= Solution
{c}
Centering gives $\sum_i z_i=0$, and hence
$$
X^{\mathsf T}X=\begin{pmatrix}n&0\\0&\sum_i z_i^2\end{pmatrix}.
$$
The least-squares <covariance> formula from part (a) therefore gives
$$
\boxed{\operatorname{Var}\begin{pmatrix}\widehat\alpha\\\widehat\beta\end{pmatrix}
=\sigma^2\begin{pmatrix}1/n&0\\0&1/\sum_i(e^{-t_i}-c)^2\end{pmatrix}.}
$$
Explicitly $\widehat\alpha=\overline Y$ and $\widehat\beta=\sum_i z_iY_i/\sum_i z_i^2$. Their zero <covariance> follows directly from the centered predictor, and their joint normality additionally makes them independent.
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