I am trying to prove that

There is no UMVUE for $\theta$ for the distribution $\text{Unif}\{\theta-1, \theta, \theta+1\}$, $\theta$ is an integer.

Here is what I have attempted.

I am trying to use the theorem 1.7 from Lehmann and Casella's book but I got stuck.

The theorem says

Let $X$ have a distribution $\mathbb{P}_\theta$, $\theta\in\Theta$, let $\delta$ be an estimator so that $\mathbb{E}_\theta\delta^2<\infty$, and let $\mathcal{U}$ denote the set of all unbiased estimator of zero, which also satisfy $\mathbb{E}_\theta U^2<\infty,\forall U\in\mathcal{U}$. Then $\delta$ is a UMVUE for its expectation $g(\theta)=\mathbb{E}(\delta)$ if and only if $$\mathbb{E}_\theta(\delta U)=0,\forall U\in\mathcal{U}\text{ and }\theta\in\Theta.$$

Suppose $\{X\}_{i=1}^n\overset{i.i.d}{\sim}\text{Unif}\{\theta-1, \theta, \theta+1\}$. Thus, $$\mathbb{P}(X=\theta-1)=\mathbb{P}(X=\theta)=\mathbb{P}(X=\theta+1)=1/3$$.

All the unbiased estimator of $0$ is of the form $$\mathcal{U}=\{a\mathbb{1}_{(X=\theta-1)}+b\mathbb{1}_{(X=\theta)}+c\mathbb{1}_{(X=\theta+1)},\text{ }\forall a,b,c\in\mathbb{R},a+b+c=1\}$$ This is because $\forall U\in\mathcal{U}$, we have $\mathbb{E}(U)=1/3(a+b+c)=0.$

Also, notice $$\mathbb{E}(X)=\theta$$

I was trying to use the theorem, but I do not know how to proceed further.

  • 3
    $\begingroup$ Your set $\mathcal U$ is not made of estimators since the expressions involve $\theta$. $\endgroup$
    – Xi'an
    Jan 25 '21 at 19:01
  • $\begingroup$ Ha, that's right. I probably need $X_{(n)}-X_{(1)}$ to get the unbiased estimator of zero. Thanks. $\endgroup$
    – Tan
    Jan 25 '21 at 20:11
  • $\begingroup$ Not sure if this helps but there is no complete sufficient statistic in this model: stats.stackexchange.com/q/61990/119261. $\endgroup$ May 5 '21 at 15:55

I think you have made the right attempt, since the method of zero estimator is the necessary and sufficient condition for a UMVUE to exist. And here is my proof:

Let's first assume there exist a UMVUE $\hat\theta $ for $\theta$. Then, by the theorem, any zero estimator $\eta$ will have:

  1. $E_{\theta}\eta = 0$, for any $𝜃∈Θ$.
  2. $E_𝜃[\eta \hat \theta]=0$

So next step is to find an estimator that breaks one of these conditions.

Let's start with the first term: $$E_𝜃[\eta ]=\frac{1}{3}[\eta(\theta-1) + \eta(\theta) + \eta(\theta+1)] = 0$$ Then any function that satisfies $\eta(x-1) + \eta(x) + \eta(x+1) = 0 $ will meet the requirement. This surely can not derive $E_{\theta}(\eta \hat\theta)= 0 $, so in this case there is no UMVUE.


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