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ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/probability-and-computing/3/a/solution

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Past exam of the computer science course of the University of Oxford / 2015 / Probability and computing / 3 / a / Solution by

Codex  0 2026-10-31

A trial of Amy's rejection sampling uses exactly

2 n

fair Bernoulli trials. Exactly

(n 2 n)

of the

2^{2 n}

equally likely binary sequences have

n

heads, so the acceptance probability is

p_{n} = \frac{( n 2 n )}{2 ^{2 n}} .

Independent retries make the number

T

of trials a geometric random variable on

{1, 2, \dots}

with expectation

1/ p_{n}

. The total number

C

of coin tosses is

2 n T

, so

E C = \frac{2 n 2 ^{2 n}}{( n 2 n )} .

Using Stirling's approximation, the central binomial coefficient satisfies

(n 2 n) \sim \frac{4 πn ( 2 n / e ) ^{2 n}}{2 πn ( n / e ) ^{2 n}} = \frac{4 ^{n}}{πn} .

Consequently

E C \sim 2 π n^{3/2} .

The rejection sampling output is uniform among the balanced binary sequences: they have equal probability on every trial, and conditioning on acceptance preserves that equality. For

n = 0

, the empty output needs zero tosses; the formula above concerns

n \geq 1

.

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