= Solution
{c}
The inverse moments in part <7/b/i> give
$$
\mathbb E[\hat\theta]=\frac{n\theta}{n-1},\qquad
\mathbb E[\hat\theta^2]=\frac{n^2\theta^2}{(n-1)(n-2)}.
$$
Thus the <bias> is positive, and direct expansion of the <mean squared error> yields
$$
\boxed{b_\theta(\hat\theta)=\frac\theta{n-1},\qquad
\operatorname{MSE}_\theta(\hat\theta)=\frac{(n+2)\theta^2}{(n-1)(n-2)}.}
$$
For comparison, the <variance> is $n^2\theta^2/[(n-1)^2(n-2)]$; adding the squared <bias> gives the same answer by the <bias-variance decomposition>.
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