Here are some notes on bounding joint probability distributions. Enjoy! This was converted from \(\LaTeX\) with pandoc, so typos, missing figures, etc., to be expected.

Consider the binary random variables \(X_1, \ldots, X_n\) following the distribution \(P\). For some collection of values, say, \(x_1, \ldots, x_n\), we are interested in computing \(P(X_1=x_1,\ldots, X_n=x_n)\).

There is rich literature on bounding joint probabilities, say, \(P(X_1,X_2,X_3)\), if one has of knowledge of the marginals, \(P(X_i),\) \(i=1,2,3\), \(P(X\_{i},X_j)\), \(i\neq j\), or of the moments of the marginal distributions. Some examples of such inequalities follow below.

When the bounds only use \(P(X_i)\), we will say that they utilize *first-order* information. Similarly, if \(P(X_i, X_j)\) are used in the bounds, they are of second-order, then third-order, etc.

We start with a classical result, inspired from the inclusion-enclusion formula, known as the *Bonferroni* inequalities [@galambos1977bonferroni]. The notation \(X^c\) corresponds to the negation of the \(X\) variable, i.e., if \(X=x\), \(X^c=1-x\) for \(x\in \{0,1\}\). First, we define:

Then, for every odd \(k\) in \(\{1,\ldots, n\}\):

\[\begin{aligned} P(X_1,\ldots, X_n)&\geq 1 -\sum_{j=1}^{k} (-1)^{j-1}S_j. \end{aligned}\]We can also get an upper bound for every even \(k\):

\[\begin{aligned} P(X_1,\ldots, X_n)&\leq 1 -\sum_{j=1}^{k} (-1)^{j-1}S_j.\end{aligned}\]By the inclusion-exclusion formula, the inequalities become equalities when \(k=n\). Thus, the inequalities can be made sharper by including more marginals. However, the upper (and lower) bounds don’t necessarily become sharper monotonically as \(k\) increases; see work by [@schwager1984bonferroni]. Also, although the inequalities are valid for all \(k\), they can be uninformative, that is, smaller than zero or greater than one.

An alternative upper bound for the joint is the Frechet-type bound:

\[\begin{aligned} \label{eq:frechet} P(X_1,\ldots, X_n)\leq \min_{i}P(X_i). \end{aligned}\]This can be simply derived by observing that, for any \(i\),

\[P(X_1,\ldots, X_n)=P(X_1,\ldots,X_{i-1},X_{i+1},\ldots, X_n|X_i)P(x_i)\leq P(X_i)\]and then picking the tightest bound. We can also include terms like \(P(X_i,X_j)\) to the upper bound, if known, to get an even tighter bound. As an upper bound, this may be more suitable than the Bonferroni bound; it is always a valid probability and can be tight when dealing with rare events. Like the Bonferroni bound, this is distribution-independent.

Now, if all we know about the \(X_i\) are the \(P(X_i)\), then the tightest bounds[^1] we can get are:

\[\begin{aligned} \label{eq:frechet-first} \max\{0,1-\sum_i(1-P(X_i))\} \leq P(X_1,\ldots, X_n)\leq \min_{i} P(X_i).\end{aligned}\]The lower bound comes from the first Bonferroni lower bound. However, it can be further sharpened by adding second-order information, that is, some of the \(P(X_i^c, X_j^c)\), as discussed by [@hochbergsome]. One example of such a sharpening is known as the *Kounias* inequality:

This can be further sharpened by replacing the max term in by

\[\sum_{i,j:(i,j)\in T} P(X_i^c, X_j^c),\]where $T$ is the maximal spanning tree, i.e., the tree that maximizes the sum of the probabilities[^2]. The new bound then is:

\[\begin{aligned} \label{eq:wolfe} 1-\sum_{i}(1-P(X_i))+\sum_{i,j:(i,j)\in T} P(X_i^c, X_j^c)\leq P(X_1,\ldots, X_n).\end{aligned}\]This bound was first derived in work by [@hunter1976upper] and has been subsequently generalized to work with more events via the construction of multi-trees; see work by [@bukszar2001upper].

In some cases, multiplicative bounds, that is,

\[P(X_1,\ldots X_n)\geq P(X_1)\ldots P(X_n),\]may also be applicable when the random variables show positive association; see work by [@esary1967association] for details on that. Those bounds are easier to apply and often tighter but may not always be correct as they are distribution dependent. Especially for Bernoulli variables, Theorem 4. in [@esary1967association] shows that association of the \(X_1,\ldots, X_n\) implies only that

\[\begin{aligned} P(X_1=1,\ldots, X_n=1)&\geq P(X_1=1)\ldots P(X_n=1),\\ P(X_1=0,\ldots, X_n=0)&\geq P(X_1=0)\ldots P(X_n=0).\end{aligned}\]