Since $n_0=qn_k$ and the $k$ experimental arms each contain $n_k$ patients,
$$
n_k=\frac{n_{\mathrm{tot}}}{k+q}.
$$
Consequently
$$
V=\frac1{n_0}+\frac1{n_k}
=\frac{(q+1)(q+k)}{qn_{\mathrm{tot}}}
=\frac{q+k+1+k/q}{n_{\mathrm{tot}}}.
$$
Differentiation gives $V'(q)=(1-k/q^2)/n_{\mathrm{tot}}$, so the stationary point is $q=\sqrt k$. Since
$$
V''(q)=\frac{2k}{q^3n_{\mathrm{tot}}}>0,
$$
this is the unique minimum. Thus a shared standard arm should be $\sqrt k$ times the size of each new-treatment arm.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 207 / 3 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-207/3/a/i/solution