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)
= $$P(-\sqrt{w} z1 < \sqrt{w})$$
= $$I (\sqrt{w} ) - I(-\sqrt{w} )$$ 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