{25}=e^{25 \cdot 5(e^t-1)}=e^{125(e^t-1)}$

And this is the mgf of a Poisson distribution with parameter 125.