Nonexistence of UMVUE

Question

In Mathematical Statistics written by Jun Shao(2003), exercise 3.22 claims that

Exercise 3.22. Let $\left(X_{1}, \ldots, X_{n}\right)$ be a random sample from $P \in \mathcal{P}$ containing all symmetric distributions with finite means and with Lebesgue densities on $\mathcal{R}$.

• (i) When $n=1$, show that $X_{1}$ is the UMVUE of $\mu=E X_{1}$.
• (ii) When $n>1$, show that there is no UMVUE of $\mu=E X_{1}$.

However, it seems that example 3.8 gives a counterexample towards 3.22.(i) when $g(\theta)=\theta$. In the solution to exercise 3.22, a uniform distribution $U(\theta_1-\theta_2,\theta_1+\theta_2),\theta_1\in \mathbb{R},\theta_2>0$ is considered as a subfamily of $\mathcal{P}$. So I think $U\left(\theta-\frac{1}{2}, \theta+\frac{1}{2}\right))$ also belongs to $\mathcal{P}$, which makes a contradiction in exercise 3.22(i). Dose anyone can answer why this happens?

Example 3.8. Let $X$ be a sample (of size 1) from the uniform distribution $U\left(\theta-\frac{1}{2}, \theta+\frac{1}{2}\right), \theta \in \mathcal{R}$. We now apply Theorem $3.2$ to show that there is no UMVUE of $\vartheta=g(\theta)$ for any nonconstant function $g$. Note that an unbiased estimator $U(X)$ of 0 must satisfy $$ \int_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} U(x) d x=0 \quad \text { for all } \theta \in \mathcal{R} \text {. } $$ Differentiating both sizes of the previous equation and applying the result of differentiation of an integral lead to $U(x)=U(x+1)$ a.e. $m$, where $m$ is the Lebesgue measure on $\mathcal{R}$. If $T$ is a UMVUE of $g(\theta)$, then $T(X) U(X)$ is unbiased for 0 and, hence, $T(x) U(x)=T(x+1) U(x+1)$ a.e. $m$, where $U(X)$ is any unbiased estimator of 0 . Since this is true for all $U, T(x)=T(x+1)$ a.e. $m$. Since $T$ is unbiased for $g(\theta)$, $$ g(\theta)=\int_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} T(x) d x \quad \text { for all } \theta \in \mathcal{R} . $$ Differentiating both sizes of the previous equation and applying the result of differentiation of an integral, we obtain that $$ g^{\prime}(\theta)=T\left(\theta+\frac{1}{2}\right)-T\left(\theta-\frac{1}{2}\right)=0 \quad \text { a.e. } m . $$

Answer 1

I think this is a very nice question and I'm surprised that no one posted any answer. My attempt here is only to invite better answers.

For the symmetric family $\mathcal{P}$ with finite means, even though the mean is the parameter of interest, the mean is not an indexing parameter, because different members of $\mathcal{P}$ may have common means. Therefore, the "$\theta$" here is not $\mu$. Rather, I would think of $\theta$ as any CDF $F$ belonging to $\mathcal{P}$. And $\mu=g(\theta)=\int_{\mathbb{R}} x\,\mathbb{d}\theta(x)$.

Next, let's consider the definition of an unbiased estimator $U(X)$. This has to satisfy $$\mathbb{E}[U(X)]=g(\theta)\ \forall\ \theta. $$ The tricky part is for all $\theta$. Since the nonparametric family $\mathcal{P}$ is very big, it is more difficult to find an estimator that remain unbiased under such many different scenarios (i.e., different $\theta$'s). I think the solution manual's argument is essentially that $X_1$ is the only unbiased estimator for this huge family.

Had the family becomes smaller, like the $\text{Unif}(\mu-0.5,\ \mu+0.5)$ parametric family, the unbiasedness of an estimator only need to be examined under a limited number of scenarios and many competitors of $X_1$ may survive the this screening test.

The same argument can be applied to unbiased estimators of $0$. In the big family $\mathcal{P}$, only a constant function $h(X)=0$ is possible; but in the smaller family $\text{Unif}(\mu-0.5,\ \mu+0.5)$, any periodic function such that $U(X)=U(X+1)$ can estimate $0$ unbiasedly. But of course, they are not necessarily unbiased for $0$ if being examined by a member of $\mathcal{P}$ outside this parametric subfamily.

Therefore, in the parametric family $\text{Unif}(\mu-0.5,\ \mu+0.5)$, it turns out that no survived unbiased estimators can beat others under all scenarios; but in the nonparametric family $\mathcal{P}$, the estimator $X_1$ faces no competition because other estimators cannot quality to join the competition (for being biased in some cases). I believe this is the reason for the seemingly paradoxical fact that an UMVUE does not exist for a smaller special family, but it does exist for a superfamily that includes the special family.

Answer 2

Here I show why, for example, $X_1$ is not a UMVUE for $g(\theta)=\theta=\mathbb{E}_{P_\theta}[X_1]$ for $P_\theta\in\mathcal{P}_u=\{\operatorname{Unif}(\theta-\tfrac12,\theta+\tfrac12): \theta\in\mathbb{R}\}$.

