91Ó°ÊÓ

Order statistics $Var\left(X_{\left(i\right)}\right),i=1,…,n$

Uniform Random variables $(0,1)$

Given function $f\left(x\right)=\frac{n!}{(i-1)!(n-i)!}{x}^{i-1}(1-x{)}^{n-i}0<x<1$

We have that,

$\mathrm{E}\left({\mathrm{X}}_{\left(\mathrm{i}\right)}\right)={∫}_0^1 \frac{n!}{(i−1)!(n−i!)}xâ‹\dots x^{i−1}(1−x{)}^{n−i}dx;0<x<1i=1,2,…,n$

$\begin{array}{r}=\frac{n!}{(i−1)!(n−i!)}{∫}_0^1 x^i(1−x{)}^{n−i}dx\end{array}$

localid="1647417641323" $=\frac{n!}{(i−1)!(n−i)!}â‹\dots \frac{\left(i\right)!(n−i)!}{(n+1)!}$

$=\frac{n!}{(i−1)!(n−i)!}â‹\dots \frac{\left(i\right)!(n−i)!}{(n+1)!}$

$=\frac{i}{n+1}$

Calculate the value of$E\left[{X^2}_{\left(i\right)}\right],$

$\mathrm{E}\left[{\mathrm{X}}_{\left(\mathrm{i}\right)}^2\right]=\frac{n!}{(i-1)!(n-i)!}{∫}_0^1{x}^{i+1}(1-x{)}^{n-i}dx$

$\begin{array}{r}=\frac{n!}{(i−1)!(n−i)!}â‹\dots \frac{(i+1)!(n−i)!}{(i+2+n−i+1−1)!}\end{array}$

localid="1647417779471" $=\frac{n!}{(i−1)!(n−i)!}â‹\dots \frac{(i+1)!(n−i)!}{(n+2)!}$

$=\frac{i(i+1)}{(n+1)(n+2)}$

Above results are calculated on the Basis of following result:

$\begin{array}{r}{∫}_0^1 x^{a−1}(1−x{)}^{b−1}dx=\frac{(a−1)!(b−1)!}{(a+b−1)!}\end{array}$

$=\frac{i(i+1)}{(n+1)(n+2)}−\frac{i^2}{(n+1{)}^2}$

$=\frac{\left[i^2+i\right][n+1]−i^2[n+2]}{(n+1{)}^2(n+2)}$

localid="1647418098408" $=\frac{i^2[n+1−n−2]+i[n+1]}{(n+1{)}^2(n+2)}$

$=\frac{i[n+1]−i^2}{(n+1{)}^2(n+2)}$

Therefore, the computing value of $Var\left(X_{\left(i\right)}\right),i=1,…,n$is $\frac{i[n+1]-i^2}{(n+1{)}^2(n+2)}$.

Order statistics $\mathrm{Var}\left(X_{\left(i\right)}\right),i=1,…,n$

Uniform Random variables $(0,1)$

Given function$f\left(x\right)=\frac{n!}{(i−1)!(n−i)!}{x}^{i−1}(1−x{)}^{n−i}0<x<1.$

Calculate the variance,

$\begin{array}{r}\mathrm{Var}\left({\mathrm{X}}_{\left(i\right)}\right)=\frac{i[n+1]−i^2}{(n+1{)}^2(n+2)}=f\left(i\right)\end{array}$

$\begin{array}{r}⇒\frac{df\left(i\right)}{di}=\frac{1}{(n+1{)}^2(n+2)}\frac{d}{di}\left[i(n+1)−i^2\right]\end{array}$

$\begin{array}{r}=\frac{1}{(n+1{)}^2(n+2)}\left(\right[n+1]−2i)\end{array}$

role="math" localid="1647527486280" $\frac{df\left(i\right)}{di}=0$

$⇒i=\frac{n+1}{2}$

$\frac{d^{2}f\left(i\right)}{di}=\frac{−2}{(n+1{)}^2(n+2)}<0$

$⇒Var\left(X_{\left(i\right)}\right)$ is maximum at$i=\frac{n+1}{2}$

Now we know that the minimum (least) value of variance can be zero.

From (1) we see that for $i=(n+1)$

Therefore, $i=(n+1)$Minimizes the $\mathrm{Var}\left({\mathrm{X}}_{\left(\mathrm{i}\right)}\right);i=1,2,…,n$

And$⇒i=\left(\frac{n+1}{2}\right)$ Maximizes the$\mathrm{Var}\left({\mathrm{X}}_{\left(\mathrm{i}\right)}\right);i=1,2,..,n$