I have some questions regarding a problem involving univariate transformations.
The problem assumes that the population is an Exponential$(\theta)$ population.
($1$) For the first part, I am asked to use MGF's to find the distribution of $Y=\sum\limits_{i = 1}^n {X_i}$, where $X_1...X_n$ is a random sample from the population.
To solve this, I noted that $Y=X_1+...+X_n$ are indeed a sum of independent random variables, and therefore reasoned that $M_Y(t) = M_{X_1+X_2+...+X_n}(t)=M_{X_1}(t)*M_{X_2}(t)*...*M_{X_n}(t)=\bigg(\dfrac{1}{1-\theta t}\bigg)^n$. Then, it was clear that $\bigg(\dfrac{1}{1-\theta t}\bigg)^n$ is the MGF of a Gamma($n,\theta$) distribution. Hence, $Y$~Gamma($n,\theta)$, so $f_y(y) = \dfrac{y^{n-1}}{\Gamma(n)\theta^n}e^\frac{-y}{\theta}$, where $y>0$.
($2$) For the second part, I was asked to use this distribution, to then apply the univariate transformation technique to find the distribution of $\bar{X}$.
I noted that since $Y=X_1+X_2+...+X_n$ and $\bar{X} = \frac{Y}{n}$, then $Y=\bar{X}n$, and $\frac{dY}{d\bar{X}} = n$. Since $n$ is positive, $|n| = n$. Thus:
$$f_\bar{X}(\bar{X}) = f_\bar{Y}(\bar{Y})\bigg|\frac{dY}{d\bar{X}}\bigg| = \dfrac{y^{n-1}}{\Gamma(n)\theta^n}e^{-Y/\theta}(n)=\dfrac{{(\bar{X}n)}^{n-1}}{\Gamma(n)\theta^n}e^{-(\bar{X}n)/\theta}(n)=\dfrac{{(\bar{X})}^{n-1}(n)^n}{\Gamma(n)\theta^n}e^{-(\bar{X}n)/\theta}$$ where $\bar{X}>0$.
So, this was the distribution I found for $\bar{X}$. However, I was a bit doubtful about this answer, since the $n^n$ part of the numerator prevents this from being a Gamma distribution.
If anyone could help point out any inaccuracies in my argument, that would be wonderful. Thank you!