Solution. A linear estimator \(\sum _i a_iY_i\) is unbiased for \(\theta \) if and only if \(\sum _i a_i = 1\), since \(E\left (\sum _i a_iY_i\right )=\theta \sum _i a_i\). Its variance is \(\sum _i a_i^{2}\sigma _i^{2}\) by independence, so the problem is to minimise \(\sum _i a_i^{2}\sigma _i^{2}\) subject to \(\sum _i a_i=1\).

With a Lagrange multiplier \(\lambda \), differentiating \(\sum _i a_i^{2}\sigma _i^{2} - \lambda \left (\sum _i a_i - 1\right )\) gives \(2a_i\sigma _i^{2}=\lambda \), so \(a_i \propto 1/\sigma _i^{2}\). Normalising to make the weights sum to one, \[\widehat {\theta } = \frac {\sum _{i=1}^{n} Y_i/\sigma _i^{2}} {\sum _{i=1}^{n} 1/\sigma _i^{2}}, \qquad \var \left (\widehat {\theta }\right ) = \frac {1}{\sum _{i=1}^{n}1/\sigma _i^{2}} .\]

Solution. The constraint \(\alpha _1+\alpha _2=0\) means \(\alpha _2=-\alpha _1\), so \[E\left (\overline {Y}_{1\cdot }\right ) = \mu +\alpha _1,\qquad E\left (\overline {Y}_{2\cdot }\right ) = \mu -\alpha _1 ,\] where \(\overline {Y}_{i\cdot }=n^{-1}\sum _{j=1}^{n}Y_{ij}\). Subtracting removes \(\mu \): \[E\left (\overline {Y}_{1\cdot }-\overline {Y}_{2\cdot }\right ) = 2\alpha _1 ,\] so \[\widehat {\alpha _1} = \frac {\overline {Y}_{1\cdot }-\overline {Y}_{2\cdot }}{2}\] is linear and unbiased. Its variance is \[\var \left (\widehat {\alpha _1}\right ) = \frac {1}{4}\left (\frac {\sigma ^{2}}{n}+\frac {\sigma ^{2}}{n}\right ) = \frac {\sigma ^{2}}{2n} .\] That it is best follows from the Gauss–Markov theorem: the model is linear with uncorrelated errors of constant variance, and \(\widehat {\alpha _1}\) is the least squares estimator of the estimable function \(\alpha _1\) under the constraint.