/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 For \(i=1,2, \ldots, k\), let \(... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 28

For \(i=1,2, \ldots, k\), let \(X_{i} \sim f_{i}\left(x \mid \theta_{i}\right)\) and suppose that \(\delta_{i}^{*}\left(x_{i}\right)\) is a unique Bayes estimator of \(\theta_{i}\) under the loss \(L_{i}\left(\theta_{i}, \delta\right)\), where \(L_{i}\) satisfies \(L_{i}(a, a)=0\) and \(L_{i}\left(a, a^{\prime}\right)>\), 0, \(a \neq a^{\prime}\). Suppose that for some \(j, 1 \leq j \leq k\), there is a value \(\theta^{*}\) such that if \(\theta_{j}=\theta^{*}\), (i) \(X_{j}=x^{*}\) with probability 1 , (ii) \(\delta_{j}^{*}\left(x^{*}\right)=\theta^{*}\). Show that \(\left(\delta_{1}^{*}\left(x_{1}\right), \delta_{2}^{*}\left(x_{2}\right), \ldots, \delta_{k}^{*}\left(x_{k}\right)\right)\) is admissible for \(\left(\theta_{1}, \theta_{2}, \cdots, \theta_{k}\right)\) under the loss \(\sum_{i} L_{i}\left(\theta_{i}, \delta\right)\); that is, there is no Stein effect.

Short Answer

Expert verified
The estimator is admissible due to its minimal expected loss property and perfect prediction for special cases.

Step by step solution

01

Understand Parameters and Conditions

We start with parameters \(i = 1, 2, \ldots, k\), where each \(X_i\) is distributed according to \(f_i(x | \theta_i)\). The \(\delta_i^*(x_i)\) is a unique Bayes estimator for \(\theta_i\) under the loss function \(L_i(\theta_i, \delta)\). The conditions given include \(L_i(a, a) = 0\) and \(L_i(a, a') > 0\) for \(a eq a'\), indicating that the loss is minimized when the estimate equals the true parameter.
02

Special Case for Parameters

For some \(j\), if \(\theta_j = \theta^*\), it's given that \(X_j = x^*\) with probability 1 and \(\delta_j^*(x^*) = \theta^*\). This means that whenever \(\theta_j\) takes the value \(\theta^*\), the Bayesian estimator perfectly predicts \(\theta^*\) due to the certainty condition on \(X_j\).
03

Loss Structure and Admissibility

We need to prove that \((\delta_1^*(x_1), \delta_2^*(x_2), \ldots, \delta_k^*(x_k))\) is an admissible decision rule under the combined loss \(\sum_i L_i(\theta_i, \delta)\). Admissibility means there is no alternative estimator providing uniformly lower expected loss.
04

Consider Bayes Estimator Properties

A Bayes estimator \(\delta_i^*(x_i)\) is by definition minimizing the expected posterior loss for each parameter \(\theta_i\). Since \(\delta_i^*(x_i)\) is given as unique and optimal under \(L_i\), and \(X_j = x^*\) gives \(\theta_j = \delta_j^*(x^*) = \theta^*\), this condition implies optimal performance for the estimator under the special case defined.
05

Conclude on Admissibility

Since each \(\delta_i^*(x_i)\) individually minimizes the posterior expected loss, and \(X_j = x^*\) leads to a perfect prediction for \(\theta_j\), any alternative estimator would not perform better across all \(i\). Additionally, aggregation of minimal losses \(\sum_i L_i(\theta_i, \delta)\) does not allow for further reduction, confirming admissibility.

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Key Concepts

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

Admissibility
In Bayesian decision theory, the concept of admissibility is central when determining whether an estimator is considered optimal or justifiable. An estimator, like \(\delta_i^*(x_i)\), is deemed admissible if there is no other estimator that performs better for every possible outcome. Here, 'better' specifically refers to minimizing the expected loss.

A decision rule becomes inadmissible if there exists an alternative rule that delivers a smaller or equal expected loss for all parameter values and has a strictly smaller expected loss for at least one parameter value. In simple terms, it's like saying you have the best solution unless someone finds another option that is equal in all ways but better in at least one.

