Q.7.21 - A First Course in Probability

Question

Let $X_{\left(i\right)},i=1,\dots ,n$denote the order statistics from a set of $n$ uniform $(0,1)$ random variables, and note that the density function of $X_{\left(i\right)}$ is given by $f\left(x\right)=\frac{n!}{(i-1)!(n-i)!}{x}^{i-1}(1-x{)}^{n-i} 0<x<1$

(a) Compute $Var\left(X_{\left(i\right)}\right),i=1,\dots ,n$

(b) Which value of $i$ minimizes, and which value maximizes, $Var\left(X_{\left(i\right)}\right)?$

Step-by-Step Solution

Answer

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

b) The value of $i=(n+1)$ Minimizes the $Var\left({\mathrm{X}}_{\left(\mathrm{i}\right)}\right);i=1,2,.,n$

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

Step 1: Given Information (Part a)

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

Uniform Random variables $(0,1)$

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

Step 2: Explanation (Part a)

We have that,

$\mathrm{E}\left({\mathrm{X}}_{\left(\mathrm{i}\right)}\right)={\int }_0^1 \frac{n!}{(i-1)!(n-i!)}x\cdot x^{i-1}(1-x{)}^{n-i}dx;0<x<1i=1,2,\dots ,n$

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

$=\frac{n!}{(i-1)!(n-i)!}\cdot \frac{\left(i\right)!(n-i)!}{(n+1)!}$

$=\frac{n!}{(i-1)!(n-i)!}\cdot \frac{\left(i\right)!(n-i)!}{(n+1)!}$

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

Step 3: Explanation (Part a)

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

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

$\begin{array}{r}=\frac{n!}{(i-1)!(n-i)!}\cdot \frac{(i+1)!(n-i)!}{(i+2+n-i+1-1)!}\end{array}$

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

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

Step 4: Explanation (Part a)

Above results are calculated on the Basis of following result:

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

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

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

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

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

Step 5: Final Answer (Part a)

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

Step 1: Given Information (Part b)

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

Uniform Random variables $(0,1)$

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

Step 2: Explanation

Calculate the variance, 

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

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

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

$\frac{df\left(i\right)}{di}=0$

$\Rightarrow i=\frac{n+1}{2}$

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

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

Step 3: Final Answer

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 $\mathrm{Var}\left({\mathrm{X}}_{\left(\mathrm{i}\right)}\right);i=1,2,\dots ,n$

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