Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid with pmf \(f(x ; p)=p^{x}(1-p)^{1-x}, x=0,1\), zero elsewhere. Show that \(C=\left\\{\left(x_{1}, \ldots, x_{n}\right): \sum_{1}^{n} x_{i} \leq c\right\\}\) is a best critical region for testing \(H_{0}: p=\frac{1}{2}\) against \(H_{1}: p=\frac{1}{3} .\) Use the Central Limit Theorem to find \(n\) and \(c\) so that approximately \(P_{H_{0}}\left(\sum_{1}^{n} X_{i} \leq c\right)=0.10\) and \(P_{H_{1}}\left(\sum_{1}^{n} X_{i} \leq c\right)=0.80\).

Short answer

First, it's proven that C is the best critical region. The Central Limit Theorem is then invoked to bring in Normal approximation. Finally, by using the standard Normal table and the given probabilities under both null and alternate hypothesis, we solve the equations to find the required n and c respective to those probabilities.

Step-by-step solution

Check critical region

We first prove that C is the best critical region. For this, it must satisfy both Neyman–Pearson conditions: The ratio \(\frac{f(x ; p_{1})}{f(x ; p_{0})}\) must be monotone increasing and the randomized test must have the form: Reject \(H_{0}\) if \(\frac{f(x ; p_{1})}{f(x ; p_{0})} > k\), do not reject \(H_{0}\) if \(\frac{f(x ; p_{1})}{f(x ; p_{0})} < k\), and random rejection when \(\frac{f(x ; p_{1})}{f(x ; p_{0})} = k\). Here, \(p_{1}=1/3\) and \(p_{0}=1/2\) and x is the sum of all random variables, \(x_{i}\). From the given pmf, we know that the likelihood ratio decreases as x increases. Hence, the first condition is satisfied.

Implement Central Limit Theorem

According to the Central Limit Theorem (CLT), the distribution of the sum (or average) of a large number of independent, identically distributed variables approaches the Normal distribution, regardless of the shape of the original distribution. Because the \(X_i\) are independent and identically distributed and because \(n\) is large, the CLT can be applied. The mean and variance under \(H_{0}\) and \(H_{1}\) can be calculated using \(E(X)=np\) and \(Var(X)=np(1-p)\).

Find n and c

Let Z denote the standardized test statistic. Our task is to find n and c so that \(P_{H_{0}}(\frac{\sum_{1}^{n} X_{i} - n*p_{0}}{\sqrt{n*p_{0}(1-p_{0})}} \leq Z) = 0.10\) and \(P_{H_{1}}(\frac{\sum_{1}^{n} X_{i} - n*p_{1}}{\sqrt{n*p_{1}(1-p_{1})}} \leq Z) = 0.80\). Using the standard Normal table, we can solve these equations for n and c. The equation under \(H_{0}\) gives Z-value corresponding to the area 0.10, and the equation under \(H_{1}\) gives Z-value corresponding to the area 0.80. The system of these two equations will yield the required n and c.

Key concepts

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental concept in statistics. It states that, for a large number of independent and identically distributed random variables, the distribution of their sum tends towards a normal distribution. This happens regardless of the original distribution's shape.

Why is this important? It allows us to make inferences about population parameters using sample statistics. In this problem, we have random variables following a Bernoulli distribution, but if our sample size is large enough, we can approximate the sum of these variables with a normal distribution. This makes calculations more manageable and enables the use of z-scores.

The CLT is crucial because it makes hypothesis testing feasible with large samples. For example, if the sample size \(n\) is big, the sum \(\sum_{i=1}^{n} X_i\) under hypothesis \(H_0\) can be approximated as a normal distribution with mean \(n \times \frac{1}{2}\) and variance \(n \times \frac{1}{2} \times \left(1 - \frac{1}{2}\right)\). This simplifies finding probabilities and determining critical values.

Neyman–Pearson Lemma

The Neyman–Pearson Lemma is a cornerstone of hypothesis testing. It provides a way to find the most powerful test (one that maximizes the probability of rejecting a false null hypothesis) for a given size \(\alpha\).

In this problem, we're dealing with hypotheses \(H_0: p=\frac{1}{2}\) and \(H_1: p=\frac{1}{3}\). The lemma states that the likelihood ratio test, expressed as \(\frac{f(x; p_1)}{f(x; p_0)}\), identifies the best critical region for rejecting \(H_0\).

The critical region \(C\) is defined such that you reject \(H_0\) when the likelihood ratio \(> k\). Here, we need to verify that the ratio is monotone, which is true because it decreases as the sum \(x\) increases. This ensures that the region \(C\) provides the highest power of the test within the set constraints.

Likelihood Ratio Test

The Likelihood Ratio Test (LRT) is a method for comparing two statistical models based on their likelihoods. For two competing hypotheses, it evaluates which hypothesis is more likely given the observed data.

To use LRT, we calculate the likelihoods under both \(H_0\) and \(H_1\). The test statistic is the ratio of these likelihoods: \(\frac{L(H_1)}{L(H_0)}\). In our specific problem, this is expressed using the probability mass function for Bernoulli variables.

A decision rule is then established based on this ratio. We reject \(H_0\) in favor of \(H_1\) if the ratio is higher than a preset threshold \(k\). Our model assumes a specific parametric form, where parameters like \(p\) change under different hypotheses. The LRT helps us decide if the observed data significantly favors one hypothesis over the other.