# What is the necessary condition for a unbiased estimator to be UMVUE?

According to the Rao-Blackwell theorem, if statistic $$T$$ is a sufficient and complete for $$\theta$$, and $$E(T)=\theta$$, then $$T$$ is a uniformly minimum-variance unbiased estimator (UMVUE).

I am wondering how to justify that an unbiased estimator is a UMVUE:

1. if $$T$$ is not sufficient, can it be a UMVUE?
2. if $$T$$ is not complete, can it be a UMVUE?
3. If $$T$$ is not sufficient or complete, can it be a UMVUE?
• Should the last one "if $T$ is not sufficient or complete" perhaps be "if T is neither sufficient nor complete" (if you mean both conditions hold simultaneously)? Commented Aug 16, 2015 at 19:43
• In 2. If $T$ is not complete, then it is an MVUE but you do need the completeness if you are to attach the letter U to it :) Commented Aug 16, 2015 at 23:41
• A necessary-sufficient condition for an unbiased estimator (with finite second moment) to be UMVUE is that it must be uncorrelated with every unbiased estimator of zero. Commented Jan 30, 2020 at 21:21

Let us show that there can be a UMVUE which is not a sufficient statistic.

First of all, if the estimator $T$ takes (say) value $0$ on all samples, then clearly $T$ is a UMVUE of $0$, which latter can be considered a (constant) function of $\theta$. On the other hand, this estimator $T$ is clearly not sufficient in general.

It is a bit harder to find a UMVUE $Y$ of the "entire" unknown parameter $\theta$ (rather than a UMVUE of a function of it) such that $Y$ is not sufficient for $\theta$. E.g., suppose the "data" are given just by one normal r.v. $X\sim N(\tau,1)$, where $\tau\in\mathbb{R}$ is unknown. Clearly, $X$ is sufficient and complete for $\tau$. Let $Y=1$ if $X\ge0$ and $Y=0$ if $X<0$, and let
$\theta:=\mathsf{E}_\tau Y=\mathsf{P}_\tau(X\ge0)=\Phi(\tau)$; as usual, we denote by $\Phi$ and $\varphi$, respectively, the cdf and pdf of $N(0,1)$.
So, the estimator $Y$ is unbiased for $\theta=\Phi(\tau)$ and is a function of the complete sufficient statistic $X$. Hence, $Y$ is a UMVUE of $\theta=\Phi(\tau)$.

On the other hand, the function $\Phi$ is continuous and strictly increasing on $\mathbb{R}$, from $0$ to $1$. So, the correspondence $\mathbb{R}\ni\tau=\Phi^{-1}(\theta)\leftrightarrow\theta=\Phi(\tau)\in(0,1)$ is a bijection. That is, we can re-parametirize the problem, from $\tau$ to $\theta$, in a one-to-one manner. Thus, $Y$ is a UMVUE of $\theta$, not just for the "old" parameter $\tau$, but for the "new" parameter $\theta\in(0,1)$ as well. However, $Y$ is not sufficient for $\tau$ and therefore not sufficient for $\theta$. Indeed, \begin{multline*} \mathsf{P}_\tau(X<-1|Y=0)=\mathsf{P}_\tau(X<-1|X<0)=\frac{\mathsf{P}_\tau(X<-1)}{\mathsf{P}_\tau(X<0)} \\ =\frac{\Phi(-\tau-1)}{\Phi(-\tau)} \sim\frac{\varphi(-\tau-1)/(\tau+1)}{\varphi(-\tau)/\tau}\sim\frac{\varphi(-\tau-1)}{\varphi(-\tau)}=e^{-\tau-1/2} \end{multline*} as $\tau\to\infty$; here we used the known asymptotic equivalence $\Phi(-\tau)\sim\varphi(-\tau)/\tau$ as $\tau\to\infty$, which follows by the l'Hospital rule. So, $\mathsf{P}_\tau(X<-1|Y=0)$ depends on $\tau$ and hence on $\theta$, which shows that $Y$ is not sufficient for $\theta$ (whereas $Y$ is a UMVUE for $\theta$).

• If the estimator $T$ always takes the value $0$, how can it be unbiased? Commented Jan 15, 2018 at 11:24
• By definition, $T$ is an unbiased estimator of a function $q(\theta)$ of the parameter $\theta$ if $E_\theta T=q(\theta)$ for all values of $\theta$. So, if $q(\theta)=0$ for all $\theta$, then of course $T=0$ will be an unbiased estimator of this $q(\theta)$. And this is what I said: that $T=0$ is clearly an unbiased estimator of the constant zero function of the parameter. Commented Jan 15, 2018 at 15:23
• OK, thanks, I had missed the fact that you were "estimating" a constant function! Commented Jan 15, 2018 at 15:26

On Uniformly Minimum Variance Unbiased Estimation when no Complete Sufficient Statistics Exist by L. Bondesson gives some examples of UMVUEs which are not complete sufficient statistics, including the following one:

Let $X_1, \ldots, X_n$ be independent observations of a random variable $X = \mu + \sigma Y$, where $\mu$ and $\sigma$ are unknown, and $Y$ is gamma distributed with known shape parameter $k$ and known scale parameter $\theta$. Then $\bar{X}$ is the UMVUE of $E(X) = \mu + k\theta\sigma$. However, when $k \neq 1$ then there is no complete sufficient statistic for $(\mu, \sigma)$.