Chapter 6, Problem 77SE

a.

To determine

Find the joint density function of $Y_{\left(j\right)}$ and $Y_{\left(k\right)}$, where j and k are integers $1\le j\le k\le n$

Answer to Problem 77SE

The joint density function of $Y_{\left(j\right)}$ and $Y_{\left(k\right)}$ is,

$g_{\left(j\right)\left(k\right)}\left(y_j,y_k\right)=\left\{\begin{array}{l}\frac{n!}{\left(j-1\right)!\left(k-1-j\right)!\left(n-k\right)!}{\left[\frac{y_j}{\theta }\right]}^{j-1}\\ {\left[\frac{y_k}{\theta }-\frac{y_j}{\theta }\right]}^{k-1-j}{\left[1-\frac{y_k}{\theta }\right]}^{n-k}\frac{1}{{\theta }^2}\end{array}\right\},0\le y_j\le \theta$ .

Explanation of Solution

Calculation:

From Theorem 6.5, the joint density function for $Y_{\left(j\right)}$ and $Y_{\left(k\right)}$ is,

$g_{\left(j\right)\left(k\right)}\left(y_j,y_k\right)=\left\{\begin{array}{l}\frac{n!}{\left(j-1\right)!\left(k-1-j\right)!\left(n-k\right)!}{\left[F\left(y_j\right)\right]}^{j-1}\times {\left[F\left(y_k\right)-F\left(y_j\right)\right]}^{k-1-j}\\ \times {\left[1-F\left(y_k\right)\right]}^{n-k}f\left(y_j\right)f\left(y_k\right)\end{array}\right\}$

From Exercise 6-74, the density function 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 joint density function for $Y_{\left(j\right)}$ and $Y_{\left(k\right)}$ is,

$\begin{array}{c}{g}_{\left(j\right)\left(k\right)}\left(y_j,y_k\right)=\frac{n!}{\left(j-1\right)!\left(k-1-j\right)!\left(n-k\right)!}{\left[\frac{y_j}{\theta }\right]}^{j-1}\times {\left[\frac{y_k}{\theta }-\frac{y_j}{\theta }\right]}^{k-1-j}\times {\left[1-\frac{y_k}{\theta }\right]}^{n-k}\frac{1}{\theta }\times \frac{1}{\theta }\\ =\frac{n!}{\left(j-1\right)!\left(k-1-j\right)!\left(n-k\right)!}{\left[\frac{y_j}{\theta }\right]}^{j-1}{\left[\frac{y_k}{\theta }-\frac{y_j}{\theta }\right]}^{k-1-j}{\left[1-\frac{y_k}{\theta }\right]}^{n-k}\frac{1}{{\theta }^2},0\le y_j\le \theta \end{array}$

b.

To determine

Find $Cov\left(Y_{\left(j\right)},Y_{\left(k\right)}\right)$, where j and k are integers $1\le j\le k\le n$.

Answer to Problem 77SE

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

Explanation of Solution

Calculation:

Consider,

$\begin{array}{c}E\left(Y_{\left(j\right)}{Y}_{\left(k\right)}\right)={\displaystyle \int {\displaystyle \int y_j{y}_k{g}_{\left(j\right)\left(k\right)}\left(y_j,y_k\right)d{y}_{j}d{y}_k}}\\ ={\displaystyle \int {\displaystyle \int y_j{y}_k\left\{\begin{array}{l}\frac{n!}{\left(j-1\right)!\left(k-1-j\right)!\left(n-k\right)!}{\left[F\left(y_j\right)\right]}^{j-1}\times \\ {\left[F\left(y_k\right)-F\left(y_j\right)\right]}^{k-1-j}\\ \times {\left[1-F\left(y_k\right)\right]}^{n-k}f\left(y_j\right)f\left(y_k\right)\end{array}\right\}d{y}_{j}d{y}_k}}\\ =\frac{n!}{\left(j-1\right)!\left(k-1-j\right)!\left(n-k\right)!}\theta {\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{v}{\int }}{u}^j}{\left[v-u\right]}^{k-1-j}v{\left[1-v\right]}^{n-k}dudv}\left[\begin{array}{c}\text{Let, }\frac{y_j}{\theta }=u\\ \frac{y_k}{\theta }=v\end{array}\right]\\ =c\theta {\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{v}{\int }}{u}^j}{\left[v-u\right]}^{k-1-j}v{\left[1-v\right]}^{n-k}dudv}\left[c=\frac{n!}{\left(j-1\right)!\left(k-1-j\right)!\left(n-k\right)!}\right]\end{array}$

Let us consider,

$\begin{array}{c}w=\frac{u}{v}\\ u=wv\\ du=vdw\end{array}$

Then, the integral becomes,

$\begin{array}{c}E\left(Y_{\left(j\right)}{Y}_{\left(k\right)}\right)=c\theta {\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{v}{\int }}{u}^j}{\left[v-u\right]}^{k-1-j}v{\left[1-v\right]}^{n-k}dudv}\\ =c{\theta }^2\left[{\displaystyle \underset{0}{\overset{1}{\int }}{u}^{k+1}{\left(1-u\right)}^{n-k}du}\right]\left[{\displaystyle \underset{0}{\overset{1}{\int }}{w}^j{\left(1-w\right)}^{k-1-j}dw}\right]\\ =\frac{\left(k+1\right)j}{\left(n+1\right)\left(n+2\right)}{\theta }^2\end{array}$

From Exercise 6-76, $E\left(Y_{\left(j\right)}\right)=\frac{j}{n+1}\theta$ and $E\left(Y_{\left(k\right)}\right)=\frac{k}{n+1}\theta$.

Therefore,

$\begin{array}{c}Cov\left(Y_{\left(j\right)},Y_{\left(k\right)}\right)=E\left(Y_{\left(j\right)}{Y}_{\left(k\right)}\right)-E\left(Y_{\left(j\right)}\right)E\left(Y_{\left(k\right)}\right)\\ =\frac{\left(k+1\right)j}{\left(n+1\right)\left(n+2\right)}{\theta }^2-\left(\frac{j}{n+1}\theta \right)\left(\frac{k}{n+1}\theta \right)\\ =\frac{\left(k+1\right)j}{\left(n+1\right)\left(n+2\right)}{\theta }^2-\frac{jk}{{\left(n+1\right)}^2}{\theta }^2\\ =\frac{n-k+1}{{\left(n+1\right)}^2\left(n+2\right)}{\theta }^2\end{array}$

c.

To determine

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

Answer to Problem 77SE

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

Explanation of Solution

Calculation:

From Exercise 6-76, $V\left(Y_{\left(k\right)}\right)=\frac{k\left(n-k+1\right)}{{\left(n+1\right)}^2\left(n+2\right)}{\theta }^2$ and $V\left(Y_{\left(j\right)}\right)=\frac{j\left(n-j+1\right)}{{\left(n+1\right)}^2\left(n+2\right)}{\theta }^2$.

Consider,

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

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

Therefore,

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