# Proving $X$ is a complete statistic to find a UMVUE

I'm learning about Stein's phenomenon. This standard problem is considered:

Let $$X_1, \dots, X_p$$ be independent random variables with $$X_i \sim N(\theta_i, 1)$$ for $$i = 1, \dots, p$$. Let $$\theta = (\theta_1, \dots, \theta_p)$$ and $$X = (X_1, \dots, X_p)$$. We wish to estimate $$\theta$$ under quadratic loss.

As part of the justification for first considering the "obvious" estimator, $$\hat{\theta} = X$$, it is noted without proof that $$\hat{\theta}$$ is the UMVUE for estimating $$\theta$$.

To show this, I think we wish to use the Lehmann-Scheffé theorem. Immediately, the estimator $$\hat{\theta}$$ is unbiased, and the statistic $$X$$ is sufficient for $$\theta$$.

How can one prove $$X$$ is a complete statistic for the underlying distribution in order to invoke the Lehmann-Scheffé Theorem please?

Many thanks

• At present the q does not make sense, probably it is a typo. $X$ is a vector so canot be an estimator of $\theta$, which is a scalar! Please correct. Commented Sep 27, 2019 at 21:28
• I believe $\theta$ is a vector? In the third line, I defined $\theta = (\theta_1, \dots, \theta_p)$. Commented Sep 27, 2019 at 21:29
• Please note that the $X_i$ are not identically distributed. Each has its own mean, $\theta_i$. Commented Sep 27, 2019 at 21:30

Completeness can be justified indirectly if you invoke results of the Exponential family.

The pdf of $$X$$ for $$\theta\in\mathbb R^p$$ is

\begin{align} f_{\theta}(x)&=\frac{1}{(\sqrt{2\pi})^p}\exp\left[-\frac{1}{2}\sum_{i=1}^p (x_i-\theta_i)^2\right] \\&=\frac{1}{(\sqrt{2\pi})^p}\exp\left[-\frac{1}{2}x^Tx-\frac{1}{2}\theta^T\theta+x^T\theta\right]\quad,\small x=(x_1,\ldots,x_p)\in\mathbb R^p \end{align}

This density is a member of a full rank exponential family, which guarantees that a complete sufficient statistic for $$\theta$$ is $$X^T$$, or simply $$X$$.

I think completeness can also be proved in the following way. Let $$g(\cdot)$$ be any function of $$x$$.

Then, $$E_{\theta}[g(X)]=0\quad\forall\,\theta\implies \int_{\mathbb R^p}e^{x^T\theta}g(x)e^{-\frac12x^Tx}\,dx=0\quad\forall\,\theta$$

The above is a (multidimensional) bilateral Laplace transform of $$g(x)e^{-\frac12x^Tx}$$, which implies $$g(x)e^{-\frac12x^Tx}=0\quad,\text{ a.e.}$$

That is, $$g(x)=0\quad,\text{ a.e.}$$

• I can see that this density is a member of the exponential family. Which result justifies that $X^T$ is a complete sufficient statistic? Commented Sep 30, 2019 at 19:25
• @JohnSmith The relevant result is Proposition 2.1 here, where exponential family is defined in this way. Commented Sep 30, 2019 at 21:05
• Thank you for this. I will accept your answer in a few days, if no other answers are given - I would like to wait a little because it would be more useful for me to have a direct proof that does not require introducing a lot more machinery and then invoking results from that. Commented Oct 1, 2019 at 8:55
• @JohnSmith Do not bother about accepting the answer. A direct proof would be great in any case. Commented Oct 1, 2019 at 15:55