Secrecy capacity of a finite degraded wiretap channel

Fix any finite $P_{U,X}$ and form the joint law $P_{U,X,Y,Z}(u,x,y,z)=P_{U,X}(u,x)W(y,z\mid x)$, which is the notation $U-X-(Y,Z)$. For each $u$ with $P_U(u)>0$, the degraded factorization [F1] makes $X-Y-Z$ under the posterior law $P_{X\mid U=u}$, so [F4] gives $I(X;Z\mid U=u)\le I(X;Y\mid U=u)$. Averaging over $u$ yields $I(X;Z\mid U)\le I(X;Y\mid U)$.

Conversely, for any input law $P_X$, take $U=X$. Then $U-X-(Y,Z)$ holds and $I(U;Y)-I(U;Z)=I(X;Y)-I(X;Z)$. Thus the supremum over auxiliaries is at least the supremum of the displayed single-letter objective.

The Markov condition $U-X-Y$ gives $I(U;Y\mid X)=0$, so the chain rule [F3] yields $I(X;Y)=I(U;Y)+I(X;Y\mid U)$. Likewise $U-X-Z$ gives $I(X;Z)=I(U;Z)+I(X;Z\mid U)$. Subtracting these identities and using step 1.1 shows $I(U;Y)-I(U;Z)=I(X;Y)-I(X;Z)-(I(X;Y\mid U)-I(X;Z\mid U))\le I(X;Y)-I(X;Z)\mathrm{.}$

By [F5], choose an auxiliary law attaining the maximum in the general formula. Step 2.1 shows that replacing this law by $U=X$ at the same marginal $P_X$ cannot decrease its value, so this input law attains the maximum of the single-letter expression. Every input law is itself an allowed auxiliary with $U=X$ by step 1.2. Thus the auxiliary maximum equals the displayed maximum, and [F6] identifies it with all four capacities in the sense of [F7].