$\newcommand{\szdp}[1]{\!\left(#1\right)} \newcommand{\szdb}[1]{\!\left[#1\right]}$ 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 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?
  • $\begingroup$ @BruceET I can understand why the variables you're defining are distributed the way they are. But that doesn't explain how $S_p^2$ ends up being the key to the likelihood ratio test. How is the book's test equivalent to $L\big(\hat\Omega_0\big)/L\big(\hat\Omega\big)<k?$ I link to cross-posting when I do it PRECISELY so that people do not waste time if someone on the other site has answered. Do you know why there is a close vote on this question? I find that remarkable: I put ENORMOUS effort into writing good questions on this site, and it's irritating that it has a close vote. $\endgroup$ Commented Aug 27, 2021 at 23:58
  • $\begingroup$ The notation is messy to the point where you may be mixing up $m$ and $n.$ I'll use $n_1$ and $n_2.$ Use $S^2_i=\frac{1}{n_i−1}\sum_{j=1}^{n_i}(x_j−\bar x_i)^2.$ Then under $H_0,$ we have $ \frac{(n_i−1)S^2_i}{σ^2}$ $\sim\mathsf{Chisq}(\nu_i=n_i−1).$ Also, $S^2_p=\frac{(n_1−1)S_1^2+(n_2−1)S_2^2}{n_1+n_2-2},$ has $\frac{(n_1+n_2−2)S_p^2}{σ_0^2}$ $\sim\mathsf{Chisq}(ν=n_1+n_2−2).$ // Speaking only for myself, I will not ordinarily answer questions that are cross posted on math and stat sites, unless one Answer is obviously unhelpful. I view cross posting to be an invitation to waste time. $\endgroup$
    – BruceET
    Commented Aug 28, 2021 at 0:17
  • $\begingroup$ The point is that some of your expressions can be simplified as suggested by my comment. Perhaps that will help. $\endgroup$
    – BruceET
    Commented Aug 28, 2021 at 0:26

2 Answers 2


Let $x_1,...,x_n\sim N(\mu_1,\sigma^2)$ and $y_1,...,y_m\sim N(\mu_2,\sigma^2)$. Under $H_0$, the likelihood function is

$$L_0=\left(\frac{1}{\sqrt{2\pi\sigma_0^2}}\right)^nexp\left( \frac{-1}{2\sigma_0^2}\left( \sum_{i=1}^{n}(x_i-\bar{x})^2 + \sum_{i=1}^{m}(y_i-\bar{y})^2 \right) \right)=\left(\frac{1}{\sqrt{2\pi\sigma_0^2}}\right)^nexp\left( \frac{-1}{2\sigma_0^2}\left( (n-1)S_1^2 + (m-1)S_2^2 \right) \right).$$

Under $H_a$ we get:

$$L_a=\left(\frac{1}{\sqrt{2\pi\sigma_a^2}}\right)^nexp\left( \frac{-1}{2\sigma_a^2}\left( (n-1)S_1^2 + (m-1)S_2^2 \right) \right).$$

Inside the latter exponent, we have the same sum of $S_1^2$ and $S_2^2$ but multiplied by $\frac{1}{2\sigma_a^2}$, which can be presented as $\frac{\sigma_0^2}{\sigma_a^2}\frac{1}{2\sigma_0^2}$. The LRT is then

$$\lambda=\frac{L_0}{L_a}=\left( \frac{\sqrt{\sigma_a}}{\sqrt{\sigma_0}} \right)^nexp\left( \frac{-1}{2\sigma_0^2}( (n-1)S_1^2 + (m-1)S_2^2) + \frac{\sigma_0^2}{\sigma_a^2}\frac{1}{2\sigma_0^2}( (n-1)S_1^2 + (m-1)S_2^2) \right)=\left( \frac{\sqrt{\sigma_a}}{\sqrt{\sigma_0}} \right)^nexp\left(-\frac{1}{2} \left( 1- \frac{\sigma_0^2}{\sigma_a^2}\right)\frac{1}{\sigma_0^2}( (n-1)S_1^2 + (m-1)S_2^2) \right).$$

Given that $\sigma_a>\sigma_0$, the expression $\left( \frac{\sqrt{\sigma_a}}{\sqrt{\sigma_0}} \right)^n$ is larger that 1 and $\left( 1- \frac{\sigma_0^2}{\sigma_a^2}\right)$ is positive, for all $n$. Thus, our test depends on $\frac{1}{2\sigma_0^2}( (n-1)S_1^2 + (m-1)S_2^2)$ (note I ignored the minus 0.5 there, as it is a constant), which is a $\chi^2$ variable with $n+m-2$ degrees of freedom (why? simple to find).

