Sampling Distribution of the Sample Mean for Wait Times With a PDF of f(x) = ce^-cx where x>=0, how does one determine the Sampling Distribution of the Sample Mean?
https://www.desmos.com/calculator/e5txrxqofw
I really don't understand the concept when the PDF has an exponential distribution.
Every resource I find magically graphs the sampling distribution then moves on to analyze the resulting graph. Isn't that skipping a step?
Is there a function? What is the process to graph the distribution?
I'm lost.
 A: Hints:
Let $S_n=\frac{1}{n}\sum_{i=1}^{n}X_i$ and assuming the samples are iid,
$$M_{S_n}(t) = \left(M_{X_i}\left(\frac{t}{n}\right)\right)^n = \left(M_{X}\left(\frac{t}{n}\right)\right)^n$$
$$M_{X}(t) = E(e^{tX}) = \frac{1}{1-ct} \implies M_{S_{n}}(t) = \left(\frac{1}{1-\frac{c}{n}t}\right)^n$$
$$Y \sim gamma(\alpha,\beta) \implies M_{Y}(t) = \left(\frac{1}{1-\beta t}\right)^\alpha$$
Edit: This way might be easier for you to understand.
Suppose you have an exponential distribution with parameter $\beta$ (i.e. $\exp(\beta)$), the probability density function (pdf) of which is,
$$f(x)=\frac{1}{\beta} e^{-\frac{x}{\beta}}\mathbb{I}(x > 0)$$
Now, you sample from the above distribution and say that your sample contains $X_1, X_2, X_3, \ldots, X_n$. Then, the sample mean will be $S_n=\frac{1}{n}\sum_{i=1}^{n}X_i$. Assume that you sample each $X_i$ independently i.e. $X_i$ is independent of $X_j$ for all $i \neq j$. Since each $X_i$ is sampled from the above distribution, this means that each $X_i$ follows $\exp(\beta)$ distribution.
Now, we are supposed to find the distribution of $S_n$. Before doing that let's find out the probability density of $X_1+X_2$.
$$\begin{align}f(X_1+X_2=x,X_2=y)  &= f(X_1=x-y,X_2=y) \\
&= \frac{1}{\beta^2}e^{-\frac{(x-y)}{\beta}}e^{-\frac{y}{\beta}}\mathbb{I}(x-y>0)\mathbb{I}(y>0) \\
&= \frac{1}{\beta^2} e^{-\frac{x}{\beta}} \mathbb{I}(y<x)\mathbb{I}(y>0)\end{align}$$
$$f(X_1+X_2=x) = \int_{0}^{x}\frac{1}{\beta^2} e^{-\frac{x}{\beta}}dy = \frac{x}{\beta^2} e^{-\frac{x}{\beta}}$$
This is the probability density function of $gamma(x; 2,\beta)$. So, you may use induction from now on to prove that $Y_n = X_1+X_2+\ldots+X_n$ will follow $gamma(x; n, \beta)$.
$$F_{S_n}(x) = P(S_n \leq x) = P(Y_n \leq nx) = F_{Y_n}(nx)$$
Take derivative wrt $x$ to get pdf of $S_n$,
$$f_{S_n}(x) = f_{Y_n}(nx)n = f_{gamma}(x; n, \frac{\beta}{n})$$
Therefore, $S_n \sim gamma(n, \frac{\beta}{n})$.
