Chapter 10.11, Problem 108E

a.

To determine

Obtain the likelihood ratio test for $H_0:{\sigma }_1^2={\sigma }_2^2={\sigma }_3^2$ against $H_a:\text{ At least one inequality}$.

Explanation of Solution

In this context, $Y_1,Y_2,\mathrm{...},Y_{n1}$ denotes a random sample from the population having normal distribution with unknown mean ${\mu }_1$ and variance ${\sigma }^2$, $X_1,X_2,\mathrm{...},X_{n2}$ denotes a random sample from the population having normal distribution with unknown mean ${\mu }_2$ and variance ${\sigma }^2$, and $W_1,W_2,\mathrm{...},W_{n3}$ denotes a random sample from the population having normal distribution with unknown mean ${\mu }_3$ and variance ${\sigma }^2$.

The likelihood ratio test has to be obtained for the following hypothesis:

$\begin{array}{l}{H}_0:{\sigma }_1^2={\sigma }_2^2={\sigma }_3^2\text{versus}\\ H_a:\text{ At least one inequality}\end{array}$

Likelihood ratio test:

A likelihood ratio test of $H_0:\Theta \in {\Omega }_0\text{versus }{H}_a:\Theta \in {\widehat{\Omega }}_a$ denotes $\lambda$ as the test statistic and the rejection region as $\lambda \le k$.

The test statistic, $\lambda$, is denoted as follows:

$\begin{array}{c}\lambda =\frac{L\left({\widehat{\Omega }}_0\right)}{L\left(\widehat{\Omega }\right)}\\ =\frac{\underset{\Theta \in {\widehat{\Omega }}_0}{\max }L\left({\widehat{\Omega }}_0\right)}{\underset{\Theta \in \widehat{\Omega }}{\max }L\left(\widehat{\Omega }\right)}\end{array}$

The likelihood function, $L\left({\Omega }_0\right)$, is given below:

$\begin{array}{c}L\left({\Omega }_0\right)={\displaystyle \prod _{i=1}^{n_1}f\left(Y_i,{\mu }_1,{\sigma }^2\right)}\times {\displaystyle \prod _{j=1}^{n_2}f\left(X_j,{\mu }_2,{\sigma }^2\right)}\times {\displaystyle \prod _{m=1}^{n_3}f\left(W_m,{\mu }_2,{\sigma }^2\right)}\\ ={\displaystyle \prod _{i=1}^{n_1}\frac{1}{\sqrt{2\pi }{\sigma }_1}{e}^{-\frac{{\left(Y_i-{\mu }_1\right)}^2}{2{\sigma }_1^2}}}\times {\displaystyle \prod _{j=1}^{n_2}\frac{1}{\sqrt{2\pi }{\sigma }_2}{e}^{-\frac{{\left(X_i-{\mu }_2\right)}^2}{2{\sigma }_2^2}}}\times {\displaystyle \prod _{m=1}^{n_3}\frac{1}{\sqrt{2\pi }{\sigma }_3}{e}^{-\frac{{\left(W_i-{\mu }_3\right)}^2}{2{\sigma }_3^2}}}\\ ={\left(\frac{1}{\sqrt{2\pi }\left({\sigma }_1+{\sigma }_2+{\sigma }_3\right)}\right)}^{n_1+n_2+n_3}\times e^{-\frac{1}{2{\sigma }_1^2}{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}}\times e^{-\frac{1}{2{\sigma }_2^2}{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-{\mu }_2\right)}^2}}\times e^{-\frac{1}{2{\sigma }_3^2}{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}}\end{array}$

Let, ${\sigma }_1={\sigma }_2={\sigma }_3=\sigma$

$\begin{array}{c}L\left({\Omega }_0\right)={\left(\frac{1}{\sqrt{2\pi }\left(\sigma \right)}\right)}^{n_1+n_2+n_3}\times e^{-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}}\times e^{-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-{\mu }_2\right)}^2}}\times e^{-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}}\\ ={\left(\frac{1}{\sqrt{2\pi }\left(\sigma \right)}\right)}^{n_1+n_2+n_3}\times e^{-\frac{1}{2{\sigma }^2}\left[{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-{\mu }_2\right)}^2}+{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}\right]}\end{array}$

Since the above model is a complete model with parameters $\left(\mu ,\sigma \right)$, just the sample size has increased to $n_1+n_2+n_3$.

Under $H_0:{\sigma }_1^2={\sigma }_2^2={\sigma }_3^2$, the likelihood function is as follows:

$L\left({\Omega }_0\right)={\left(\frac{1}{\sqrt{2\pi }\left(\sigma \right)}\right)}^{n_1+n_2+n_3}\times e^{-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-\overline{Y}\right)}^2}}\times e^{-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-\overline{X}\right)}^2}}\times e^{-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-\overline{W}\right)}^2}}$

MLE of $\theta$ is obtained as given below:

$l=-\frac{\left(n_1+n_2+n_3\right)}{2}\ln \left({\sigma }^2\right)-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-{\mu }_2\right)}^2}-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}$

Differentiate MLE with respect to $\theta$.

