Bounding mutual information given bounds on pointwise mutual information

Suppose I have two sets X and Y and a joint probability distribution over these sets p(x,y). Let p(x) and p(y) denote the marginal distributions over X and Y respectively.

The mutual information between X and Y is defined to be:

i.e. it is the average value of the pointwise mutual information pmi(x,y)log(p(x,y)p(x)p(y)).

Suppose I know upper and lower bounds on pmi(x,y): i.e. I know that for all x,y the following holds:

What upper bound does this imply on I(X;Y). Of course it implies I(X;Y)k, but I would like a tighter bound if possible. This seems plausible to me because p defines a probability distribution, and pmi(x,y) cannot take its maximum value (or even be non-negative) for every value of x and y.


My contribution consists of an example. It illustrates some limits on how the mutual information can be bounded given bounds on the pointwise mutual information.

Take X=Y={1,,n} and p(x)=1/n for all xX. For any m{1,,n/2} let k>0 be the solution to the equation
Then we place point mass ek/n2 in nm points in the product space {1,,n}2 in such a way that there are m of these points in each row and each column. (This can be done in several ways. Start, for instance, with the first m points in the first row and then fill out the remaining rows by shifting the m points one to the right with a cyclic boundary condition for each row). We place the point mass ek/n2 in the remaining n2nm points. The sum of these point masses is
so they give a probability measure. All the marginal point probabilities are
so both marginal distributions are uniform.

By the construction it is clear that pmi(x,y){k,k}, for all x,y{1,,n}, and (after some computations)
with the mutual information behaving as k2/2 for k0 and as k for k.

Source : Link , Question Author : Florian , Answer Author : NRH

Leave a Comment