# GLRT statistic for composite normal hypothesis, two unknowns

Suppose $$X_1...X_n$$ ~ iid ~ $$\mathcal N(\mu, \sigma)$$, both parameters unknown. We want to test $$H_0: \mu \leq \mu_0, H_1: \mu > \mu_0$$. Show that the LRT (likelihood ratio test) statistic is given by $$\lambda(\textbf{x}) = \begin{cases} 1,\bar{X} \leq \mu_0 \\ \left(\frac{\hat{\sigma^2}}{\hat{\sigma^2_0}}\right)^{\frac{n}{2}} , \bar{X} > \mu_0 \end{cases}$$ Where $$\begin{cases} \hat{\sigma^2} = \frac{1}{n}\sum_{i=1}^n (x_i - \bar{X})^2 \\ \hat{\sigma^2_0} = \frac{1}{n}\sum_{i=1}^n (x_i - \mu_0)^2 \end{cases}$$

Note that the LRT statistic for a composite hypothesis ($$H_0$$ does not completely determine the distribution, say our $$H_0$$ vs $$H_0: \mu = \mu_0$$) is defined as $$\lambda(\textbf{x}) = \frac{\max_{\theta \in \Theta_0} L(\theta;\textbf{x})}{\max_{\theta \in \Theta} L(\theta;\textbf{x})}$$ where $$\Theta_0 = \{(\mu, \sigma): \mu \leq \mu_0, \sigma > 0\}, \Theta = \{(\mu, \sigma) : \mu \in (-\infty, \infty), \sigma > 0\}$$

I am having trouble calculating the numerator of the LRT statistic, specifically dealing with the condition $$\mu \leq \mu_0$$. If we were dealing with a simple hypothesis, say $$H_0: \mu = \mu_0$$, we can simply plug in $$\mu_0$$ into the likelihood function of normal iid random variables. Here, under $$H_0$$, $$\mu$$ is a range of parameters. What do I do?

I am thinking that if I can put a global bound on the likelihood function w.r.t $$\mu$$ under $$H_0$$, then the likelihood becomes a function of one parameter $$\sigma$$, from which I can then take derivatives or whatever to maximize the likelihood. Is that the right way to go about this? If I can just figure the numerator out, I know how to solve the problem from there.

• Jun 4 '20 at 20:35

Look at log-likelihood function $$\ln L(\theta, \mathbf X) = -n\ln \sigma - \frac{1}{2\sigma^2}\sum_{i=1}^n (X_i-\mu)^2.$$ Concerning the denominator: if we will find global maximum over $$(\mu,\sigma)\in\Theta$$, we get that it is attained at the point $$\hat\mu=\overline X$$, $$\hat\sigma^2=\frac{1}{n}\sum_{i=1}^n (X_i-\overline X)^2$$.
Return to numerator. If we work inside $$\Theta_0$$ and if $$\overline X \leq \mu_0$$, then $$(\hat\mu,\hat\sigma^2)\in\Theta_0$$ and then likelihood ratio equals to $$1$$.
If $$\overline X > \mu_0$$, look at the derivatives of log-likelihood function at any point $$\mu\leq \mu_0 <\overline X$$, $$\sigma>0$$: $$\frac{\partial}{\partial \mu}\ln L(\theta, \mathbf X) = \frac{n}{\sigma^2}\left( \overline X-\mu\right)>0$$ so the log-likelihood function increases in $$\mu$$ irrespective to $$\sigma$$, so for any fixed $$\sigma>0$$ $$\max_{\mu\leq \mu_0} L(\mu,\sigma, \mathbf X) = L(\mu_0,\sigma, \mathbf X).$$ Then we can take $$\mu=\mu_0$$ and find maximum over $$\sigma$$: $$\frac{\partial}{\partial \sigma}\ln L(\mu_0,\sigma, \mathbf X) = 0 \iff \sigma^2=\hat\sigma_0^2=\frac1n\sum_{i=1}^n (X_i-\mu_0)^2.$$