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Show that the gamma distribution is a conjugate prior for the exponential distribution. Suppose that the waiting time in a queue is modeled as an exponential random variable with unknown parameter \(\lambda,\) and that the average time to serve a random sample of 20 customers is 5.1 minutes. A gamma distribution is used as a prior. Consider two cases: (1) the mean of the gamma is 0.5 and the standard deviation is \(1,\) and (2) the mean is 10 and the standard deviation is \(20 .\) Plot the two posterior distributions and compare them. Find the two posterior means and compare them. Explain the differences.

Short Answer

Expert verified
Scenario 1's posterior mean is 0.8168, while Scenario 2's is 1.4286, driven by differing prior beliefs.

Step by step solution

01

Understand the Problem

We need to demonstrate that the gamma distribution is a conjugate prior for the exponential distribution when modeling the parameter \( \lambda \). We have an average serving time from a sample and two scenarios with different gamma priors to consider.
02

Definitions and Prior Setup

The exponential distribution's parameter \( \lambda \) represents the rate, and our prior for \( \lambda \) is a gamma distribution. For a gamma distribution, the mean \( \mu \) is given by \( \mu = \frac{\alpha}{\beta} \), and variance \( \sigma^2 = \frac{\alpha}{\beta^2} \).
03

Solve for Prior Parameters

For Scenario 1, the gamma prior mean \( 0.5 \) gives us \( \alpha = 0.5\beta \). Given \( \sigma = 1 \), \( \sigma^2 = 1 \) gives \( \alpha = \beta \). Hence, \( \beta = 1 \) and \( \alpha = 0.5 \). For Scenario 2, mean \( 10 \) (\( \alpha = 10\beta \)) and \( \sigma = 20 \) (\( \alpha = 0.025\beta^2 \)), solve for \( \alpha = 0.025 \times 400 = 10 \) and \( \beta = 1 \).
04

Posterior Distribution Derivation

For an exponential likelihood with sample size \( n \, = \, 20 \) and average time \( \bar{x} = 5.1 \), the likelihood is based on \( n \bar{x} \). Their conjugate is gamma prior: posterior \( \sim \text{Gamma}(\alpha + n, \beta + n\bar{x}) \). For both scenarios: Scenario 1 gives posterior parameters \( \alpha = 20.5 \) and \( \beta = 25.1 \). Scenario 2: \( \alpha = 30 \), and \( \beta = 21 \).
05

Calculate Posterior Means

From \( \text{Gamma}(\alpha, \beta) \), the posterior mean is \( \frac{\alpha}{\beta} \). For Scenario 1, mean = \( \frac{20.5}{25.1} \approx 0.8168 \). For Scenario 2, mean = \( \frac{30}{21} \approx 1.4286 \).
06

Comparison and Explanation

Scenario 1 yields a lower mean (0.8168) because the prior was more concentrated around the lower value. Scenario 2's mean (1.4286) is higher, reflecting a prior belief of a higher rate. The variation in posterior means indicates how the prior influences post-service rate estimation.
07

Plot and Visualize Posterior Distributions

Plot the gamma distributions for both scenarios using the derived parameters; visually compare each. Notice how prior beliefs shape the posteriors, reflecting in different shapes and peaks for each scenario.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gamma Distribution
The gamma distribution is a continuous probability distribution often used in Bayesian statistics as a prior distribution, particularly when expressing uncertainty about a rate parameter. It is defined by two parameters: \( \alpha \) (shape) and \( \beta \) (rate). These parameters give us useful properties:
  • The mean of the gamma distribution is \( \frac{\alpha}{\beta} \).
  • The variance is \( \frac{\alpha}{\beta^2} \).
One reason why the gamma distribution is favored as a prior is its role as a conjugate prior for the exponential distribution. A conjugate prior means that the resulting posterior distribution has the same functional form as the prior, making calculations simpler when updating beliefs with new data. This property is particularly valuable as it allows for efficient analytical solutions in Bayesian inference. By using this distribution as a prior, you can easily update your understanding of a rate parameter as new data is observed.
Exponential Distribution
The exponential distribution is widely used to model the time between independent events that happen at a constant average rate. This makes it a natural choice for modeling scenarios like the time between customer arrivals or the duration of service. The exponential distribution is defined by a single parameter \( \lambda \), which is the rate at which events occur. If you know that the average waiting time for an event is 5.1 minutes, the average rate \( \lambda \) would be \( \frac{1}{5.1} \) events per minute. The exponential distribution is memoryless, meaning the probability of an event occurring in the future is independent of any past events. This property makes it suitable for systems where past events provide no additional information about future ones, enhancing its utility in queueing theory and reliability analysis. In this context, it pairs well with the gamma distribution in Bayesian analysis, as the update process becomes straightforward when data about these independent events is collected.
Bayesian Statistics
Bayesian statistics is a method of statistical inference where Bayes' theorem is used to update the probability of a hypothesis as more evidence becomes available. It emphasizes the use of prior distributions, which represent your initial beliefs about parameters before observing any data. As new data is incorporated, these beliefs are updated to form a posterior distribution, which reflects our updated understanding. Here are the basics of how Bayesian statistics works:
  • **Prior:** Your initial belief about a parameter, expressed as a probability distribution.
  • **Likelihood:** How probable your observed data is, given your hypotheses or model parameters.
  • **Posterior:** The updated belief after considering the observed data, which is proportional to the product of the likelihood and the prior.
Bayesian statistics allows for a flexible, coherent framework that incorporates uncertainty and prior knowledge directly into the analysis. This approach is especially beneficial when data is scarce or expensive to obtain, as it provides a powerful way to harness prior information.
Posterior Distribution
The posterior distribution is the core of Bayesian inference. It represents the updated beliefs about a parameter after including new evidence through observed data. Mathematically, the posterior distribution is derived using Bayes' theorem:\[P(\theta | X) = \frac{P(X | \theta) \times P(\theta)}{P(X)}\]Where \( \theta \) represents the parameter of interest and \( X \) is the observed data. For the given problem, after observing the average service time, the posterior distribution of the rate parameter \( \lambda \) is calculated by combining the gamma prior with the exponential likelihood. The resulting posterior remains a gamma distribution because of the conjugate prior relationship. The posterior parameters are adjusted by adding the data influence: the prior sample information and the observed sample sum. This allows us effectively to incorporate both our prior beliefs and new data to make more informed decisions about the parameter. When we plot the posterior distributions for different scenarios, we can observe how initial assumptions influence the updated probability distributions, seen in their shapes and locations of peaks.

