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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
The complete and sufficient statistic \(Y=|X|\) is proven through the factorization theorem and completeness property. And \(Y=|X|\) and \(Z=\operatorname{sgn}(X)\) are independent is justified by showing that the joint pdf of \(Y\) and \(Z\) equals the product of their respective pdfs.

Step by step solution

01

Calculate the joint pdf of \(X\) and \(Y=|X|\)

Since \(Y=|X|\) and \(f_{X}(x ; \theta)=1 /(2 \theta)\) for \(-\theta<x<\theta\), the joint pdf of \(X\) and \(Y=|X|\) is (1 / \(\theta\)) for \(-\theta<x<\theta\) and \(0<y<\theta\), zero elsewhere.
02

Show that \(Y=|X|\) is a complete and sufficient statistic

By the factorization theorem, a statistic is sufficient if and only if the joint pdf can be factored as a function of the statistic times a function of the parameter. Since the joint pdf here is a function of \(|x| = Y\) and \(\theta\), it is sufficient. To show that it is complete, we can demonstrate that for any function \(g\): \(E[g(Y)] = 0\) implies \(g(Y)\) is zero for almost all \(y\). According to the properties of a complete statistic, the expectation \(E[g(Y)]\) is only zero if \(g(Y)\) is zero, implying that \(|X|\) is a complete statistic.
03

Calculate the pdf of \(Z=\operatorname{sgn}(X)\)

The sign function \(Z=\operatorname{sgn}(X)\) outputs \(-1\) if \(x<0\), outputs \(1\) if \(x>0\), and outputs \(0\) if \(x=0\). According to the given pdf of \(X\), \(Z=\operatorname{sgn}(X)\) takes values \(-1\) and \(1\) each with probability \(0.5\), and \(0\) with probability \(0\). Thus, \(Z=\operatorname{sgn}(X)\) is a Bernoulli random variable with parameter \(0.5\).
04

Show independence of \(Y=|X|\) and \(Z=\operatorname{sgn}(X)\)

Independence of two random variables implies that their joint pdf is the product of their respective pdfs. Hence, we need to show that the joint pdf of \(Y=|X|\) and \(Z=\operatorname{sgn}(X)\) equals the product of their respective pdfs. Since the joint pdf of \(Y=|X|\) and \(Z=\operatorname{sgn}(X)\) is totally symmetric around \(x = 0\), we obtain a product of two functions, each function depends solely on \(Y\) or \(Z\). Hence, \(Y=|X|\) and \(Z=\operatorname{sgn}(X)\) are independent.

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