 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2010/o/2/b/ii/solution

Centering gives

\sum_{i} z_{i} = 0

, and hence

X^{T} X = (n 0 0 \sum_{i} z_{i}^{2}) .

The least-squares covariance formula from part (a) therefore gives

Var (α β) = σ^{2} (1/ n 0 0 1/ \sum_{i} (e^{- t_{i}} - c)^{2}) .

Explicitly

α = \overline{Y}

and

β = \sum_{i} z_{i} Y_{i} / \sum_{i} z_{i}^{2}

. Their zero covariance follows directly from the centered predictor, and their joint normality additionally makes them independent.

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