/*! 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 8 What is the sufficient statistic... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the sufficient statistic for \(\theta\) if the sample arises from a beta distribution in which \(\alpha=\beta=\theta>0 ?\)

Short Answer

Expert verified
The sample independent observations arising from the Beta distribution in which \(\alpha=\beta=\theta>0\) are the sufficient statistic for \(\theta\).

Step by step solution

01

Understanding a Beta Distribution

A Beta distribution is defined by two parameters, \(\alpha\) and \(\beta\). Given that the problem mentions \(\alpha=\beta=\theta>0\), we have a symmetric Beta distribution with parameter \(\theta\). The pdf of a Beta distribution is given by: \(f(x;\alpha,\beta)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}\), where \(B(\alpha,\beta)\) is the Beta function. Here, since \(\alpha=\beta=\theta\), the pdf simplifies to \(f(x;\theta,\theta)=\frac{x^{\theta-1}(1-x)^{\theta-1}}{B(\theta,\theta)}\).
02

Deriving the Sufficient Statistic

A statistic \(T(X)\) is said to be sufficient for \(\theta\) if the conditional probability distribution of the data \(X\), given the statistic \(T(X)\), does not depend on \(\theta\). In the case of a Beta distribution, the sufficient statistic is generally the order statistics. Now, given \(n\) i.i.d Beta random variables \(X_i\), \(i=1,2,...,n\), with parameters \(\theta\) (i.e., \(\alpha=\beta\)=\(\theta\)), the joint density function of \(X_i\) is as follows: \(f(X_1,...,X_n;\theta)=\prod_{i=1}^{n}\frac{x_i^{\theta-1}(1-x_i)^{\theta-1}}{B(\theta,\theta)}\). We can see that for the i.i.d sample \(X_i\), the joint density function can be factorized into a function of \(\theta\) and a function of data \(X_i\). This shows that the data does not provide any other information about \(\theta\) beyond the sample observations, thus all i.i.d observations \(X_i\) are a complete sufficient statistics for \(\theta\) on its own.
03

Final Answer

Hence, the sample independent observations arising from the Beta distribution in which \(\alpha=\beta=\theta>0\) are the sufficient statistic for \(\theta\), as no other values derived from the sample data could provide any additional information about \(\theta\) beyond what these observations provide.

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 91Ó°ÊÓ!

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 \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample with the common pdf \(f(x)=\) \(\theta^{-1} e^{-x / \theta}\), for \(x>0\), zero elsewhere; that is, \(f(x)\) is a \(\Gamma(1, \theta)\) pdf. (a) Show that the statistic \(\bar{X}=n^{-1} \sum_{i=1}^{n} X_{i}\) is a complete and sufficient statistic for \(\theta\). (b) Determine the MVUE of \(\theta\). (c) Determine the mle of \(\theta\). (d) Often, though, this pdf is written as \(f(x)=\tau e^{-\tau x}\), for \(x>0\), zero elsewhere. Thus \(\tau=1 / \theta\). Use Theorem \(6.1 .2\) to determine the mle of \(\tau\). (e) Show that the statistic \(\bar{X}=n^{-1} \sum_{i=1}^{n} X_{i}\) is a complete and sufficient statistic for \(\tau\). Show that \((n-1) /(n \bar{X})\) is the MVUE of \(\tau=1 / \theta\). Hence, as usual, the reciprocal of the mle of \(\theta\) is the mle of \(1 / \theta\), but, in this situation, the reciprocal of the MVUE of \(\theta\) is not the MVUE of \(1 / \theta\). (f) Compute the variances of each of the unbiased estimators in parts (b) and (e).

Show that the first order statistic \(Y_{1}\) of a random sample of size \(n\) from the distribution having pdf \(f(x ; \theta)=e^{-(x-\theta)}, \theta

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 \(\theta\).

Let \(X\) have the pdf \(f_{X}(x ; \theta)=1 /(2 \theta)\), for \(-\theta0\) (a) Is the statistic \(Y=|X|\) a sufficient statistic for \(\theta ?\) Why? (b) Let \(f_{Y}(y ; \theta)\) be the pdf of \(Y\). Is the family \(\left\\{f_{Y}(y ; \theta): \theta>0\right\\}\) complete? Why?

We consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a distribution with pdf \(f(x ; \theta)=(1 / \theta) \exp (-x / \theta), 0

See all solutions

Recommended explanations on Math Textbooks

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.