Chapter 8, Problem 132SE

a.

To determine

Verify that the distribution function of $Y_{\left(n\right)}=\text{max}\left(Y_1,Y_2,\dots ,Y_n\right)$ is $F_{Y_{\left(n\right)}}\left(y\right)=\{\begin{array}{l}0,y<0,\\ {\left(\frac{y}{\theta }\right)}^{nc},0\le y\le \theta \\ 1,y>\theta \end{array}$.

Explanation of Solution

The distribution for $Y_{\left(n\right)}=\text{max}\left(Y_1,Y_2,\dots ,Y_n\right)$ is given as follows:

$\begin{array}{c}{F}_{Y_{\left(n\right)}}\left(y\right)=P\left(Y_{\left(n\right)}\le y\right)\\ =P\left(\text{max}\left\{Y_1,Y_2,\dots ,Y_n\right\}\le y\right)\\ =P\left(Y_1\le y\right)\cdot \dots \cdot P\left(Y_n\le y\right)\\ ={\left(F_{Y_1}\left(y\right)\right)}^n,y\in R\end{array}$

For $y<0$,

$\begin{array}{c}{F}_{Y_{\left(n\right)}}\left(y\right)={\left({\left(\frac{y}{\theta }\right)}^{\alpha }\right)}^n\\ =0^n\\ =0\end{array}$

For $0\le y\le \theta$,

$\begin{array}{c}{F}_{Y_{\left(n\right)}}\left(y\right)={\left({\left(\frac{y}{\theta }\right)}^{\alpha }\right)}^n\\ ={\left(\frac{y}{\theta }\right)}^{n\alpha }\end{array}$

For $y>\theta$,

$\begin{array}{c}{F}_{Y_{\left(n\right)}}\left(y\right)=1^n\\ =1\end{array}$

Substitute $\alpha =c$. Thus, the distribution function of $Y_{\left(n\right)}=\text{max}\left(Y_1,Y_2,\dots ,Y_n\right)$ is $F_{Y_{\left(n\right)}}\left(y\right)=\{\begin{array}{l}0,y<0,\\ {\left(\frac{y}{\theta }\right)}^{nc},0\le y\le \theta \\ 1,y>\theta \end{array}$.

b.

To determine

Verify that the distribution function of $\frac{Y_{\left(n\right)}}{\theta }$ is a pivotal quantity.

Explanation of Solution

From Part (a), the distribution function of $Y_{\left(n\right)}=\text{max}\left(Y_1,Y_2,\dots ,Y_n\right)$ is $F_{Y_{\left(n\right)}}\left(y\right)=\{\begin{array}{l}0,y<0,\\ {\left(\frac{y}{\theta }\right)}^{nc},0\le y\le \theta \\ 1,y>\theta \end{array}$.

Consider the following:

$\begin{array}{c}{F}_{\frac{Y_{\left(n\right)}}{\theta }}\left(y\right)=P\left(\frac{Y_{\left(n\right)}}{\theta }\le y\right)\\ =P\left(Y_{\left(n\right)}\le y\theta \right)\\ =F_{Y_{\left(n\right)}}\left(y\theta \right)\end{array}$

For $\theta y<0$,

$F_{Y_{\left(n\right)}}\left(y\right)=0$

For $0\le \theta y\le \theta$,

$\begin{array}{c}{F}_{\frac{Y_{\left(n\right)}}{\theta }}\left(y\right)={\left(\frac{y\theta }{\theta }\right)}^{nc}\\ =y^{nc}\end{array}$

For $\theta y>\theta$,

$F_{\frac{Y_{\left(n\right)}}{\theta }}\left(y\right)=1$

The distribution function for $\frac{Y_{\left(n\right)}}{\theta }$ is given below:

$F_{\frac{Y_{\left(n\right)}}{\theta }}\left(y\right)=\{\begin{array}{l}0,y<0,\\ y^{nc},0\le y\le 1,\\ 1,y>1\end{array}$

It is said to be pivotal quantity for $\theta$, if its probability is independent of $\theta$. Now, check whether the probability is independent on $\theta$ as shown below:

$\begin{array}{c}P\left(k<\frac{Y_{\left(n\right)}}{\theta }\le 1\right)=P\left(\frac{Y_{\left(n\right)}}{\theta }\le 1\right)-P\left(\frac{Y_{\left(n\right)}}{\theta }<k\right)\\ =1^{nc}-k^{nc}\\ =1-k^{nc}\end{array}$

Since the above expression is independent of $\theta$, the distribution function of $\frac{Y_{\left(n\right)}}{\theta }$ is a pivotal quantity.

c.

To determine

i Calculate the value of $k$ so that $P\left(k<\frac{Y_{\left(5\right)}}{\theta }\le 1\right)=0.95$.

ii Construct 95% confidence interval for $\theta$.

Answer to Problem 132SE

i The value of $k$ is 0.78.

ii The 95% confidence interval for $\theta$ is $\frac{Y_{\left(5\right)}}{1},\frac{Y_{\left(5\right)}}{0.78}$.

Explanation of Solution

The distribution function for $\frac{Y_{\left(n\right)}}{\theta }$ is $F_{\frac{Y_{\left(n\right)}}{\theta }}\left(y\right)=\{\begin{array}{l}0,y<0,\\ y^{nc},0\le y\le 1\\ 1,y>1\end{array}$.

i The value of k is calculated as follows:

$\begin{array}{c}P\left(\frac{Y_{\left(5\right)}}{\theta }\le y\right)=0.95\\ P\left(k<\frac{Y_{\left(n\right)}}{\theta }\le 1\right)=1-k^{nc}\\ P\left(k<\frac{Y_{\left(5\right)}}{\theta }\le 1\right)=1-k^{5\left(2.4\right)}\\ 1-k^{12}=0.95\\ k^{12}=0.05\\ k={0.05}^{\frac{1}{12}}\\ =0.78\end{array}$

Therefore, the value of $k$ is 0.78.

ii

The 95% confidence interval for $\theta$ is calculated as follows:

$\begin{array}{c}0.95=P\left(0.78<\frac{Y_{\left(5\right)}}{\theta }\le 1\right)\\ =P\left(0.78<\frac{Y_{\left(5\right)}}{\theta },\frac{Y_{\left(5\right)}}{\theta }\le 1\right)\\ =P\left(\frac{Y_{\left(5\right)}}{0.78}>\theta ,\frac{Y_{\left(5\right)}}{1}\le \theta \right)\\ =P\left(\frac{Y_{\left(5\right)}}{1}\le \theta <\frac{Y_{\left(5\right)}}{0.78}\right)\end{array}$

The 95% confidence interval for $\theta$ is $\frac{Y_{\left(5\right)}}{1},\frac{Y_{\left(5\right)}}{0.78}$.