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Chapter 7: Problem 4

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

Short Answer

Expert verified
The family of probability density functions is complete when \(\Omega = \{\theta:0<\theta<\infty\}\), and is not complete when \(\Omega = \{\theta:1<\theta<\infty\}\).

Step by step solution

01

Part (a): Show completeness with hint

Use the given density function and the fact that \(u(z)\) is continuous to find the derivative of \(E[u(Z)]\) with respect to \(\theta\). Calculate the expectation \(E[u(Z)]\). For the derivative of \(E[u(Z)]\) with respect to \(\theta\) to be zero, it implies that the expectation does not depend on \(\theta\). Thus, this will show that for any function \(u(z)\) such that \(E[u(Z)]=0\), the function \(u(z)=0\) a.e. on the support of the distribution. This shows completeness of the family under \(\Omega = \{\theta:0<\theta<\infty\}\).
02

Part (b): Show non-completeness with hint

Concentrate on the interval \(01\). A function like \(u(z) = z\) can work. In this case, you can compute the expectation \(E[u(Z)]\) and show that it equals zero for all \(\theta>1\). This shows that the family is not complete under \(\Omega = \{\theta:1<\theta<\infty\}\), as found a non-zero function \(u(z)\) such that \(E[u(Z)]=0\).

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Most popular questions from this chapter

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