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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 164 / 4 / ii / Solution

{c}

If $U,V,X$ are independent random variables in $\mathbb F_2^n$, then
$$
d_R(U\mid U+V;X)
\leq\frac12\bigl(d_R(U;X)+d_R(V;X)+d_R(U;V)\bigr).
$$
Indeed, <conditioning reduces entropy> and independence give
$$
d_R(U\mid U+V;X)
\leq H(U+X)-\frac12H(U)-\frac12H(V)
+\frac12H(U+V)-\frac12H(X).
$$
In $\mathbb F_2^n$, $V=U+(U+V)$, so <conditional entropy under a deterministic change of variables> shows that the left-hand side is unchanged when $U$ and $V$ are exchanged. Averaging the displayed bound with its exchanged version gives the result.

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