 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/5l/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Put

K = nY

. Apart from a factor independent of

p

, the likelihood is

L (p) = p^{nY} (1 - p)^{n (1 - Y)} .

S (p) = ℓ^{'} (p) = n (\frac{Y}{p} - \frac{1 - Y}{1 - p}) = \frac{n ( Y - p )}{p ( 1 - p )}, I (p) = \frac{n}{p ( 1 - p )} .

p^{(r + 1)} = p^{(r)} - \frac{ℓ ^{'} ( p ^{(r)} )}{ℓ ^{''} ( p ^{(r)} )}, p^{(r + 1)} = p^{(r)} + I (p^{(r)})^{- 1} S (p^{(r)}) .

Since

p = Y

, initialization at

p^{(0)} = Y

requires zero steps for either algorithm. Moreover,

p + I (p)^{- 1} S (p) = p + \frac{p ( 1 - p )}{n} \frac{n ( Y - p )}{p ( 1 - p )} = Y,

so binomial-proportion Fisher scoring reaches the MLE in one step from every interior

p^{(0)} \neq = Y

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