/*! 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 13 Consider any closed sequential b... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider any closed sequential binomial sampling plan with a set \(B\) of stopping points, and let \(B^{\prime}\) be the set \(B \cup\left\\{\left(x_{0}, y_{0}\right)\right\\}\) where \(\left(x_{0}, y_{0}\right)\) is a point not in \(B\) that has positive probability of being reached under plan \(B\). Show that the sufficient statistic \(T=(X, Y)\) is not complete for the sampling plan which has \(B^{\prime}\) as its set of stopping points. [Hint: For any point \((x, y) \in B\), let \(N(x, y)\) and \(N^{\prime}(x, y)\) denote the number of paths to \((x, y)\) when the set of stopping points is \(B\) and \(B^{\prime}\), respectively, and let \(N\left(x_{0}, y_{0}\right)=0, N^{\prime}\left(x_{0}, y_{0}\right)=1\). Then, the statistic \(1-\left[N(X, Y) / N^{\prime}(X, Y)\right]\) has expectation 0 under \(B^{\prime}\) for all values of \(p .]\)

Short Answer

Expert verified
The statistic is not complete due to the introduction of additional stopping points, altering path counts without affecting expectations.

Step by step solution

01

Understanding the Problem

The task is to show that the statistic \( T = (X, Y) \) is not a complete sufficient statistic for a certain binomial sampling plan. We analyze the closed sequential binomial sampling plan characterized by a set of stopping points \( B \), and we extend it by adding a new point \((x_0, y_0)\) to form \( B' \). Our goal is to demonstrate that this changes the completeness of the sufficient statistic.
02

Analyzing Stopping Points

Initially, consider the stopping points in \( B \) and how paths reach each point \((x, y) \in B\). We denote \( N(x, y) \) as the number of different paths that can lead to \((x, y)\) using stopping points in \( B \). When introducing a new point \((x_0, y_0)\), counted as part of \( B' \), the number of paths to \((x, y)\) now becomes \( N'(x, y) \), reflecting the additional paths possible with \( (x_0, y_0) \).

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Key Concepts

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

Sufficient Statistic
In the context of statistical analysis, a sufficient statistic provides all the necessary information from your data that is needed to calculate a particular parameter. In sequential binomial sampling, we often encounter situations where we want to summarize data efficiently. The statistic \( T = (X, Y) \) is traditionally used as a sufficient statistic because it captures the outcomes of binomial trials. A sufficient statistic like \( T \) allows us to perform inference without dealing directly with individual data points, which simplifies calculations. However, when we modify our sampling plan by adding a new stopping point, \( (x_0, y_0) \), this changes the dynamics. Although \( T \) remains sufficient, it may no longer maintain completeness, as will be discussed further. Understanding the role and limitations of sufficient statistics is crucial. Especially when dealing with data where the plans or observations change, resulting in different paths or complexities.
Completeness
Completeness of a statistic is an advanced concept that ensures there are no unobserved or hidden distributions within a statistic that provide no additional information. When we add this new stopping point \( (x_0, y_0) \) to our set \( B \) to create \( B' \), we face a challenge regarding completeness. Even though \( T = (X, Y) \) is a sufficient statistic, it may fail the completeness test. This means there exists another function of \( T \) that has an expected value of zero, yet it is not a trivial function (non-zero elsewhere). The formula given in the original exercise, \(1 - [N(X, Y) / N'(X, Y)]\), shows such a function where even though its expected value is zero, its existence implies that the distribution is not completely captured by \( T \).When a statistic is not complete, it forces us to consider additional information or a change in our analysis approach, as there might be unidentified patterns or data that are not captured by \( T \).
Stopping Points
Stopping points in sequential sampling plans designate the moments when sampling should cease. They are crucial as they define when a decision can be made accurately based on accumulated data. In the case of a binomial sampling plan characterized by the set of stopping points \( B \), these points guide the process until a conclusion is reached.By introducing a new stopping point \( (x_0, y_0) \), forming the set \( B' \), we alter the sampling landscape. This additional point provides different scenarios with which observations can stop, thereby affecting the number of potential paths to each existing point. This change can complicate the traditional use of a statistic as well, impacting its completeness as discussed earlier.Recognizing and analyzing these stopping points help in understanding the complexity in sampling plans and how they influence both data collection and subsequent analysis results.
Probability Paths
Probability paths refer to the various sequences or ways by which a trial can reach a particular stopping point. In our sequential binomial sampling plan, each path to a point \((x, y)\) represents a potential outcome for the sampling process.Originally, with the set \( B \), the number of paths to any stopping point \((x, y)\) is given by \( N(x, y) \). When we expand our stopping points set to \( B' \) by adding \((x_0, y_0)\), we need to count new paths as well. This total becomes \( N'(x, y) \), reflecting increased complexity and variability due to additional stopping scenarios.Understanding probability paths helps us calculate probabilities accurately and also critically determines whether a statistic remains complete. Changes in these paths indicate shifts in underlying probabilities, necessitating reevaluation of inference strategies.

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Most popular questions from this chapter

Curtailed single sampling. Let \(a, b

Let \(X_{1}, \ldots, X_{n}\) be iid with distribution \(F\). (a) Characterize the totality of functions \(f\left(X_{1}, \ldots, X_{n}\right)\) which are unbiased estimators of zero for the class \(\mathcal{F}_{0}\) of all distributions \(F\) having a density. (b) Give one example of a nontrivial unbiased estimator of zero when (i) \(n=2\) and (ii) \(n=3\)

(a) Any two random variables \(X\) and \(Y\) with finite second moments satisfy the covariance inequality \([\operatorname{cov}(X, Y)]^{2} \leq \operatorname{var}(X) \cdot \operatorname{var}(Y)\). (b) The inequality in part (a) is an equality if and only if there exist constants \(a\) and \(b\) for which \(P(X=a Y+b)=1\).

Verify the following statements, asserted by Basu (1988, Chapter 1 ), which illustrate the relationship between information, sufficiency, and ancillarity. Suppose that we let \(I(\theta)=E_{\theta}\left[-\partial^{2} / \partial \theta^{2} \log f(x \mid(\theta)]\right.\) be the information in \(X\) about \(\theta\) and let \(J(\theta)=E_{\theta}\left[-\partial^{2} / \partial \theta^{2} \log g(T \mid \theta)\right]\) be the information about \(\theta\) contained in a statistic \(T\), where \(g(\cdot \mid \theta)\) is the density function of \(T\). Define \(\lambda(\theta)=I(\theta)-J(\theta)\), a measure of information lost by using \(T\) instead of \(X .\) Under suitable regularity conditions, show that (a) \(\lambda(\theta) \geq 0\) for all \(\theta\) (b) \(\lambda(\theta)=0\) if and only if \(T\) is sufficient for \(\theta\). (c) If \(Y\) is ancillary but \((T, Y)\) is sufficient, then \(I(\theta)=E_{\theta}[J(\theta \mid Y)]\), where $$ J(\theta \mid y)=E_{\theta}\left[-\frac{\partial^{2}}{\partial \theta^{2}} \log h(T \mid y, \theta) \mid Y=y\right] $$ and \(h(t \mid y, \theta)\) is the conditional density of \(T\) given \(Y=y\).

Let \(X, Y\), and \(g\) be such that \(E[g(X, Y) \mid y]\) is independent of \(y\). Then, \(E[f(Y) g(X, Y)]=\) \(E[f(Y)] E[g(X, Y)]\), and hence \(f(Y)\) and \(g(X, Y)\) are uncorrelated, for all \(f\).

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