Under independent assignment,
$$
\mathbb P(Z=z)=\prod_{i=1}^n\pi_{z_i}
=\prod_{r=0}^2\pi_r^{N_r(z)}.
$$
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 $\prod_r\pi_r^{n_r}$, so
$$
\mathbb P(Z=z\mid N_0=n_0,N_1=n_1,N_2=n_2)
=\frac{n_0!n_1!n_2!}{n!}
$$
for compatible $z$, and zero otherwise. Conditioning independent assignment on its arm sizes therefore recovers <complete randomization>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-221/1/i/b/solution