# How to get the LRT statistic of $n(\theta, \sigma^2)$

This is from George Casella textbook question 8.37 (c). Let $$X_1,...,X_n$$ be a random sample from a $$n(\theta, \sigma^2)$$ population. Consider testing $$H_0:\theta\leq \theta_0$$ versus $$H_1:\theta> \theta_0$$. If $$\sigma^2$$ is unknown, what is the LRT statistic? The solution is $$\lambda=(\hat{\sigma}^2/\hat{\sigma_0}^2)^{n/2}$$. I didn't get this.

My attempt: the LRT of denimoator is easier and reached when $$\theta=\bar{x}$$ and $$\sigma^2=\hat{\sigma}^2$$ below in the solution. The numerator is $$\sigma^2=\hat{\sigma}^2$$ and $$\theta=\theta_0$$.(We don't need to consider when $$\lambda=1$$ case.) I didn't get the LRT statistic in the solution.

According to Casella & Berger's notation (Definition 8.2.1), the likelihood ratio test statistic for testing $$H_0: \theta \in \Theta_0$$ versus $$H_1: \theta \in \Theta_0^c$$ is $$\lambda(\boldsymbol{x}) = \frac{\text{sup}_{\Theta_0}L(\theta \mid \boldsymbol{x})}{\text{sup}_{\Theta}L(\theta \mid \boldsymbol{x})}.$$ A likelihood ratio test (LRT) is any test that has a rejection region of the form $$\{\boldsymbol{x}: \lambda(\boldsymbol{x}) \leq c\}$$, where $$c$$ is any number satisfying $$0 \leq c \leq 1$$.

In the problem of testing $$H_0: \theta \leq \theta_0$$ versus $$H_1: \theta > \theta_0$$, where $$X_1, \dots, X_n \overset{i.i.d.}{\sim} n(\theta, \sigma^2)$$ and $$\sigma^2$$ is unknown, we have $$\Theta_0 = \{(\theta, \sigma^2): \theta \leq \theta_0, \sigma^2 > 0\}$$ and $$\Theta = \{(\theta, \sigma^2): \theta \in \mathbb{R}, \sigma^2 > 0\}$$, and the likelihood is given by $$L(\theta, \sigma^2 \mid \boldsymbol{x}) = \left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^n \exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^n(x_i - \theta)^2\right).$$

The supremum in the denominator of $$\lambda(\boldsymbol{x})$$ is attained at the unrestricted MLE of $$(\theta, \sigma^2)$$, that is, $$\hat{\theta} = \bar{x}$$ and $$\hat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^n(x_i - \bar{x})^2$$ (See Section 4.1 if this is unclear).

The supremum in the numerator of $$\lambda(\boldsymbol{x})$$ needs to be a bit more careful, since it is "restricted" to the parameter space $$\Theta_0$$. Specifically, if we observe $$\bar{x} > \theta_0$$ ("$$>$$" is more precise than "$$\geq$$" the solution), then $$L(\theta, \sigma^2)$$ cannot be maximized at the unrestricted MLE ($$\hat{\theta} = \bar{x}, \hat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^n(x_i - \bar{x})^2$$) since $$(\bar{x}, \hat{\sigma}^2) \notin \Theta_0$$. In this case, the restricted MLE (with respect to $$\Theta_0$$) of $$(\theta, \sigma^2)$$ is taken at the boundary of the parameter space, that is, $$\hat{\theta}_0 = \theta_0$$ and $$\hat{\sigma}_0^2 = \frac{1}{n}\sum_{i=1}^n(x_i - \theta_0)^2$$. On the other hand, if we observe $$\bar{x} \leq \theta_0$$, then $$L(\theta, \sigma^2)$$ can be maximized at the unrestricted MLE ($$\hat{\theta} = \bar{x}, \hat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^n(x_i - \bar{x})^2$$) since $$(\bar{x}, \hat{\sigma}^2) \in \Theta_0$$. In this case, the numerator and denominator coincides thus $$\lambda(\boldsymbol{x}) = 1$$.

To sum up, for $$n(\theta, \sigma^2)$$ where $$\sigma^2$$ is unknown, the LRT statistic of testing $$H_0: \theta \leq \theta_0$$ versus $$H_1: \theta > \theta_0$$ is given by $$\lambda(\boldsymbol{x}) = \begin{cases} \dfrac{\text{sup}_{\Theta_0}L(\theta, \sigma^2 \mid \boldsymbol{x})}{\text{sup}_{\Theta}L(\theta, \sigma^2 \mid \boldsymbol{x})} = \dfrac{L(\theta_0, \hat{\sigma}_0^2 \mid \boldsymbol{x})}{L(\bar{x}, \hat{\sigma}^2 \mid \boldsymbol{x})} = \left(\dfrac{\hat{\sigma}^2}{\hat{\sigma}_0^2}\right)^{n/2} & \text{if } \bar{x} > \theta_0; \\ 1 & \text{if } \bar{x} \leq \theta_0. \end{cases}$$

Remark: It is worth to note the fact that $$\exp\left(-\frac{1}{2\hat{\sigma}_0^2}\sum_{i=1}^n(x_i - \theta_0)^2\right) = \exp\left(-\frac{1}{2\hat{\sigma}^2}\sum_{i=1}^n(x_i - \bar{x})^2\right) = \exp\left(-\frac{n}{2}\right),$$ which cancels in the derivation of the LRT statistic $$\lambda(\boldsymbol{x})$$ for $$\bar{x} > \theta_0$$.

• Thank you so much! We can also refer to Example 7.2.11. Apr 4 at 14:27
• I asked another relevant question here. Could you help me? math.stackexchange.com/questions/4089371/… Apr 4 at 20:09
• @Yichuan No problem! Glad that it is helpful. Apr 4 at 21:00
• @Yichuan Sure, I can take a look. Apr 4 at 21:00