# moment generating functions for independent var.

• Nov 14th 2009, 01:46 PM
moment generating functions for independent var.
let X~N(mu,sigma^2),Y~Gamma(alpha,Beta) . X and Y are independent , the moment generating function is e^(ut + sigma^2*t^2)/2 for X and (1-Beta)^-alpha for Y. Find mgf of X+Y....I know if you have two indp. var. you multiply them to get the answer, but what form do you leave it in. My answer is e^(ut + sigma^2*t^2)/2 ((1-Beta)^-alpha . Can this answer be broken down anymore?

2)Also how do you compute mgf for 3X? Is it just 3(e^(ut + sigma^2*t^2)/2) or do you find the differiate 3 times?

3) last question computing mgf for these types of problems 2+ Y or 2 + 2X + 3Y. Is it just 2 + (1-Beta)^ -alpha and 2 + (3X)*(3Y)?
I just need to make sure I understand how to compute the mgf.
• Nov 15th 2009, 12:22 AM
Moo
Hello,
Quote:

let X~N(mu,sigma^2),Y~Gamma(alpha,Beta) . X and Y are independent , the moment generating function is e^(ut + sigma^2*t^2)/2 for X and (1-Beta)^-alpha for Y. Find mgf of X+Y....I know if you have two indp. var. you multiply them to get the answer, but what form do you leave it in. My answer is e^(ut + sigma^2*t^2)/2 ((1-Beta)^-alpha . Can this answer be broken down anymore?

It's rather
Quote:

the moment generating function is e^(ut + sigma^2*t^2/2) for X
Quote:

2)Also how do you compute mgf for 3X? Is it just 3(e^(ut + sigma^2*t^2)/2) or do you find the differiate 3 times?
The definition of the mgf is \$\displaystyle E(e^{Xt})\$
So the mgf of 3X is \$\displaystyle E(e^{3Xt})=E(e^{X(3t)})\$, which is the mgf of X taken at the point 3t.

Quote:

3) last question computing mgf for these types of problems 2+ Y or 2 + 2X + 3Y. Is it just 2 + (1-Beta)^ -alpha and 2 + (3X)*(3Y)?
I just need to make sure I understand how to compute the mgf.
Same method as question 2) : write it in the form of the exponential.
Try it (Wink)