Discussion
In a linear model \(Y_{4 \times 1} = X_{4 \times 3}\,\theta_{3 \times 1} + \epsilon_{4 \times 1}\) \(E(Y_{1}) = E(Y_{3}) = \theta_{1} + \theta_{2}\) and \(E(Y_{2}) = E(Y_{4}) = \theta_{1} - \theta_{3}\). Hence which of the following statement is true ?
Answer: D
The estimable function is \(\theta_{2} + \theta_{3}\).
A linear function is estimable exactly when it is a linear combination of the expected observations. Here those are \(\theta_{1} + \theta_{2}\) and \(\theta_{1} - \theta_{3}\), so estimable functions have the form
\[ \begin{aligned} &a(\theta_{1} + \theta_{2}) + b(\theta_{1} - \theta_{3}) \\ &= (a + b)\theta_{1} + a\theta_{2} - b\theta_{3} \end{aligned} \]
For \(\theta_{2} + \theta_{3}\): \(a = 1\), \(b = -1\), \(a + b = 0\). It works, and \(E(Y_{1} - Y_{2}) = \theta_{2} + \theta_{3}\).
For \(\theta_{1} + \theta_{2} + \theta_{3}\): \(a = 1\), \(b = -1\) gives coefficient 0 for \(\theta_{1}\), not 1. For \(\theta_{1} + \theta_{3}\): \(a = 0\), \(b = -1\) gives \(-1\) for \(\theta_{1}\). The individual \(\theta\)'s are not estimable, since the design matrix has rank 2.
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