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Most popular questions from this chapter

Let \(X_{1}, \ldots, X_{n}\) be an i.i.d. sample from an exponential distribution with the density function $$f(x | \tau)=\frac{1}{\tau} e^{-x / \tau}, \quad 0 \leq x<\infty$$ a. Find the mle of \(\tau.\) b. What is the exact sampling distribution of the mle? c. Use the central limit theorem to find a normal approximation to the sampling distribution. d. Show that the mle is unbiased, and find its exact variance. (Hint: The sum of the \(X_{i}\) follows a gamma distribution.) e. Is there any other unbiased estimate with smaller variance? f. Find the form of an approximate confidence interval for \(\tau\) g. Find the form of an exact confidence interval for \(\tau\)

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. \(N\left(\mu_{0}, \sigma_{0}^{2}\right)\) and \(\mu\) and \(\sigma^{2}\) are estimated by the method of maximum likelihood, with resulting estimates \(\hat{\mu}\) and \(\hat{\sigma}^{2} .\) Suppose the bootstrap is used to estimate the sampling distribution of \(\hat{\mu}\) a. Explain why the bootstrap estimate of the distribution of \(\hat{\mu}\) is \(N\left(\hat{\mu}, \frac{\hat{\sigma}^{2}}{n}\right)\). b. Explain why the bootstrap estimate of the distribution of \(\hat{\mu}-\mu_{0}\) is \(N\left(0, \frac{\hat{\sigma}^{2}}{n}\right)\). c. According to the result of the previous part, what is the form of the bootstrap confidence interval for \(\mu,\) and how does it compare to the exact confidence interval based on the \(t\) distribution?

In Example A of Section \(8.4,\) we used knowledge of the exact form of the sampling distribution of \(\hat{\lambda}\) to estimate its standard error by $$s_{\hat{\lambda}}=\sqrt{\frac{\hat{\lambda}}{n}}$$ This was arrived at by realizing that \(\sum X_{i}\) follows a Poisson distribution with parameter \(n \lambda_{0} .\) Now suppose we hadn't realized this but had used the bootstrap, letting the computer do our work for us by generating \(B\) samples of size \(n=23\) of Poisson random variables with parameter \(\lambda=24.9,\) forming the mle of \(\lambda\) from each sample, and then finally computing the standard deviation of the resulting collection of estimates and taking this as an estimate of the standard error of \(\hat{\lambda}\) Argue that as \(B \rightarrow \infty,\) the standard error estimated in this way will tend to \(s_{\hat{\lambda}}\).

The Poisson distribution has been used by traffic engineers as a model for light traffic, based on the rationale that if the rate is approximately constant and the traffic is light (so the individual cars move independently of each other), the distribution of counts of cars in a given time interval or space area should be nearly Poisson (Gerlough and Schuhl 1955 ). The following table shows the number of right turns during 300 3-min intervals at a specific intersection. Fit a Poisson distribution. Comment on the fit by comparing observed and expected counts. It is useful to know that the 300 intervals were distributed over various hours of the day and various days of the week. $$\begin{array}{cc} \hline n & \text { Frequency } \\ \hline 0 & 14 \\ 1 & 30 \\ 2 & 36 \\ 3 & 68 \\ 4 & 43 \\ 5 & 43 \\ 6 & 30 \\ 7 & 14 \\ 8 & 10 \\ 9 & 6 \\ 10 & 4 \\ 11 & 1 \\ 12 & 1 \\ 13+ & 0 \\ \hline \end{array}$$

Suppose that a random sample of size 20 is taken from a normal distribution with unknown mean and known variance equal to \(1,\) and the mean is found to be \(\bar{x}=10 .\) A normal distribution was used as the prior for the mean, and it was found that the posterior mean was 15 and the posterior standard deviation was 0.1. What were the mean and standard deviation of the prior?

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