Moment generating function from p.d.f

X has chi square distribution with 4 degrees of freedom, with pdf

$\displaystyle f_{X}(x) = \frac{xe^-\frac{x}{2}}{4}, x>0$

(i) Find the moment generating function.

Now I did moment generating function = E($\displaystyle e^{tx})$ so for continuous r.v =

$\displaystyle \int e^{tx}\frac{xe^-\frac{x}{2}}{4}$.

Do I leave this here? I do I have to integrate and what happens with the t? Should I isolate the t outside of the integral first?

(ii) If X1...Xn are independent, identically distributed random variables, with chi square distribution, with 4 degrees of freedom, show that the moment generating function $\displaystyle Y = \sum X_{i}$ is $\displaystyle (1-2t)^{-2n}$ and find the expecation and variance of Y.

Re: Moment generating function from p.d.f

Writing it out in full notation can provide hints on what the answer should look like:

$\displaystyle M_x(t) = \int_0^\infty e^{tx} \frac{ x e^{-0.5x}}{4} dx $

Written in full its clear that the next step is to complete the integration with respect to x. Its also clear from the LHS that the final answer should be a function of t (ie, you do not have to isolate or solve for t).

Re: Moment generating function from p.d.f

Quote:

Originally Posted by

**SpringFan25** Writing it out in full notation can provide hints on what the answer should look like:

$\displaystyle M_x(t) = \int_0^\infty e^{tx} \frac{ x e^{-0.5x}}{4} dx $

Written in full its clear that the next step is to complete the integration with respect to x. Its also clear from the LHS that the final answer should be a function of t (ie, you do not have to isolate or solve for t).

I used partical integration.

I got:

$\displaystyle \frac{1}{4}(\frac{xe^{(t-\frac{1}{2})x}}{t-\frac{1}{2}}-\frac{e^{(t-\frac{1}{2})x}}{(t-\frac{1}{2})^2})$

is this correcT? I also have to specify the range of t. Would this just be t not equal 1/2? or is there another stipulation? also, moving onto the second part? any starters?

(ii) If X1...Xn are independent, identically distributed random variables, with chi square distribution, with 4 degrees of freedom, show that the moment generating function $\displaystyle Y = \sum X_{i}$ is $\displaystyle (1-2t)^{-2n}$ and find the expecation and variance of Y.

Re: Moment generating function from p.d.f

this is a gamma density

the idea is to combine the exponent of e and use the fact that a density integrates to one

hence find your new BETA