Example 3.2: Finding The Most Powerful Test For Exponential Distribution
Introduction
In this article, we will delve into Example 3.2, which focuses on finding the most powerful test for a random sample drawn from an Exponential distribution. Specifically, we aim to determine the most powerful test of size α for testing the null hypothesis H₀: θ = θ₀ versus the alternative hypothesis H₁: θ = θ₁, where θ₀ > θ₁. Additionally, we will calculate β, the probability of a Type II error, when the sample size n is equal to 1. This example provides a comprehensive illustration of applying the Neyman-Pearson Lemma in the context of exponential distributions, a fundamental concept in statistical hypothesis testing. Understanding how to construct the most powerful test is crucial for making informed decisions based on data, especially in scenarios where minimizing the probability of false negatives is paramount. This article breaks down the mathematical steps and provides clear explanations to enhance comprehension.
Problem Statement
Let X₁, X₂, ..., Xₙ be a random sample from an Exponential(θ) distribution. Our goal is to address the following questions:
(a) Find the most powerful test of size α for testing the null hypothesis H₀: θ = θ₀ versus the alternative hypothesis H₁: θ = θ₁, where θ₀ > θ₁.
(b) If n = 1, find β, the probability of a Type II error.
(a) Finding the Most Powerful Test
To find the most powerful test, we will employ the Neyman-Pearson Lemma. This lemma states that the most powerful test for testing a simple null hypothesis against a simple alternative hypothesis is given by the likelihood ratio test. The likelihood ratio is the ratio of the likelihood function under the alternative hypothesis to the likelihood function under the null hypothesis. In our case, the likelihood function for a random sample from an Exponential(θ) distribution is given by:
L(θ; x₁, x₂, ..., xₙ) = ∏ᵢ₌₁ⁿ f(xᵢ; θ) = ∏ᵢ₌₁ⁿ θ⁻¹e⁻ˣⁱ/θ = θ⁻ⁿ exp(-∑ᵢ₌₁ⁿ xᵢ / θ)
where f(xᵢ; θ) is the probability density function of the Exponential(θ) distribution, which is θ⁻¹e⁻ˣ/θ for x > 0 and 0 otherwise. The likelihood ratio, denoted by Λ, is then:
Λ = L(θ₁; x₁, x₂, ..., xₙ) / L(θ₀; x₁, x₂, ..., xₙ)
Substituting the likelihood function, we get:
Λ = (θ₁⁻ⁿ exp(-∑ᵢ₌₁ⁿ xᵢ / θ₁)) / (θ₀⁻ⁿ exp(-∑ᵢ₌₁ⁿ xᵢ / θ₀))
Simplifying, we have:
Λ = (θ₀ / θ₁)ⁿ exp(-∑ᵢ₌₁ⁿ xᵢ (1/θ₁ - 1/θ₀))
Since θ₀ > θ₁, then 1/θ₁ > 1/θ₀, and thus (1/θ₁ - 1/θ₀) > 0. The Neyman-Pearson Lemma states that the most powerful test rejects H₀ if Λ > k, where k is a constant chosen to achieve the desired significance level α. Equivalently, we can reject H₀ if:
(θ₀ / θ₁)ⁿ exp(-∑ᵢ₌₁ⁿ xᵢ (1/θ₁ - 1/θ₀)) > k
Taking the natural logarithm of both sides:
n ln(θ₀ / θ₁) - ∑ᵢ₌₁ⁿ xᵢ (1/θ₁ - 1/θ₀) > ln(k)
Rearranging the terms:
∑ᵢ₌₁ⁿ xᵢ (1/θ₁ - 1/θ₀) < n ln(θ₀ / θ₁) - ln(k)
Let C = n ln(θ₀ / θ₁) - ln(k), which is a constant. Since (1/θ₁ - 1/θ₀) > 0, we can divide both sides by (1/θ₁ - 1/θ₀) without changing the inequality direction:
∑ᵢ₌₁ⁿ xᵢ < C / (1/θ₁ - 1/θ₀) = k'
where k' is another constant. Let's denote the sum of the sample by T = ∑ᵢ₌₁ⁿ Xᵢ. The test then rejects H₀ if T < k'.
