3
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ – kjetil b halvorsen Sep 27 '19 at 21:28
  • 2
    $\begingroup$ I believe $\theta$ is a vector? In the third line, I defined $\theta = (\theta_1, \dots, \theta_p)$. $\endgroup$ – John Smith Sep 27 '19 at 21:29
  • 2
    $\begingroup$ Please note that the $X_i$ are not identically distributed. Each has its own mean, $\theta_i$. $\endgroup$ – John Smith Sep 27 '19 at 21:30
3
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ I can see that this density is a member of the exponential family. Which result justifies that $X^T$ is a complete sufficient statistic? $\endgroup$ – John Smith Sep 30 '19 at 19:25
  • 1
    $\begingroup$ @JohnSmith The relevant result is Proposition 2.1 here, where exponential family is defined in this way. $\endgroup$ – StubbornAtom Sep 30 '19 at 21:05
  • $\begingroup$ 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. $\endgroup$ – John Smith Oct 1 '19 at 8:55
  • 1
    $\begingroup$ @JohnSmith Do not bother about accepting the answer. A direct proof would be great in any case. $\endgroup$ – StubbornAtom Oct 1 '19 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.