= Solution
{c}
Put $D_j=X_j-\overline X$. The supplied linear-transformation result shows that $(\overline X,D_2,\ldots,D_n)$ is jointly <multivariate normal>. Its marginal <distributions> are
$$
\boxed{\overline X\sim N(\mu,\sigma^2/n),\qquad D_j\sim N(0,\sigma^2(1-1/n))\quad(2\leq j\leq n).}
$$
Indeed, $\operatorname{Cov}(X_j,\overline X)=\sigma^2/n$, so
$$
\operatorname{Cov}(\overline X,D_j)=\sigma^2/n-\sigma^2/n=0.
$$
For a jointly <Gaussian vector>, zero cross-covariance between blocks implies their <independence>. Thus $\overline X$ is independent of the entire vector $(D_2,\ldots,D_n)$, not merely of each coordinate separately. The residual coordinates themselves need not be independent: for $j\ne k$, their <covariance> is $-\sigma^2/n$.
Back to article page
