Talk:Characteristic function (probability theory)

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I think the formula for calculating the n-th moment based on the characteristic function was not correct. It was:

E(X^n)=-i^n \varphi^{(n)}_X(0).

I changed it to

E(X^n)=(-i)^n \varphi^{(n)}_X(0).

Please compare with [1].

Contents

[edit] Inversion theorem

I have a bit of a different definition for the inversion theorem that seems to contradict the one in this article. From "probability and random processes" by Grimmet I have:

\overline{F}(x) = \frac{1}{2}[F(x) + \lim_{y \to x^-}F(y)]

then

\overline{F}_X(y) - \overline{F}_X(x) = \lim_{\tau \to +\infty} \frac{1} {2\pi} \int_{-\tau}^{+\tau} \frac{e^{-itx} - e^{-ity}} {it}\, \varphi_X(t)\, dt.

--Faisel Gulamhussein 22:17, 20 October 2006 (UTC)

Probably you're correct. I don't think there is a contradiction exactly. Just many authors assume their random variables have a density, if this is the case the statements would be the same. Thenub314 13:07, 21 October 2006 (UTC)

Would someone like to write a brief explanation of the meaning of a characteristic function, for those of us without such strong statistics and mathematics backgrounds? I have no idea of its concept or application. An example that appeals to intuition would be great, thanks! BenWilliamson 10:28, 21 February 2007 (UTC)

[edit] Bochner-Khinchin theorem

There is something strange as it is formulated.

The would-be characteristic function

\varphi(t) = e^{- c t^4}

satisfies all the requested properties but it implies that the corresponding distribution has vanishing second moment

E[X^2] \propto \varphi''(0)=0

--MagnusPI (talk) 09:25, 24 April 2008 (UTC)

I believe the problem lies in condition: " \varphi(t) = e^{- c t^4} is a positive definite function". I suggest that this doesn't hold, and there are results that indirectly prove it isn't ... but I don't know of a direct approach to showing this. Note that "positive definite function" is a non-straightforward math thing if you are not familiar with it. Melcombe (talk) 09:34, 25 April 2008 (UTC)

On any compact \varphi(t) = e^{- c t^4} is positive definite, the only points where it is not are \pm \infty but in this case I do not understand why the characteristic function for some Levy stable \varphi(t)= e^{ - | c t |^\alpha } can be accepted since they show the same behaviour at \infty

--MagnusPI (talk) 08:44, 28 April 2008 (UTC)

I suggest you follow the link to positive definite function to see what is required .. it is not just \varphi(t) > 0 . Melcombe (talk) 08:56, 28 April 2008 (UTC)

Interesting but this is not what a physicist would call a positive definite function. I will therefore add a note to the main page

--MagnusPI (talk) 08:55, 29 April 2008 (UTC)

[edit] History of Characteristic Functions

Historia Matematica has a nice little thread on the history of characteristic functions which could be used to give some background on where c.f.'s came from. --Michael Stone 19:22, 10 April 2007 (UTC)

Please check your work.

E(X^n)=i^n\varphi^{(n)}_X (0).

[edit] Inverse fourier transform

The article mentions that you can calculate the characteristic function taking the conjugate of the Fourier transform of the pdf, but isn't this simply the inverse Fourier transform?... This should be made more clear. We tend to think about transforming using only the "direct" transform first, and then transforming back again, but you can perfectly say that you can find the characteristic function taking the inverse Fourier transform, then come back again taking the direct Fourier transform...

It's like the pdf is already the "transform", and then the characteristic function is the "original" function, that we could see as a signal to be transformed...

The idea is just that we save some words using the idea of "inverse fourier transform" instead of "conjugate of the transform", and also help to see the inverse transform as something as natural as the "direct" transform. No transform domain is intrinsically better! there is no reason to insist in seeing the pdf as the correlate of a signal to be transformed by the direct fourier transform, and not the inverse one... —Preceding unsigned comment added by Nwerneck (talk • contribs) 22:53, 17 November 2007 (UTC)