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Chapter 10: Problem 7

Minimax-Tests. Betrachten Sie ein einfaches Alternativ-Standardmodell \(\left(\not x, \mathscr{F} ; P_{0}, P_{1}\right) .\) Ein Test \(\varphi\) von \(P_{0}\) gegen \(P_{1}\) hei脽t ein Minimax-Test, wenn das Maximum der Irrtumswahrscheinlichkeiten erster und zweiter Art minimal ist. Zeigen Sie: Es gibt einen Neyman-Pearson Test \(\varphi\) mit \(\mathbb{E}_{0}(\varphi)=\mathbb{E}_{1}(1-\varphi)\), und dieser ist ein Minimax-Test.

Short Answer

Expert verified
A Neyman-Pearson test with equal Type I and Type II error probabilities is a minimax test.

Step by step solution

01

Understand the Definitions

A minimax test is a test where we want to minimize the worst-case maximum of the probabilities of making Type I and Type II errors. A Neyman-Pearson test is a test that maximizes the likelihood ratio, subject to a constraint on the Type I error.
02

Recall Neyman-Pearson Lemma

The Neyman-Pearson Lemma states that the most powerful test for simple hypotheses involves rejecting the null hypothesis in favor of the alternative hypothesis when the likelihood ratio exceeds a certain threshold.
03

Define the Test Significance Level

For a test \(\varphi\), define the expected values:\(\mathbb{E}_{0}(\varphi) = \alpha\) as the probability of Type I error and \(\mathbb{E}_{1}(1-\varphi) = \beta\) as the probability of Type II error under respective hypotheses.
04

Minimize Maximum Error Probability

Choose a test function \(\varphi\) that minimizes the maximum of \(\alpha\) and \(\beta\). The Neyman-Pearson test criterion \(\mathbb{E}_{0}(\varphi) = \mathbb{E}_{1}(1-\varphi)\) satisfies this because it equalizes the Type I and Type II error probabilities.
05

Conclusion from Equivalence

Since both error probabilities are equal under this test \(\varphi\), the maximum error probability is minimized. Therefore, such a \(\varphi\) is both a Neyman-Pearson test and a minimax test.

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Key Concepts

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

Neyman-Pearson Lemma
The Neyman-Pearson Lemma is a fundamental concept in statistical hypothesis testing. It provides a framework for deciding between two simple hypotheses: the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \). The lemma states that the most powerful test to distinguish between these two hypotheses is one that uses a likelihood ratio.Here鈥檚 how it works:
  • You calculate the likelihood ratio \( \frac{P_1(x)}{P_0(x)} \), where \( P_1(x) \) is the probability of observing data \( x \) under the alternative hypothesis, and \( P_0(x) \) under the null hypothesis.
  • You reject the null hypothesis \( H_0 \) in favor of \( H_1 \) when this ratio exceeds a certain level. This level is determined by the desired significance level or Type I error \( \alpha \).
By following this rule, you achieve the highest possible power (1-尾) for the test given a specific Type I error rate.
Type I Error
A Type I error occurs in hypothesis testing when you mistakenly reject the null hypothesis \( H_0 \) when it is actually true. It is essentially a false positive error.For example, if you were testing whether a new drug is effective, a Type I error would mean concluding the drug works when, in fact, it doesn鈥檛.The probability of making a Type I error is represented by \( \alpha \), the significance level of the test. Key points include:
  • You decide \( \alpha \) before conducting the test. Common values are 0.05 or 0.01.
  • A lower \( \alpha \) reduces the risk of a Type I error but can increase the chance of a Type II error.
Balancing \( \alpha \) with the risk of making a Type II error is critical in designing hypothesis tests.
Type II Error
A Type II error happens when a test fails to reject a null hypothesis \( H_0 \) that is actually false. It's like a false negative, where you miss detecting an effect that is truly present.In the context of a new drug, this error would mean concluding the drug is not effective when it truly is. Several points are crucial to understand about Type II errors:
  • The probability of making a Type II error is denoted by \( \beta \).
  • Its complement, 1-\( \beta \), is the power of the test, which is the probability of correctly rejecting a false null hypothesis.
  • Power increases with larger sample sizes or more pronounced effects.
Effective testing should aim to minimize both Type I and Type II errors in a balanced manner.
Likelihood Ratio
In hypothesis testing, the likelihood ratio is a crucial statistic for comparing two competing hypotheses. The likelihood ratio, denoted as \( \frac{P_1(x)}{P_0(x)} \), indicates how much more likely the observed data \( x \) are under one hypothesis compared to the other. Here鈥檚 a breakdown of its use:
  • If the ratio is greater than a set threshold, the data are more consistent with the alternative hypothesis \( H_1 \).
  • If less than the threshold, it supports the null hypothesis \( H_0 \).
  • The threshold is determined based on the acceptable level of Type I error (\( \alpha \)).
Using the likelihood ratio simplifies making clear and objective decisions in hypothesis testing.
Hypothesis Testing
Hypothesis testing is a method used to determine the validity of a claim about a population based on sample data. It's essential in scientific research and decision-making.This process involves several key steps:
  • First, form two hypotheses: the null hypothesis \( H_0 \) (a statement of no effect) and the alternative hypothesis \( H_1 \) (a statement indicating the presence of an effect).
  • Next, choose a significance level \( \alpha \), which is the probability threshold for rejecting \( H_0 \).
  • Conduct a test using sample data to compute a statistic.
  • Compare this statistic to a critical value or use a p-value to decide whether to reject \( H_0 \).
Hypothesis testing helps to make data-driven decisions, often being a trade-off between Type I and Type II errors. Achieving this balance is critical for reliable conclusions.

