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For a nonnegative weakly self-bounding function $Z=f(X_1,\dots ,X_n)$ of independent random variables, a Gaussian lower-tail conclusion requires a further coordinate-increment hypothesis in the standard theorem. Suppose $m=\mathbb{E}Z\in (0,\infty )$ and the measurable coordinate-deleted functions satisfyThenThis is the weakly $(a,b)$ self-bounding lower-tail theorem with $a=1,b=0$. The bound on each increment is a separate hypothesis and must not be dropped when invoking that theorem.^[ref]

For example, for $Z={\sum }_i{X}_i$ with independent Bernoulli distribution variables, deleting coordinate $i$ changes $Z$ by $X_i\in \{0,1\}$. Both displayed hypotheses hold since ${\sum }_i{X}_i^2=Z$. With $m=100$, a downward deviation of $20$ has probability at most $e^{-2}$. If $m=0$, nonnegativity gives $Z=0$ almost surely; if $t>m$, the lower-tail event is empty.

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.

Let $Z_i$ be the longest increasing subsequence length after deleting coordinate $i$. Then $0\le Z-Z_i\le 1$. Choose one longest increasing subsequence $I$ of length $Z$. If deleting $i$ reduces the optimum, then $i$ must belong to $I$; consequently at most $Z$ coordinates can satisfy $Z-Z_i=1$. Henceso $Z$ is a weakly self-bounding function. The variance bound for a weakly self-bounding function gives

For a weakly self-bounding function $Z=f(X_1,\dots ,X_n)$ of independent random variables, the coordinatewise variance inequality giveswhere $Z_i=f_i(X^{(i)})$.