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.
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.