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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).]

Short Answer

Expert verified
The estimator is inadmissible under both conditions as the risk decreases with a better choice of \(\varepsilon\).

Step by step solution

01

Define Squared Error Loss

The squared error loss is defined as the expectation of the square of the difference between the estimator and the true value. Therefore, for our estimator \(T = \left[\frac{1}{1+\lambda} + \varepsilon\right] X\), the squared error loss function is:\[ L(T, E_{\theta}(X)) = \left(T - E_{\theta}(X)\right)^2 = \left(\left[\frac{1}{1+\lambda} + \varepsilon\right] X - E_{\theta}(X)\right)^2. \]
02

Calculate the Risk Function

The risk function \(R(\varepsilon)\) is the expected value of the squared error loss. For the estimator \(T = \left[\frac{1}{1+\lambda} + \varepsilon\right] X\), the risk function is:\[ R(\varepsilon) = E_{\theta}\left( \left(\left[\frac{1}{1+\lambda} + \varepsilon\right] X - E_{\theta}(X)\right)^2 \right). \]
03

Differentiate the Risk Function

According to the hint, differentiate \(R(\varepsilon)\) with respect to \(\varepsilon\) to examine its behavior:\[ \frac{d}{d\varepsilon} R(\varepsilon). \] This derivative will show us how the risk changes as \(\varepsilon\) is adjusted.
04

Analyze Condition (a)

Under condition (a), where \(\operatorname{var}_{\theta}(X) / E_{\theta}^{2}(X) > \lambda > 0\) and \(\varepsilon > 0\), the risk function \(R(\varepsilon)\) decreases as \(\varepsilon\) decreases. This means that making \(\varepsilon\) smaller reduces the risk, suggesting the estimator is not optimal. Since we can achieve lower risk by decreasing \(\varepsilon\), the estimator is inadmissible in this case.
05

Analyze Condition (b)

Under condition (b), where \(\operatorname{var}_{\theta}(X) / E_{\theta}^{2}(X) < \lambda\) and \(\varepsilon < 0\), the opposite happens—\(R(\varepsilon)\) increases as \(\varepsilon\) increases. Again, this implies the estimator is inadmissible because we can improve our estimator's performance by choosing a different \(\varepsilon\).
06

Conclude with the Estimator's Inadmissibility

For both conditions, adjusting \(\varepsilon\) in the right direction (decreasing it in case (a) and increasing it in case (b)) consistently improves the estimator, proving its inadmissibility as we can always find a better choice of \(\varepsilon\) that reduces the risk.

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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 context of statistics, the squared error loss is a measure of accuracy for an estimator. It quantifies how far off an estimation is from the actual value. The formula for squared error loss is the squared difference between the estimator and the true parameter value. Specifically, for an estimator denoted by \( T = \left[\frac{1}{1+\lambda} + \varepsilon\right] X \), the squared error loss is given by: \[ L(T, E_{\theta}(X)) = \left(T - E_{\theta}(X)\right)^2 = \left(\left[\frac{1}{1+\lambda} + \varepsilon\right] X - E_{\theta}(X)\right)^2. \] This loss function is crucial because it helps us understand how effective an estimator is. A lower squared error loss indicates that the estimator is closer to the true value and is therefore more accurate. This concept ensures that statistical methods focus on minimizing errors to yield better and more reliable estimations.
Risk Function
The risk function is a statistical tool used to evaluate the overall performance of an estimator. It is essentially the expected value of the squared error loss. When considering an estimator \( T = \left[\frac{1}{1+\lambda} + \varepsilon\right] X \), the risk function quantifies how much "risk" or error is expected, on average, when using the estimator: \[ R(\varepsilon) = E_{\theta}\left( \left(\left[\frac{1}{1+\lambda} + \varepsilon\right] X - E_{\theta}(X)\right)^2 \right). \] Risk is a useful metric as it accounts for both variability and bias in an estimator. By minimizing the risk function, statisticians aim to find the most reliable estimator, ideally one with the lowest possible expectation of error. Minimizing the risk function ensures that the estimator we choose not only fits the current data well but also performs adequately on new data.
Differentiation of Risk
To understand the behavior of the risk function, it is beneficial to differentiate it with respect to the parameter of interest, in this case, \( \varepsilon \). By taking the derivative, we can assess how changes in \( \varepsilon \) impact the risk. This mathematical exercise reveals whether the risk increases or decreases, helping us decide whether a particular change in \( \varepsilon \) leads to a better estimator: \[ \frac{d}{d\varepsilon} R(\varepsilon). \] Differentiation informs us if adjustments to \( \varepsilon \) improve or worsen our estimator's performance. In practical terms, it means observing whether the risk increases when \( \varepsilon \) changes. If the derivative shows that risk decreases as \( \varepsilon \) decreases under certain conditions, an alternative \( \varepsilon \) like that results in an inadmissible estimator, meaning a better one can be found. This concept is decisive in choosing the optimal parameters for effective estimation.
Variability and Expectation Ratio
The ratio of variability to expectation squared, expressed as \( \operatorname{var}_{\theta}(X) / E_{\theta}^{2}(X) \), is an important consideration when assessing estimators under different conditions. This ratio contrasts the degree of spread or dispersion in the data (variability) with the square of its mean (expectation). It is a standalone factor in determining whether the estimator is suitable or can be improved: - **For Condition (a):** If \( \operatorname{var}_{\theta}(X) / E_{\theta}^{2}(X) > \lambda > 0 \) with \( \varepsilon > 0 \), it suggests that by reducing \( \varepsilon \), the risk function \( R(\varepsilon) \) tends to decrease. Hence, the estimator can be improved, proving it to be inadmissible under these conditions.- **For Condition (b):** If \( \operatorname{var}_{\theta}(X) / E_{\theta}^{2}(X) < \lambda \) and \( \varepsilon < 0 \), increasing \( \varepsilon \) reduces risk, indicating unfounded estimation stability. Again, the estimator is inadmissible here. This ratio is a crucial determinant of an estimator's admissibility, assisting in selecting whether the estimator should be modified for improved accuracy.

