91Ó°ÊÓ

Suppose

$U=\max \left(X_1,…,X_n\right)$

This means that

$F_U\left(u\right)=P(U≤u)$

Obviously, if the maximum is bounded above, then every element in the set is similarly bounded. So:

$F_U\left(u\right)=P(U≤u)=P\left(X_1≤u,…,X_n≤u\right)$

And by assumptions, we know that the joint distribution is just the product of the marginal distributions, so:

$F_U\left(u\right)={\left(\frac{x-0}{1-0}\right)}^n=x^n$

$⇒f_U\left(u\right)=n{x}^{n-1}$

Now it is relatively simple to calculate$E\left[U\right]$

$E\left[U\right]={∫}_0^{1}x{f}_U\left(u\right)dx={∫}_0^{1}xn{x}^{n-1}dx={∫}_0^{1}n{x}^{n}dx=\frac{n}{n+1}$

Suppose

$L=\min \left(X_1,…,X_n\right)$

This means that

$F_L\left(l\right)=P(L≤l)=1-P(Lâ‰\yen l)$

If the minimum $L$is bounded below by $l$, then every element is the set is similarly bounded. So:

$F_L\left(l\right)=1-P\left(X_1â‰\yen l,…,X_nâ‰\yen l\right)$

And by assumptions, we know that the join distribution is just the product of the marginal distributions, so:

$F_L\left(l\right)=1-P\left(X_1â‰\yen l\right)…P\left(X_nâ‰\yen l\right)=1-{\left(1-\frac{x-0}{1-0}\right)}^n$

$⇒F_L\left(l\right)=1-(1-x{)}^n⇒f_L(l)=n(1-x{)}^{n-1}$

Now it is relatively simple to calculate$E\left[L\right]$

$E\left[L\right]={∫}_0^{1}x{f}_L\left(l\right)dx={∫}_0^{1}xn(1-x{)}^{n-1}dx$

This can be simplified using integration by parts where

$u=x,dv=n(1-x{)}^{n-1}⇒du=dx,v=(1-x{)}^n$

So we end up with

$E\left[L\right]=-{∫}_0^1(1-x{)}^{n}dx=-{\left[\frac{(1-x{)}^{n+1}}{n+1}\right]}_0^1$

$⇒E\left[L\right]=\frac{1}{n+1}$