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Consider the Bayes model $$ \begin{aligned} X_{i} \mid \theta & \sim \operatorname{iid} \Gamma\left(1, \frac{1}{\theta}\right) \\ \Theta \mid \beta & \sim \Gamma(2, \beta) \end{aligned} $$ By performing the following steps, obtain the empirical Bayes estimate of \(\theta\). (a) Obtain the likelihood function $$ m(\mathbf{x} \mid \beta)=\int_{0}^{\infty} f(\mathbf{x} \mid \theta) h(\theta \mid \beta) d \theta $$ (b) Obtain the mle \(\widehat{\beta}\) of \(\beta\) for the likelihood \(m(\mathbf{x} \mid \beta)\). (c) Show that the posterior distribution of \(\Theta\) given \(\mathbf{x}\) and \(\widehat{\beta}\) is a gamma distribution. (d) Assuming squared-error loss, obtain the empirical Bayes estimator.

Short Answer

Expert verified
The empirical Bayes estimate of \(\theta\) can be obtained through these steps. First calculate the likelihood function. Then calculate the maximum likelihood estimate (MLE) of \(\beta\). Next step is to show that the posterior distribution of \(\Theta\) given \(\mathbf{x}\) and \(\widehat{\beta}\) follows a gamma distribution under given condition of Bayesian model. Finally, by considering the squared-error loss, the desired empirical Bayes estimator can be obtained from the expectation of the posterior distribution.

Step by step solution

01

Obtain the Likelihood Function

The likelihood function \(m(\mathbf{x} \mid \beta)\) can be found by integrating the joint PDF of \(f(\mathbf{x} \mid \theta) h(\theta \mid \beta)\). With given distribution functions, the integration from 0 to \(\infty\) gives the likelihood function.
02

Obtain MLE of \(\beta\)

To obtain the maximum likelihood estimate (MLE) \(\widehat{\beta}\) of \(\beta\), take the derivative of \(m(\mathbf{x} \mid \beta)\) and set it equal to zero, and then solve for \(\beta\). From the result, the estimate \(\widehat{\beta}\) will be found.
03

Show Posterior Distribution of \(\Theta\) as Gamma Distribution

The posterior distribution of \(\Theta\) given \(\mathbf{x}\) and \(\widehat{\beta}\) can be obtained according to Bayes' theorem. It combines the likelihood function and the prior distribution. For the given conditions, after normalization, it will be confirmed that the posterior distribution follows a gamma distribution.
04

Obtain Empirical Bayes Estimator

Assuming squared-error loss, according to Bayes rule, the empirical Bayes estimator is the expected value of the posterior distribution derived in step 3, which is the product of the shape and scale parameter of the gamma distribution obtained from the posterior.

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Most popular questions from this chapter

Consider the Bayes model \(X_{i} \mid \theta, i=1,2, \ldots, n \sim\) iid with distribution \(b(1, \theta), 0<\theta<1\) (a) Obtain the Jeffreys' prior for this model. (b) Assume squared-error loss and obtain the Bayes estimate of \(\theta\).

In Example 11.1.2, let \(n=30, \alpha=10\), and \(\beta=5\), so that \(\delta(y)=(10+y) / 45\) is the Bayes estimate of \(\theta\). (a) If \(Y\) has a binomial distribution \(b(30, \theta)\), compute the risk \(E\left\\{[\theta-\delta(Y)]^{2}\right\\}\). (b) Find values of \(\theta\) for which the risk of part (a) is less than \(\theta(1-\theta) / 30\), the risk associated with the maximum likelihood estimator \(Y / n\) of \(\theta\).

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?

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 \(\theta\).

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