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Suppose we have a finite sample space $S$ and some events $A_1, \dots, A_k \subseteq S$. We would like to put a probability distribution on $S$ so that no element has probability greater than a parameter $\rho$, while each event $A_i$ has probability at least $1 - \epsilon_i$ for some parameter $\epsilon_i$. When is this possible?

For example, with $k = 1$ it's pretty easy to see that such a distribution exists iff $|S| \ge 1 / \rho$ and $|A_1| \ge (1 - \epsilon_1) / \rho$. With $k = 2$ you also need $| A_1 \cap A_2 | \ge (1 - \epsilon_1 - \epsilon_2) / \rho$, and in general you need lower bounds on all pairwise, 3-wise, etc. intersections of the events. I have a general set of conditions extending these which I can prove are necessary, but I can't show they're sufficient for all $k$ (only for $k \le 3$).

This problem seems so basic that I think it's likely it has already been studied, so if anyone can point me to anything relevant that would be great.

In case it helps suggest anything, let me mention a connection with linear programming: if all the parameters are fixed, then the existence of an appropriate distribution is equivalent to the feasibility of a system of linear inequalities (which is in a sense an answer to my question, but I'm looking for much simpler conditions like the inequalities I gave above). I've tried to prove sufficiency of my conditions by constructing the dual system and proving it's bounded, but I can't seem to do that other than by an enormous case analysis for each $k$.

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  • $\begingroup$ The book "Bonferroni-type Inequalities with Applications" by Galombos and Simonelli might have something you can use. $\endgroup$ Commented May 12, 2016 at 0:34
  • $\begingroup$ Thanks for the reference, but I'm not sure how it's applicable. As far as I can tell the book is concerned with deriving inequalities that must hold for all probability distributions, whereas I want conditions under which a certain type of distribution exists. Maybe Bonferroni-type bounds could yield necessary conditions, but I don't see how they could yield sufficient ones. Am I missing something? $\endgroup$
    – Daniel
    Commented May 16, 2016 at 19:01

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