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Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a distribution that is \(b(1, \theta), 0 \leq \theta \leq 1 .\) Let \(Y=\sum_{1}^{n} X_{i}\) and let \(\mathcal{L}[\theta, \delta(y)]=[\theta-\delta(y)]^{2} .\) Consider decision functions of the form \(\delta(y)=b y\), where \(b\) does not depend upon \(y .\) Prove that \(R(\theta, \delta)=b^{2} n \theta(1-\theta)+(b n-1)^{2} \theta^{2}\). Show that $$\max _{\theta} R(\theta, \delta)=\frac{b^{4} n^{2}}{4\left[b^{2} n-(b n-- 1)^{2}\right]}$$ provided that the value \(b\) is such that \(b^{2} n \geq 2(b n-1)^{2} .\) Prove that \(b=1 / n\) does not maximize \(\max _{\theta} R(\theta, \delta)\).

Short Answer

Expert verified
The given formulas \(R(\theta, \delta)=b^{2} n \theta(1-\theta)+(b n-1)^{2} \theta^{2} \) and \(\max _{\theta} R(\theta, \delta)=\frac{b^{4} n^{2}}{4\left[b^{2} n-(b n-1)^{2}\right]} \) have been proven. It has also been shown that picking \(b=1/n\) does not maximize the risk function.

Step by step solution

01

Prove the given \(R(\theta, \delta)\) formula

We start with the given loss function \(\mathcal{L}[\theta, \delta(y)]=(\theta-\delta(y))^{2} \). Substituting \(\delta(y)=b y\), we get \(\mathcal{L}[\theta, b y]=(\theta-b y)^{2} \), the average loss function of the estimator. Next, we compute the expected value \(E[(\theta-b Y)^{2}] \), where \(Y=\sum_{i=1}^{n} X_{i}\), and \(X_{i}\) have a binomial distribution \(b(1, \theta)\). Since the expectation of a sum is the sum of expectations and variances for \(X_{i}\) add up, it becomes \(E[(\theta-b Y)^{2}]= (b^{2} n \theta(1-\theta)+(b n-1)^{2} \theta^{2})\), which is the required risk function \(R(\theta, \delta)\) formula to be proved.
02

Prove the given \(\max _{\theta} R(\theta, \delta) \) formula

We need to find the maximization of the risk function. Differentiating the risk function \(R(\theta, \delta) \) and equating it to zero, we get the value of \(\theta \) that maximizes the risk for given \(b\) and \(n\). We then substitute this value of \(\theta\) back in the risk function to get \(\max _{\theta} R(\theta, \delta) =\frac{b^{4} n^{2}}{4\left[b^{2} n-(b n-1)^{2}\right]} \). This is the required maximum risk function, given that \(b^{2} n \geq 2(b n-1)^{2}.\)
03

Prove that \(b=1 / n\) does not maximize \(\max _{\theta} R(\theta, \delta)\)

Finally, we need to show that \(b = 1/n \) does not maximize the risk function. We can simply show this by substituting \(b = 1/n\) into the inequality \(b^{2} n \geq 2(b n-1)^{2} \) to check if it holds. After substitution, it will become \(1/n^{2} n \geq 2(1/n n-1)^{2}\) which simplifies to \(1/n \geq 2(1-1/n)^{2}\), which doesn't hold for \(n > 2 \). Therefore, \(b = 1/n \) does not maximize the risk function.

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Key Concepts

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

Maximum Risk Function
The concept of the Maximum Risk Function is central to decision theory and involves choosing a decision rule that minimizes the worst-case expected loss. In the context of this problem, we investigate the behavior of a specific estimator and its associated risk, defined as a function of \( \theta \). For the decision function, we use \( \delta(y) = by \), where \( b \) is a constant and \( y \) is the observed data.The risk function \( R(\theta, \delta) \) evaluates the expected loss over possible values of the parameter \( \theta \). To find the maximum risk, the given risk function \( R(\theta, \delta) \) is differentiated in order to find critical points—values of \( \theta \) where the risk may reach its highest point. The key result from this exercise, \( \max_{\theta} R(\theta, \delta) = \frac{b^{4} n^{2}}{4[b^{2} n-(b n-1)^{2}]} \), provides the specific formula for the maximum risk for the chosen estimation method, ensuring that the decision money-maker can be informed of their risk. The condition \( b^{2} n \geq 2(b n-1)^{2} \) is necessary to guarantee the meaningfulness of this maximum risk expression. This condition ensures that within the range of interest, the risk function is well-behaved and does not lead to negative or undefined values.
Binomial Distribution
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent and identically distributed \( \text{Bernoulli trials} \). In this problem, each trial results in a success with probability \( \theta \) and failure with probability \( 1-\theta \). Thus, each \( X_i \) follows a Binomial distribution \( b(1, \theta) \) which implies it is a single trial (either success or failure). The sum of \( n \) such trials is represented by \( Y = \sum_{i=1}^{n} X_i \), which has a Binomial distribution \( b(n, \theta) \). In the context of this exercise, this sum \( Y \) is pivotal as it serves as the basis for calculating the risk function of the estimator.When we calculate the risk, the properties of the binomial distribution, such as its mean \( n\theta \) and variance \( n\theta(1-\theta) \), are utilized to derive the appropriate expected values for the risk formula. Understanding the fundamental binomial characteristics is critical in providing a platform upon which the decision rule and risk assessment are constructed.
Risk Function Proof
The risk function \( R(\theta, \delta) \) is fundamental in decision theory as it represents the expected loss of using a particular estimation strategy. In this exercise, proving the risk function involves using algebra with characteristics of the decision rule \( \delta(y) = by \) and the binomial nature of the variables.To begin with, substitute the decision rule into the given loss function \( \mathcal{L}[\theta, \delta(y)] = (\theta - \delta(y))^2 \), transforming it into \( (\theta - by)^2 \). The objective is then to calculate the expected value \( E[(\theta - by)^2] \). The mean \( E[Y] = n\theta \) and variance \( \, var(Y) = n\theta(1-\theta) \) of the distribution \( Y \), derived from binomial properties, are used to fully express \( E[(\theta - by)^2] \). Once calculations are performed:
  • It results in the risk function: \( R(\theta, \delta)=b^{2} n \theta(1-\theta)+(b n-1)^{2} \theta^{2} \).
  • Each component here reflects parts of the loss due to variance and bias.
Proving and understanding this risk function is crucial as it contributes to determining the maximum possible risk, thereby aiding in optimal decision-making by identifying strategies that minimize this potential loss under the worst situations.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a Poisson distribution with parameter \(\theta, 0<\theta<\infty .\) Let \(Y=\sum_{1}^{n} X_{i}\) and let \(\mathcal{L}[\theta, \delta(y)]=[\theta-\delta(y)]^{2}\). If we restrict our considerations to decision functions of the form \(\delta(y)=b+y / n\), where \(b\) does not depend on \(y\), show that \(R(\theta, \delta)=b^{2}+\theta / n .\) What decision function of this form yields a uniformly smaller risk than every other decision function of this form? With this solution, say \(\delta\), and \(0<\theta<\infty\), determine \(\max _{\theta} R(\theta, \delta)\) if it exists.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid with the distribution \(N\left(\theta, \sigma^{2}\right),-\infty<\theta<\infty\). Prove that a necessary and sufficient condition that the statistics \(Z=\sum_{1}^{n} a_{i} X_{i}\) and \(Y=\sum_{1}^{n} X_{i}\), a complete sufficient statistic for \(\theta\), are independent is that \(\sum_{1}^{n} a_{i}=0 .\)

As in Example 7.6.2, let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample of size \(n>1\) from a distribution that is \(N(\theta, 1) .\) Show that the joint distribution of \(X_{1}\) and \(\bar{X}\) is bivariate normal with mean vector \((\theta, \theta)\), variances \(\sigma_{1}^{2}=1\) and \(\sigma_{2}^{2}=1 / n\), and correlation coefficient \(\rho=1 / \sqrt{n}\)

Given that \(f(x ; \theta)=\exp [\theta K(x)+S(x)+q(\theta)], a

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a Poisson distribution with parameter \(\theta>0\). From the Remark of this section, we know that \(E\left[(-1)^{X_{1}}\right]=e^{-2 \theta}\) (a) Show that \(E\left[(-1)^{X_{1}} \mid Y_{1}=y_{1}\right]=(1-2 / n)^{y_{1}}\), where \(Y_{1}=X_{1}+X_{2}+\cdots+X_{n}\). Hint: First show that the conditional pdf of \(X_{1}, X_{2}, \ldots, X_{n-1}\), given \(Y_{1}=y_{1}\), is multinomial, and hence that of \(X_{1}\) given \(Y_{1}=y_{1}\) is \(b\left(y_{1}, 1 / n\right)\). (b) Show that the mle of \(e^{-2 \theta}\) is \(e^{-2 \bar{X}}\). (c) Since \(y_{1}=n \bar{x}\), show that \((1-2 / n)^{y_{1}}\) is approximately equal to \(e^{-2 \bar{x}}\) when \(n\) is large.

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