Chapter 6, Problem 76SE

a.

To determine

Find the density function of $Y_{\left(k\right)}$, k^th-order statistic.

Answer to Problem 76SE

The density function of $Y_{\left(k\right)}$, k^th-order statistic is,

$g_{\left(k\right)}\left(y\right)=\frac{n!}{\left(k-1\right)!\left(n-k\right)!}\frac{y^{k-1}{\left(\theta -y\right)}^{n-k}}{{\theta }^n},0\le y_k\le \theta$.

Explanation of Solution

Calculation:

From Theorem 6.5, the density function for k^th order statistic, $Y_{\left(k\right)}$ is,

$g_{\left(k\right)}\left(y_k\right)=\frac{n!}{\left(k-1\right)!\left(n-k\right)!}{\left[F\left(y_k\right)\right]}^{k-1}\times {\left[1-F\left(y_k\right)\right]}^{n-k}f\left(y_k\right),-\infty <y_k<\infty$.

It is known that, the density function of Uniform distribution over the interval $\left[0,\theta \right]$ is $f\left(y\right)=\frac{1}{\theta },0\le y\le \theta$ and the distribution function is $F\left(y\right)=\frac{y}{\theta }$.

The density function for $Y_{\left(k\right)}$ is,

$\begin{array}{c}{g}_{\left(k\right)}\left(y\right)=\frac{n!}{\left(k-1\right)!\left(n-k\right)!}{\left[\frac{y}{\theta }\right]}^{k-1}\times {\left[1-\frac{y}{\theta }\right]}^{n-k}\frac{1}{\theta },-\infty <y_k<\infty \\ =\frac{n!}{\left(k-1\right)!\left(n-k\right)!}\frac{y^{k-1}{\left(\theta -y\right)}^{n-k}}{{\theta }^n},0\le y_k\le \theta \end{array}$

b.

To determine

Find the value of $E\left(Y_{\left(k\right)}\right)$.

Answer to Problem 76SE

The value of $E\left(Y_{\left(k\right)}\right)$ is $\frac{k}{n+1}\theta$.

Explanation of Solution

Calculation:

Consider,

$\begin{array}{c}E\left(Y_{\left(k\right)}\right)={\displaystyle \underset{0}{\overset{\theta }{\int }}y{g}_{\left(k\right)}}\left(y\right)dy\\ ={\displaystyle \underset{0}{\overset{\theta }{\int }}y\frac{n!}{\left(k-1\right)!\left(n-k\right)!}\frac{y^{k-1}{\left(\theta -y\right)}^{n-k}}{{\theta }^n}}dy\\ =\frac{k}{n+1}\frac{\Gamma \left(n+2\right)}{\Gamma \left(k+1\right)\Gamma \left(n-k+1\right)}{\displaystyle \underset{0}{\overset{\theta }{\int }}{\left(\frac{y}{\theta }\right)}^k{\left(1-\frac{y}{\theta }\right)}^{n-k}}dy\\ =\frac{k}{n+1}\frac{\Gamma \left(n+2\right)}{\Gamma \left(k+1\right)\Gamma \left(n-k+1\right)}{\displaystyle \underset{0}{\overset{\theta }{\int }}{\left(z\right)}^k{\left(1-z\right)}^{n-k}}\theta dz\left[\begin{array}{c}\text{Let, }\frac{y}{\theta }=z\\ dy=\theta dz\end{array}\right]\end{array}$

This function  ${\displaystyle \underset{0}{\overset{\theta }{\int }}{\left(\frac{y}{\theta }\right)}^k{\left(1-\frac{y}{\theta }\right)}^{n-k}}dy$ representing the beta density function with $\alpha =k+1$ and $\beta =n-k+1$.

Therefore,

$E\left(Y_{\left(k\right)}\right)=\frac{k}{n+1}\theta$

c.

To determine

Find the value of $V\left(Y_{\left(k\right)}\right)$.

Answer to Problem 76SE

The value of $V\left(Y_{\left(k\right)}\right)$ is $\frac{k\left(n-k+1\right)}{{\left(n+1\right)}^2\left(n+2\right)}{\theta }^2$.

Explanation of Solution

Calculation:

Consider,

$\begin{array}{c}E\left(Y^2{}_{\left(k\right)}\right)={\displaystyle \underset{0}{\overset{\theta }{\int }}{y}^2{g}_{\left(k\right)}}\left(y\right)dy\\ ={\displaystyle \underset{0}{\overset{\theta }{\int }}{y}^2\frac{n!}{\left(k-1\right)!\left(n-k\right)!}\frac{y^{k-1}{\left(\theta -y\right)}^{n-k}}{{\theta }^n}}dy\\ =\frac{k\theta }{n+1}\frac{\Gamma \left(n+2\right)}{\Gamma \left(k+1\right)\Gamma \left(n-k+1\right)}{\displaystyle \underset{0}{\overset{\theta }{\int }}{\left(\frac{y}{\theta }\right)}^{k+1}{\left(1-\frac{y}{\theta }\right)}^{n-k}}dy\\ =\frac{k\theta }{n+1}\frac{\Gamma \left(n+2\right)}{\Gamma \left(k+1\right)\Gamma \left(n-k+1\right)}{\displaystyle \underset{0}{\overset{\theta }{\int }}{\left(z\right)}^{k+1}{\left(1-z\right)}^{n-k}}\theta dz\left[\begin{array}{c}\text{Let, }\frac{y}{\theta }=z\\ dy=\theta dz\end{array}\right]\end{array}$

                   $=\frac{k\left(k+1\right)}{\left(n+1\right)\left(n+2\right)}{\theta }^2$

Therefore,

$\begin{array}{c}V\left(Y_{\left(k\right)}\right)=E\left(Y_{\left(k\right)}^2\right)-{\left[E\left(Y_{\left(k\right)}\right)\right]}^2\\ =\frac{k\left(k+1\right)}{\left(n+1\right)\left(n+2\right)}{\theta }^2-{\left(\frac{k}{n+1}\theta \right)}^2\\ =\frac{k\left(n-k+1\right)}{{\left(n+1\right)}^2\left(n+2\right)}{\theta }^2\end{array}$

d.

To determine

Find the mean difference between two successive order statistics, $E\left(Y_{\left(k\right)}-Y_{\left(k-1\right)}\right)$.

Answer to Problem 76SE

The mean difference between two successive order statistics, $E\left(Y_{\left(k\right)}-Y_{\left(k-1\right)}\right)$ is $\frac{1}{n+1}\theta$.

The expected order statistics are equally spaced.

Explanation of Solution

Calculation:

Consider,

$\begin{array}{c}E\left(Y_{\left(k\right)}-Y_{\left(k-1\right)}\right)=\frac{k}{n+1}\theta -\frac{k-1}{n+1}\theta \\ =\left(\frac{k-k+1}{n+1}\right)\theta \\ =\frac{1}{n+1}\theta \end{array}$

This function is a constant for all k. Thus, the expected order statistics are equally spaced.