 Solution

Put $D^{\ast }=D-A$. By (i), $D^{\ast }X=0$. Thus $D^{\ast }{A}^T=D^{\ast }X(X^{T}X{)}^{-1}=0$, and its transpose vanishes too. Expanding $D{D}^T=(D^{\ast }+A)(D^{\ast }+A{)}^T$ yieldsThe covariance difference is positive semidefinite. In particular its $i$th diagonal entry is ${\sigma }^2{\sum }_j(D_{ij}^{\ast }{)}^2\ge 0$, proving the requested componentwise variance inequalities. Prefer $\boxed{{\hat{\theta }}_1}$ within the class of linear unbiased estimators, since it has the least variance and the same expectation. A tie is possible if the first row of $D^{\ast }$ is zero. This is the Gauss-Markov theorem conclusion; no normal-distribution assumption on the errors is required.