Let $U(x)=\sin(2\pi x)$. As $U$ is $1$-periodic, $\int^1_0\sin(2\pi x)\,dx=0$, and $\int^1_0\sin^2(2\pi x)\,dx=\frac12$, we have $T_c(X_1)=X_1+cU(X_1)$ is also an unbiased estimator for $g(\theta)$ and $$\operatorname{var}_\theta[T_c(X_1)]=\operatorname{var}_\theta[X_1]+c\big(\frac1{2\pi}\cos(2\pi \theta)+ \frac{c}{2}\big)$$ For each $\theta\in\mathbb{R}\setminus\{\frac{2k+1}{4}:k\in\mathbb{Z}\}$, all $c$ in iside the interval with endpoints $0$ and $-\frac{\pi}{\cos(2\pi \theta)}$ yield estimators $T_c$ of $\theta$ that have smaller variance than that of $X_1$.

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The general argument to answer the first part of the OP is based on measure theory.

The issue is that the population $\mathcal{P}$ of all symmetric distributions dominated by Lebesgue's measure $m$ that have finite moments up-to-order 2 (which it is meant to be the collection of all the absolutely continuous probability measures $P$ on $\mathbb{R}$ for which there is $a\in\mathbb{R}$ such that $\frac{dP}{dm}(x+a)=\frac{dP}{dm}(-x+a)$ $m$-a.s. and $\int x^2 \frac{dP}{dm}(x)\,dm<-\infty$) is large enough so that the sample $(X_1)$ of $\mathcal{P}$ of size $1$ admits a complete sufficient statistic: $T(X_1)=X_1$.

Notice for example that the subfamily $\mathcal{P}_n=\{\operatorname{Norma}(\mu;1):\mu\in\mathbb{R}\}\subset \mathcal{P}$ is of the exponential type and that $T(X_1)=X_1$ is complete sufficient for $\mathcal{P}_n$. Hence, for any measurable function $h$ such that $h\in L_1(P)$ for all $P\in\mathcal{P}$, if $\mathbb{E}_{P}[h(X_1)]=0$ for $P\in\mathcal{P}$, then $\mathbb{E}_{P}[h(X_1)]=0$ for all $P\in\mathcal{P}_n$ and so, $h(X_1)\equiv0$ $m$-a.s. This means that IF $U(X_1)$ is another unbiased estimator for $\phi(P)=\int x \frac{dP}{dm}$, then $h(X_1):=X_1-U(X_1)=0$ $m$-a.s. In other words, $T(X_1)=X_1$ is the unique UMVUE of $\mathcal{P}$.

In contrast, the subfamily $P_u=\{\operatorname{Unif}(\theta-\frac12,\theta+\frac12): \theta\in\mathbb{R}\}\subset\mathcal{P}$, does not admit a complete sufficient statistic. In fact, for any function $g$ on $\mathbb{R}$ that is not constant, there is no UMVUE for $g(\theta)$. Suppose there is one such UMVUE $T(X_1)$. Then for unbiased estimator $U(X_1)\not\equiv0$ of $0$, that is any measurable function $U\not\equiv0$ on $\mathbb{R}$ such that \begin{align} \mathbb{E}_\theta[U(X_1)]=\int^{\theta+\tfrac12}_{\theta-\tfrac12}U(x)\,dx=0, \qquad \theta\in\mathbb{R},\tag{0}\label{zero} \end{align} the statistic $T_c(X_1)=T(X_1)+c U(X_1)$ is also an unbiased estimator for $g(\theta)$ and so, $$\operatorname{var}_\theta[T_c(X_1)]\geq\operatorname{var}_\theta[T(X_1)],\qquad \theta\in\mathbb{R},\,c\in\mathbb{R}.$$ This is equivalent to \begin{align} c^2\mathbb{E}_\theta[U^2(X_1)]+2c\mathbb{E}_\theta[(T(X_1)-g(\theta))U(X_1)]\geq0\tag{1}\label{one} \end{align} for all $c\in\mathbb{R}$ and $\theta\in\mathbb{R}$. Fixing $\theta$ for the moment, inequality~\eqref{one} implies that $$\mathbb{E}_\theta[(T(X)-g(\theta))U(X_1)]=\mathbb{E}_\theta[T(X_1)U(X_1)]=0, \qquad\theta\in\mathbb{R}.$$

Application of Lebesgue's differentiation shows that condition \eqref{zero} implies that $U$ is $1$-periodic $m$-a.s. In particular, $V(x)=T(x)U(x)$ must be a $1$-periodic function $m$-a.s. Hence $T$ is also $1$-periodic a.s. for $T(x+1)U(x+1)=T(x)U(x)$ and $U(x+1)=U(x)$ a.s. From this, it follows that $$g(\theta)=\mathbf{E}_\theta[T(X_1)]=\int^{\theta-\tfrac12}_{\theta-\tfrac12}T(x)\,dx=\int^{-\tfrac12}_{-\tfrac12}T(x)\,dx$$ This all means that $g(\theta)$ is constant!