# Chi-square problem

1. Aug 29, 2009

### 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 dont understand what the W is refering 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 :)

2. Aug 29, 2009

### statdad

Re: Chi-square

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

3. Aug 29, 2009

### gerv13

Re: Chi-square

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). Im not sure if that question relates to the question that I'm supposed to be answering.

But I dont 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 dont 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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook