The entropy method combines a coordinatewise certificate with <tensorization of entropy> to bound the <moment-generating function> of a certifiable random quantity and hence its tails.

Entropy method for certifiable functions

Let $d_i=f(X)-f_i(X^{(i)})$. The <self-bounding function> assumptions say $0\leq d_i\leq1$ and $\sum_i d_i\leq Z$. Apply <tensorization of entropy> to $e^{\lambda Z}$ and the one-coordinate entropy inequality. Since $\varphi(u)=e^u-u-1$ is convex and $\varphi(td)\leq d\varphi(t)$ for $0\leq d\leq1$, the resulting bound is
$$
\operatorname{Ent}(e^{\lambda Z})
\leq\varphi(-\lambda)\mathbb E[Ze^{\lambda Z}].
$$

Writing $\psi(\lambda)=\log\mathbb E e^{\lambda(Z-\mu)}$ and dividing by the moment-generating function reduces this to
$$
\left(\frac{\psi(\lambda)}{e^\lambda-1}\right)'
\leq\mu\left(\frac{-\lambda}{e^\lambda-1}\right)'.
$$
Both sides have finite limits at zero and $\psi(0)=\psi'(0)=0$. Integrating from zero to $\lambda$, with the direction interpreted correctly when $\lambda<0$, yields
$$
\psi(\lambda)\leq\mu(e^\lambda-\lambda-1)=\mu\varphi(\lambda),
$$
which is the required inequality.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 208 / 2 / b / Solution

For a nonnegative <function> $Z$ of <independent random variables>, entropy is at most the sum of its conditional coordinate entropies. This tensorization is a basic step of the <entropy method for certifiable functions>[entropy method].

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 Incoming links: Tensorization of entropy