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I'm trying to find theorems regarding random variables derived from sampling permutations, specifically concentration bounds.

As an example, let $X_i$ be the $\{0,1\}$-random variable that represents whether a uniformly random permutation $\sigma \in S_m$, has that $\sigma(i)$ is even (assuming permutations in $S_m$ are taken as bijections from $\{1, \ldots, m\}$ to itself)

Note that then the $X_i$ are identically distributed, but not independent, because each $X_i$ essentially 'samples without replacement' from the set $\{1,\ldots,m\}$

Then it's easy to see that (say) for $n = \frac{m}{2}$, $Pr[X_1 + X_2 + \ldots + X_n = n] = \frac{n!}{m!}$ which is asymptotically small.

More generally, given some set of functions $f_i: S_m \rightarrow \{0,1\}$, let $Y_i$ be a $\{0,1\}$-random variable that represents whether a uniformly random permutation $\sigma \in S_m$ has $f_i(\sigma) = 1$.

Then it seems plausible, given some reasonable conditions on the $f_i$ (for example, maybe $Pr[Y_i = 1] = 1/c$ for some constant $c$, or $Pr[Y_i + Y_j = 2] < 1/c$) that we can get some kinds of concentration bound for $Y_1 + \ldots + Y_n$

The best I can think of is to use the above two conditions on the $f_i$'s to calculate the expected value and variance of $Y_1 + \ldots + Y_n$ then invoke Chebyshev's inequality.

However, I can't believe that there's not some additional conditions that would lead to a better bound. Considering the "$\sigma(i)$ is even" example above, even though the variables are not mutually independent (or even pairwise independent) we still have that $X_1 + \ldots X_n$ is strongly concentrated around its expectation (in some vague sense).

Is there some way to interpret random variables like $X_1, \ldots, X_n, Y_1, \ldots, Y_n$, as 'nice' samples of a permutation distribution, as retaining some of the properties of independence that might lead to concentration?

This isn't my area so I might not be asking in the best way, so part of my question is also "what's the right way to ask this question?" :) Thanks a lot.

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3 Answers 3

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One useful trick that comes in handy sometimes (I originally saw it in this paper of Talagrand, though it may go further back): We can view a random permutation in $S_n$ as being generated as follows: Start with the identity permutation, and successively perform transpositions of the form $(n, a_n)$, then $(n-1, a_{n-1})$, and so on down to $(2, a_2)$, where each $a_j$ is uniformly chosen from the numbers between $1$ and $j$ (with no transposition if $a_j=j$).

The point is that the $a_j$ are now independent. So if we have a situation where each individual $a_j$ has little impact on the sum of the $X_i$ or $Y_i$, then we can apply concentration bounds for independent variables. For example, changing an individual $a_j$ only impacts at most $3$ positions in the final permutation (whereever $j$, the old $a_j$, and the new $a_j$ end up), changing an individual $a_j$ can only change $X_1+\dots+X_n$ by $3$ (or at most $1$, if you're more careful), so McDiarmid's inequality immediately gives exponential concentration on the sum.

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  • $\begingroup$ Thanks, I think that set of notes is exactly the sort of thing I was looking for. $\endgroup$
    – Joe Bebel
    Feb 10, 2013 at 11:14
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Note that the variables $X_i$ can be permuted without changing their statistics. There is a paper by Persi Diaconis that discusses this situation. Essentially one has bounds on the distance from the distribution of IID variables: what that means for concentrations specifically I do not know.

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The paper

Chao, C., Bai, Z., & Liang, W. (1993). Asymptotic Normality for Oscillation of Permutation. Probability in the Engineering and Informational Sciences, 7(2), 227-235. doi:10.1017/S0269964800002886

gives asymptotic normality results for the random variable $\sum_{i=1}^n|\sigma(i)-\sigma(i+1)|$ (with $\sigma(i+1)=\sigma(1)$).

That type of results is pretty far from the example given in the question, but "random variables derived from sampling permutations, specifically concentration bounds" could encompass the above result.

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