The generalized likelihood ratio test of $H_0: \sigma^2=\sigma_0^2$ v.s. $H_1: \sigma^2\neq \sigma_0^2$ with $\mu$ unknown

Let $$X_i$$ be an iid sample from $$X\sim N(\mu,\sigma^2)$$. I try to find the generalized likelihood ratio test of $$H_0: \sigma^2=\sigma_0^2$$ v.s. $$H_1: \sigma^2\neq \sigma_0^2$$ with $$\mu$$ unknown.

My work:

I try to find the likelihood ratio statistic: \begin{align} \lambda(x) &= \frac{\sup_{\theta=\theta_0}L(\theta\mid X)}{\sup_{\theta\neq\theta_0}L(\theta\mid X)} \end{align}

For the global MLE case, I know that $$\sup_{\theta\neq\theta_0}L(\theta\mid X)=L(\hat{\mu},\hat{\sigma}^2)=(\frac{1}{\sqrt{2\pi\hat{\sigma}^2}})^{n} \exp[-\frac{1}{2\hat{\sigma}^2}\sum_{i=1}^n (X_i-\bar{X})^2]$$ where $$\hat{\mu}$$ is the sample mean and $$\hat{\sigma}^2=\frac{1}{n}\sum_i (X_i-\bar{X})^2$$.

But for the restricted MLE, I am a little bit confused. Since $$\theta=\theta_0$$ means $$(\mu,\sigma^2)=(\mu, \sigma_0^2)$$, then $$\sup_{\theta=\theta_0}L(\theta\mid X)=\sup L(\mu_0, \sigma_0^2)?$$

Is $$\mu_0=\bar{X}$$ and $$\sigma_0^2=\frac{1}{n}\sum_i (X_i-\bar{X})^2$$? So this will be the same as in the global case...

No, under $$H_0$$ you know that $$\sigma^2$$ equals $$\sigma^2_0$$ thus you don't have to estimate $$\sigma^2$$ but only $$\mu$$. Thus under $$H_0$$, the maximum likelihood is $$\sup_{\theta=\theta_0}L(\theta\mid X)= L(\bar X, \sigma_0^2\mid X).$$

Don't get distracted by the fact that the estimators of $$\mu$$ in the two hypotheses coincide. It is the estimator of $$\theta = (\mu,\sigma^2)$$ that matters here.

As per request, the likelihood ratio is

\begin{align} \lambda(X) &= \frac{\sup_{\theta=\theta_0}L(\theta\mid X)}{\sup_{\theta\neq\theta_0}L(\theta\mid X)} = \frac{L(\bar X, \sigma_0^2|X)}{L(\bar X, \hat\sigma^2|X)}\\ & = \left(\frac{\hat\sigma^2}{\sigma_0^2}\right)^{n/2}\exp\left(-\frac{n\hat\sigma^2}{2\sigma_0^2} + \frac{n}{2}\right). \end{align}

The rejection region has "shape" $$\left\{(X_1,\ldots X_n): \left(\frac{\hat\sigma^2}{\sigma_0^2}\right)^{n/2}\exp\left(-\frac{n\hat\sigma^2}{2\sigma_0^2}\right) \leq \exp(-n/2)\right\}.$$

After some algebra, you will find that the likelihood ratio test of level $$\alpha$$ is to reject $$H_0$$ whenever $$T_n >\chi_{n-1, 1-\alpha/2}^2\,\text{ or } T_n <\chi^2_{n-1,\alpha/2},$$

where $$\chi_{n-1, p}^2$$ denotes the $$p$$th quantile of the $$\chi_{n-1}^2$$ distribution and $$T_n = \frac{n\hat\sigma^2}{\sigma_0^2} \sim \chi_{n-1}^2$$.

• But how to simply the likelihood ratio test? And find the cut point? Commented Apr 6, 2023 at 15:07
• that's another question not related to the post :-) Commented Apr 6, 2023 at 15:27
• @Hermi I've added some more details. Since this is a self-study type of question I cannot provide you with a full solution. Can you fill in the missing details yourself? Commented Apr 6, 2023 at 19:17