Question

A random sample \(X_{1}, X_{2}, \ldots, X_{n}\) arises from a distribution given by $$H_{0}: f(x ; \theta)=\frac{1}{\theta}, \quad 0

Short answer

The likelihood ratio (\(\Lambda\)) for the given hypotheses \(H_{0}\) and \(H_{1}\) is \(\Lambda = e^{\sum_{i=1}^{n}x_{i}/\theta}\).

Step-by-step solution

Compute the Likelihoods

First, it is necessary to calculate the likelihoods under each hypothesis. For a given sample \(X_{1}, X_{2}, \ldots, X_{n}\), the likelihoods are given by: \[L(\theta|H_{0}) = \prod_{i=1}^{n} f(x_{i}; \theta|H_{0}) = \prod_{i=1}^{n} \frac{1}{\theta} = \theta^{-n}\] for \(H_0\), and \[L(\theta|H_{1}) = \prod_{i=1}^{n} f(x_{i}; \theta|H_{1}) = \prod_{i=1}^{n} \frac{1}{\theta} e^{-x_{i}/\theta} = \theta^{-n} e^{-\sum_{i=1}^{n}x_{i}/\theta}\] for \(H_1\).

Calculate the Likelihood Ratio

Next, compute the likelihood ratio, which is given by the ratio of the likelihoods under the two hypotheses. Therefore, \[\Lambda = \frac{L(\theta|H_{0})}{L(\theta|H_{1})} = \frac{\theta^{-n}}{\theta^{-n} e^{-\sum_{i=1}^{n}x_{i}/\theta}} = e^{\sum_{i=1}^{n}x_{i}/\theta}\]

Simplify the Ratio

Finally, simplify the ratio. As there is no information given on which will accept or reject the hypothesis, the solution will remain as likelihood ratio. In other scenarios, you may need to compare this ratio to a threshold to decide whether to accept or reject.

Key concepts

Hypothesis Testing

Hypothesis testing is an essential statistical tool used to decide between two competing hypotheses based on sample data. In this exercise, we examine two statistical hypotheses, known as the null hypothesis \(H_0\) and the alternative hypothesis \(H_1\).

• The null hypothesis \(H_0\) assumes that data is uniformly distributed between \(0\) and \(\theta\).
• The alternative hypothesis \(H_1\) suggests that the data follows an exponential distribution with rate \(\frac{1}{\theta}\).

The goal is to determine which hypothesis is more likely given the sample data. A common approach is the likelihood ratio test, which compares the likelihoods of the data under both hypotheses.
The test statistic, the likelihood ratio \(\Lambda\), helps us make this decision. A high or low value (based on a threshold) indicates which hypothesis is more plausible, though setting this threshold often depends on the context and desired confidence level.

Exponential Distribution

The exponential distribution is a continuous probability distribution commonly used to model the time between events in a process that occur continuously and independently.
It is characterized by a single parameter \(\theta\), which is both the mean and standard deviation of the distribution. The probability density function (PDF) is given by:
\[ f(x ; \theta) = \frac{1}{\theta} e^{-x / \theta} \quad \text{for} \; x > 0 \]
This distribution is widely used in scenarios like modeling the lifespan of electronic components or time until a radioactive particle decays.
Its memoryless property is particularly interesting, meaning the future probability of an event occurring does not depend on past events. In our exercise, \(H_1\) uses this exponential distribution to describe data that can occur over an infinite range, providing a contrast to \(H_0\) and aiding in hypothesis testing.

Uniform Distribution

The uniform distribution, also known as the rectangular distribution, is perhaps the simplest continuous probability distribution. It describes a scenario where every outcome in a finite range is equally likely. For this exercise, \(H_0\) represents the uniform distribution with a PDF defined as:
\[ f(x ; \theta) = \frac{1}{\theta} \quad \text{for} \; 0 < x < \theta \]
It is zero elsewhere, indicating the data can only take on values within \(0\) and \(\theta\).
When using the uniform distribution in hypothesis testing, the likelihood function under \(H_0\) is computed as the product of densities at the observed data points. This distribution is applied in scenarios like generating random numbers where each number in a given range should have an equal chance of being chosen. The simplicity and bounded nature of the uniform distribution make it a useful baseline hypothesis when contrasting with more complex alternatives, such as the exponential distribution in this case.