Expected value of distance between min and max of independent events with uniform distribution Let $U_1, \cdots , U_5$ be independent, each with uniform distribution on (0, 1).  Let R be the distance between the minimum and maximum of the $U_i's$.  Find E(R).
I think R might have a Beta distribution.  I really don't know how to solve this.
Thanks!
 A: Hint: Let $A$ be the minimum and $B$ the maximum.  Then 
if $0 \le a \le b \le 1$, $P(a \le A \le B \le b) = (b-a)^5$.  Use this to find the joint density of $A$ and $B$.
A: Let $A$ be the minimum, $B$ be the maximum. 
Then $R=B-A$. 
Thus $E(R)=E(B)-E(A)$. 
It is easy to find $E(A)=\int_0^1 5y(1-y)^4 dy=1/6$,  and $E(B)=\int_0^1 5y^5 dy=5/6$
Hence $E(R)=2/3$. 
A: You want the difference between the maximum and the minimum.
Both follow a Beta distribution as you can see in this page of Wikipedia. This is:


*

*max = $X_{(5)}$ follows $Beta(5, 1)$

*min = $X_{(1)}$ follows $Beta(1, 5)$


then you can look at your distribution table to know the expected value:


*

*$E[X_{(5)}]=\frac{\alpha}{\alpha+\beta}=\frac{5}{6} $

*$E[X_{(1)}]=\frac{\alpha}{\alpha+\beta}=\frac{1}{6} $


Then just get there expected values and substracted since $E(X-Y) = E(X) -E(Y)$ (by linearity property of Expected value):
$$E[X_{(5)}-X_{(1)}] = E[X_{(5)}] -E[X_{(1)}] = \frac{4}{6} = \frac{2}{3}$$
Additionally:
If you want the distribution of $X_{(n)} - X_{(1)}$.
You may want to use $R=X_{(5)} - X_{(1)}$ and $C=X_{(5)}+X_{(1)}$ to ease the transformation and obtain $f(r,c)$.
Notes:


*

*The range, R, won't be beta distributed.

*You can find more information here.

