If $X$ is exponential distributed with parameter $\lambda$, how is $\bar{X}/\lambda$ distributed?

Question

In Davison A.C., Hinkley D.V. - Bootstrap methods and their application It say's $\bar{X}/\lambda$ "has the Gamma distribution with index n and unit mean". I don't understand the sentence. Isn't gamma distribution a generell distribution, including the exponential distribution, Erlang distribution, and chi-squared distribution? And which one would it be here with which parameters?

Thank you.

Answer 1

Let's $X\sim Exp(\lambda)$, say, for $x>0$

$$f_X(x)=\frac{1}{\lambda} e^{-\frac{x}{\lambda}}$$

That is equivalent to a Gamma distribution $\Gamma(1;\lambda)$

Let's have $X_1,...,X_n$ iid rv's with the same denisity,

$$\frac{\sum_i X_i}{n \lambda}\sim \Gamma(n;n)$$

That is a Gamma distribution with index $n$ and mean $\frac{n}{n}=1$

Due to the fact that $n$ is integer, this distribution is a.k.a. Erlang

The proof is trivial using Moment Generating Function and its properties

Answer 2

Comment continued: Let $n=10; \lambda =\mathrm{scale} = 2.$ Vector a has 100,000 sample averages. Simulation in R.

set.seed(2020);  n = 10;  scale=2
a = replicate(10^5, mean(rexp(n,1/scale)))  
 # In R 2nd param = rate
mean(a); var(a)
[1] 1.998913    # aprx E(A) = 2
[1] 0.3998701   # aorx Var(A) = 4/10
mean(a/scale)
[1] 0.9994563   # aprx 1
var(a/scale)
[1] 0.09996752  # aprx 1/10


hist(a/scale, prob=T, col="skyblue2")
 curve(dgamma(x, n, n), add=T, col="red")
 # density of gamma with shape=n, rate=n