How do you transform Gamma to Chi-squared distribution

Question

Here is the question not sure how to turn a Gamma into a Chi-Squared:

Suppose $X_1....X_n$ is a sample from the distribution Gamma($\alpha=3,\ \lambda=\theta$) with unknown $\theta > 0$. We wish to eventually construct a 98% Confidence Interval for the scale parameter $\theta$

a) Show first that $2\theta\sum_{1 \le \ i \le n} X_i$ has Chi-squared distribution, identify the number of degree of freedom

b) Using $2\theta\sum_{1 \le \ i \le n} X_i$ as a piviot, explain what tables you would need to use in order to find an $a(x_1,...x_n)$ and a $b(x_1,...x_n)$ so that: $$P(a(x_1,...x_n) < \theta < b(x_1,...x_n))=.98$$ The 98% CI for $\theta$ is $a(x_1,...x_n),b(x_1,...x_n)$

Answer 1

This question likely needs a homework or self study tag and the OP, who has asked several similar kinds of questions before, should know better than to skip this.

Here are some hints:

• The sum of independent Gamma random variables with identical scale parameter is a Gamma random variable with the same scale parameter and order parameter equal to the sum of the order parameters. So you should be able to show that $\sum_i X_i$ is a Gamma random variable.

• If $E[Y] = \mu$, then $E[\mu^{-1}Y] = 1$. Furthermore, if $Y$ is a Gamma random variable, then so is $aY$ a Gamma random variable. You should be able to figure out the distribution of $2\theta\sum_i X_i$ from this.

• A Gamma random variable with order parameter $\frac{n}{2}$ and mean $n$ is a $\chi^2$ random variable with $n$ degrees of freedom.