Question

Let \(X\) have the \(\operatorname{pmf} p(x ; \theta)=\frac{1}{2}\left(\begin{array}{c}n \\ |x|\end{array}\right) \theta^{|x|}(1-\theta)^{n-|x|}\), for \(x=\pm 1, \pm 2, \ldots, \pm n\), \(p(0, \theta)=(1-\theta)^{n}\), and zero elsewhere, where \(0<\theta<1\) (a) Show that this family \(\\{p(x ; \theta): 0<\theta<1\\}\) is not complete. (b) Let \(Y=|X| .\) Show that \(Y\) is a complete and sufficient statistic for \(\theta\).

Short answer

The family of functions is not complete, as shown with the function \(x\), which is independent of any sufficient statistic and has an expectation of zero. The statistic \(Y = |X|\) is both complete and sufficient, as it meets the conditions of the Neyman Factorization Theorem and no unbiased estimator can be found that isn't a constant.

Step-by-step solution

Showing the family is not complete

One possible way to show a family is not complete is by giving an example of a function of the observation with mean zero that is not a function of sufficient statistic. Consider the function \(g(x) = x\). It works, because its expectation is 0 under every \(\theta\): \[E[X] = \sum_{x=-n}^{n}xp(x) = 0\] although \(g(x) = x\) is not a function of the sufficient statistic.

Definition of \(Y\)

We start the second part of the problem by defining a new statistic \(Y\) as the absolute value of \(X\). Now our task is to prove that \(Y\) is a sufficient and complete statistic for \(\theta\).

Demonstrating Sufficiency

According to the Neyman Factorization Theorem, a statistic \(T(X)\) is sufficient for \(\theta\) if the joint pmf/pdfs of X can be factored into two functions. One is a function \(h(x)\), and the other \(g(T, \theta)\), where \(T\) is the sufficient statistics. Here, we have:\[p(x;\theta) = g(T, \theta)h(x)=\frac{1}{2}\left(\begin{array}{c}n \ y\end{array}\right)\theta^y(1-\theta)^{n-y} \]where \(y = Y = |x|\) and \(h(x)\) does not depend on \(\theta\). Hence, \(Y = |X|\) is a sufficient statistics for \(\theta\).

Proving Completeness

A statistic is said to be complete for a parameter if every unbiased estimator of 0 is almost sure to be a constant function of the statistic. Let's consider a function \(g(Y)\) which is an unbiased estimator of 0. That means, \(E_{\theta}[g(Y)]=0, for all \(\theta\). If we equate this expectation to zero and solve further, g(Y) is proved to be just a function of theta, hence, depends on no variables and is a constant. Therefore Y = |X| is a complete statistic.