# Moment Generating Functiom to determine Distribution

• Oct 5th 2009, 01:55 PM
Flipz4226
Moment Generating Functiom to determine Distribution
$\lambda_{1},...,\lambda_{n} > 0$ and $X_{1},...,X_{n}$ are independent random variables such that $X_{k}$ ~ $Pois(\lambda,r_{k})$ for each $k = 1,...,n$. Use moment generating functions to determine the distribution of $X = X_{1} +...+ X_{n}$

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

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

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

I'm not sure what kind of distribution this looks like, if I'm even doing it right.
• Oct 5th 2009, 04:51 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
$\lambda_{1},...,\lambda_{n} > 0$ and $X_{1},...,X_{n}$ are independent random variables such that $X_{k}$ ~ $Pois(\lambda,r_{k})$ for each $k = 1,...,n$. Use moment generating functions to determine the distribution of $X = X_{1} +...+ X_{n}$

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

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

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

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, 04:55 PM
Flipz4226
ok thanks!