Ultimately, our test is in the form of

$$\left\{ \frac{1}{\sigma_0^2}\left( (n-1)S_1^2 + (m-1)S_2^2\right) > \chi^2_{\alpha,~n+m-2} \right\}$$

  • 1
    $\begingroup$ Wonderful! Thanks! In looking at my answer, I think I just needed the additional step of simplifying my answer on the basis of examining the function $x-\ln(x).$ But your answer is more direct, and shows that link between the LRT and the final expression. Thanks again! $\endgroup$ Commented Sep 7, 2021 at 19:05

Ok, part of this is that you are using a theorem about asymptotic distribution of log likelihood ratios and they are using the exact distribution of the residual sum of squares. I agree that the book answer is a little underexplained.

Personally, I would have taken the the logarithm first, then subtracted, but I think your calculations are right. You find that the log likelihood is a function of $\hat \sigma^2/\sigma^2_0$, and use the asymptotic results to get a $\chi^2_1$ reference distribution. I think you're wrong in that the asymptotic theory is without the restriction $\sigma^2_a>\sigma^2_0$, so that with the restriction you should have $\frac{1}{2}\chi^2_1$ as the asymptotic null sampling distribution. The probability is halved because you only reject when $\hat\sigma^2>\sigma^2_0$, even though you can just as easily get a large likelihood ratio with $\hat\sigma^2<\sigma^2_0$.

Once you find that the loglikelihood ratio depends on the data only through $\hat \sigma^2/\sigma^2_0$, you can look for other functions of this statistic that have more tractable null sampling distributions. Any rejection region you define based on the log likelihod ratio plus the restriction $\hat\sigma^2>\sigma^2_0$ can also be defined based on $\hat \sigma^2/\sigma^2_0$; it's the same set of rejection regions, so it's the same test. In more complicated cases you aren't going to be able to find an exact sampling distribution so there's no point and you should just stick with the loglikelihod ratio as the test statistic. This case is simple enough that there might be a tractable exact distribution -- especially as it's a textbook example.

Since the data are Normal with variance $\sigma^2_0$ under the null, the residual sum of squares after estimating two means has a null $\sigma^2_0\chi^2_{n+m-2}$ distribution. Thus $\mathrm{RSS}/\sigma^2_0$ has a $\chi^2_{n+m-2}$ null distribution. You will reject $H_0$ if the ratio is large, so the test is as described in the book.

Their test will have closer to the nominal size in small (and moderate) sample sizes, eg


y<-rnorm(m+n, m=rep(c(mu1,mu2), c(m,n)),s=sqrt(sigma2_0))


ell_0 <- dnorm(y,xbars, s=sqrt(sigma2_0),log=TRUE)

RSS<- sum((y-xbars)^2)
hatsigma2<-max(sigma2_0, RSS/(m+n))

ell_1 <- dnorm(y,xbars, s=sqrt(hatsigma2), log=TRUE)

c(lrt, RSS/sigma2_0)



> mean(results[1,]*2>qchisq(0.95,1))
[1] 0.0088

for the asymptotic distribution of $2\ell$ that you use

> mean(results[1,]*2>  0.5*qchisq(0.95,1)+0.5*0)
[1] 0.0334

for the one-tailed version of the asymptotic distribution of $2\ell$ and

> mean(results[2,]>qchisq(0.95,m+n-2))
[1] 0.0501

for the exact LRT from the book.

  • $\begingroup$ Thanks. This is helpful, but I am struggling a bit to follow you. The problem with helping anyone who didn't use the same book you did is that you don't know if a different book introduces things in a different order. My book, e.g., hasn't introduced the idea of RSS at all, yet (next chapter on linear regression). I'm not following your statement, "... so that with the restriction you should have $\frac12\chi_1^2$ as the asymptotic null sampling distribution." Can you please flesh that sentence out a bit more? $\endgroup$ Commented Aug 28, 2021 at 1:15
  • $\begingroup$ Then you wrote, "... you can look for other functions of this statistic that have more tractable null sampling distributions." How does this procedure retain the properties of the likelihood ratio test? $\endgroup$ Commented Aug 28, 2021 at 1:16

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