/*! 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 11 Consider a random sample \(X_{1}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 11

Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a distribution with pdf \(f(x ; \theta)=\theta(1-x)^{\theta-1}, 00\) (a) Find the form of the uniformly most powerful test of \(H_{0}: \theta=1\) against \(H_{1}: \theta>1\) (b) What is the likelihood ratio \(\Lambda\) for testing \(H_{0}: \theta=1\) against \(H_{1}: \theta \neq 1 ?\)

Short Answer

Expert verified
For part (a), the rule will have the form \(Reject H_0: \theta =1 if \frac{1-n}{s} > k\), where \(s\) is the sample mean. The likelihood ratio for part (b) would be \(\Lambda = \frac{1}{(\theta(1-x)^{\theta-1})}\).

Step by step solution

01

Writing Down the Likelihood Function

The likelihood function can be given as \(L(\theta; x) = \prod_{i=1}^{n} \theta(1-x_i)^{\theta-1}\). Notice that likelihood function is a function of \(\theta\) given the observed sample.
02

Deriving the Uniformly Most Powerful Test

The Neyman-Pearson lemma states that the most powerful test for testing \(H_0: \theta = \theta_0\) against \(H_1: \theta = \theta_1\) has a critical region defined by \{x : \frac{f(x|\theta_1)}{f(x|\theta_0)} > k \} where \(k\) a constant depending on the size of the test. Applying this to our hypotheses, the rejection rule becomes \(\frac{(1-x)^0}{(1-x)} > k\). Solve for \(k\), this gives us the uniformly most powerful test.
03

Calculating the Likelihood Ratio

For the likelihood ratio test, we need to compute the ratio \( \Lambda = \frac{L(\theta_0)}{L(\theta_1)} \). This can be computed by substituting \(\theta = 1\) and \(\theta \neq 1\) in the derived likelihood function.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a distribution having pdf of the form \(f(x ; \theta)=\theta x^{\theta-1}, 0

Consider a normal distribution of the form \(N(\theta, 4)\). The simple hypothesis \(H_{0}: \theta=0\) is rejected, and the alternative composite hypothesis \(H_{1}: \theta>0\) is accepted if and only if the observed mean \(\bar{x}\) of a random sample of size 25 is greater than or equal to \(\frac{3}{5}\). Find the power function \(\gamma(\theta), 0 \leq \theta\), of this test.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a distribution with pdf \(f(x ; \theta)=\) \(\theta x^{\theta-1}, 0

Let the random variable \(X\) have the pdf \(f(x ; \theta)=(1 / \theta) e^{-x / \theta}, 0

Let \(X_{1}, \ldots, X_{n}\) and \(Y_{1}, \ldots, Y_{m}\) follow the location model $$\begin{aligned} X_{i} &=\theta_{1}+Z_{i}, \quad i=1, \ldots, n \\ Y_{i} &=\theta_{2}+Z_{n+i}, \quad i=1, \ldots, m\end{aligned}$$ where \(Z_{1}, \ldots, Z_{n+m}\) are iid random variables with common pdf \(f(z)\). Assume that \(E\left(Z_{i}\right)=0\) and \(\operatorname{Var}\left(Z_{i}\right)=\theta_{3}<\infty\) (a) Show that \(E\left(X_{i}\right)=\theta_{1}, E\left(Y_{i}\right)=\theta_{2}\), and \(\operatorname{Var}\left(X_{i}\right)=\operatorname{Var}\left(Y_{i}\right)=\theta_{3}\). (b) Consider the hypotheses of Example \(8.3 .1\); i.e, $$H_{0}: \theta_{1}=\theta_{2} \text { versus } H_{1}: \theta_{1} \neq \theta_{2}$$ Show that under \(H_{0}\), the test statistic \(T\) given in expression \((8.3 .5)\) has a limiting \(N(0,1)\) distribution. (c) Using Part (b), determine the corresponding large sample test (decision rule) of \(H_{0}\) versus \(H_{1}\). (This shows that the test in Example \(8.3 .1\) is asymptotically correct.)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.