Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 12 / 2

Use the uniform probability measure on the symmetric group $S_n$ and the unnormalized Hamming distanceLet $f:S_n\to \mathbb{R}$ have Lipschitz constant $K>0$ for this distance. We will prove the one-sided concentration inequalitiesConsequently $\mathbb{P}(|f-\mathbb{E}f|\ge t)\le 2e^{-t^2/(2n{K}^2)}$ for $t>0$. A zero Lipschitz constant makes $f$ constant, so that case is immediate.

Let $\sigma$ be a uniform permutation, reveal $\sigma (1),\dots ,\sigma (k)$ successively, and define the Doob exposure martingaleFix the first $k-1$ images. For an available value $a$, let $h(a)$ be the conditional mean when $\sigma (k)=a$. Given two available values $a,b$, left-composition with the transposition exchanging $a,b$ is a bijection from completions with next image $a$ to completions with next image $b$. It fixes the previously exposed images and changes exactly two coordinates of each complete permutation. Since conditional completions are uniform, averaging the Lipschitz condition over this pairing gives $|h(a)-h(b)|\le 2K$.

Thus, conditionally on the first $k-1$ images, $M_k-M_{k-1}$ has mean zero and takes values in an interval of length at most $2K$. We need the range-length version of the Hoeffding lemma: if a real random variable $Z$ has mean zero and lies in an interval of length $\ell$, thenHere is a proof. For $\psi (\lambda )=\log \mathbb{E}{e}^{\lambda Z}$, differentiation gives ${\psi }^{\prime \prime }(\lambda )={\mathrm{Var}}_{\lambda }(Z)$ under the exponentially tilted probability distribution. Any distribution on an interval of length $\ell$ has variance at most ${\ell }^2/4$, because its squared distance from the interval midpoint is at most ${\ell }^2/4$. Since $\psi (0)={\psi }^{\prime }(0)=0$, integrating this second-derivative bound gives the lemma for both signs of $\lambda$. The same argument holds for every conditional distribution.

Apply it conditionally to each martingale difference and use the tower property of conditional expectation repeatedly:The exponential Markov bound now gives, for $\lambda >0$,Choose $\lambda =t/(n{K}^2)$ to obtain the upper tail in the box, and apply the argument to $-f$ for the lower tail. This proves concentration of Lipschitz functions on the symmetric group. Using only the bound $|M_k-M_{k-1}|\le 2K$ in the absolute-increment Azuma-Hoeffding inequality would lose a factor four in the exponent; the conditional range length is what yields the constant needed in part (ii).

Put $d=d_H(A,B)=\min \{d_H(\alpha ,\beta ):\alpha \in A,\beta \in B\}$, and define $f(\sigma )=d_H(\sigma ,A)$. The triangle inequality shows that $f$ is Lipschitz continuous with Lipschitz constant one. It vanishes on $A$, and $f\ge d$ on $B$. Let $\mu =\mathbb{E}f$ for a uniform permutation.

If $\mu \ge d/2$, apply the lower-tail bound proved in part (i):If $\mu <d/2$, apply its upper-tail bound:For $d=0$ the desired inequality is simply $\min \{|A|,|B|\}\le n!$. Hence in all casesThis uses only the precise one-sided concentration inequality from part (i) and the Hamming distance triangle inequality. The normalization of the distance matters: it counts mismatching coordinates, rather than dividing their number by $n$.