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The question is Let $Y$ be a random variable with a Gamma distribution with parameters $\alpha > 0$ and $\theta > 0$. Show that $2Y/\theta$ has a chi-square distribution. What is the number of degrees of freedom?

I am confused about what the $2Y/\theta$ represents. How does one show something is a chi-square distribution. I know that if $X$ follows a normal distribution than $X^2$ is a chi-square distribution?

If someone could show work and explain how all the pieces work together I would appreciate that. Trying to understand the concept and relation between Gamma and chi-square distribution.

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closed as off-topic by Did, Namaste, Shailesh, астон вілла олоф мэллбэрг, user91500 Nov 21 '16 at 5:32

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    $\begingroup$ Is there something that you are leaving out, such as $\alpha = \frac n2$ for some positive integer $n$? A $\chi^2$ random variable is indeed a Gamma random variable, but its order parameter $\alpha$ must be of the form $\frac n2$ for some integer $n$. The scaling of $Y$ by $\frac{2}{\theta}$ can adjust the mean parameter appropriately, but will not affect the order parameter at all. $\endgroup$ – Dilip Sarwate Nov 13 '15 at 4:09
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The moment generating function will uniquely determine the distribution of the random variable in this case. The mgf of $2Y/\theta$ is calculated using the law of the unconscious statistician:

$$M_{2Y/\theta}(t) = \mathbb{E}[e^{2tY/\theta}] = \int_{0}^{\infty}\frac{e^{\frac{2ty}{\theta}}}{\Gamma(\alpha)\theta^{\alpha}}y^{\alpha-1}e^{-\frac{y}{\theta}}\operatorname{d}\!y.$$

It should be very easy from here to show that

$$\int_{0}^{\infty}\frac{e^{\frac{2ty}{\theta}}}{\Gamma(\alpha)\theta^{\alpha}}y^{\alpha-1}e^{-\frac{y}{\theta}}\operatorname{d}\!y =(1-2t)^{-\alpha} \int_{0}^{\infty}\frac{1}{\Gamma(\alpha)(\theta(1-2t)^{-1})^{\alpha}}y^{\alpha-1}e^{-\frac{y}{\theta(1-2t)^{-1}}}\operatorname{d}\!y.$$

Conclude using your knowledge of the MGF of the $\chi^{2}$-distribution and noting that the final integrand above is a probability density function.

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