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Most popular questions from this chapter

Zweiseitiger Chiquadrat-Test. Betrachten Sie im zweiparametrigen Gau脽'schen Produktmodell das zweiseitige Testproblem \(H_{0}: v=v_{0}\) gegen \(H_{1}: v \neq v_{0}\) mit folgender Entscheidungsvorschrift: \(H_{0}\) werde akzeptiert, falls \(c_{1} \leq \frac{n-1}{u_{0}} V^{*} \leq c_{2}\) (a) Berechnen Sie die G眉tefunktion \(G\) dieses Tests und zeigen Sie: Es gilt $$ \frac{\partial G}{\partial v}\left(m, v_{0}\right) \geqq 0 \text { je nachdem, ob }\left(\frac{c_{2}}{c_{1}}\right)^{(n-1) / 2} \geqq e^{\left(c_{2}-c_{1}\right) / 2} $$ (b) Naiv w眉rde man \(c_{1}, c_{2}\) ja so w盲hlen, dass $$ P_{m, v_{o}}\left(\frac{n-1}{v_{0}} V^{*}c_{2}\right)=\alpha / 2 $$ Zeigen Sie im Fall \(\alpha=0.02, n=3\), dass dieser Test verf盲lscht ist, und skizzieren Sie \(G\) (c) Wie kann man einen unverf盲lschten Test der obigen Bauart konstruieren? (d) Welche Gestalt hat der zugeh枚rige Likelihood-Quotienten-Test?

Zeigen Sie, dass im zweiparametrigen GauB'schen Produktmodell kein gleichma脽ig bester Test f眉r das einseitige Testproblem \(H_{0}: m \leq 0\) gegen \(H_{1}: m>0\) existiert.

Konstruieren Sie einen zweiseitigen Binomialtest zum Niveau \(\alpha\), d. h. einen Test im Binomialmodell f眉r die Nullhypothese \(H_{0}: \vartheta=\vartheta_{0}\) gegen die Alternative \(H_{1}: \vartheta \neq \vartheta_{0}\), \(0<\vartheta_{0}<1\). Leiten Sie au脽erdem mit Hilfe des Satzes von de Moivre-Laplace eine asymptotische Version des Tests her.

Bei einem Preisratsel wird der Gewinner dadurch ermittelt, dass aus der Menge aller eingegangenen Postkarten solange (mit Zur眉cklegen) gezogen wird, bis man eine Karte mit der richtigen Losung in der Hand hallt. Da bei der letzten Auslosung dazu 7 Karten gezogen werden mussten, argw枚hnt der verantwortliche Redakteur, dass der Anteil \(p\) der eingegangenen richtigen L枚sungen weniger als \(50 \%\) betragen habe, die Quizfrage also zu schwierig gewesen sei. Liegt er mit dieser Entscheidung richtig? F眉hren Sie anhand des vorliegenden Ergebnisses in einem geeigneten statistischen Modell einen Test f眉r \(H_{0}: p \geq 0.5\) gegen \(H_{1}: p<0.5\) zum Niveau \(\alpha=0.05\) durch.

Zusammenhang von Konfidenzbereichen und Tests. Sei \(\left(\mathfrak{x}, \mathscr{F}, P_{\vartheta}: \vartheta \in \Theta\right.\) ) ein statistisches Modell. Zeigen Sie: (a) Ist \(C: X \rightarrow \mathfrak{P}(\Theta)\) ein Konfidenzbereich zum Irrtumsniveau \(\alpha\) und \(\vartheta_{0} \in \Theta\), so ist \(\left\\{\vartheta_{0} \notin C(\cdot)\right\\}\) der Ablehnungsbereich eines Tests von \(H_{0}: \vartheta=\theta_{0}\) gegen \(H_{1}: \vartheta \neq \vartheta_{0}\) zum Niveau \(\alpha\). (b) Ist umgekehrt fur jedes \(\vartheta_{0} \in \Theta\) ein nichtrandomisierter Test fur \(H_{0}: \vartheta=\vartheta_{0}\) gegen \(H_{1}: \vartheta \neq \vartheta_{0}\) zum Niveau \(\alpha\) gegeben, so l盲sst sich daraus ein Konfidenzbereich zum Irrtumsniveau \(\alpha\) gewinnen.
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