# Expansion of Taylor series for statistical functionals

1. Aug 26, 2011

### Testguy

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. Aug 26, 2011

### Stephen Tashi

Does "distribution function" mean a probability distribution function? If so, wouldn't the argument of $\theta$ have to be $\frac {f_1 + f_2}{2}$ 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 $T(x,h)$ acts on a set X, it also acts (for a given value of the parameters h) to take a function $f(x)$ to another function $g(x) = f(T(x,h))$. There are many things known about that scenario. I don't know whether you could arrange for a group to take $f_1$ to $f_1 + f_2$.

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 $T(f_1,f_2,h) = (1-h) f_1(x) + h (f_1(x) + f_2(x) )$, you might make sense of a derivative with respect to $h$.