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

Question

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.

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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...

Answer 1

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 $T_n = \frac{n\hat\sigma^2}{\sigma_0^2} \sim \chi_{n-1}^2$.