Let $X_1,X_2,Y$ be independent, with $X_1,X_2$ distributed as $X$. Apply part (b) to $X_1,-Y,X_2$:
$$
H(X_1+X_2-Y)+H(Y)\leq2H(X-Y).
$$
Adding an <independent random variable> cannot decrease <information entropy>, so
$$
H(X_1+X_2)+H(Y)\leq2H(X-Y).
$$
In terms of <Entropic Ruzsa distance>, this is $d_R(X,-X)\leq2d_R(X,Y)$. The <Entropic Ruzsa triangle inequality> and invariance under simultaneous negation now give
$$
d_R(X,-Y)
\leq d_R(X,-X)+d_R(-X,-Y)
\leq2d_R(X,Y)+d_R(X,Y),
$$
which is the <Entropic Ruzsa sum-difference inequality>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 224 / 1 / d / Solution

For independent $X,Y$, expand the <Entropic Ruzsa sum-difference inequality> from part (d):
$$
H(X+Y)-\frac12H(X)-\frac12H(Y)
\leq3\left(H(X-Y)-\frac12H(X)-\frac12H(Y)\right).
$$
Collecting the <information entropy> terms gives
$$
H(X+Y)+H(X)+H(Y)\leq3H(X-Y).
$$
Replacing $Y$ by $-Y$ interchanges sum and difference and preserves $H(Y)$, yielding the requested orientation
$$
H(X-Y)+H(X)+H(Y)\leq3H(X+Y).
$$

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 224 / 1 / e / Solution

{c}

For finitely supported random variables,
$$
d_R(X,-Y)\leq3d_R(X,Y).
$$
Equivalently, independent $X,Y$ satisfy
$$
H(X+Y)+H(X)+H(Y)\leq3H(X-Y).
$$


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