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Expansion of Taylor series for statistical functionals

  1. Aug 26, 2011 #1
    Hi

    By some googling it seems like there exist some kind of expansion of the Taylor series for statistical functionals. I can however, not sort out how it is working and what the derivative-equivalent of the functional actually is.

    My situation is that I have a functional, say \theta which depend on the distribution function given to it. I want to expand \theta(f_1 + f_2) around the distribution function f_1. I think it should look like

    \theta(f_1(y) + f_2(y)) = \theta(f_1(y)) + \frac{d \theta}{d (something)} *f_2(y) + o(1),

    but I do not know what the derivative-equivalent actually is.

    What is the definition of this and how can it be calculated in a given situation? Or am I totally out of bounds with such a formula?

    Can someone help me or at least point me in the right direction?

    Any help is appreciated.
     
  2. jcsd
  3. Aug 26, 2011 #2

    Stephen Tashi

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    Does "distribution function" mean a probability distribution function? If so, wouldn't the argument of [itex] \theta [/itex] have to be [itex] \frac {f_1 + f_2}{2} [/itex] or some other combination that produced a probability density or cumulative probability function?


    I'm not an expert on functionals, so I can't say whether you are out of bounds. When a Lie group of transformations [itex] T(x,h) [/itex] acts on a set X, it also acts (for a given value of the parameters h) to take a function [itex] f(x) [/itex] to another function [itex] g(x) = f(T(x,h)) [/itex]. There are many things known about that scenario. I don't know whether you could arrange for a group to take [itex] f_1 [/itex] to [itex] f_1 + f_2 [/itex].

    It isn't clear what variable's powers appear in the Taylor series that you want and it isn't clear what variable the "o(1)" applies to.

    If you define a set of parameterized transformations ( such as [itex] T(f_1,f_2,h) = (1-h) f_1(x) + h (f_1(x) + f_2(x) ) [/itex], you might make sense of a derivative with respect to [itex] h [/itex].
     
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