To understand why the given Bayesian estimator is admissible under the loss function \(\sum_i L_i(\theta_i, \delta)\), consider the uniqueness of how it performs for every parameter \(\theta_i\), in line with the special case given for \(\theta_j = \theta^*\). This specificity implies that no alternative estimator can consistently outperform this decision rule without encountering greater loss elsewhere.
Loss Functions
Loss functions are essential in evaluating the performance of an estimator in statistical decision theory. They assign a numerical value, often a penalty, to the error or discrepancy between the estimated values and the true values. In the given problem, each \(L_i(\theta_i, \delta)\) denotes the loss associated with the decision rule \((\delta_i(x_i))\) for estimating \(\theta_i\).

The structure of the loss function here is such that the loss is zero if the estimate equals the true parameter (i.e., \(L_i(a, a) = 0\)) and positive if they differ (i.e., \(L_i(a, a') > 0\) for \(a eq a'\)). This reflects a fundamental property of a good loss function: it rewards accuracy and penalizes deviation.

In practice, choosing an appropriate loss function can significantly affect the behavior of an estimator. For Bayesian estimation, a common choice is the squared error loss function, which emphasizes precision. However, other forms like absolute error or asymmetric loss functions are used depending on the context and desired properties of the estimation process.
Bayes Estimator
A Bayes estimator is a statistical estimator that minimizes the expected value of a loss function under the Bayesian probability framework, considering prior beliefs. Throughout this exercise, \(\delta_i^*(x_i)\) represents such an estimator for each parameter \(\theta_i\). Its goal is to achieve the smallest possible average loss or risk, incorporating both data evidence and prior information.

The uniqueness of a Bayes estimator plays a vital role in ensuring that no other estimator can achieve better performance in minimizing expected loss 鈥� a property we discussed under admissibility. This uniqueness guarantees that the estimator not only aligns with the true parameter under prior knowledge but also effectively handles the data

When considering characteristics like the special condition \(X_j = x^*\) with probability 1 and \(\delta_j^*(x^*) = \theta^*\), the Bayes estimator is shown to perfectly anticipate this scenario due to its optimal usage of information. This alone underscores its efficacy and suitability in complex probabilistic models.

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Most popular questions from this chapter

For the most part, the risk function of a Stein estimator increases as \(|\theta|\) moves away from zero (if zero is the shrinkage target). To guarantee that the risk function is monotone increasing in \(|\theta|\) (that is, that there are no "dips" in the risk as in Berger's 1976 a tail minimax estimators) requires a somewhat stronger assumption on the estimator (Casella 1990). Let \(X \sim N_{r}(\theta, I)\) and \(L(\theta, \delta)=|\theta-\delta|^{2}\), and consider the Stein estimator $$ \delta(\mathbf{x})=\left(1-c\left(|\mathbf{x}|^{2}\right) \frac{(r-2)}{|\mathbf{x}|^{2}}\right) \mathbf{x} $$ (a) Show that if \(0 \leq c(\cdot) \leq 2\) and \(c(-)\) is concave and twice differentiable, then \(\delta(\mathbf{x})\) is minimax. [Hint: Problem 1.7.7.] (b) Under the conditions in part (a), the risk function of \(\delta(\mathbf{x})\) is nondecreasing in \(|\theta|\). [Hint: The conditions on \(c(\cdot)\), together with the identity $$ (d / d \lambda) E_{\dot{\lambda}}\left[h\left(\chi_{p}^{2}(\lambda)\right)\right]=E_{\lambda}\left\\{\left[\partial / \partial \chi_{p+2}^{2}(\lambda)\right] h\left(\chi_{p+2}^{2}(\lambda)\right)\right\\} $$ where \(\chi_{\rho}^{2}(\lambda)\) is a noncentral \(\chi^{2}\) random variable with \(p\) degrees of freedom and noncentrality parameter \(\lambda\), can be used to show that \(\left(\partial / \partial|\theta|^{2}\right) R(\theta, \delta)>0\).]

Prove the following (equivalent) version of Blyth's Method (Theorem 7.13). Theorem \(8.7\) Suppose that the parameter space \(\Omega \in \Re^{r}\) is open, and estimators with continuous risks are a complete class. Let \(\delta\) be an estimator with a contimous risk function, and let \(\left[\pi_{e}\right\\}\) be a sequence of (possibly improper) prior measures such that (i) \(r\left(\pi_{n}, \delta\right)<\infty\) for all \(n\), (ii) for any nonempty open set \(\Theta_{0} \in \Omega\), $$ \frac{r\left(\pi_{n}, \delta\right)-r\left(\pi_{N}, \delta^{\pi_{x}}\right)}{\int_{\Theta_{0}} \pi_{n}(\theta) d \theta} \rightarrow 0 \quad \text { as } n \rightarrow \infty $$ Then, \(\delta\) is an admissible estimator.

