Question

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

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

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.

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.

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 \(0.5\).

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.