Under H₀, T follows a Gamma distribution with parameters n and θ₀. Therefore, T ~ Gamma(n, θ₀). The significance level α is given by:
α = P(T < k' | H₀) = P(∑ᵢ₌₁ⁿ Xᵢ < k' | θ = θ₀)
The constant k' is chosen such that:
α = ∫₀ᵏ' f_T(t; n, θ₀) dt
where f_T(t; n, θ₀) is the probability density function of the Gamma distribution with parameters n and θ₀. This integral represents the cumulative distribution function (CDF) of the Gamma distribution evaluated at k'. Solving this integral equation for k' gives us the critical value for our test.
The most powerful test of size α rejects H₀ if ∑ᵢ₌₁ⁿ Xᵢ < k', where k' is chosen to satisfy the equation above. This test minimizes the probability of a Type II error while maintaining the desired significance level α. The derivation underscores the importance of the Neyman-Pearson Lemma in constructing optimal hypothesis tests, particularly in statistical inference where the power of a test is crucial for reliable decision-making.
(b) Calculating β when n = 1
Now, let's calculate β, the probability of a Type II error, when n = 1. The probability of a Type II error is the probability of failing to reject H₀ when H₁ is true. In our case, it is the probability of observing X₁ ≥ k' when θ = θ₁.
β = P(X₁ ≥ k' | H₁) = P(X₁ ≥ k' | θ = θ₁)
Since X₁ follows an Exponential(θ₁) distribution under H₁, its probability density function is f(x₁; θ₁) = θ₁⁻¹e⁻ˣ¹/θ₁. Thus, we need to compute the integral:
β = ∫ₖ'^{∞} θ₁⁻¹e⁻ˣ¹/θ₁ dx₁
The cumulative distribution function (CDF) of the Exponential(θ) distribution is given by F(x; θ) = 1 - e⁻ˣ/θ for x ≥ 0. Therefore, the integral simplifies to:
β = 1 - ∫₀ᵏ' θ₁⁻¹e⁻ˣ¹/θ₁ dx₁ = 1 - (1 - e⁻ᵏ'/θ₁) = e⁻ᵏ'/θ₁
To find the value of k' when n = 1, we use the significance level α condition from part (a):
α = P(X₁ < k' | H₀) = ∫₀ᵏ' θ₀⁻¹e⁻ˣ¹/θ₀ dx₁ = 1 - e⁻ᵏ'/θ₀
Solving for k':
1 - α = e⁻ᵏ'/θ₀
ln(1 - α) = -k'/θ₀
k' = -θ₀ ln(1 - α)
Now we substitute this expression for k' into the equation for β:
β = e⁻⁽⁻θ₀ ln(1 - α)⁾/θ₁ = exp(θ₀ ln(1 - α) / θ₁) = (1 - α)^(θ₀/θ₁)
Therefore, the probability of a Type II error β when n = 1 is given by:
β = (1 - α)^(θ₀/θ₁)
This result shows how β depends on the significance level α and the parameters θ₀ and θ₁. Since θ₀ > θ₁, the exponent θ₀/θ₁ is greater than 1, and β will decrease as α decreases (as expected). This calculation provides a clear understanding of the trade-off between Type I and Type II errors in hypothesis testing.
Conclusion
In summary, we have successfully found the most powerful test of size α for testing H₀: θ = θ₀ versus H₁: θ = θ₁ for an Exponential(θ) distribution using the Neyman-Pearson Lemma. We showed that the test rejects H₀ when the sum of the sample is less than a critical value k', which is determined by the significance level α and the parameters of the null distribution. Additionally, we calculated the probability of a Type II error β when the sample size n is 1, expressing it in terms of α and the parameters θ₀ and θ₁. This comprehensive analysis provides a solid understanding of hypothesis testing in the context of exponential distributions and illustrates the practical application of the Neyman-Pearson Lemma.
The ability to construct and analyze the power of statistical tests is a crucial skill in statistical inference, particularly in fields such as engineering, medicine, and economics, where data-driven decisions are paramount. The detailed solution presented in this article aims to enhance the understanding of these concepts and provide a valuable resource for students and practitioners alike. By mastering these techniques, one can make more informed and reliable decisions based on statistical evidence.
Keywords
Most powerful test, Neyman-Pearson Lemma, Exponential distribution, Hypothesis testing, Type II error, Significance level, Likelihood ratio, Gamma distribution, Statistical inference, Probability of error.
