The binary XOR MAC exposes the independent private-input law

For independent Bernoulli inputs with parameters $p,q$, given $X_2$ the XOR is a bijective relabeling of $X_1$, and given $(X_1,X_2)$ the output is fixed. Thus [F4]–[F5] give $I(X_1;Y\mid X_2)=H(X_1\mid X_2)=h_b(p)$; interchanging users gives $h_b(q)$. The deterministic output and [F3], [F6] give $I(X_1,X_2;Y)=H(Y)=h_b(p+q-2pq)$.

For any finite time-sharing variable, the sum bound is $I(X_1,X_2;Y\mid Q)=H(Y\mid Q)\le \ln 2$, because each conditional output is binary. With $Q$ constant and $p=q=1\mathrm{/}2$, both individual informations and the sum information equal $\ln 2$. Hence every nonnegative pair with sum at most $\ln 2$ is in the closed region, and no pair with larger sum is.

The diagonal joint law has probability $1\mathrm{/}2$ on $(0,0)$ and $(1,1)$, so it is not a product law: its marginals are uniform but their product puts positive mass on $(0,1)$ and $(1,0)$. More generally, suppose a finite mixture $P_Q{P}_{X_1\mathrm{\mid }Q}{P}_{X_2\mathrm{\mid }Q}$ equals this diagonal law. For every positive-mass $q$, nonnegativity and zero total off-diagonal mass force $P(X_1=0\mathrm{\mid }q)P(X_2=1\mathrm{\mid }q)=P(X_1=1\mathrm{\mid }q)P(X_2=0\mathrm{\mid }q)=0$. If $X_1$ had positive conditional mass on both bits, these equalities would force $P(X_2=0\mathrm{\mid }q)=P(X_2=1\mathrm{\mid }q)=0$, impossible; hence $X_1$ is deterministic given $q$, and the two equalities then force $X_2$ to be the same deterministic bit. Thus any such public time-sharing representation uses deterministic diagonal product components. With $Q$ selecting $(0,0)$ or $(1,1)$, [F5] gives $I(X_1;Y\mid X_2,Q)=I(X_2;Y\mid X_1,Q)=0$; the mixture adds no private-message rate.