 Solution

The pivot $Z=\sqrt{n}{\Sigma }^{-1/2}(\overline{X}-\theta )$ has independent standard normal coordinates, so $Z^{T}Z$ has the chi-squared distribution with $d$ degrees of freedom. Let $q_{d,1-\alpha }$ be its $(1-\alpha )$ quantile. The confidence ellipsoidtherefore has exactly $1-\alpha$ coverage for every $\theta$. No large-sample approximation is required when the covariance is known.