# Moment Generating Functiom to determine Distribution

• Oct 5th 2009, 12:55 PM
Flipz4226
Moment Generating Functiom to determine Distribution
$\displaystyle \lambda_{1},...,\lambda_{n} > 0$ and $\displaystyle X_{1},...,X_{n}$ are independent random variables such that $\displaystyle X_{k}$ ~ $\displaystyle Pois(\lambda,r_{k})$ for each $\displaystyle k = 1,...,n$. Use moment generating functions to determine the distribution of $\displaystyle X = X_{1} +...+ X_{n}$

So far I have $\displaystyle M_{X_1 + X_2 +...+X_n}(t)=M_{x_1}(t)M_{x_2}(t)...M_{x_n}(t)$

=$\displaystyle e^{\lambda_{1}(e^t-1)} \star e^{\lambda_{2}(e^t-1)} \star ... \star e^{\lambda_{n}(e^t-1)}$

=$\displaystyle e^{e^t-1 [\lambda_{1}+\lambda_{2}+...+\lambda_{n}}]$

I'm not sure what kind of distribution this looks like, if I'm even doing it right.
• Oct 5th 2009, 03:51 PM
matheagle
you're missing a ( )
I fixed it below, I don't know how to do the red highlight stuff here.

Quote:

Originally Posted by Flipz4226
$\displaystyle \lambda_{1},...,\lambda_{n} > 0$ and $\displaystyle X_{1},...,X_{n}$ are independent random variables such that $\displaystyle X_{k}$ ~ $\displaystyle Pois(\lambda,r_{k})$ for each $\displaystyle k = 1,...,n$. Use moment generating functions to determine the distribution of $\displaystyle X = X_{1} +...+ X_{n}$

So far I have $\displaystyle M_{X_1 + X_2 +...+X_n}(t)=M_{x_1}(t)M_{x_2}(t)...M_{x_n}(t)$

=$\displaystyle e^{\lambda_{1}(e^t-1)} \star e^{\lambda_{2}(e^t-1)} \star ... \star e^{\lambda_{n}(e^t-1)}$

=$\displaystyle e^{(e^t-1) [\lambda_{1}+\lambda_{2}+...+\lambda_{n}]}$

I'm not sure what kind of distribution this looks like, if I'm even doing it right.

so the sum of indep poissons is a poisson (as I said in class today)
clearly the mean of the sum is the sum of the means
• Oct 5th 2009, 03:55 PM
Flipz4226
ok thanks!