# Is maximizing the Shannon differential entropy equivalent to minimizing the predictability and/or minimizing the maximum density?

For a real-valued, 1-dimensional, continuous random variable X with density f(x), I am trying to determine if maximizing the Shannon differential entropy of f(x) is mathematically equivalent to finding the f(x) with the least predictive density (as defined below) for a given set of constraints:

Least predictive density: The density that minimizes the maximum probabiliy associated with each value of the Lebesgue measure on the domain of f(x).

For example, let X be an RV with domain (-inf,+inf) and specified mean M and variance V and I be a set of intervals on the real line with total Lebesgue measure L (i.e., sum of the measure of each interval in the set. The actual intervals not specified, only their total Lebesgue measure is given). The least predictive distribution would minimize the maximum probability that can be contained in I, for each value of L, while also having mean M and variance V.

I don't know if this problem can be formulated generally as a functional so that calculus of variations can be applied. At this point, I have formulated it as a functional for the specific case of a fixed interval for unimodal symmetric unbounded densities, but generalizing this has been problematic.

Related to the above, this problem appears to boil down to minimizing the maximum value of the density subject to any constraints (such as specified mean and variance).

Given the above "preamble", the two issues I am hoping to get guidance on are:

1) Is maximum Shannon entropy equivalent to minimal predictability (as defined above for a particular set of constraints)?

2) Is minimal predictability equivalent to minimizing the maximum density (subject to a set of constraints)?

I hope I have specified this precisely enough and given enough background on my current thinking and work on this. Any pointers, references, or research results related to the above would be much appreciated.

Thanks :)

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