Likelihood ratio test confusion

Question

Let's say I have a sample $X_1, X_2, \cdots ,X_n$ from a Bernoulli distribution, and I want to test the following: $$ H_0 \colon p=p_{0} \quad\text{vs}\quad H_1 \colon p= p_{1} $$ where $p$ is the parameter of a Bernoulli distribution. The likelihood ratio test is then: $$\Lambda(x) = \left(\frac{1 - p_0}{1 - p_1}\right)^n \left(\frac{p_0 (1 - p_1)}{p_1 (1 - p_0)}\right)^{\sum x_{i}} $$ Suppose we want to determine the exact distribution of $\Lambda$. In order to do that, do we suppose the $x_{i}$ considered a Bernoulli with parameter $p_0$ or $p_1$?.

EDIT: In other words, is the distribution of $\Lambda$ determined under $H_0$?

Answer 1

As the likelihood ratio is a function of the data $x$, it is a statistic. You calculate the statistic and test a hypothesis under $H_0$, so here we assume $p_0$ is the true value of the parameter.

Regarding the exact distribution of $\Lambda(x)$:

Say $H_0$ and $H_1$ have parameter spaces $\Theta_0$ (of dimension $d_0$) and $\Theta_1$(of dimension $d_1$) respectively. If $\Theta_0$ is a subspace of $\Theta_1$ (i.e. testing against a general alternative $H_1$), then the distribution of $\Lambda(x)$ is given by the Wilks' Theorem. Here $H_0$ is either simple or composite, but the general alternative $H_1$ is composite.

The theorem states that, as the sample size $n\rightarrow \infty$, then the statistic $-2 \operatorname{log}\Lambda(x)$ converges in distribution to the chi-squared distribution $\chi_{d_1-d_0}^2$ i.e. with degrees of freedom the difference in dimensions in parameter spaces.

Here the null and alternative hypothesis are both simple, so we can't use Wilks' theorem to find the distribution. Instead, after finding the likelihood ratio $\Lambda(x)$ for the observed data, we can use it to apply the Neyman-Pearson lemma in order to find the most powerful test among all tests with a fixed significance level $\alpha$. Let me know if the theorem doesn't make sense.

Hope this helps.