Use moment generating function to show that $f_Y(y)$ has chi squared distrubution Problem in my Stat class
Let $Y_1, Y_2,\ldots, Y_n$ be arandom sample of size $n$ from 
$$f_Y(y)= \frac{1}{\theta } e^{-y/\theta}, \qquad y> 0,\theta > 0$$
Use moment-generating functions to show that the ratio $2n\bar{y}/\theta$ has a chi square distribution with $2n$ degrees of freedom.
First question do they mean that $2n\bar{y}$ is the $y$ variable and $\theta$ is the $\theta$ variable?
secondly how do you show that anything has a chi squared distribution?
Small hints to guide me in the right direction would also be helpful. 
 A: Note that your given distribution for $Y_1,\dots,Y_n$ is that of an exponential family. In particular,
$$
M_Y(t) = E[e^{tY}] = \int_{0}^\infty e^{ty} \left( \dfrac{1}{\theta} e^{-y / \theta} \right)  \; dy = \cdots = \dfrac{1}{1-\theta t} \text{ for } \dfrac{1}{\theta} > t.
$$
Let $Z = \dfrac{2  n \overline{Y}}{\theta} = \dfrac{2}{\theta} \displaystyle\sum_{k=1}^n Y_i$. It follows that
\begin{align*}
M_Z(t) = E[e^{tZ}] &= E \left[ \exp \left( \dfrac{2t}{\theta} \displaystyle\sum_{k=1}^n Y_i \right) \right] \\
&= E \left[ \displaystyle\prod_{k=1}^n \exp \left( \dfrac{2t}{\theta}  Y_i \right) \right] \\
&= \displaystyle\prod_{k=1}^n E \left[ \exp \left( \dfrac{2t}{\theta}  Y_i \right) \right] \quad \text{(Independence)} \\
&= \left( E \left[  e^{2tY/\theta} \right] \right)^n \quad \text{(Identically Distributed)} \\
&= (M_Y(2t/\theta))^n \\
&= \left( \dfrac{1}{1-\theta (2t/\theta)} \right)^n,  \quad \text{ with } \frac{1}{\theta}>\left( \dfrac{2t}{\theta} \right) \iff t < \frac{1}{2} \\
&= \left( \dfrac{1}{1-2t} \right)^n, \quad t < \frac{1}{2}
\end{align*}
which is precisely the mgf of a chi-square distribution with $2n$ degrees of freedom.
