 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iii/7/b/ii/solution

The inverse moments in part i give

E [\hat{θ}] = \frac{n θ}{n - 1}, E [\hat{θ}^{2}] = \frac{n ^{2} θ ^{2}}{( n - 1 ) ( n - 2 )} .

Thus the bias is positive, and direct expansion of the mean squared error yields

b_{θ} (\hat{θ}) = \frac{θ}{n - 1}, MSE_{θ} (\hat{θ}) = \frac{( n + 2 ) θ ^{2}}{( n - 1 ) ( n - 2 )} .

For comparison, the variance is

n^{2} θ^{2} / [(n - 1)^{2} (n - 2)]

; adding the squared bias gives the same answer by the bias-variance decomposition.

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