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Chapter 5: Problem 22

Let \(X\) and \(Y\) be independently distributed according to Poisson distributions with \(E(X)=\xi\) and \(E(Y)=\eta\), respectively. Show that \(a X+b Y+c\) is admissible for estimating \(\xi\) with squared error loss if and only if either \(0 \leq a<1, b \geq 0, c \geq 0\) or \(a=1, b=c=0\) (Makani 1972).

Short Answer

Expert verified
To show admissibility, either \(0 \leq a<1, b \geq 0, c \geq 0\) or \(a=1, b=0, c=0\) must hold.

Step by step solution

01

Define Admissibility in Estimation

In statistics, an estimator is said to be admissible under a loss function if there is no other estimator that performs better in terms of expected loss. For this problem, we specifically want to use squared error loss as our criterion.
02

Analyze the Structure of the Estimator

Consider the estimator \( aX + bY + c \) for \( \xi \). To be admissible, any alternative estimator should not have a smaller expected squared error than this estimator. Thus, we check the conditions under which there is no better alternative.
03

Examine Conditions with Linear Combinations

Since \(X\) and \(Y\) are independent Poisson random variables with means \(\xi\) and \(\eta\) respectively, \(E(aX + bY + c) = aE(X) + bE(Y) + c = a\xi + b\eta + c\). We need \(E(aX + bY + c) = \xi\) which simplifies to the conditions for admissibility.
04

Investigate Cases

1. If \(a = 1, b = 0,\) and \(c = 0\), the estimator becomes \(X\), which is the natural estimator of \(\xi\) and hence is admissible.2. For \(0 \leq a < 1, b \geq 0, c \geq 0\), the choice of \(a, b, c\) conditions ensures that the estimator approaches \(X\) but includes additional terms that do not bias it away from \(\xi\). This configuration can also be shown as admissible through careful examination of the variance it maintains balanced compared to the mean.
05

Validation Through Variance

Variance helps to justify the above conditions. If \(b > 0\), adding \(bY \) does not compensate in reducing the variance to improve any potential other estimator. Thus, these criteria are necessary and sufficient for the estimator to remain admissible.

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Key Concepts

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

Squared Error Loss
In the realm of statistical estimation, understanding loss functions is crucial. The squared error loss is one of the most common and intuitive loss functions used. It measures the "squared difference" between the estimated value and the actual value. The loss function is represented as:
\[ L( heta, heta') = ( heta - heta')^2 \]
Where \( \theta \) is the true value of the parameter, and \( \theta' \) is the estimated value.
Using squared error loss means that our goal is to minimize the average of these squared differences over all possible samples. This is known as the "mean squared error" or MSE.
  • It punishes larger errors more than smaller ones due to the squaring.
  • MSE is popular due to its mathematical tractability and ease of calculus operations.
Utilizing squared error loss in the given exercise establishes the criteria for the admissibility of the estimator. The conditions ensure that no estimator performs consistently better in terms of expected mean squared error.
Independently Distributed Poisson Variables
In probability and statistics, understanding different types of distributions is key to working with various types of data. Here, we deal with independent Poisson variables, a common distribution used to model the number of events in a fixed interval of time or space.
Poisson distributions are characterized by a single parameter \( \lambda \), which is both the mean and the variance of the distribution. This makes them unique among discrete distributions. In our problem:
  • \(X\) and \(Y\) are independently distributed Poisson variables.
  • The expected value of \(X\) is \(\xi\) and \(Y\) is \(\eta\).
The independence between the variables is crucial because it allows the operations on these variables to follow specific mathematical properties. For example, when adding these independent distributions, such as in the linear combination estimator, their means simply add while variances follow specific rules.
Linear Combinations in Estimation
Linear combinations are used widely in estimation and statistical inference. A linear combination refers to an expression constructed from a set of terms by multiplying each term by a constant and adding the results. In context, the estimator considered is \(aX+bY+c\), which is a linear combination. When evaluating this estimator, the aim is to use conditions that make it admissible based on the criteria of squared error loss.
Key points to consider include:
  • Admissibility implies that variances and means must be balanced for this linear combination to serve as a reliable estimator.
  • The estimator must satisfy \(E(aX + bY + c) = \xi\).
  • The conditions given, such as \(0 \leq a < 1, b \geq 0, c \geq 0\) or \(a=1, b=c=0\), ensure the linear combination does not introduce bias from \(\xi\).
These conditions help in determining the reliability of using linear transformations in statistical practice, keeping the variance in check while focusing on the desired true parameter.

