 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/28k/f/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For an exponential proposal of rate

λ > 0

h_{λ} (x) = λ e^{- λ x} 1_{x \geq 0},

and the smallest valid rejection constant is

M (λ) = x \geq 0 sup \frac{f ( x )}{h _{λ} ( x )} = \frac{1}{λ} \frac{2}{π} x \geq 0 sup exp (- \frac{x ^{2}}{2} + λ x) = \frac{2}{π} \frac{e ^{λ^{2} /2}}{λ},

because the exponent is maximized at

x = λ

. Differentiating its logarithm gives

\frac{d}{d λ} lo g M (λ) = λ - \frac{1}{λ} .

The unique minimum occurs at

λ = 1

. Since rejection efficiency is

1/ M (λ)

, choosing any

λ \neq = 1

makes the algorithm less efficient:

the rate-one proposal is optimal within the exponential family .

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