Question

The incidence of a rare disease seems to be increasing. In successive years the numbers of new cases have been \(y_{1}, \ldots, y_{n}\). These may be assumed to be independent observations from Poisson distributions with means \(\lambda \theta, \ldots, \lambda \theta^{n}\). Show that there is a family of tests each of which, for any given value of \(\lambda\), is a uniformly most powerful test of its size for testing \(\theta=1\) against \(\theta>1\).

Short answer

For any given \(\lambda\), the test comparing \(\theta=1\) against \(\theta>1\) using the likelihood ratio is uniformly most powerful.

Step-by-step solution

Understand the Problem Statement

We have a set of data points \( y_1, ..., y_n \) which are independently observed from Poisson distributions with means \( \lambda \theta, ..., \lambda \theta^n \). Our goal is to construct a test that can determine if \( \theta \) is greater than 1, while assuming it is equal to 1 at the beginning. This test must be uniformly most powerful (UMP) for every given value of \( \lambda \).

Review the Poisson Distribution

In a Poisson distribution, the probability of observing \( y_i \) given a mean \( \lambda_i \) is given by \( P(Y_i = y_i) = \frac{e^{-\lambda_i} \lambda_i^{y_i}}{y_i!} \). Here, \( \lambda_i = \lambda \theta^i \). The likelihood function considering all observations is the product of these Poisson probabilities.

Construct the Likelihood Function

The likelihood function for observations \( y_1, ..., y_n \) is expressed as: \[ L(\theta; y_1, ..., y_n) = \prod_{i=1}^{n} \frac{e^{-\lambda \theta^i} (\lambda \theta^i)^{y_i}}{y_i!} = e^{-\lambda \sum_{i=1}^{n} \theta^i} \prod_{i=1}^{n} \frac{(\lambda \theta^i)^{y_i}}{y_i!} \].

Formulate the Hypothesis Test

We need to test the null hypothesis \( H_0: \theta = 1 \) against the alternative \( H_a: \theta > 1 \). The likelihood ratio test statistic is given by \( \Lambda = \frac{L(1; y_1, ..., y_n)}{L(\hat{\theta}; y_1, ..., y_n)} \), where \( \hat{\theta} \) is the MLE under the alternative hypothesis.

Derive the Likelihood Ratio Test

Since \( \theta = 1 \) under the null hypothesis, the likelihood ratio \( \Lambda \) simplifies to comparing the occurrence of higher observations (more cases of disease) than expected when \( \theta = 1 \). A UMP test is one that decides in favor of \( \theta > 1 \) by evaluating this statistic and comparing it against a critical value.

Maximize the Power of the Test

The power of a test is maximized when it accounts for all possible values of \( y_1, ..., y_n \) under \( H_a \) while minimizing Type I error. Each test compares the likelihood of observations under both hypotheses at each \( y_i \) and determines if \( H_0 \) should be rejected or not.

Conclude with Uniformly Most Powerful Test

Given any \( \lambda \), this test maintains its power because it specifically targets the exponential growth factor \( \theta^i \) included in the likelihood function's exponent. The reliance on individual \( y_i \), which is within a Poisson relationship contingent on \( \lambda \) and \( \theta^i \), ensures the test's optimality.

Key concepts

Poisson Distribution

The Poisson distribution is a probabilistic model used to describe the occurrence of rare events. If an event occurs with a constant mean rate in a fixed interval, the number of events in that interval follows a Poisson distribution.
This model is especially useful when analyzing the number of new cases in epidemiological studies, like the rare disease example given. In our context, if we have observed new cases in successive years as data points\( y_1, y_2, \ldots, y_n \), these can be independently modeled as Poisson distributions with respective means \( \lambda \theta, \lambda \theta^2, \ldots, \lambda \theta^n \).
This implies that each year, the number of new cases increases exponentially with \( \theta \), representing a potential growth factor.Using the formula for the Poisson probability:

• \( P(Y_i = y_i) = \frac{e^{-\lambda_i} \lambda_i^{y_i}}{y_i!} \)
• where \( \lambda_i = \lambda \theta^i \).

By understanding how the data is distributed, one can better assess the potential trends and shifts in the disease's occurrence and then decide on the parameters to test.

Uniformly Most Powerful Test

A Uniformly Most Powerful (UMP) test is designed to maximize its power across all possible alternatives of a parameter. In hypothesis testing, the power of a test is its ability to correctly reject a false null hypothesis.In this exercise, we are analyzing if the incidence of disease cases is growing, driven by the parameter \( \theta \). The hypothesis in question is:

• \( H_0: \theta = 1 \)
• versus \( H_a: \theta > 1 \).

The goal here is to determine if \( \theta \) is indeed greater than 1, which would indicate an increase in the disease cases. When establishing a UMP test, one considers all possible values for \( \lambda \) to ensure the test's effectiveness regardless of this parameter.In this context:

• We utilize the observed data \( y_1, \ldots, y_n \)
• To assess each possible \( \lambda \)-dependent test's rejection capability against \( H_0 \).

Therefore, a UMP test effectively focuses on maximizing test power without dependence on unknown parameters.

Likelihood Ratio Test

The Likelihood Ratio Test (LRT) is a statistical technique used to compare the fit of two models and is a building block for establishing significance in hypothesis testing.For the exercise at hand, the aim is to test \( \theta \) being 1 against \( \theta > 1 \). Here, the likelihood ratio \( \Lambda \) is a function comparing the likelihood of the data under the null hypothesis \( H_0 \) to the likelihood under the alternative hypothesis \( H_a \).Define \( \Lambda \) as:

• \( \Lambda = \frac{L(1; y_1, \ldots, y_n)}{L(\hat{\theta}; y_1, \ldots, y_n)} \)
• Where \( L(\cdot) \) denotes the likelihood function

This statistic provides a decision metric such that:

• A smaller \( \Lambda \) suggests the alternative hypothesis is more likely, indicating \( \theta > 1 \).
• The MLE, \( \hat{\theta} \), optimizes the likelihood under \( H_a \).

By comparing this test statistic to a critical threshold, one can choose to reject \( H_0 \) in favor of \( H_a \).
The LRT framework simplifies evaluating whether the observed increase in disease matches expected growth, effectively determining if changes in the disease pattern are statistically significant.