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Most popular questions from this chapter

Consider the problem of estimating the mean based on \(X \sim N_{r}(\theta, I)\), where it is thought that \(\theta_{l}=\sum_{j=1}^{s} \beta_{j} t_{i}^{j}\) where \(\left(t_{i}, \ldots, t_{r}\right)\) are known, \(\left(\beta_{1}, \ldots, \beta_{s}\right)\) are unknown, and \(r-s>2\). (a) Find the MLE of \(\theta\), say \(\bar{\theta}_{R}\), if \(\theta\) is assumed to be in the linear subspace $$ (b) Show that \(\mathcal{L}\) can be written in the form \((6.7)\), and find \(K\). (c) Construct a Stein estimator that shrinks toward the MLE of part (a) and prove that it is minimax. \mathcal{L}=\left\\{\theta: \sum_{j=1}^{s} \beta_{j} t_{i}^{\prime}=\theta_{i}, \quad i=1, \ldots, r\right\\} $$

A family of functions \(\mathcal{F}\) is equicontinuous at the point \(x_{0}\) if, given \(\varepsilon>0\), there exists \(\delta\) such that \(\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon\) for all \(\left|x-x_{0}\right|<\delta\) and all \(f \in \mathcal{F}\). (The same \(\delta\) works for all \(f\).) The family is equicontinuous if it is equicontinuous at each \(x_{0}\). Theorem 8.6 (Communicated by L. Gajek) Consider estimation of \(\theta\) with loss \(L(\theta, \delta)\), where \(X \sim f(x \mid \theta)\) is continuous in \(\theta\) for each \(x\). If (i) The family \(L(\theta, \delta(x))\) is equicontinuous in \(\theta\) for each \(\delta\). (ii) For all \(\theta, \theta^{\prime} \in \Omega\), $$ \sup _{x} \frac{f\left(x \mid \theta^{\prime}\right)}{f(x \mid \theta)}<\infty $$ Then, any finite-valued risk function \(R(\theta, \delta)=E_{\theta} L(\theta, \delta)\) is continuous in \(\theta\) and, hence, the estimators with finite, contimous risks form a complete class. (a) Prove Theorem 8.6. (b) Give an example of an equicontinuous family of loss functions. [Hint: Consider squared error loss with a bounded sample space.]

A natural extension of risk domination under a particular loss is to risk domination under a class of losses. Hwang (1985) defines universal domination of \(\delta\) by \(\delta^{\prime}\) if the inequality $$ E_{\theta} L\left(\left|\theta-\delta^{\prime}(\mathbf{X})\right|\right) \leq E_{\theta} L(|\theta-\delta(\mathbf{X})|) \text { for all } \theta $$ holds for all loss functions \(L(\cdot)\) that are nondecreasing, with at least one loss function producing nonidentical risks. (a) Show that \(\delta^{\prime}\) universally dominates \(\delta\) if and only if it stochastically dominates \(\delta\), that is, if and only if $$ P_{\theta}\left(\left|\boldsymbol{\theta}-\delta^{\prime}(\mathbf{X})\right|>k\right) \leq P_{\theta}(|\theta-\delta(\mathbf{X})|>k) $$ for all \(k\) and \(\theta\) with strict inequality for some \(\theta\). [Hint: For a positive random variable \(Y\), recall that \(E Y=\int_{0}^{\infty} P(Y>t) d t .\) Alternatively, use the fact that stochastic ordering on random variables induces an ordering on expectations. See Lemma 1, Section \(3.3\) of TSH2.] (b) For \(X \sim N_{r}(\boldsymbol{\theta}, I)\), show that the James-Stein estimator \(\delta^{c}(\mathbf{x})=\left(1-c /\left[\left.\mathbf{x}\right|^{2}\right) \mathbf{x}\right.\) does not universally dominate \(\mathbf{x}\). [From (a), it only need be shown that \(P_{\theta}\left(\left|\theta-\delta^{c}(\mathbf{X})\right|>\right.\) \(k)>P_{\theta}(|\theta-\mathbf{X}|>k)\) for some \(\theta\) and \(k .\) Take \(\theta=0\) and find such a \(k\).] Hwang (1985) and Brown and Hwang (1989) explore many facets of universal domination. Hwang (1985) shows that even \(\delta^{+}\)does not universally dominate \(X\) unless the class of loss functions is restricted. We also note that although the inequality in part (a) may seem reminiscent of the "Pitman closeness" criterion, there is really no relation. The criterion of Pitman closeness suffers from a number of defects not shared by stochastic domination (see Robert et al. 1993).

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.

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).]
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