Because $\sigma_{RQE}$ is pure, $S(RQ)=S(E)$ and $S(RE)=S(Q)$. Therefore
$$
\begin{aligned}
I(R:Q)-I(R:E)
&=[S(R)+S(Q)-S(RQ)]\\
&\quad-[S(R)+S(E)-S(RE)]\\
&=2[S(Q)-S(E)].
\end{aligned}
$$
Thus
$$
\boxed{I_c(\Lambda,\rho)=\frac12[I(R:Q)-I(R:E)]}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 323 / 5 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-323/5/d/solution