Question

If $X_1,X_2,\dots ,X_n$are independent and identically distributed random variables having uniform distributions over $(0,1)$, find

(a) $E\left[\max \left(X_1,\dots ,X_n\right)\right]$;

(b) $E\left[\min \left(X_1,\dots ,X_n\right)\right]$.

Short answer

$E\left[U\right]=\frac{n}{n+1}$

$E\left[L\right]=\frac{1}{n+1}$

Step-by-step solution

Given Information (part a)

Suppose

$U=\max \left(X_1,\dots ,X_n\right)$

This means that

$F_U\left(u\right)=P(U\le u)$

Calculation (part a)

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

$F_U\left(u\right)=P(U\le u)=P\left(X_1\le u,\dots ,X_n\le u\right)$

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

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

$\Rightarrow f_U\left(u\right)=n{x}^{n-1}$

Final Answer (part a)

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

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

Given Information (part b)

Suppose

$L=\min \left(X_1,\dots ,X_n\right)$

This means that

$F_L\left(l\right)=P(L\le l)=1-P(L\ge l)$

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

$F_L\left(l\right)=1-P\left(X_1\ge l,\dots ,X_n\ge l\right)$

Calculation (part b)

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

$F_L\left(l\right)=1-P\left(X_1\ge l\right)\dots P\left(X_n\ge l\right)=1-{\left(1-\frac{x-0}{1-0}\right)}^n$

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

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

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

This can be simplified using integration by parts where

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

Final Answer (part b)

So we end up with

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

$\Rightarrow E\left[L\right]=\frac{1}{n+1}$