Show that an estimator \([1 /(1+\lambda)+\varepsilon] X\) of \(E_{\theta}(X)\) is inadmissible (with squared error loss) under each of the following conditions: (a) if \(\operatorname{var}_{\theta}(X) / E_{\theta}^{2}(X)>\lambda>0\) and \(\varepsilon>0\) (b) if \(\operatorname{var}_{\theta}(X) / E_{\theta}^{2}(X)<\lambda\) and \(\varepsilon<0\) [ Hint: (a) Differentiate the risk function of the estimator with respect to \(\varepsilon\) to show that it decreases as \(\varepsilon\) decreases (Karlin 1958).]

Let the distribution of \(X\) depend on parameters \(\theta\) and \(\vartheta\), let the risk function of an estimator \(\delta=\delta(x)\) of \(\theta\) be \(R(\theta, \vartheta ; \delta)\), and let \(r(\theta, \delta)=\int R(\theta, \vartheta ; \delta) d P(\vartheta)\) for some distribution \(P\). If \(\delta_{0}\) minimizes \(\sup _{\theta} r(\theta, \delta)\) and satisfies sup \(_{\theta} r\left(\theta, \delta_{0}\right)=\sup _{\theta, b} R\left(\theta, \vartheta ; \delta_{0}\right)\), show that \(\delta_{0}\) minimizes sup \(_{\theta, b} R(\theta, \vartheta ; \delta)\)

For \(X \mid \theta \sim N_{r}(\theta, I)\), George (1986a, 1986b) looked at multiple shrinkage estimators, those that can shrink to a number of different targets. Suppose that \(\theta \sim \pi(\theta)=\) \(\sum_{j=1}^{k} \omega_{i} \pi_{i}(\theta)\), where the \(\omega_{i}\) are known positive weights, \(\sum \omega_{i}=1\). (a) Show that the Bayes estimator against \(\pi(\theta)\), under squared error loss, is given by \(\delta^{*}(\mathbf{x})=\mathbf{x}+\nabla \log m^{*}(\mathbf{x})\) where \(m^{*}(\mathbf{x})=\sum_{j=1}^{k} \omega_{j} m_{j}(\mathbf{x})\) and $$ m_{i}(\mathbf{x})=\int_{\Omega} \frac{1}{(2 \pi)^{p / 2}} e^{-(1 / 2)|\mathbf{x}-\boldsymbol{\theta}|^{2}} \pi_{i}(\theta) d \theta $$ (b) Clearly, \(\delta^{*}\) is minimax if \(m^{*}(\mathbf{x})\) is superharmonic. Show that \(\delta^{*}(\mathbf{x})\) is minimax if either (i) \(m_{i}(\mathbf{x})\) is superharmonic, \(i=1, \ldots, k\), or (ii) \(\pi_{i}(\theta)\) is superharmonic, \(i=1, \ldots, k_{.}[\)Hint: Problem 1.7.16] (c) The real advantage of \(\delta^{*}\) occurs when the components specify different targets. For \(\rho_{j}=\omega_{j} m_{j}(x) / m^{*}(x)\), let \(\delta^{*}(\mathbf{x})=\sum_{j=1}^{k} \rho_{j} \delta_{j}^{+}(\mathbf{x})\) where $$ \delta_{j}^{+}(\mathbf{x})=\mu_{j}+\left(1-\frac{r-2}{\left|\mathbf{x}-\mu_{j}\right|^{2}}\right)^{+}\left(\mathbf{x}-\mu_{j}\right) $$ and the \(\mu_{j}\) 's are target vectors. Show that \(\delta^{*}(\mathbf{x})\) is minimax. [Hint: Problem 5.19] [George (1986a, 1986b) investigated many types of multiple targets, including multiple points, subspaces, and clusters and subvectors. The subvector problem was also considered by Berger and Dey (1983a, 1983b). Multiple shrinkage estimators were also investigated by \(\mathrm{Ki}\) and Tsui (1990) and Withers (1991).]
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