## Monday, 27 April 2015

### Parametric Inference: Likelihood Ratio Test by Example

Hypothesis testing have been extensively used on different discipline of science. And in this post, I will attempt on discussing the basic theory behind this, the Likelihood Ratio Test (LRT) defined below from Casella and Berger (2001), see reference 1.
Definition. The likelihood ratio test statistic for testing $H_0:\theta\in\Theta_0$ versus $H_1:\theta\in\Theta_0^c$ is $$\label{eq:lrt} \lambda(\mathbf{x})=\frac{\displaystyle\sup_{\theta\in\Theta_0}L(\theta|\mathbf{x})}{\displaystyle\sup_{\theta\in\Theta}L(\theta|\mathbf{x})}.$$ A likelihood ratio test (LRT) is any test that has a rejection region of the form $\{\mathbf{x}:\lambda(\mathbf{x})\leq c\}$, where $c$ is any number satisfying $0\leq c \leq 1$.
The numerator of equation (\ref{eq:lrt}) gives us the supremum probability of the parameter, $\theta$, over the restricted domain (null hypothesis, $\Theta_0$) of the parameter space $\Theta$, that maximizes the joint probability of the sample, $\mathbf{x}$. While the denominator of the LRT gives us the supremum probability of the parameter, $\theta$, over the unrestricted domain, $\Theta$, that maximizes the joint probability of the sample, $\mathbf{x}$. Therefore, if the value of $\lambda(\mathbf{x})$ is small such that $\lambda(\mathbf{x})\leq c$, for some $c\in [0, 1]$, then the true value of the parameter that is plausible in explaining the sample is likely to be in the alternative hypothesis, $\Theta_0^c$.

Example 1. Let $X_1,X_2,\cdots,X_n\overset{r.s.}{\sim}f(x|\theta)=\frac{1}{\theta}\exp\left[-\frac{x}{\theta}\right],x>0,\theta>0$. From this sample, consider testing $H_0:\theta = \theta_0$ vs $H_1:\theta<\theta_0$.

Solution:
The parameter space $\Theta$ is the set $(0,\Theta_0]$, where $\Theta_0=\{\theta_0\}$. Hence, using the likelihood ratio test, we have $$\lambda(\mathbf{x})=\frac{\displaystyle\sup_{\theta=\theta_0}L(\theta|\mathbf{x})}{\displaystyle\sup_{\theta\leq\theta_0}L(\theta|\mathbf{x})},$$ where, \begin{aligned} \sup_{\theta=\theta_0}L(\theta|\mathbf{x})&=\sup_{\theta=\theta_0}\prod_{i=1}^{n}\frac{1}{\theta}\exp\left[-\frac{x_i}{\theta}\right]\\ &=\sup_{\theta=\theta_0}\left(\frac{1}{\theta}\right)^n\exp\left[-\displaystyle\frac{\sum_{i=1}^{n}x_i}{\theta}\right]\\ &=\left(\frac{1}{\theta_0}\right)^n\exp\left[-\displaystyle\frac{\sum_{i=1}^{n}x_i}{\theta_0}\right], \end{aligned} and \begin{aligned} \sup_{\theta\leq\theta_0}L(\theta|\mathbf{x})&=\sup_{\theta\leq\theta_0}\prod_{i=1}^{n}\frac{1}{\theta}\exp\left[-\frac{x_i}{\theta}\right]\\ &=\sup_{\theta\leq\theta_0}\left(\frac{1}{\theta}\right)^n\exp\left[-\displaystyle\frac{\sum_{i=1}^{n}x_i}{\theta}\right]=\sup_{\theta\leq\theta_0}f(\mathbf{x}|\theta). \end{aligned} Now the supremum of $f(\mathbf{x}|\theta)$ over all values of $\theta\leq\theta_0$ is the MLE (maximum likelihood estimator) of $f(x|\theta)$, which is $\bar{x}$, provided that $\bar{x}\leq \theta_0$.

