= Solution
Write $\delta=d_R(X,Y)$, $A=U_1+U_2$, $B=V_1+V_2$, and $S=A+B$. Distances depend only on distributions, so all variables used in any one application may be realized as <independent random variables>.
First, the <entropy submodularity for three independent sums> implies
$$
d_R(U_1+U_2,X)
\leq\frac12\bigl(2d_R(U,X)+d_R(U,U)\bigr),
$$
and analogously for $V$. The <Entropic Ruzsa triangle inequality> gives $d_R(U,U)\leq2d_R(U,X)$ and $d_R(V,V)\leq2d_R(V,Y)$. Consequently the <relevance of independent self-sums> gives
$$
p+q:=d_R(A,X)+d_R(B,Y)\leq2C\delta.
$$
Apply the <Conditioned entropic Ruzsa distance of a summand> first to $(A,B,X)$ and then to $(B,A,Y)$:
$$
\begin{aligned}
d_R(A\mid S;X)&\leq\frac12\bigl(d_R(A,X)+d_R(B,X)+d_R(A,B)\bigr),\\
d_R(B\mid S;Y)&\leq\frac12\bigl(d_R(B,Y)+d_R(A,Y)+d_R(A,B)\bigr).
\end{aligned}
$$
Three applications of the <Entropic Ruzsa triangle inequality> give
$$
d_R(B,X)\leq q+\delta,
\qquad
d_R(A,Y)\leq p+\delta,
\qquad
d_R(A,B)\leq p+\delta+q.
$$
Adding all these bounds yields
$$
\begin{aligned}
d_R(A\mid S;X)+d_R(B\mid S;Y)
&\leq2(p+q+\delta)\\
&\leq(4C+2)\delta.
\end{aligned}
$$
Thus the required absolute constants may be taken as $a=4$ and $b=2$.
Solved by gpt-5.6-sol high.
Back to article page
