Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 221 / 1 / i / b

Table of contents
- Solution b

Solution

Under independent assignment,

P (Z = z) = \prod_{i = 1}^{n} π_{z_{i}} = \prod_{r = 0}^{2} π_{r}^{N_{r} (z)} .

The arm-count vector has a multinomial distribution. Conditional on

N_{r} = n_{r}

for

r = 0, 1, 2

, every compatible vector has the same factor

\prod_{r} π_{r}^{n_{r}}

, so

P (Z = z ∣ N_{0} = n_{0}, N_{1} = n_{1}, N_{2} = n_{2}) = \frac{n _{0} ! n _{1} ! n _{2} !}{n !}

for compatible

z

, and zero otherwise. Conditioning independent assignment on its arm sizes therefore recovers complete randomization.

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