Properties of the Exponential Distribution
Suppose that $X_1,X_2,…,X_n$ is a random sample from the exponential distribution with rate $\lambda\gt0$.
We know that
\[\sum_{i-1}^{n}X_i\sim\Gamma(n,\lambda)\]and therefore that
\[2\lambda\sum_{i-1}^{n}X_i\sim\Gamma\bigg(\frac{2n}{2},\frac{1}{2}\bigg)=\chi^2(2n)\]We also know that
\[\overline{X}=\frac{1}{n}\sum_{i-1}^{n}X_i\sim\Gamma(n,n\lambda)\]and there for that
\[2n\lambda\overline{X}\sim\Gamma\bigg(n,\frac{1}{2}\bigg)=\Gamma\bigg(\frac{2n}{2},\frac{1}{2}\bigg)=\chi^2(2n)\]Suppose that $X_1,X_2,…,X_n$ is a random sample from the exponential distribution with rate $\lambda\gt0$.
What is the distribution of minimum?
Let $Y_n=\text{min}(X_1,X_2,…,X_n)$.
The cdf for each $X_i$ is
\[F(x)=P(X_i\le x)=1-e^{-\lambda x}\]Let $Y_n=min(X_1,X_2,…,X_n)$.
The cdf for each $X_i$ is
\[F(x) = P(X_i\le x)=1-e^{-\lambda x}\]The cdf for $Y_n$ is
\[\begin{align} F_{Y_n}(y)&=P(Y_n\le y)\\ &=P(min(X_1,X_2,...,X_n)\le y) \end{align}\]If we choose $x$’s like the way below, there’s a lot of way for $x$’s to fall so that the minimum is less that or equal to $y$ it’s going to be hard to translate this about the minimum to a statement involving where the individual $x$’s fall with respect to this fixed number $y$.
But it will be easier if we look at the complement of this event. The minimum in a sample is greater than a fixed number $y$, if and only if every number in the sample is greater than $y$.
If we take the derivative and we get the pdf of an exponential with paramenter $n\lambda$
\[f(Y_n(y) = \frac{d}{dx}F_{Y_n}(y)=n\lambda e^{-n\lambda y}\]The minimum of $n$ iid exponential with rate $\lambda$ is exponential with rate $n\lambda$!
\[Y_n\sim \text{exp}(\text{rate}=n\lambda)\]