Discussion
Consider the model \(E(Y_{1}) = \beta_{1} + \beta_{2}\), \(E(Y_{2}) = \beta_{1} - \beta_{2}\), \(E(Y_{3}) = \beta_{1} + 2\beta_{2}\) with the assumption that \(Y_{i}\)'s are uncorrelated and \(\text{var}(Y_{i}) = \sigma^{2}\) for all \(i = 1, 2, 3\). Then the best linear unbiased estimator of \(\beta_{2}\) is :
Answer: C
The BLUE of \(\beta_{2}\) is \(\dfrac{Y_{1} - 5Y_{2} + 4Y_{3}}{14}\).
The design matrix has rows (1, 1), (1, -1), (1, 2). By the Gauss-Markov theorem the BLUE is the least squares estimator \((X'X)^{-1}X'Y\):
\[ \begin{aligned} X'X &= \begin{pmatrix} 3 & 2 \\ 2 & 6 \end{pmatrix},\quad (X'X)^{-1} = \frac{1}{14}\begin{pmatrix} 6 & -2 \\ -2 & 3 \end{pmatrix} \\ X'Y &= (Y_{1} + Y_{2} + Y_{3},\ Y_{1} - Y_{2} + 2Y_{3})' \end{aligned} \]
\[ \begin{aligned} \hat{\beta}_{2} &= \frac{-2(Y_{1} + Y_{2} + Y_{3}) + 3(Y_{1} - Y_{2} + 2Y_{3})}{14} \\ &= \frac{Y_{1} - 5Y_{2} + 4Y_{3}}{14} \end{aligned} \]
Check unbiasedness: the coefficients give \(\beta_{1}(1 - 5 + 4)/14 + \beta_{2}(1 + 5 + 8)/14 = \beta_{2}\). The first option has expectation \(\beta_{1} + 2\beta_{2}/3\), the second has expectation \(\beta_{1}\), and the last has expectation \(\frac{4 - 8 + 4}{3}\beta_{1} + \frac{4 + 8 + 8}{3}\beta_{2}\), which is \(\frac{20}{3}\beta_{2}\).
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