= Solution
The <multivariate Wald statistic> for a proposed value $\theta$ is
$$
W_n(\theta)
=n(\overline X-\theta)^T\widehat i_n(\overline X-\theta).
$$
Because $\widehat i_n\to I_p$ and
$$
\sqrt n(\overline X-\theta_0)\sim N_p(0,I_p),
$$
<Slutsky theorem> gives
$$
\boxed{W_n(\theta_0)\xrightarrow{\mathrm d}\chi_p^2.}
$$
If $q_{p,1-\alpha}$ is the $(1-\alpha)$-quantile of the <chi-squared distribution> with $p$ degrees of freedom, an asymptotic $(1-\alpha)$ confidence region is
$$
\boxed{\left\{\theta:
n(\overline X-\theta)^T\widehat i_n(\overline X-\theta)
\leq q_{p,1-\alpha}\right\}.}
$$
In this known-identity-covariance model one may replace $\widehat i_n$ by $I_p$; then the confidence ball is exact. For an individual coordinate, the corresponding exact <confidence interval> is
$$
\overline X_j\mathbin\pm z_{1-\alpha/2}/\sqrt n.
$$
Back to article page
