For priors $p_1,p_2$,
$$
p_{\rm succ}
=p_2+\operatorname{Tr}[E(p_1\rho_1-p_2\rho_2)].
$$
If $\Delta=p_1\rho_1-p_2\rho_2$, the maximum is $p_2+\operatorname{Tr}\Delta_+$. Since
$$
\operatorname{Tr}\Delta_+
=\frac12(\|\Delta\|_1+\operatorname{Tr}\Delta)
=\frac12(\|\Delta\|_1+p_1-p_2),
$$
one obtains
$$
\boxed{p_{\rm succ}^{\max}
=\frac12\left(1+\|p_1\rho_1-p_2\rho_2\|_1\right)}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 4 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-323/4/b/ii/solution