= Solution
The transmitted branch is the <Kraus operator>
$$
\boxed{E=\sqrt{1-R}\,|0\rangle\langle0|+|1\rangle\langle1|,\qquad E^\dagger E=\operatorname{diag}(1-R,1).}
$$
For nonzero loss it is not trace preserving: $\operatorname{tr}(E\rho E^\dagger)$ is the transmission probability. A complete channel includes an orthogonal vacuum or erasure flag, with a loss operator $L=\sqrt R\,|\mathrm{vac}\rangle\langle0|$ and $E^\dagger E+L^\dagger L=I$. Postselecting only the first outcome gives the single-operator description in the question.
For $|\pm\rangle$, the unnormalized transmitted vector is $(\sqrt{1-R}|0\rangle\pm|1\rangle)/\sqrt2$. Its norm squared and normalized conditional state are
$$
\boxed{P_{\mathrm{tr}}=1-\frac R2,\qquad|\pm_R\rangle=\frac{\sqrt{1-R}|0\rangle\pm|1\rangle}{\sqrt{2-R}},\qquad P_{\mathrm{loss}}=\frac R2.}
$$
The reduced density matrix of either member of the initial pair has eigenvalues $\alpha,1-\alpha$. Since the joint state is pure, its <Von Neumann entropy> is zero. With logarithms to base two and $h_2$ the <binary entropy function>,
$$
\boxed{S(AB)=0,\qquad S(A)=S(B)=h_2(\alpha),\qquad I(A:B)=2h_2(\alpha).}
$$
The two nonzero <Schmidt coefficients> show that this pure pair is entangled exactly when $0<\alpha<1$. Its mutual information rises from zero to two bits at $\alpha=1/2$ and returns to zero.
For this known pure family, every positive mutual information signals entanglement. For an otherwise unspecified two-qubit state, positive mutual information can also be classical. The <mutual information bound for separable states> gives the stronger general witness $I(A:B)>1$ bit. On this curve it holds for approximately $0.110028<\alpha<0.889972$; the rest of the open interval is still entangled, but that generic one-bit witness alone would not certify it.
\Image[/past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2016-filtered-pair-mutual-information.png]
{title=Mutual information of the pure pair, the generic one-bit entanglement witness, and the optimal lossy-filter compensation}
If Bob's qubit passes through the channel, the unnormalized retained state is $\sqrt\alpha|11\rangle+\sqrt{(1-\alpha)(1-R)}|00\rangle$. Thus
$$
\boxed{P_s=\alpha+(1-\alpha)(1-R),\qquad|\psi_s\rangle=\frac{\sqrt\alpha|11\rangle+\sqrt{(1-\alpha)(1-R)}|00\rangle}{\sqrt{P_s}}.}
$$
For $0\le R<1$, equal <Schmidt coefficients> require $\alpha=(1-\alpha)(1-R)$, giving
$$
\boxed{\alpha_{\mathrm{opt}}=\frac{1-R}{2-R},\qquad P_s=\frac{2(1-R)}{2-R},\qquad\text{pair rate}=N\frac{2(1-R)}{2-R}.}
$$
The retained state is then $(|00\rangle+|11\rangle)/\sqrt2$. At $R=1$, this limiting compensation has zero transmission probability and yields no maximally entangled pairs. Conditioning can increase the entanglement of retained pairs, while the discarded fraction accounts for the cost.
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