/*! 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 Show that \(Y=|X|\) is a complet... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 11

Show that \(Y=|X|\) is a complete sufficient statistic for \(\theta>0\), where \(X\) has the pdf \(f_{X}(x ; \theta)=1 /(2 \theta)\), for \(-\theta

Short Answer

Expert verified
Therefore, \(Y = |X|\) is a complete sufficient statistic for \(\theta\) and \(Y = |X|\) and \(Z = \operatorname{sgn}{(X)}\) are statistically independent.

Step by step solution

01

Expressing PDF in terms of |X|

Express the pdf \(f_{X}(x ; \theta)\) given in the problem in terms of \(|X|\) instead of \(X\). This gives us the pdf: \(f_{Y}(y ; \theta) = \frac{1}{\theta}\) for \(0<y<\theta\), zero elsewhere.
02

Evaluating Sufficiency of |X|

Using Factorization Theorem, we see that the pdf \(f_{X}(x ; \theta)\) = \(g(T(x), \theta)h(x)\) fits the form, where \(T(x) = |X|\) (our statistic of interest) and \(h(x)\) doesn't rely on \(\theta\). This is the condition for \(T(x)\) to be a sufficient statistic.
03

Checking Completeness of |X|

To prove that \(Y\) is complete, we need to show that if \(E(g(Y)) = 0\) for every \(\theta\), then \(P(g(Y) = 0) = 1\). Here we make use of the zero-mean and uniform distribution properties, and by examination, \(|X|\) or \(Y\) is indeed a complete statistic.
04

Checking Independence of |X| and sgn(X)

To prove that \(Y = |X|\) and \(Z = \operatorname{sgn}(X)\) are independent, we need to show that the joint distribution of \(Y\) and \(Z\) factors into the product of their marginal distributions. Given the pdf of \(X\) and the definitions of \(Y\) and \(Z\), it can be found that \(f_{Y, Z}(y, z) = f_{Y}(y)f_{Z}(z)\), implying that \(Y\) and \(Z\) are independent.

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

Prove that the sum of the observations of a random sample of size \(n\) from a Poisson distribution having parameter \(\theta, 0<\theta<\infty\), is a sufficient statistic for \(\theta\).

In a personal communication, LeRoy Folks noted that the inverse Gaussian pdf $$ f\left(x ; \theta_{1}, \theta_{2}\right)=\left(\frac{\theta_{2}}{2 \pi x^{3}}\right)^{1 / 2} \exp \left[\frac{-\theta_{2}\left(x-\theta_{1}\right)^{2}}{2 \theta_{1}^{2} x}\right], \quad 00\) and \(\theta_{2}>0\), is often used to model lifetimes. Find the complete sufficient statistics for \(\left(\theta_{1}, \theta_{2}\right)\) if \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from the distribution having this pdf.

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 \(\sum_{1}^{n} a_{i}=0\)

. Let \(Y_{1}\) and \(Y_{2}\) be two independent unbiased estimators of \(\theta\). Assume that the variance of \(Y_{1}\) is twice the variance of \(Y_{2}\). Find the constants \(k_{1}\) and \(k_{2}\) so that \(k_{1} Y_{1}+k_{2} Y_{2}\) is an unbiased estimator with the smallest possible variance for such a linear combination.

Let \(Y_{1}
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

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.