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. Let \(Y_{4}\) be the largest order statistic of a sanple of size \(n=4\) from a distribution with uniform pdf \(f(x ; \theta)=1 / \theta, 0

Short Answer

Expert verified
The Bayesian estimator for \(\theta\) given \(Y_4\) and the loss function \(|\delta(Y_4) - \theta|\) is \(\delta(Y_4) = 2Y_{4}\).

Step by step solution

01

Form the posterior distribution

Bayesian analysis starts with a prior distribution and then updates this with the likelihood function to form the posterior distribution. Our prior for this problem is \(g(\theta)=2 / \theta^{3}, 1<\theta<\infty\), and the likelihood function is \(f(x ; \theta)=1 / \theta, 0<x<\theta\). Therefore, the posterior distribution is proportional to the product of the prior and the likelihood: \(p(\theta|Y_{4}) \propto g(\theta)f(Y_{4}; \theta) = (2 / \theta^{3})*(1 / \theta) = 2 / \theta^{4}\).
02

Normalize the posterior distribution

For the posterior distribution to be a true probability distribution, it must integrate (sum) to 1. We need to find the normalization constant, say C, such that \(\int_{1}^{\infty}C * p(\theta|Y_{4})d\theta=1\). Computing this integral we get \(C = 1 / Y_{4}\). Thus, the normalized posterior distribution is \(p(\theta|Y_{4}) = 2 / (\theta^{4} * Y_{4})\).
03

Derive the Bayesian estimator

In Bayesian analysis, the estimator is the value that minimizes the expected loss. In this case, the loss function is absolute error loss, given by \(L(\theta, \delta) = |\delta - \theta|\). By taking the derivative of this loss function with respect to \(\delta\) and setting it to zero, we find that the Bayesian estimator is the median of the posterior distribution. It turns out the Bayesian estimator for this problem, given this loss function, is \(\delta(Y_{4}) = 2Y_{4}\).

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Key Concepts

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

Order Statistic
The largest order statistic from a sample refers to the maximum value within that sample. In our exercise, we consider a sample of size 4 from a uniform distribution. This order statistic is denoted as \(Y_4\), representing the largest observed value among four draws from the same distribution.

Order statistics are crucial in statistics as they help identify extremes like minimum, maximum, and various percentiles of a dataset. In a Bayesian context, they provide valuable information for estimating unknown parameters, such as \(\theta\) in our case. Knowing \(Y_4\) allows us to better understand which values of \(\theta\) are more plausible, given the observed data.
Uniform Distribution
A uniform distribution is a probability distribution where every outcome in the range is equally likely. For this exercise, our uniform distribution is defined such that the probability density function (pdf) is \(f(x; \theta) = 1/\theta\) for \(0 < x < \theta\), and zero elsewhere. This means any value between 0 and \(\theta\) is just as likely as any other value within this range.

Uniform distributions are simple yet powerful tools in statistics, representing randomness without favoring any outcome within the specified bounds. They serve as a foundation for more complex probabilistic models and analyses, such as the one we're dealing with in forming the posterior distribution.
Posterior Distribution
The posterior distribution reflects our updated knowledge about a parameter after considering new evidence or data. This update is based on Bayes' theorem, a cornerstone of Bayesian statistics. Our problem involves determining the posterior distribution of \(\theta\) given \(Y_4\).
  • The prior distribution here is \(g(\theta) = 2/\theta^3\) for \(1 < \theta < \infty\).
  • The likelihood function is based on the uniform distribution, given as \(f(Y_4; \theta) = 1/\theta\).
After forming the product of the prior and likelihood, the posterior becomes \(p(\theta|Y_4) \propto 2/\theta^4\). Normalizing this gives the posterior distribution as \(p(\theta|Y_4) = 2/(\theta^4 \cdot Y_4)\), ensuring the distribution sums to 1 over all possible values of \(\theta\). This allows us to learn the Bayesian estimator by focusing on the area where the posterior distribution is centered.
Absolute Error Loss
Absolute error loss is a simple yet effective way to measure the accuracy of an estimator. This loss function is defined as \(L(\theta, \delta) = |\delta - \theta|\), highlighting the absolute difference between our estimate \(\delta\) and the true parameter \(\theta\).

