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(a) Let \(X \sim N(\boldsymbol{\theta}, \Sigma)\) and consider the estimation of \(\theta\) under the loss \(L(\theta, \delta)=\) \((\theta-\delta)(\theta-\delta)\). Show that \(R(\theta, \mathbf{X})=\) tr \(\Sigma\), the minimax risk. Hence, \(\mathbf{X}\) is a minimax estimator. (b) Let \(X \sim N(\theta, I)\) and consider estimation of \(\theta\) under the loss \(L(\theta, \delta)=(\theta-\) \delta) \(Q(\theta-\delta)\), where \(Q\) is a known positive definite matrix. Show that \(R(\theta, \mathbf{X})=\operatorname{tr} Q\), the minimax risk. Hence, \(\mathbf{X}\) is a minimax estimator. (c) Show that the calculations in parts (a) and (b) are equivalent.

Short Answer

Expert verified
For both setups, the risk is trace of a matrix, proving X is a minimax estimator.

Step by step solution

01

Understand the Setup for Part (a)

We are given that \(X \sim N(\boldsymbol{\theta}, \Sigma)\), meaning \(X\) is a multivariate normal random variable with mean \(\theta\) and covariance matrix \(\Sigma\). We are estimating \(\theta\) under the squared error loss function \(L(\theta, \delta) = (\theta - \delta)^2\).
02

Calculate the Risk for Part (a)

The risk function \(R(\theta, \delta)\) is defined as the expected value of the loss: \(E[(\theta - \delta(X))^2]\). When \(\delta(X) = X\), the estimator is unbiased with variance \(\Sigma\). Therefore, the risk can be expressed as the trace of covariance, \(R(\theta, X) = \text{tr}(\Sigma)\).
03

Conclude Minimax Property for Part (a)

Since the risk is constant and independent of \(\theta\), \(R(\theta, X) = \text{tr}(\Sigma)\) is indeed the smallest maximum risk, showing \(X\) is a minimax estimator.
04

Understand the Setup for Part (b)

In this case, \(X \sim N(\theta, I)\), where \(I\) is the identity matrix. The loss function is given as \((\theta - \delta)Q(\theta - \delta)\), with \(Q\) a positive definite matrix.
05

Calculate the Risk for Part (b)

The risk is \(R(\theta, \delta) = E[(\theta - X)Q(\theta - X)]\). Since \(X\) is an unbiased estimator of \(\theta\), \(R(\theta, X) = E[(X - \theta)^T Q (X - \theta)] = \operatorname{tr}(Q)\).
06

Conclude Minimax Property for Part (b)

The risk \(R(\theta, X) = \operatorname{tr}(Q)\) is constant, independent of \(\theta\), demonstrating \(X\) is also a minimax estimator.
07

Show Equivalence for Part (c)

In both parts, the risk is derived from the unbiased nature of the estimator and results in the trace of a matrix (\(\Sigma\) in part a and \(Q\) in part b). This risk calculation structure shows equivalence between parts (a) and (b), confirming \(R(\theta, X)\) as the trace of the respective covariance structure.

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Key Concepts

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

Risk Function
In statistical estimation, the risk function is crucial to understanding an estimator's performance. It measures the expected loss when estimating a parameter. We denote it as:
  • \( R(\theta, \delta) = E[L(\theta, \delta)] \)
Here, \( L(\theta, \delta) \) is the loss function reflecting the error in estimation. Evaluating the risk function allows us to compare how different estimators perform under a given loss.
In the scenario of minimax estimation, the goal is to minimize the maximum risk across all possible values of \(\theta\). This 'smallest' maximum risk represents the least risky estimator when you don't know the exact value of the parameter you're estimating.
For example, in our exercise, the estimation under squared error loss results in a risk function that is constant and independent of \(\theta\). This characteristic shows that such estimators fall into the category of minimax as they effectively manage the worst-case scenario.
Multivariate Normal Distribution
A multivariate normal distribution is an extension of the normal distribution to multiple dimensions. It describes a distribution where each component follows a normal distribution, but components can be correlated. Some key properties are:
  • The distribution is defined by a mean vector \( \boldsymbol{\theta} \) and a covariance matrix \( \Sigma \).
  • Each component \(X_i\) has a normal distribution with its corresponding mean and variance from \( \Sigma \).
In our problem, \(X\) follows a multivariate normal distribution with mean \(\boldsymbol{\theta}\) and covariance \(\Sigma\). This setup is common in application areas like finance or biology.
The multivariate setup becomes particularly interesting for estimating parameters (like \(\theta\)) because the dependencies captured by \( \Sigma \) affect how we interpret the data.
For example, the covariance matrix \(\Sigma\) not only contributes to the understanding of data variability but directly influences the risk function's outcome (as the trace of the covariance).
Squared Error Loss
Squared error loss is a common loss function used in statistics. It measures the cost of an error as the square of its magnitude. We express it as:
  • \( L(\theta, \delta) = (\theta - \delta)^2 \)
The squared error loss penalizes larger errors more severely compared to smaller ones. This property makes it particularly useful for problems where it's crucial to avoid large discrepancies.
In our context, using squared error loss, we evaluate estimators via the risk function. The risk for an unbiased estimator (whose expectation matches the true value \(\theta\)) often relates to its variance.
This concept plays out in our exercise where minimizing this squared error across all \(\theta\) values led us to define the risk in terms of tr\( (Q) \).
Understanding how loss functions feed into the risk function gives us deeper insights into selecting the optimal estimator. It shows why the 'squaring' aspect of the loss influences how we measure performance—critical for minimax strategy.

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Most popular questions from this chapter

Let \(X_{1}, \ldots, X_{m}\) and \(Y_{1}, \ldots, Y_{n}\) be independently distributed as \(N\left(\xi, \sigma^{2}\right)\) and \(N\left(\eta, \tau^{2}\right)\), respectively, and consider the problem of estimating \(\Delta=\eta-\xi\) with squared error loss. (a) If \(\sigma\) and \(\tau\) are known, \(\bar{Y}-\bar{X}\) is minimax. (b) If \(\sigma\) and \(\tau\) are restricted by \(\sigma^{2} \leq A\) and \(\tau^{2} \leq B\), respectively \((A, B\) known and finite), \(\bar{Y}-\bar{X}\) continues to be minimax.

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).]

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\\} $$

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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