Question

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?

Short answer

Yes, the statistic \(Y = |X|\) is a sufficient statistic for \(\theta\) and the family \(\left\{f_{Y}(y ; \theta): \theta>0\right\}\) is complete.

Step-by-step solution

Determine sufficiency of Y

To decide whether or not \(Y = |X|\) is a sufficient statistic for \(\theta\), it should contain all the information about \(\theta\). The pdf of X is given by \(f_X(x;\theta)=1/(2 \theta)\), for \(-\theta<x<\theta\) and zero elsewhere. To compute Y given X, applying the transformation \(Y = |X|\) to the pdf of X,we can see that the pdf of Y is \(f_{Y}(y;\theta) = 2/(2 \theta) = 1/\theta\), for \(0 < y < \theta\) and zero elsewhere. Hence, the observed value of the absolute value of \(X\), which is \(Y\), varies directly based on \(\theta\). As such, \(Y = |X|\) is a sufficient statistic for \(\theta\).

Determine completeness of the family

To determine whether a family of distributions is complete, it is required to demonstrate that, for this family, there's no non-zero function such that its expected value equals to zero for all \(\theta\). The family \(\left\{f_{Y}(y ; \theta): \theta>0\right\}\) can be rewritten as \(\left\{1/\theta: 00\right\}\). Now, if there exists a function \(h(y)\) such as \(E[h(Y)]=0\), we have the equation: \(\int_{0}^{\theta} h(y) \cdot \frac{1}{\theta} dy = 0\). This integral equals zero only if \(h(y) = 0\) for all \(y\) in the interval \(0 < y < \theta\). Therefore, the family of distributions \(\left\{f_{Y}(y ; \theta): \theta>0\right\}\) is complete.