# Proving the nonexistence of UMVUE for $\text{Unif}\{\theta-1, \theta, \theta+1\}$

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.

• Your set $\mathcal U$ is not made of estimators since the expressions involve $\theta$. Jan 25, 2021 at 19:01
• Ha, that's right. I probably need $X_{(n)}-X_{(1)}$ to get the unbiased estimator of zero. Thanks.
– Tan
Jan 25, 2021 at 20:11
• Not sure if this helps but there is no complete sufficient statistic in this model: stats.stackexchange.com/q/61990/119261. May 5, 2021 at 15:55

Let's approach this from first principles.

An estimator $$t$$ assigns some guess of $$\theta$$ to any possible outcome $$X.$$ Since the possible outcomes are integers, write this guess as $$t_i$$ when $$X=i$$ for any integer $$i.$$

We are hoping to find estimators that tend to be close to $$\theta$$ when $$\theta$$ is the parameter. Clearly, when $$i$$ is observed the possible values of $$\theta$$ are limited (for sure) to the set $$\{i+1,i,i-1\}.$$ Thus, a good estimator is likely to guess $$\theta$$ is close to the observation $$i.$$ Let us therefore express $$t$$ in terms of how far it departs from the observation; namely, let

$$t_i = i + \delta_i.$$

When $$\theta$$ is the parameter, the outcomes $$\theta-1,$$ $$\theta,$$ and $$\theta+1$$ have equal probabilities of $$1/3$$ and all other integers have zero probability. Consequently,

1. The expectation of $$t$$ when $$\theta$$ is the parameter is $$E(t\mid \theta=i) = \frac{1}{3}\left(t_{i-1} + t_i + t_{i+1}\right) = i + \frac{1}{3}\left(\delta_{i-1} + \delta_i + \delta_{i+1}\right).$$ Because $$t$$ must be unbiased, this quantity equals $$i$$ no matter what $$i$$ might be, showing that for all $$i,$$ $$\delta_{i-1} + \delta_{i} + \delta_{i+1}=0.$$ Already this is a huge restriction, because if we specify (say) $$\delta_0$$ and $$\delta_1,$$ this relation recursively requires $$\delta_{-1} = \delta_2 = -(\delta_0 + \delta_1),$$ *etc., thereby completely determining the estimator.

2. The variance of $$t$$ is $$\operatorname{Var}(t\mid \theta=i) = \frac{1}{3}\left((t_{i-1}-i)^2 + (t_i-i)^2 + (t_{i+1}-i)^2\right) \\= \frac{1}{3}\left((\delta_{i-1}-1)^2 + \delta_i^2 + (\delta_{i+1}+1)^2\right).$$ Among all unbiased estimators, this must have the smallest variance for all $$i.$$

It is a straightforward exercise in algebra (or Calculus, using a Lagrange multiplier) to show that for a specific $$i,$$ a minimum of $$(2)$$ can be obtained subject to the constraint $$(1)$$ and implies $$\delta_{i-1}=\delta_i=\delta_{i+1}.$$ Since this must hold for all $$i,$$ clearly the $$\delta_i$$ are all equal, whence they must all equal $$0$$ (because $$\delta_1 = \delta_2 = -(\delta_0+\delta_1) = -2\delta_1$$ has the unique solution $$\delta_1=0,$$ etc.).

Consequently, if an UMVUE exists, its variance is a constant given by $$(2),$$ equal to $$2/3.$$ Unfortunately, there are unbiased estimators that achieve smaller variances for specific values of $$\theta.$$

For instance, suppose you had a strong belief that $$\theta=0.$$ You might then adjust your estimator to guess $$\theta=0$$ whenever an outcome consistent with that guess showed up. That is, you would set $$t_0=t_1=t_{-1}=0.$$ That is equivalent to $$\delta_{-1}=1,$$ $$\delta_0=0,$$ and $$\delta_1=-1.$$ As we have remarked earlier, these initial conditions determine $$t$$ completely from the recursion $$(1).$$ Its variance when $$\theta=0$$ is zero, because it always guesses the correct value of $$\theta.$$ You can't do any better than that! Moreover, $$0 \ll 2/3$$ is a huge improvement. But compensating for that is a larger variance for certain other values of $$\theta.$$ For instance, since $$\delta_2 = \delta_{-1} = 1,$$ when $$\theta=1$$ the possible outcomes are $$0,1,2,$$ for which $$t$$ guesses $$0,$$ $$0,$$ and $$3,$$ respectively, for a variance of

$$\frac{1}{3}\left((0-1)^2 + (0-1)^2 + (3-1)^2\right) = 2 \gg \frac{2}{3}.$$

This contradiction--obtaining a lower variance for certain values of $$\theta$$--shows no UMVUE exists.

You might enjoy re-interpreting $$\delta$$ as an estimator of 0 ;-).

• Setting partial derivatives of the Lagrangian equal to zero gives $2/3(\delta_{i-1}-1)+\lambda = 2/3 \delta_i + \lambda = 2/3(\delta_{i+1} + 1) + \lambda = \delta_{i-1}+\delta_i+\delta_{i+1}=0$. Why does this imply $\delta_{i-1}=\delta_i=\delta_{i+1}$? I thought it implies $\delta_{i-1}-1=\delta_i=\delta_{i+1}+1$, which also satisfies the constraint that they sum to zero. Jul 18 at 18:06
• @Daniel The equations have to hold for all $i.$ See the text immediately above that result.
– whuber
Jul 18 at 18:22
• Not sure I understand- if they hold for all $i$, then they hold for $i=0$. Then the first equation reads $\delta_{-1}-1=\delta_0$, which isn't satisfied by $\delta_{-1}=\delta_0=0$. Jul 18 at 20:35
• @Daniel The integers include the negative numbers.
– whuber
Jul 18 at 20:37

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.