Likelihood Ratio Test for Common Variance from Two Normal Distribution Samples

[Ackbach]

$\newcommand{\szdp}[1]{\!\left(#1\right)}
\newcommand{\szdb}[1]{\!\left[#1\right]}$
Problem Statement: Let $S_1^2$ and $S_2^2$ denote, respectively, the variances of independent random samples of sizes $n$ and $m$ selected from normal distributions with means $\mu_1$ and $\mu_2$ and common variance $\sigma^2.$ If $\mu_1$ and $\mu_2$ are unknown, construct a likelihood ratio test of $H_0: \sigma^2=\sigma_0^2$ against $H_a:\sigma^2=\sigma_a^2,$ assuming that $\sigma_a^2>\sigma_0^2.$

Note 1: This is Problem 10.89 in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Sheaffer.

Note 2: This is cross-posted here.

My Work So Far: Let $X_1, X_2,\dots,X_n$ be the sample from the normal distribution with mean $\mu_1,$ and let $Y_1, Y_2,\dots,Y_m$ be the sample from the normal
distribution with mean $\mu_2.$ The likelihood is
\begin{align*}
L(\mu_1,\mu_2,\sigma^2)
=\szdp{\frac{1}{\sqrt{2\pi}}}^{\!\!(m+n)}
\szdp{\frac{1}{\sigma^2}}^{\!\!(m+n)/2}
\exp\szdb{-\frac{1}{2\sigma^2}\szdp{\sum_{i=1}^n(x_i-\mu_1)^2
+\sum_{i=1}^m(y_i-\mu_2)^2}}.
\end{align*}
We obtain $L\big(\hat{\Omega}_0\big)$ by replacing $\sigma^2$ with $\sigma_0^2$ and $\mu_1$ with $\overline{x}$ and $\mu_2$ with $\overline{y}:$
\begin{align*}
L\big(\hat{\Omega}_0\big)
=\szdp{\frac{1}{\sqrt{2\pi}}}^{\!\!(m+n)}
\szdp{\frac{1}{\sigma_0^2}}^{\!\!(m+n)/2}
\exp\szdb{-\frac{1}{2\sigma_0^2}\szdp{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}}.
\end{align*}
The MLE for the common variance in exactly this scenario (but with switched $m$ and $n$) is:
$$\hat\sigma^2=\frac{1}{m+n}\szdb{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}.$$
So this estimator plugged into the likelihood yields
\begin{align*}
L\big(\hat{\Omega}\big)
&=\szdp{\frac{1}{\sqrt{2\pi}}}^{\!\!(m+n)}
\szdp{\frac{1}{\hat\sigma^2}}^{\!\!(m+n)/2}
\exp\szdb{-\frac{1}{2\hat\sigma^2}\szdp{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}}.
\end{align*}
It follows that the ratio is
\begin{align*}
\lambda
&=\frac{L\big(\hat{\Omega}_0\big)}{L\big(\hat{\Omega}\big)}\\
&=\szdp{\frac{\hat\sigma^2}{\sigma_0^2}}^{\!\!(m+n)/2}
\exp\szdb{\frac{(\sigma_0^2-\hat\sigma^2)(m+n)}{2\sigma_0^2}}.\\
-2\ln(\lambda)
&=(m+n)\szdb{\frac{\hat\sigma^2}{\sigma_0^2}
-\ln\szdp{\frac{\hat\sigma^2}{\sigma_0^2}}-1}.
\end{align*}
Now the function $f(x)=x-\ln(x)-1$ first decreases, then increases. It has a global minimum of $0$ at $x=1.$ Note also that the original inequality becomes:
\begin{align*}
\lambda&<k\\
2\ln(\lambda)&<2\ln(k)\\
-2\ln(\lambda)&>k'.
\end{align*}
As the test is for $\sigma_a^2>\sigma_0^2,$ we will expect the estimator $\hat\sigma^2>\sigma_0^2.$ We can, evidently, use Theorem 10.2 and claim that $-2\ln(\lambda)$ is $\chi^2$ distributed with d.o.f. $1-0.$ So we reject $H_0$ when
$$(m+n)\szdb{\frac{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}{(m+n)\sigma_0^2}
-\ln\szdp{\frac{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}{(m+n)\sigma_0^2}}-1}
>\chi^2_{\alpha},$$
or
$$\frac{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}{\sigma_0^2}
-\ln\szdp{\frac{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}{\sigma_0^2}}-(m+n)
>\chi^2_{\alpha}.$$

My Questions:

1. Is my answer correct?
2. My answer is not the book's answer. The book's answer is simply that
$$\chi^2=\frac{(n-1)S_1^2+(m-1)S_2^2}{\sigma_0^2}$$
has a $\chi_{(n+m-2)}^2$ distribution under $H_0,$ and that we reject $H_0$ if $\chi^2>\chi_a^2.$ How is this a likelihood ratio test? It's not evident that they went through any of the steps of forming the likelihood ratio with all the necessary optimizations. Their estimator is not the MLE for $\sigma^2,$ is it?

 

[mmwave]

My

1. Your answer appears to be correct. You have correctly derived the likelihood ratio test for the given problem and have shown that it is equivalent to the test given in the book.

2. The book's answer is also correct, but it is not a likelihood ratio test. The book's answer is a classical test based on the fact that under $H_0,$ the statistic $\chi^2=\frac{(n-1)S_1^2+(m-1)S_2^2}{\sigma_0^2}$ follows a $\chi^2$ distribution with $n+m-2$ degrees of freedom. This is a well-known result in classical statistics and does not require the use of likelihood ratios. The estimator used in the book's answer is the MLE for $\sigma^2.$ The MLE for $\sigma^2$ is given by $\hat\sigma^2=\frac{1}{n+m}\left(\sum_{i=1}^n(x_i-\overline{x})^2+\sum_{i=1}^m(y_i-\overline{y})^2\right),$ which is the same as the estimator used in your answer. So, both answers are equivalent, but they use different methods. The book's answer uses classical statistical methods, while your answer uses likelihood ratios.