Question

Let \(X\) have the pmf \(p(x ; \theta)=\frac{1}{2}\left(\begin{array}{l}n \\\ x\end{array}\right) \theta^{|x|}(1-\theta)^{n-|x|}\), 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

(a) The family \( p(x; \theta) \) is not complete because we found a function g(X) = X, which yields \( E_{\theta}[X] = 0 \) for all \( \theta \), but is not always zero. (b) |X| is a sufficient statistic for \( \theta \), as shown by the factorization theorem. Moreover, |X| is also a complete statistic, because we proved that any unbiased estimator based on |X| is constant.

Step-by-step solution

Prepare for Proof of Non-Completeness

We will show that there exists a function g(X) such that \( E_{\theta}[ g(X) ] = 0 \) for all \( \theta \), but g(X) is not equal to zero. Let's take g(X) = X as a trial function to prove this point.

Calculate Expectation

Find the expectation: \( E_{\theta}[X] = \sum_{x=-n}^{n} x*p(x; \theta) \) . As the function is symmetric (|x|), this expectation is equal to 0, for all \( \theta \).

Show Non-Zero of g(X)

Although \( E_{\theta}[X] = 0 \), X is not always equal to zero. Thus, we have shown that the family \( p(x; \theta) \) is not complete.

Prepare for Proof of Sufficiency

We now show that |X| is a sufficient statistic for \( \theta \). We can use the factorization theorem.

Use Factorization Theorem

Observe the pmf: \( p(x; \theta) = \frac{1}{2} {n \choose {|x|}} \theta^{|x|}(1-\theta)^{n-|x|} \), which can be factored into \( h(x)g(t(x), \theta) \) where \( h(x) = \frac{1}{2} \) and \( g(t(x), \theta) = {n \choose t} \theta^{t}(1-\theta)^{n-t} \) with t = |X|. This factoring proves |X| is a sufficient statistic.

Prepare for Proof of Completeness

We now show that |X| is a complete statistic. We have to show that if an unbiased estimator based on |X| has expectation zero, it must be a constant.

Assume E(g(Y))=0

Assume there exists a function of |X|, say g(|X|), such that \( E_{\theta}[ g(|X|) ] = 0 \) for all \( \theta \). Then we have \( \sum_{y=0}^{n} g(y) {n \choose y} \theta^{y} (1-\theta)^{n-y} = 0 \).

Prove uniqueness of solution

The above equation, which is a polynomial of degree n, equals to zero for all \( \theta \) in (0,1). It can only be true if coefficients of \( \theta^y \) are all zero. That means, \( g(y) = 0 \) for y = 0,...,n. This indicates that g(|X|) is a constant, proving |X| to be a complete statistic.

Key concepts

Complete Statistic

Understanding a 'Complete Statistic' is key when diving into the field of mathematical statistics. A statistic, by definition, is a function of random sample data. Now, a statistic is called 'complete' for a parameter if no other statistic that is not almost surely a constant can have an expectation of zero for all values of the parameter.

In simpler terms, think of it as a unique representation of data. If a statistic is complete, any attempt to estimate the parameter based only on that statistic will lead to a unique estimator. In the exercise provided, we look at the absolute value of a random variable, \(Y=|X|\), and we show that it is complete by proving that any function of \(Y\), say \(g(|X|)\), that has an expectation of zero for all possible values of \(\theta\), must be a constant function - indicating that no other information or variation could represent the parameter \(\theta\) except for \(Y\).

This is pivotal because a complete statistic helps to ensure the best possible estimations within the framework of unbiased estimators, avoiding any redundant or irrelevant information that could mislead the estimation process.

Sufficient Statistic

A 'Sufficient Statistic' is another fundamental concept in statistics. This kind of statistic provides all necessary information to perform inference about a parameter. Linking closely to the notion of data reduction, a sufficient statistic essentially captures all the information needed about the parameter contained in the sample.

The exercise demonstrates, using the Factorization Theorem, that \(Y=|X|\) is a sufficient statistic for the parameter \(\theta\). To understand this better, suppose you're given a jigsaw puzzle that's already completed – that's what a sufficient statistic does. It compiles the entire essence of the dataset with reference to estimating the parameter, rendering further details from the data redundant.

In technical terms, the pmf of \(X\), when factored appropriately, leaves us with a component that only involves the sufficient statistic \(Y\) and the parameter \(\theta\). This supports the argument that \(Y\) contains all the 'necessary' information about the parameter in question.

Probability Mass Function (pmf)

The 'Probability Mass Function (pmf)' is a basic concept in probability theory for discrete random variables. Simply put, it defines the probability that a discrete random variable is exactly equal to some value. The pmf serves as a bridge between outcomes and their probabilities, often denoted as \(p(x)\) for a discrete random variable \(X\).

In our exercise, the pmf describes the probabilities associated with a random variable \(X\), taking on values from \(-n\) to \(n\), including zero. The function \(p(x; \theta)\) is parameterized by \(\theta\) and is what forms the backbone of our statistical inferences. It tells us how likely each outcome is and consequently is used in the proofs to show completeness and sufficiency of the statistic \(Y\). The pmf, therefore, is not just a listing of probabilities, it frames the entire statistical enquiry we are conducting on the variable \(X\).

Expectation of a Random Variable

The concept of 'Expectation of a Random Variable' is akin to what is commonly known as the average or mean in everyday language. Mathematically, for discrete random variables, it is calculated as the sum of all values that the variable can take, each weighted by its probability of occurrence - similar to calculating a weighted average.

In practice, the expectation tells us the 'center' or the expected value of the random variable's distribution if we were to repeat the random process numerous times. In the given exercise, when we calculate the expectation of \(X\), denoted as \(E_{\theta}[X]\), and find it to be zero for all \(\theta\), we use this result in the context of proving non-completeness. This zero expectation shows that though we have an average outcome of zero, the statistic \(X\) carries information about \(\theta\) that varies with different possible outcomes - reinforcing that \(X\) is not complete. Conversely, the expectation of \(g(Y)\) being zero would mean \(Y\) suffices to capture all necessary information to estimate \(\theta\), representing the idea of a complete statistic.