Find the best unbiased estimator for $\mu^T \mu + 1^T \mu$.

Question

Let $X_1, X_2, \dots, X_n$ independent $n$-dimensional vectors with the same distribution $N(\mu, I)$. Find the best unbiased estimator for

$$ \mu^T \mu + 1^T\mu $$

where $1^T = (1, 1, \dots, 1)$.

How to solve this problem?

Thx.

Answer 1

As the joint pdf of $X_1,\ldots,X_n$ is a member of a regular (full rank) exponential family, it follows that a complete sufficient sufficient statistic for $\mu$ is $\sum\limits_{i=1}^n X_i$, or equivalently the sample mean vector $\overline X=\frac1n\sum\limits_{i=1}^n X_i$. By Lehmann-Scheffé, the uniformly minimum variance unbiased estimator of $g(\mu)=\mu^T\mu+1^T\mu$ is that unbiased estimator of $g$ which is based on $\overline X$.

Since $X_i\sim N_n(\mu,I_n)$ independently for all $i$, we have $$\overline X=(\overline X_1,\ldots,\overline X_n)^T\sim N_n\left(\mu,\frac1n I_n\right)$$

So for every $\mu=(\mu_1,\ldots,\mu_n)^T\in \mathbb R^n$,

$$E_{\mu}\left[\overline X^T\overline X\right]=E_{\mu}\left[\sum_{i=1}^n \overline X_i^2\right]=\sum_{i=1}^n\left(\frac1n+\mu_i^2\right)=1+\mu^T\mu$$

And $$E_{\mu}\left[1^T \overline X\right]=1^T\mu$$

Hence, $$E_{\mu}\left[\overline X^T\overline X+1^T\overline X-1\right]=g(\mu)$$