# Show that $2Y/\theta$ has a chi-square distribution [closed]

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.

## closed as off-topic by Did, Namaste, Shailesh, астон вілла олоф мэллбэрг, user91500Nov 21 '16 at 5:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Shailesh, астон вілла олоф мэллбэрг, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – Dilip Sarwate Nov 13 '15 at 4:09

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:
$$M_{2Y/\theta}(t) = \mathbb{E}[e^{2tY/\theta}] = \int_{0}^{\infty}\frac{e^{\frac{2ty}{\theta}}}{\Gamma(\alpha)\theta^{\alpha}}y^{\alpha-1}e^{-\frac{y}{\theta}}\operatorname{d}\!y.$$
$$\int_{0}^{\infty}\frac{e^{\frac{2ty}{\theta}}}{\Gamma(\alpha)\theta^{\alpha}}y^{\alpha-1}e^{-\frac{y}{\theta}}\operatorname{d}\!y =(1-2t)^{-\alpha} \int_{0}^{\infty}\frac{1}{\Gamma(\alpha)(\theta(1-2t)^{-1})^{\alpha}}y^{\alpha-1}e^{-\frac{y}{\theta(1-2t)^{-1}}}\operatorname{d}\!y.$$
Conclude using your knowledge of the MGF of the $\chi^{2}$-distribution and noting that the final integrand above is a probability density function.