So that, \begin{aligned} \lambda(\mathbf{x})&=\frac{\left(\frac{1}{\theta_0}\right)^n\exp\left[-\displaystyle\frac{\sum_{i=1}^{n}x_i}{\theta_0}\right]} {\left(\frac{1}{\bar{x}}\right)^n\exp\left[-\displaystyle\frac{\sum_{i=1}^{n}x_i}{\bar{x}}\right]},\quad\text{provided that}\;\bar{x}\leq \theta_0\\ &=\left(\frac{\bar{x}}{\theta_0}\right)^n\exp\left[-\displaystyle\frac{\sum_{i=1}^{n}x_i}{\theta_0}\right]\exp[n]. \end{aligned} And we say that, if $\lambda(\mathbf{x})\leq c$, $H_0$ is rejected. That is, \begin{aligned} \left(\frac{\bar{x}}{\theta_0}\right)^n\exp\left[-\displaystyle\frac{\sum_{i=1}^{n}x_i}{\theta_0}\right]\exp[n]&\leq c\\ \left(\frac{\bar{x}}{\theta_0}\right)^n\exp\left[-\displaystyle\frac{\sum_{i=1}^{n}x_i}{\theta_0}\right]&\leq c',\quad\text{where}\;c'=\frac{c}{\exp[n]}\\ n\log\left(\frac{\bar{x}}{\theta_0}\right)-\frac{n}{\theta_0}\bar{x}&\leq \log c'\\ \log\left(\frac{\bar{x}}{\theta_0}\right)-\frac{\bar{x}}{\theta_0}&\leq \frac{1}{n}\log c'\\ \log\left(\frac{\bar{x}}{\theta_0}\right)-\frac{\bar{x}}{\theta_0}&\leq \frac{1}{n}\log c-1. \end{aligned} Now let $h(x)=\log x - x$, then $h'(x)=\frac{1}{x}-1$. So the critical point of $h'(x)$ is $x=1$. And to test if this is maximum or minimum, we apply second derivative test. That is, $$h''(x)=-\frac{1}{x^2}<0,\forall x.$$ Thus, $x=1$ is a maximum. Hence, $\log\left(\frac{\bar{x}}{\theta_0}\right)-\frac{\bar{x}}{\theta_0}$ is maximized if $\frac{\bar{x}}{\theta_0}=1\Rightarrow\bar{x}=\theta_0$. To see this consider the following plot,
Above figure is the plot of $h(\bar{x})$ function with $\theta_0=1$. Given the assumption that $\bar{x}\leq \theta_0$ then assuming $R=\frac{1}{n}\log c-1$ designates the orange line above, then we reject $H_0$ if $h(\bar{x})\leq R$, if and only if $\bar{x}\leq k$. In practice, $k$ is specified to satisfy, $$\mathrm{P}(\bar{x}\leq k|\theta=\theta_0)\leq \alpha,$$ where $\alpha$ is called the level of the test.

It follows that $X_i|\theta = \theta_0\overset{r.s.}{\sim}\exp[\theta_0]$, then $\mathrm{E}X_i=\theta_0$ and $\mathrm{Var}X_i=\theta_0^2$. If $\bar{x}=\frac{1}{n}\sum_{i=1}^{n}X_i$ and if $G_n$ is the distribution of $\frac{(\bar{x}_n-\theta_0)}{\sqrt{\frac{\theta_0^2}{n}}}$. By CLT (central limit theorem) $\lim_{n\to\infty}G_n$ converges to standard normal distribution. That is, $\bar{x}|\theta = \theta_0\overset{r.s.}{\sim}AN\left(\theta_0,\frac{\theta_0^2}{n}\right)$. $AN$ - assymptotically normal.

Thus, $$\mathrm{P}(\bar{x}\leq k|\theta=\theta_0)=\Phi\left(\frac{k-\theta_0}{\theta_0/\sqrt{n}}\right),\quad\text{for large }n.$$ So that, $$\mathrm{P}(\bar{x}\leq k|\theta=\theta_0)=\Phi\left(\frac{k-\theta_0}{\theta_0/\sqrt{n}}\right)\leq \alpha.$$ Plotting this gives us,
with corresponding PDF given by,
Implying, $$\frac{k-\theta_0}{\theta_0/\sqrt{n}}=z_{\alpha}\Rightarrow k=\theta_0+z_{\alpha}\frac{\theta_0}{\sqrt{n}}.$$ Therefore, a level-$\alpha$ test of $H_0:\theta=\theta_0$ vs $H_1:\theta<\theta_0$ is the test that rejects $H_0$ when $\bar{x}\leq\theta_0+z_{\alpha}\frac{\theta_0}{\sqrt{n}}$.

### Plot's Python Codes

In case you might ask how above plots were generated: