= Solution
{c}
\b[They need not be <independent>.] For a counterexample take $n=2$, $x_1=-1$, $x_2=1$, and independent errors taking $-1,1$ with equal <probability>. Then
$$
\widehat\alpha-\alpha=\frac{\epsilon_1+\epsilon_2}{2},\qquad
\widehat\beta-\beta=\frac{\epsilon_2-\epsilon_1}{2}.
$$
Each centered <estimator> is zero with <probability> $1/2$, but they are never both zero. Their zero <events> therefore violate <independence>.
A sufficient additional assumption is that the errors have a common <normal distribution>. The two <estimators> are then <Jointly Gaussian>, being linear combinations of the independent <Gaussian random variables>. Their zero <covariance>, established in <7/b/iii>, implies <independence of uncorrelated jointly normal variables>. Normal errors are a sufficient condition here, not a claim that no other error <probability distribution> could ever work.
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