Sufficient Statisitics and Discrete Distributions

Question

I am trying to master minimal/complete sufficient statistics, however I am having trouble when the distributions are discrete and involve indicator functions. Here is my 3 part question:

Let $X$ be a random variable from the following distribution: $$f(x;\theta) = \left\{\begin{array}{ccc} \theta & , & x = -1 \\ 1 - 2\theta & , & x = 0 \\ \theta & , & x = 1\end{array}\right.$$

where $0\leq \theta \leq\frac{1}{2}$.

1. Find a minimal sufficient statistic for parameter $\theta$.

Answer: Define the indicator functions: $$I_1(x) = \left\{\begin{array}{ccc} 1 & , & x = -1 \\ 0 & , & {\rm otherwise}\end{array}\right.$$ $$I_2(x) = \left\{\begin{array}{ccc} 1 & , & x = 0 \\ 0 & , & {\rm otherwise}\end{array}\right.$$ $$I_3(x) = \left\{\begin{array}{ccc} 1 & , & x = 1 \\ 0 & , & {\rm otherwise}\end{array}\right.$$

Then, \begin{eqnarray*} f(x;\theta) & = & (\theta)^{I_1(x)}(1 - 2\theta)^{I_2(x)}(\theta)^{I_3(x)} \\ & = & \theta^{I_1(x) + I_3(x)}(1 - 2\theta)^{I_2(x)}. \end{eqnarray*}

Therefore, by the Factorization Theorem, a sufficient statistic for $\theta$ is: $$T(\underline{X}) = (I_1(x) + I_3(x), I_2(x))$$

To show that this is a minimum: $$\frac{f(x;\theta)}{f(y;\theta)} = \frac{\theta^{I_1(x) + I_3(x)}(1 - 2\theta)^{I_2(x)}}{\theta^{I_1(y) + I_3(y)}(1 - 2\theta)^{I_2(y)}}$$

is independent of $\theta$ if and only if $I_1(x) + I_3(x) = I_1(y) + I_3(y)$ and $I_2(x) = I_2(y)$. Therefore, $T(\underline{X})$ is a minimal sufficient statistic.

2. Is $X$ a complete statistic?

Answer: By definition I know I need to show $E[u(z)] = 0$ implies $u(z) = 0$ for all $\theta$. However, I am unsure how to show $X$ is complete.

3. From Part 1, is the statistic complete? Why?

Answer: Once again I understand the definition of complete, but am unsure on how to reach the conclusion based on: \begin{eqnarray*} E[u(z)] & = & 0 \\ \sum\limits_{z = -1}^1 u(z)\theta^{I_1(z) + I_3(z)}(1 - 2\theta)^{I_2(z)} & = & 0 \end{eqnarray*}

Answer 1

The minimal sufficient partition is

$$\{ \{-1,1\}, \{0\} \} $$

(since $\frac{f_\theta(1)}{f_\theta(-1)}$ does not depend on $\theta$).

So the minimal sufficient statistics(MSS) is

\begin{eqnarray} S(x)=\left\{ \begin{array}{cc} a & x=0 \\ b & x=-1,1 \end{array} \right. \end{eqnarray}

Is it a complete statistics?

$\forall \theta\in(0,\frac{1}{2})$

\begin{eqnarray} 0=E(g(S(X))) &=& g(b)\theta +g(a) (1-2\theta )+g(b)\theta \\ &=& g(a)+\left(g(b)-2g(a)+g(b)\right) \theta \\ &=& g(a)+2\left(g(b)-g(a)\right) \theta \end{eqnarray}

So you can easily check it.