Chapter 6, Problem 91SE

a.

To determine

Show that $T_j$ for $j=1,2,\mathrm{3...},r$ has an Exponential distribution with mean $\frac{\theta }{\left(n-j+1\right)}$.

Answer to Problem 91SE

The random variable $T_j$ for $j=1,2,\mathrm{3...},r$ has an Exponential distribution with mean $\frac{\theta }{\left(n-j+1\right)}$.

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 the given information, $Y_1,Y_2,\mathrm{..},Y_n$ be the independent random sample from the Exponential distribution $f\left(y\right)=\frac{1}{\theta }{e}^{\frac{-y}{\theta }}$ and the distribution function is $F\left(y\right)=1-e^{\frac{-y}{\theta }}$.

The joint density function for $W_{\left(j-1\right)}$ and $W_{\left(j\right)}$ is,

$g_{\left(j-1\right)\left(j\right)}\left(w_{j-1},w_j\right)=\frac{n!}{\left(j-2\right)!\left(n-j\right)!}{\left[1-e^{\frac{-w_{j-1}}{\theta }}\right]}^{j-2}\left[e^{\frac{-\left(w_{j-1}+w_j\right)}{\theta }}\right]{\left[e^{\frac{-w_j}{\theta }}\right]}^{n-j}\frac{1}{{\theta }^2},0\le w_{j-1}\le w_j\le \infty$

Let us consider, $S=W_{\left(j-1\right)}$ and $T_{\left(j\right)}=W_{\left(j\right)}-W_{\left(j-1\right)}$.

This implies that, $w_{j-1}=s$ and $w_j=t_j+s$.

Consider the Jacobian,

$\begin{array}{c}J=\left|\begin{array}{l}\frac{\partial w_{j-1}}{\partial s}\frac{\partial w_{j-1}}{\partial t_j}\\ \frac{\partial w_j}{\partial s}\frac{\partial w_j}{\partial t_j}\end{array}\right|\\ =\left|\begin{array}{l}\text{1 0}\\ \text{1 1}\end{array}\right|\\ =1\end{array}$

The joint density for S and $T_j$ is,

$\begin{array}{c}f\left(s,t_j\right)=g_{\left(j-1\right)\left(j\right)}\left(w_{j-1},w_j\right)\left|J\right|\\ =\frac{n!}{\left(j-2\right)!\left(n-j\right)!}{\left[1-e^{\frac{-s}{\theta }}\right]}^{j-2}\left[e^{\frac{-\left(2s+t_j\right)}{\theta }}\right]{\left[e^{\frac{-\left(t_j+s\right)}{\theta }}\right]}^{n-j}\frac{1}{{\theta }^2}\times \left|1\right|\\ =\frac{n!}{\left(j-2\right)!\left(n-j\right)!}\frac{1}{{\theta }^2}{e}^{\frac{-\left(n-j+1\right)t_j}{\theta }}{\left[1-e^{\frac{-s}{\theta }}\right]}^{j-2}{e}^{\frac{-\left(n-j+2\right)s}{\theta }},s\ge 0,t_j\ge 0\end{array}$

The marginal density function for $T_j$ is,

$\begin{array}{c}{f}_{T_j}\left(t_j\right)={\displaystyle \underset{s}{\int }f\left(s,t_j\right)ds}\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{n!}{\left(j-2\right)!\left(n-j\right)!}\frac{1}{{\theta }^2}{e}^{\frac{-\left(n-j+1\right)t_j}{\theta }}{\left[1-e^{\frac{-s}{\theta }}\right]}^{j-2}{e}^{\frac{-\left(n-j+2\right)s}{\theta }}ds}\\ =\frac{n!}{\left(j-2\right)!\left(n-j\right)!}\frac{1}{{\theta }^2}{e}^{\frac{-\left(n-j+1\right)t_j}{\theta }}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left[1-e^{\frac{-s}{\theta }}\right]}^{j-2}{e}^{\frac{-\left(n-j+2\right)s}{\theta }}ds}\end{array}$

Let us consider $u=e^{\frac{-s}{\theta }}$, then the integral function becomes as beta density with marginal density function $f_{T_j}\left(t_j\right)=\frac{n-j+1}{\theta }{e}^{\frac{-\left(n-j+1\right)t_j}{\theta }},t_j\ge 0$

This is the Exponential density function with parameters mean $\frac{\theta }{n-j+1}$.

b.

To determine

Show that $U_r={\displaystyle \sum _{j=1}^r{W}_j+\left(n-r\right)W_r={\displaystyle \sum _{j=1}^r\left(n-j+1\right)T_j}}$ and hence, $E\left(U_r\right)=r\theta$.

Explanation of Solution

Calculation:

Consider,

$\begin{array}{c}{\displaystyle \sum _{j=1}^r\left(n-j+1\right)T_j}=n{T}_1+\left(n-1\right)T_2+\mathrm{...}+\left(n-r+1\right)T_r\\ =n{W}_1+\left(n-1\right)\left(W_2-W_1\right)+\mathrm{...}+\left(n-r+1\right)\left(W_r-W_{r-1}\right)\left[\text{Since, }{T}_j=W_j-W_{j-1}\right]\\ =W_1+W_2+\mathrm{...}+W_{r-1}+\left(n-r+1\right)W_r\\ ={\displaystyle \sum _{j=1}^r{W}_j+\left(n-r\right)W_r}\\ =U_r\end{array}$

Hence,

$\begin{array}{c}E\left(U_r\right)=E\left({\displaystyle \sum _{j=1}^r\left(n-j+1\right)T_j}\right)\\ ={\displaystyle \sum _{j=1}^r\left(n-j+1\right)E\left(T_j\right)}\\ ={\displaystyle \sum _{j=1}^r\left(n-j+1\right)\times \frac{\theta }{\left(n-j+1\right)}}\\ ={\displaystyle \sum _{j=1}^r\theta }\end{array}$

$E\left(U_r\right)=r\theta$