For a given pdf function f(x) defined in [0,1], which is monotonically increasing and convex, I need to find a probability generation (polynomial) function Q(x) of a maximum degree D, where its derivative is always larger than or equal to the pdf function in [0,1-e], for every e>0, Q'(x)>f(x). It can be solved by numerical optimization algorithms, but I am looking for some theorems to prove the existence of such a PGF for every e.
Please find here more details on the question:
http://math.stackexchange.com/questions/1400918/finding-a-polynomial-approximation-of-a-pdf
http://math.stackexchange.com/questions/1400918/finding-a-polynomial-approximation-of-a-pdf
Hi Yahaya. Can you explain a bit more or give me some references?
1. The derivative of a polynomial of degree D is a polynomial of degree D-1, so you are looking for a polynomial (with bounded degree by a prespecified bound) that dominates your pdf.
2. Dominating f on [0,1-e] for every e>0 is the same as dominating f on [0,1).
3. As your f is convex, what is wrong with using the cord (a linear function = polynomial of degree (at most) 1) stretching from the point (0, f(0)) to (1, f(1)) on the graph of f ?
Thanks David. The problem is that f is unbounded when x goes to 1. But the integral of f over [0,1] is exactly 1. That is why I cannot use linear functions. Please find more details on the question here:
http://math.stackexchange.com/questions/1400918/finding-a-polynomial-approximation-of-a-pdf
Thanks Mahyar. I looked at the reference you supplied and now your question makes sense. What initially confused me is your statement that f is defined on the closed unit interval [0,1], where in fact what you meant was that it is defined only on [0,1) and f(x) tends to infinity as x--->1. Sorry, I don't know the answer.
Function $f$ is unbounded on [0,1), but any polynomial (or its derivative) is always bounded there. So, there is no polynomial that is greater or equal to $f$ on [0,1).
Andrey, the polynomial should only dominate from zero to one minus epsilon. I think the question is wheter for each epsilon there exists such a polynomial.
Andrey is right, so if - as Peter says - the required polynomial, for each e, depends on e and dominates f only on [0,1-e], then the linear polynomial joining the points [0,f(0)] and [1-e,f(1-e)] on the graph of f should work since f is assumed convex.
Peter, I know what are generating functions of discrete probability distributions on the integers, but what do you mean by a generating function of a non-atomic distribution on the unit interval, even one that is absolutely continuous w.r.t. Lebesgue measure (i.e. with density)?
David, find a polynomial Q(x) such that the coefficients are non-negative and add up to 1 and such that Q dominates f on the interval [0,1-e]. This seems to be the problem.
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