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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.

The Herbst argument integrates a logarithmic Sobolev bound on $\lambda H^{\prime }(\lambda )-H(\lambda )$ to control the centered cumulant-generating function $H$, then applies a Chernoff bound to obtain concentration.

A modified logarithmic Sobolev inequality bounds the entropy of $e^{\lambda Z}$ by a one-sided coordinate-difference energy. For independent coordinates and arbitrary coordinate-deleted functions $f_i$, tensorization givesIf that energy is at most $v$, one common normalization is

Let $X_1,\dots ,X_n$ be independent random variables, let $f=f(X_1,\dots ,X_n)$, and let each $f_i$ depend on every coordinate except $X_i$. The modified logarithmic Sobolev inequality states that, for every real $\lambda$ for which the expectations exist,

To prove it, first apply tensorization of entropy:Condition on all coordinates except $X_i$ and use the stated variational formula with the admissible constant $u=e^{\lambda f_i}$. The $i$th conditional entropy is at mostSumming and taking the remaining expectations proves the inequality.

Put $Z_i=f_i(X^{(i)})$ and ${\Delta }_i=Z-Z_i$. The weakly self-bounding function assumptions give ${\Delta }_i\ge 0$ and ${\sum }_i{\Delta }_i^2\le Z$. For $0\le \lambda <2$, the bound $\varphi (-x)\le x^2/2$ and the modified logarithmic Sobolev inequality implyLet $H(\lambda )=\log \mathbb{E}{e}^{\lambda Z}$. Dividing by $\mathbb{E}{e}^{\lambda Z}$ turns this intoand thereforeSince $H(\lambda )\sim \lambda \mathbb{E}Z$ as $\lambda \downarrow 0$, integration givesSubtracting $\lambda \mathbb{E}Z$ from both sides yieldsas required. This is a Herbst argument with a variance proxy controlled by $Z$ itself.

One useful form of the modified logarithmic Sobolev inequality is the following. For a function $F$ of independent coordinates, letand $V^{+}(x)={\sum }_i(F(x)-F_i(x^{(i)}){)}^2$. If $V^{+}\le v$ and $\lambda \ge 0$, the inequality givesand the Herbst argument yields

Talagrand's one-sided bounded differences inequality gives the complementary tail under the same one-sided bounded-difference condition:Equivalent versions use an independent coordinate replacement and its conditional positive-part variance proxy.

For convex $f$, the subgradient inequality givesTaking the positive supremum over $z\in [0,1]$ and summing squares showsThe modified logarithmic Sobolev inequality from part a with $v=1$ therefore gives