$\frac{dl}{d{\sigma }^2}=-\frac{\left(n_1+n_2+n_3\right)}{2{\sigma }^2}-\frac{1}{2{\sigma }^4}{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}-\frac{1}{2{\sigma }^4}{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-{\mu }_2\right)}^2}-\frac{1}{2{\sigma }^4}{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}$

Equate $\frac{dl}{d\theta }=0$,

$\begin{array}{c}-\frac{\left(n_1+n_2+n_3\right)}{2{\sigma }^2}-\frac{1}{2{\sigma }^4}{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}-\frac{1}{2{\sigma }^4}{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-{\mu }_2\right)}^2}-\frac{1}{2{\sigma }^4}{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}=0\\ \frac{1}{2{\sigma }^4}\left[{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-{\mu }_2\right)}^2}+{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}\right]=\frac{\left(n_1+n_2+n_3\right)}{2{\sigma }^2}\\ \left[{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-{\mu }_2\right)}^2}+{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}\right]=\frac{2{\sigma }^4\left(n_1+n_2+n_3\right)}{2{\sigma }^2}\\ \left[{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-{\mu }_2\right)}^2}+{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}\right]={\sigma }^2\left(n_1+n_2+n_3\right)\end{array}$

Thus, the value of ${\sigma }^2$ is obtained as follows:

$\begin{array}{c}{\sigma }^2=\frac{1}{n_1+n_2+n_3}\left[{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-\overline{Y}\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-\overline{X}\right)}^2}+{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-\overline{W}\right)}^2}\right]\\ {\sigma }^2=T^2\end{array}$

The likelihood function is as follows:

$\begin{array}{c}L\left({\Omega }_0\right)=\left(\frac{1}{{\left(\sqrt{2\pi }T\right)}^{n_1+n_2+n_3}}\right)\times e^{-\frac{1}{2T^2}{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-\overline{Y}\right)}^2}}\times e^{-\frac{1}{2T^2}{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-\overline{X}\right)}^2}}\times e^{-\frac{1}{2T^2}{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-\overline{W}\right)}^2}}\\ =\left(\frac{1}{{\left(\sqrt{2\pi }T\right)}^{n_1+n_2+n_3}}\right)\times e^{-\frac{1}{2}\left[\frac{{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-\overline{Y}\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-\overline{X}\right)}^2}+{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-\overline{W}\right)}^2}}{T^2}\right]}\\ =\left(\frac{1}{{\left(\sqrt{2\pi }T\right)}^{n_1+n_2+n_3}}\right)\times e^{-\frac{1}{2}\left[n_1+n_2+n_3\right]}\end{array}$

The likelihood function, $L\left(\Omega \right)$, is given below:

$\begin{array}{c}L\left(\Omega \right)={\displaystyle \prod _{i=1}^{n_1}f\left(Y_i,{\mu }_1,{\sigma }^2\right)}\times {\displaystyle \prod _{j=1}^{n_2}f\left(X_j,{\mu }_2,{\sigma }^2\right)}\times {\displaystyle \prod _{m=1}^{n_3}f\left(W_i,{\mu }_2,{\sigma }^2\right)}\\ ={\displaystyle \prod _{i=1}^{n_1}\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{\left(Y_i-{\mu }_1\right)}^2}{2{\sigma }^2}}}\times {\displaystyle \prod _{j=1}^{n_2}\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{\left(X_i-{\mu }_2\right)}^2}{2{\sigma }^2}}}\times {\displaystyle \prod _{m=1}^{n_3}\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{\left(W_i-{\mu }_3\right)}^2}{2{\sigma }^2}}}\\ ={\left(\frac{1}{\sqrt{2\pi {\sigma }^2}}\right)}^{n_1+n_2+n_3}\times e^{-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}}\times e^{-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{j=1}^{n_2}{\left(X_i-{\mu }_2\right)}^2}}\times e^{-\frac{1}{2{\sigma }^2}{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}}\end{array}$

Since the above model is a complete model with parameters $\left(\mu ,\sigma \right)$, just the sample size has increased to $n_1+n_2+n_3$.

Define:

$\begin{array}{l}{T}_2^2=\frac{{\displaystyle \sum _{j=1}^{n_2}{\left(X_j-\overline{X}\right)}^2}}{n_1}\\ \left(\text{Note}:T_2^2\text{ is the MLE's of }{\sigma }_2^2\right)\end{array}$

Similarly, one can write for $T_1,T_3$.