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Most popular questions from this chapter

Classes of priors for \(\Gamma\)-minimax estimation have often been specified using moment restrictions. (a) For \(X \sim b(p, n)\), find the \(\Gamma\)-minimax estimator of \(p\) under squared error loss, with $$ \Gamma_{\mu}=\left\\{\pi(p): \pi(p)=\operatorname{beta}(a, b), \mu=\frac{a}{a+b}\right\\} $$ where \(\mu\) is considered fixed and known. (b) For \(X \sim N(\theta, 1)\), find the \(\Gamma\)-minimax estimator of \(\theta\) under squared error loss, with $$ \Gamma_{\mu, \tau}=\left\\{\pi(\theta): E(\theta)=\mu, \operatorname{var} \theta=\tau^{2}\right\\} $$ where \(\mu\) and \(\tau\) are fixed and known. [Hint: In part (b), show that the \(\Gamma\)-minimax estimator is the Bayes estimator against a normal prior with the specified moments (Jackson et al. 1970; see Chen, EichenhauerHerrmann, and Lehn 1990 for a multivariate version). This somewhat nonrobust \(\Gamma\) minimax estimator is characteristic of estimators derived from moment restrictions and shows why robust Bayesians tend to not use such classes. See Berger 1985 , Section \(4.7 .6\) for further discussion.]

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.]

Let \(X=1\) or 0 with probabilities \(p\) and \(q\), respectively, and consider the estimation of \(p\) with loss \(=1\) when \(|d-p| \geq 1 / 4\), and 0 otherwise. The most general randomized estimator is \(\delta=U\) when \(X=0\), and \(\delta=V\) when \(X=1\) where \(U\) and \(V\) are two random variables with known distributions. (a) Evaluate the risk function and the maximum risk of \(\delta\) when \(U\) and \(V\) are uniform on \((0,1 / 2)\) and \((1 / 2,1)\), respectively. (b) Show that the estimator \(\delta\) of (a) is minimax by considering the three values \(p=0\), \(1 / 2,1\) [Hint: (b) The risk at \(p=0,1 / 2,1\) is, respectively, \(P(U>1 / 4), 1 / 2[P(U<1 / 4)+\) \(P(V>3 / 4)]\), and \(P(V<3 / 4)]\).

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).

Efron (1990), in a discussion of Brown's (1990a) ancillarity paradox, proposed an alternate version. Suppose \(\mathbf{X} \sim N_{r}(\mu, I), r>2\), and with probability \(1 / r\), independent of \(\mathbf{X}\), the value of the random variable \(J=j\) is observed, \(j=1,2, \ldots, r\). The problem is to estimate \(\theta_{j}\) using the loss function \(L\left(\theta_{j}, d\right)=\left(\theta_{j}-d\right)^{2}\). Show that, conditional on \(J=j, X_{j}\) is a minimax and admissible estimator of \(\theta_{j}\). However, unconditionally, \(X_{f}\) is dominated by the \(j\) th coordinate of the James-Stein estimator. This version of the paradox may be somewhat more transparent. It more clearly shows how the presence of the ancillary random variable forces the problem to be considered as a multivariate problem, opening the door for the Stein effect.

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