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

Question

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

Answer 1

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