The likelihood function is as follows:

$\begin{array}{c}L\left(\Omega \right)=\left(\frac{1}{{\left(\sqrt{2\pi }\right)}^{n_1+n_2+n_3}\left(T_1^{{}^{n_1}}{T}_2^{{}^{n_2}}{T}_3^{{}^{n_3}}\right)}\right)\times e^{-\frac{1}{2T_1^2}{\displaystyle \sum _{i=1}^{n_1}{\left(Y_i-{\mu }_1\right)}^2}}\times e^{-\frac{1}{2T_2^2}{\displaystyle \sum _{j=1}^{n_2}{\left(X_j-\overline{X}\right)}^2}}\times e^{-\frac{1}{2T_3^2}{\displaystyle \sum _{m=1}^{n_3}{\left(W_i-{\mu }_3\right)}^2}}\\ =\left(\frac{1}{{\left(\sqrt{2\pi }\right)}^{n_1+n_2+n_3}\left(T_1^{{}^{n_1}}{T}_2^{{}^{n_2}}{T}_3^{{}^{n_3}}\right)}\right)\times e^{-\frac{n_1}{2}}\times e^{-\frac{n_2}{2}}\times e^{-\frac{n_3}{2}}\\ =\left(\frac{1}{{\left(\sqrt{2\pi }\right)}^{n_1+n_2+n_3}\left(T_1^{{}^{n_1}}{T}_2^{{}^{n_2}}{T}_3^{{}^{n_3}}\right)}\right)\times e^{-\frac{1}{2}\left[n_1+n_2+n_3\right]}\end{array}$

The likelihood ratio test for testing $H_0:{\sigma }_1^2={\sigma }_2^2={\sigma }_3^2$ versus $H_a:\text{Atleast one inequality}$ is obtained as given below:

$\begin{array}{c}\lambda =\frac{L\left({\widehat{\Omega }}_0\right)}{L\left(\widehat{\Omega }\right)}\\ =\frac{\left(\frac{1}{{\left(\sqrt{2\pi }T\right)}^{n_1+n_2+n_3}}\right)\times e^{-\frac{1}{2}\left[n_1+n_2+n_3\right]}}{\left(\frac{1}{{\left(\sqrt{2\pi }\right)}^{n_1+n_2+n_3}\left(T_1^{{}^{n_1}}{T}_2^{{}^{n_2}}{T}_3^{{}^{n_3}}\right)}\right)\times e^{-\frac{1}{2}\left[n_1+n_2+n_3\right]}}\\ =\left(\frac{T_1^{{}^{n_1}}{T}_2^{{}^{n_2}}{T}_3^{{}^{n_3}}}{T^{n_1+n_2+n_3}}\right)\end{array}$

Now, the rejection region becomes as follows:

 $\begin{array}{c}\lambda \le k\\ \left(\frac{T_1^{{}^{n_1}}{T}_2^{{}^{n_2}}{T}_3^{{}^{n_3}}}{T^{n_1+n_2+n_3}}\right)\le k\\ \left(n_1\ln T_1+n_2\ln T_2+n_3\ln T_3\right)-\left(n_1+n_2+n_3\right)\ln T\le \ln k\\ \left(n_1\ln T_1+n_2\ln T_2+n_3\ln T_3\right)-\left(n_1+n_2+n_3\right)\ln T>k_1\end{array}$.

Here, $\ln k=k_1$, where $k_1$ is some constant.

Therefore, the rejection region will be in the form of $RR:\left\{\left(n_1\ln T_1+n_2\ln T_2+n_3\ln T_3\right)-\left(n_1+n_2+n_3\right)\ln T>k_1\right\}$ for some constant $k_1$.

b.

To determine

Identify the approximate critical region for the test in Part (a), when $n_1,n_2\text{ and }{n}_3$ and $\alpha =0.05$.

Explanation of Solution

Based on the theorem 10.2, let $Y_1,Y_2,\mathrm{...}{Y}_n$ have a joint likelihood function $L\left(\Theta \right)$. Denote $r_0$ as the number of free parameter that specifies $H_0:\Theta \in {\Omega }_0$ and denote r as the number of free parameter that specifies $\Theta \in \Omega$. Then, for large sample of size n, $-2\ln \left(\lambda \right)$ has approximately a ${\chi }^2$ distribution with $r_0-r$ df.

In this context, the sample sizes of $n_1,n_2\text{, and }{n}_3$ are large and $\alpha =0.05$.

Then, the rejection region for $\lambda$ is as follows:

$RR:\left\{\lambda <k\right\}$

If $\lambda <k$, then the $-2\ln \left(\lambda \right)>-2\ln \left(k\right)$

Based on the theorem of large n, the rejection region for $-2\ln \left(\lambda \right)$ is as follows:

$RR:\left\{-2\ln \left(\lambda \right)>-2\ln \left(k\right)\approx {\chi }_{\alpha }^2\right\}$

For large n, $-2\ln \left(\lambda \right)$ has approximately a ${\chi }^2$ distribution.

In this situation, the degrees of freedom of ${\chi }^2$ distribution is equal to 2 because the dimensionality of $\Omega$ is $6\left({\mu }_1,{\mu }_2,{\mu }_3,{\sigma }_1^2,{\sigma }_2^2,{\sigma }_3^2\right)$.

According to the above explanation, the rejection region will be in the form of $RR:\left\{-2\ln \left(\lambda \right)>{\chi }_{0.05}^2\right\}$.

From Appendix 3, Table 6: Percentage Points of the ${\chi }^2$ Distributions, the value corresponding to ${\chi }_{0.05}^2$ and 2 degrees of freedom is 5.991.

Therefore, the rejection region is $RR:\left\{-2\ln \left(\lambda \right)>5.991\right\}$.