 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

Each residual has mean zero and <variance> $\sigma^2(1-1/n)$. By <linearity of expectation>, without needing independence of residuals,
$$
\mathbb E[S^2]=\frac1{n-1}\sum_{i=1}^n\mathbb E[D_i^2]
=\frac n{n-1}\sigma^2(1-1/n)=\boxed{\sigma^2}.
$$
Hence $S^2$ is an <unbiased estimator> of $\sigma^2$.

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