Chi-square problem

  • Thread starter gerv13
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  • #1
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Hi, For this question:
If Z1 ~ [tex]\Gamma(\alpha1, \beta)[/tex] and Z2 ~ [tex]\Gamma(\alpha2, \beta)[/tex]; Z1 and Z2 are independent, then Z = Z1 + Z2 ~[tex]\Gamma(\alpha1+\alpha2,\beta)[/tex]. Hence show that [tex]W [/tex]~ [tex]\Gamma(k/2,1/2)[/tex]

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 ~ [tex]\Gamma(1/2,1/2)[/tex]?

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

Any guidance would be appreciated :)
 

Answers and Replies

  • #2
statdad
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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
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well in my lecture notes we answered the question "show that Chi^2 ~ Gamma(1/2,1/2)" and then they end with [tex]\therefore W_1[/tex]~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)
= [tex]P(-\sqrt{w} z1 < \sqrt{w}) [/tex]
= [tex]I (\sqrt{w} ) - I(-\sqrt{w} )[/tex] Note: the I is has a 0 through it
= [tex]2I(\sqrt{w} ) - 1[/tex]

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

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

I dont understand where the [tex]\frac{1}{\sqrt{2\Pi}} e^{\frac{-w}{2}} [/tex] 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
 

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