How to Show That W Follows a Gamma Distribution

[gerv13]

Hi, For this question:
If Z1 ~ \Gamma(\alpha1, \beta) and Z2 ~ \Gamma(\alpha2, \beta); Z1 and Z2 are independent, then Z = Z1 + Z2 ~\Gamma(\alpha1+\alpha2,\beta). Hence show that W~ \Gamma(k/2,1/2)

Well i know how to do the first part, by just multiplying the moment generating function of the gamma. But i don't understand what the W is referring to in the second part of the question, am i supposed to use the fact that Chi^2 ~ \Gamma(1/2,1/2)?

If so, do i just somehow put the k instead of a 1?

Any guidance would be appreciated :)

 

[statdad]

There must be some other part of the problem, or something to which it makes a reference, that defines W. Can you find it?

 

[gerv13]

well in my lecture notes we answered the question "show that Chi^2 ~ Gamma(1/2,1/2)" and then they end with \therefore W_1~Gamma(1/2,1/2). I am not sure if that question relates to the question that I'm supposed to be answering.

But I don't really understand what they did. Maybe if someone can explain it to me it would help me to answer this question:


The cumulative distribution function of W1:
P(W1<w)
= P(-\sqrt{w} z1 &lt; \sqrt{w})
= I (\sqrt{w} ) - I(-\sqrt{w} ) Note: the I is has a 0 through it
= 2I(\sqrt{w} ) - 1

Density Function of W1:
2\frac{d I (\sqrt{w} )}{dw}
= 2\frac{1}{\sqrt{2\Pi}} e^{\frac{-w}{2}}\frac{1}{2}w^{-1/2} (*)
= \frac{1}{\sqrt{2\Pi}} e^{\frac{-w}{2}} w^{-1/2}

\therefore W_1 ~ Gamma(1/2,1/2).

I don't understand where the \frac{1}{\sqrt{2\Pi}} e^{\frac{-w}{2}} came from in lines (*)

So would this help me to answer my question? Coz this is the only thing in my notes before my original question

thanks guys