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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-43/1/a/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 43 / 1 / a / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

For any measurable set

A

, integrating the joint acceptance event in rejection sampling gives

P (Y \in A, accept) = \int_{A} g (y) \frac{f ( y )}{C g ( y )} d y = \frac{1}{C} \int_{A} f (y) d y .

Setting

A

equal to the whole support yields

P (accept) = 1/ C

. Thus

P (Y \in A ∣ accept) = \int_{A} f (y) d y .

To verify repeated attempts directly, the probability that the first accepted observation falls in

A

is

\sum_{r = 1}^{\infty} (1 - 1/ C)^{r - 1} \frac{1}{C} \int_{A} f (y) d y = \int_{A} f (y) d y .

With finite

C

, eventual acceptance has probability one. Applying the same argument after each success proves that the retained sequence consists of independent draws from the target.

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