Discussion
\(y_{1}, y_{2}, y_{3}\) are three uncorrelated random variables with common variance \(\sigma^{2}\) and \(E(y_{1}) = \theta_{1} + \theta_{3}\), \(E(y_{2}) = \theta_{2} + \theta_{3}\), \(E(y_{3}) = \theta_{1} - \theta_{3}\). Hence :
Answer: B
The correct statement is the BLUE of \(\theta_{2}\) is \(y_{2} - (y_{1} - y_{3})/2\) with variance \(\frac{3}{2}\sigma^{2}\).
Three observations and three parameters give an exactly determined system: \(y_{1} - y_{3}\) estimates \(2\theta_{3}\), \(y_{1} + y_{3}\) estimates \(2\theta_{1}\), and then \(y_{2} - \hat{\theta}_{3}\) estimates \(\theta_{2}\). Each estimable function has a unique linear unbiased estimator, which is therefore the BLUE.
\[ \begin{aligned} \mathrm{Var}\left(y_{2} - \tfrac{y_{1} - y_{3}}{2}\right) &= \sigma^{2} + \tfrac{1}{4}(2\sigma^{2}) \\ &= \tfrac{3}{2}\sigma^{2} \end{aligned} \]
\(y_{2}\) alone estimates \(\theta_{2} + \theta_{3}\), not \(\theta_{2}\). The BLUE of \(\theta_{1}\), \((y_{1} + y_{3})/2\), has variance \(\sigma^{2}/2\), not \(\sigma^{2}\), and \((y_{1} + y_{2})/2\) is biased for \(\theta_{1}\).
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