chi square degrees of freedom

**mathh** · February 17th 2010, 07:40 AM

If X and Y are independent random variables, X having a chi-square distribution with 3 degrees of freedom, and X+Y having a chi-square dist with 7 degrees of freedom, prove that Y has chi-square dist with 4 degrees of freedom

Thank you for any help that can be provided

**matheagle** · February 17th 2010, 03:30 PM

I would use MGFs.

use $X\sim\chi^2_3$ and $W=X+Y\sim\chi^2_7$

Then $M_W(t)=M_X(t)M_Y(t)$

Plug in and by division you have the MGF of Y.

**Statistik** · February 17th 2010, 06:33 PM

How do you know (aside from practice?) when to use mgf vs. anything else? We just looked at a getting the limiting distribution of

$<br /> \frac {X_r-r*\beta} {\sqrt r*\beta} <br />$

where

$<br /> X\sim\gamma(r,\beta)<br />$

and the solution was best obtained using mgf. Thanks!

PS: I was able to do the problem, but had the hint do use mgf, so just trying to get a better sense of how to tackle these problems without hints.

**matheagle** · February 17th 2010, 09:28 PM

That's just the central limit theorem

$\Gamma(\alpha,\beta)$ is equal in distribution to

$\sum_{i=1}^{\alpha} \Gamma_i(1,\beta)$

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