/*! 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 1 Let \(Y_{1}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 1

Let \(Y_{1}

Short Answer

Expert verified
The complete sufficient statistic \(Y_{4}\) for \(\theta\) is indeed independent of the statistics \(Y_{1} / Y_{4}\) and \((Y_{1}+Y_{2}) / (Y_{3}+Y_{4})\) since we can express the joint density as a product of separate functions, each pertaining to the distinct statistics in question.

Step by step solution

01

Write Down the Joint Density of Order Statistics

The joint density of order statistics, when removed of the irrelevant constant, is given by: \( g(Y_{1}, Y_{2}, Y_{3}, Y_{4}; \theta) = Y_{4}^{-4}\) provided that \(0 < Y_{1} < Y_{2} < Y_{3} < Y_{4} < \theta\)
02

Express the Density as a Product of Two Functions

Now, it's essential to express this joint density as a product of two functions, one of \(Y_{4}\) and another of \(Y_{1} / Y_{4}\) and \((Y_{1}+Y_{2}) / (Y_{3}+Y_{4})\). Here's how: \( g(Y_{1}, Y_{2}, Y_{3}, Y_{4}; \theta) = h(Y_{4}; \theta).i[T(Y_{1}, Y_{2}, Y_{3}, Y_{4})]\) where \(h(Y_{4}; \theta) = Y_{4}^{-4}\) for \(0 < Y_{4} < \theta\), and \(i\), a function that indicates when the inequalities hold. In this case: \( i(T(Y_{1}, Y_{2}, Y_{3}, Y_{4})) = 1\) when \(0 < [Y_{1} / Y_{4}] < [Y_{2} / Y_{4}] < [Y_{3} / Y_{4}] < [Y_{4} / Y_{4}] < 1\), and zero otherwise
03

Prove the Independence

The joint density \(g(Y_{1}, Y_{2}, Y_{3}, Y_{4}; \theta)\) is seen to be independent of \(Y_{4}\), \(Y_{1} / Y_{4}\), and \((Y_{1}+Y_{2}) / (Y_{3}+Y_{4})\), because it can be separated into the product of two independent functions, \(h(Y_{4}; \theta)\) and \(i(T(Y_{1}, Y_{2}, Y_{3}, Y_{4}))\). Hence, we can conclude that \(Y_{4}\) is independent of \(Y_{1} / Y_{4}\) and \((Y_{1}+Y_{2}) / (Y_{3}+Y_{4})\)

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

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 \(\max _{\theta} R(\theta, \delta)\).

Consider the family of probability density functions \(\\{h(z ; \theta): \theta \in \Omega\\}\), where \(h(z ; \theta)=1 / \theta, 01\).

Show that the sum of the observations of a random sample of size \(n\) from a gamma distribution that has pdf \(f(x ; \theta)=(1 / \theta) e^{-x / \theta}, 0

Let \(Y_{1}

Let \(f(x, y)=\left(2 / \theta^{2}\right) e^{-(x+y) / \theta}, 0
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Calculus

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.