In Bayesian estimation, this loss function helps determine the best estimate by seeking the value of \(\delta\) that minimizes the expected error. Often, this leads us to choosing the median of the posterior distribution as the most accurate estimator under absolute error loss. This aligns with our exercise, where the Bayesian estimator for \(\theta\) given \(Y_4\) is \(\delta(Y_4) = 2Y_4\). This approach ensures that we do not heavily penalize outlying errors and instead find a central, balanced estimate.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a distribution that is \(N\left(\theta, \sigma^{2}\right)\), where \(-\infty<\theta<\infty\) and \(\sigma^{2}\) is a given positive number. Let \(Y=\bar{X}\) denote the mean of the random sample. Take the loss function to be \(\mathcal{L}[\theta, \delta(y)]=|\theta-\delta(y)|\). If \(\theta\) is an observed value of the random variable \(\Theta\), that is, \(N\left(\mu, \tau^{2}\right)\), where \(\tau^{2}>0\) and \(\mu\) are known numbers, find the Bayes' solution \(\delta(y)\) for a point estimate \(\theta\).

. Consider the following mixed discrete-continuous pdf for a random vector \((X, Y)\), (discussed in Casella and George, 1992): $$ f(x, y) \propto\left\\{\begin{array}{ll} \left(\begin{array}{l} n \\ x \end{array}\right) y^{x+\alpha-1}(1-y)^{n-x+\beta-1} & x=0,1, \ldots, n, 00\) and \(\beta>0\). (a) Show that this function is indeed a joint, mixed discrete continuous pdf by finding the proper constant of proportionality. (b) Determine the conditional pdfs \(f(x \mid y)\) and \(f(y \mid x)\). (c) Write the Gibbs sampler algorithm to generate random samples on \(X\) and \(Y\). (d) Determine the marginal distributions of \(X\) and \(Y\).

Let \(Y\) have a binomial distribution in which \(n=20\) and \(p=\theta\). The prior probabilities on \(\theta\) are \(P(\theta=0.3)=2 / 3\) and \(P(\theta=0.5)=1 / 3 .\) If \(y=9\), what are the posterior probabilities for \(\theta=0.3\) and \(\theta=0.5 ?\)

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample of size \(n=10\) from a gamma distribution with \(\alpha=3\) and \(\beta=1 / \theta\). Suppose we believe that \(\theta\) has a gamma distribution with \(\alpha=10\) and \(\beta=2\) (a) Find the posterior distribution of \(\theta\). (b) If the observed \(\bar{x}=18.2\), what is the Bayes point estimate associated with square error loss function. (c) What is the Bayes point estimate using the mode of the posterior distribution? (d) Comment on an HDR interval estimate for \(\theta\). Would it be easier to find one having equal tail probabilities? Hint: Can the posterior distribution be related to a chi-square distribution?

The following amounts are bet on horses \(A, B, C, D, E\) to win. \begin{tabular}{cr} Horse & Amount \\ \hline\(A\) & \(\$ 600,000\) \\ \(B\) & \(\$ 200,000\) \\ \(C\) & \(\$ 100,000\) \\ \(D\) & \(\$ 75,000\) \\ \(E\) & \(\$ 25,000\) \\ \hline Total & \(\$ 1,000,000\) \end{tabular}Suppose the track wants to take \(20 \%\) off the top, namely \(\$ 200,000\). Determine the payoff for winning with a two dollar bet on each of the five horses. (In this exercise, we do not concern ourselves with "place" and "show.") Hint: Figure out what would be a fair payoff so that the track does not take any money, (that is, the track's take is zero), and then compute \(80 \%\) of those payoffs.

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