# On concave functions over spaces of probabilty distributions

1. May 9, 2012

### bayesian

Given two (dependent) random variables $X$ and $Y$ with joint PDF $p(x,y)$ $=p(x|y)p(y)$ $=p(y|x)p(x)$, let $H[X]$ be real-valued concave function of $p(x)$, and $H[X|Y]$ the expectation of $H$ of $p(x|y)$ with respect to $p(y)$.

Examples of possible functions $H$ include the entropy of $X$, or its variance.

The concavity of $H$ implies that $H[X]-H[X|Y]≥0$ (through Jensen's inequality).

Question:

What additional conditions (if any) on $H$ are imposed if we in addtion require that $H[X]-H[X|Y]$ should also be concave with respect to $p(x)$, if $p(y|x)$ remains fixed?

2. May 9, 2012

### chiro

Hey bayesian and welcome to the forums.

If you want to do a rigorous proof on this, then the way I would approach is to show what needs to hold for concavity of h when h = f*g. You can use the definition of a concave function for this, simplify and then collect terms.

We also know that p(x,y) and q(x,y) are separable since if p(y|x) is constant between two distributions, then if we have two different distributions corresponding to your two cases where the first corresponds to p(y) being concave, and the second corresponds to p(x) as well as p(y) being concave then this implies that p(x,y)/p(x) = q(x,y)/q(x) which implies q(x)/p(x) = q(x,y)/p(x,y) which means both p(x,y) and q(x,y) are separable.

So once you show what is needed for concavity to hold you can show what conditions p(x,y) and q(x,y) need and thus what the joint distributions need (or vice-versa) for concavity and hopefully answer your question.

This is only a suggestion, but I think those steps would be more rigorous, although it probably would be complicated.

3. May 9, 2012

### bayesian

chiro, I do not understand your suggestion, why should I consider products of functions that are concave? I cannot see how such products arise from the problem that I stated.

Just to make things clear: in my problem, H is a function of the probability distribution p(x) (e.g. the variance) and concave with respect to the space of probability distributions.

4. May 9, 2012

### viraltux

By a "function of the probability distribution" you mean that H(X) could be V(X) where X is a random variable? Because if so V(X) is not a function but a number. I don't think yo mean that, but I am not sure what you mean either.

5. May 10, 2012

### chiro

I'm a little confused like viraltux as well, but if i understand correctly then H[X] will be a number, but H[X|Y] will be a function and this will have to concave. Is this right?

If this is concave and you end up getting H[X|Y] = Z(X) for some Y, then you want to show that this is concave as well. For H[X|Y], we get an integral in the form of p(x|y)p(y)ydy and the result of this will be concave (a function of x).

Now you can show that using your condition, the probability density function corresponding to keeping p(y|x) constant between changes will result in p(x,y) = b(x)c(y) [Separable].

From this you can take b(x) outside the integral (only with respect to y as in the dy) and then this implies that b(x) is concave, which implies H[X|Y] as a function is concave.