How to find an unbiased estimator?

Question

Suppose $X_1, X_2, ...,X_n$ are samples from a uniform discrete distribution with probability 1/3 on each of the points $\theta-1, \theta, \theta+1$, where $\theta\in\mathbb{Z}.$ From "Theory of point Estimation" by Lehman and Casella. Page130 Problem 1.9

I want to find all unbiased estimators of zero. How should I do that?

My understanding:
The definition of unbiased estimator says, $E[\delta(x)]=g(\theta)$.

which means, I want anything that has $bias=E[\delta(x)]-g(\theta)=0$

But how exactly do I find this guy, $\delta(x)$? And what is the parameter of interest here?

My work:
$f(x;3)= \frac{1}{3}, \text{for } \theta-1\le x \le \theta+1, \text{where }\theta\in\mathbb{Z}$

Could someone point out what I am missing here? I feel that I am pretty close to the answer.
Thanks in advance!

Answer 1

If the assignment asks for all estimators, then probably the correct answer would be to list a property which the desired estimator must satisfy. The estimator is simply the function $\delta:\mathbb{R}^n\to \mathbb{R}$. The requirement is that it should be unbiased:

$$E_\theta\delta(X_1,...,X_n)=0.$$

Now vector $(X_1,...,X_n)$ assumes value $(x_{i_1},...,x_{i_n})$ with probability $(1/3)^n$, where $(x_{i_1},...,x_{i_n})\subset \{\theta-1,\theta,\theta+1\}^n$. So you can write down how the expectation looks like and you'll get a new condition of what $\delta$ must satisfy. Maybe it will give you some additional ideas how to proceed.

Answer 2

The answer is simple, any statistic $\delta(X)$ satisfying $$\delta(\theta-1)+\delta(\theta)+\delta(\theta+1)=0, \forall\theta\in\mathbb{Z}$$ is an unbiased estimator of zero. Thus, the totality of the unbiased estimators of zero is given by

$$\left\{\delta\in\{\text{measurable functions from }\mathbb{Z}\text{ to }\mathbb{R}\},\delta(\theta-1)+\delta(\theta)+\delta(\theta+1)=0, \forall\theta\in\mathbb{Z}\right\}$$

Answer 3

Actually you need to solve the following equation: $$Ef(X_1,\dots,X_n)=\theta.$$ For example if $f(x_1,\dots,x_2)=\sum_{i=1}^n \lambda_i x_i$ then $$E\sum_{i=1}^n \lambda_i X_i=\sum_{i=1}^n \lambda_i EX_i=\theta.$$ Then $EX_1=\dots=EX_n=1/3\cdot((\theta-1)+(\theta)+(\theta+1))=\theta$, and mind that $EX_i$ only equals to $\theta$ if all three probabilities are the same! Then we insert $EX_i=\theta$ and get $$\sum_{i=1}^n \lambda_i \theta= \theta \sum_{i=1}^n \lambda_i=\theta.$$ Now we see that if $\sum_{i=1}^n \lambda_i=1$ we have unbiased estimator of $\theta$, for example $X_1$ is unnbiased estimator yielding $\lambda_1=1$ and $\lambda_i=0, \forall i \neq 1$.

So these are all linear unbiased estimators and what about non linear? Like median? Is there some additional info that we could be missing in you assignment?