We have $n$ i.i.d. samples $X_1,X_2,\dots,X_n$, which have exponential probability density with mean $\mu$, so pdf is $f(x)=(1/\mu)e^{-x/\mu}$ for $x>0$ and $0$ otherwise. Now how to calculate the expectations $$E\left[\min{\{X_1,X_2,\ldots,X_n}\}\right]$$
and $$E\left[\max{\{X_1,X_2,\ldots,X_n}\}\right]$$
The one thing that I get from the density function is that probability of minimum value of $x$ would be the highest and probability of maximum value of $x$ would be the lowest. So from intuition, as we increase the value of $n$ the probability of highest number will move close to $0$ and that of lowest one will move to $1$.
How is this going to help me out?
Best Answer
Hint ahead: first do it for standard case $\mu=1$. Then realize that $X_i=\mu Y_i$ where $Y_i$ has standard exponential distribution.
In general if $Z$ is a nonnegative random variable with continuous distribution then:$$\mathbb EZ=\int_0^{\infty}P(Z>z)dz$$
For $Z:=\min(X_1,\dots,X_n)$ we find $P(Z>z)=P(X_1>z)^n$.
For $Z:=\max(X_1,\dots,X_n)$ we find $P(Z>z)=1-P(Z\leq z)=1-P(X_1\leq z)^n$
addendum
$\int_{0}^{\infty}xdF_Z\left(x\right)=\int_{0}^{\infty}\int_{0}^{\infty}1_{\left(0,x\right)}\left(z\right)dzdF_Z\left(x\right)=\int_{0}^{\infty}\int_{0}^{\infty}1_{\left(0,x\right)}\left(z\right)dF_Z\left(x\right)dz=\int_{0}^{\infty}P\left(Z>z\right)dz$