Question

If x1, x2, . . . , xn are independent and identically distributed random variables having uniform distributions over (0, 1), find (a) e[max(x1, . . . , xn)]; (b) e[min(x1, . . . , xn)].

Answer 1

Denote by $X_{(n)}$ the maximum order statistic, with $X_{(n)}=\max\{X_1,\ldots,X_n\}$, and similarly denote by $X_{(1)}$ the minimum order statistic. Then the CDF for $X_{(n)}$ is

$F_{X_{(n)}}(x)=\mathbb P(X_{(n)}\le x)$

In order for there to be some $x$ that exceeds the value of $X_{(n)}$, it must be true that $x$ exceeds the value of all the $X_i$, so the above is equivalent to the joint probability


$F_{X_{(n)}}(x)=\mathbb P(X_1\le x,\ldots,X_n\le x)$

and since the $X_i$ are i.i.d., we have

$F_{X_{(n)}}(x)=\mathbb P(X_1\le x)\cdots\mathbb P(X_n\le x)=\mathbb P(X_1\le x)^n$
$\implies F_{X_{(n)}}(x)=F_X(x)^n$

where $X\sim\mathrm{Unif}(0,1)$. We have


$F_X(x)=\begin{cases}0&\text{for }x<0\\x&\text{for }0\le x\le1\\1&\text{for }x>1\end{cases}$

and so

$F_{X_{(n)}}(x)=\begin{cases}0&\text{for }x<0\\x^n&\text{for }0\le x\le1\\1&\text{for }x>1\end{cases}$
$\implies f_{X_{(n)}}(x)=\begin{cases}nx^{n-1}&\text{for }0<x<1\\0&\text{otherwise}\end{cases}$
$\implies\mathbb E[X_{(n)}]=\displaystyle\int_0^1xnx^{n-1}\,\mathrm dx=n\int_0^1x^n\,\mathrm dx=\frac n{n+1}$

Using similar reasoning, we can find the CDF for $X_{(1)}$. We have

$F_{X_{(1)}}(x)=\mathbb P(X_{(1)}\le x)=1-\mathbb P(X_{(1)}>x)$
$F_{X_{(1)}}(x)=1-\mathbb P(X_1>x,\ldots,X_n>x)=1-\mathbb P(X_1>x)^n$
$F_{X_{(1)}}(x)=1-(1-\mathbb P(X\le x))^n=1-(1-F_X(x))^n$
$\implies F_{X_{(1)}}(x)=\begin{cases}0&\text{for }x<0\\1-(1-x)^n&\text{for }0\le x\le1\\1&\text{for }x>1\end{cases}$
$\implies f_{X_{(1)}}(x)=\begin{cases}n(1-x)^{n-1}&\text{for }0<x<1\\0&\text{otherwise}\end{cases}$
$\implies\mathbb E[X_{(1)}]=\displaystyle\int_0^1xn(1-x)^{n-1}\,\mathrm dx=\frac1{n+1}$