Talk:Characteristic function (probability theory)

message: 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)
thread_title:  Inversion theorem 

message: 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)
replies: text: 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)
thread_title:  Bochner-Khinchin theorem 

message: 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)
thread_title:  History of Characteristic Functions 

message: 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)
thread_title:  Inverse fourier transform 

message: The article gives the impression that every random variable can be described by a characteristic function; I am almost sure (no, that's too strong to say in a statistics page - let me rephrase it: I have more than 95% confidence) that a pathological example can be designed that does not have a characteristic function... Something along the lines of the perforated interval in page standard probability space. Darth Albmont (talk) 19:06, 18 November 2008 (UTC)
replies: text: I suggest you see page 11 of the Lukacs reference. He cites a theorem of Cramer (1946) which he says ensures the existence of the cf for "every distribution function". The previous context makes it clear that this includes even dfs with singular components. See also page 20, 36 and 64 for other results on the relation between the cf and whether the df has a singular component.  Melcombe (talk) 10:01, 19 November 2008 (UTC)
replies: text:  Does every random variable defined in the real numbers have a distribution function? Darth Albmont (talk) 12:25, 19 November 2008 (UTC)
replies: text: You linked to density function but wrote distribution function; they are not equivalent.  Every real valued RV has a distribution function; not all have density functions (e.g., Cantor distribution).  Your original question was about characteristic functions; they always exist for any distribution with a Lebesgue measure due to the dominated convergence theorem.  I cannot say for sure about distributions with more pathological measures, but I suspect it's true there too.  I cleaned up the wikilinks in your original post. Baccyak4H (Yak!) 15:02, 19 November 2008 (UTC)
replies: text:  You linked to density function but wrote distribution function; they are not equivalent - Baccyak4H, yes, but the page distribution function is about physics, and it gives both cumulative and density as the probability concept. I don't know if there are pathological real valued random variables so pathological that it's not possible to compute its characteristic function - but let's try a counting argument. Let X be any set that is dense in [0,1]; is it possible to define a real-valued random variable whose sample space is X? If we accept the Axiom of Choice (the psychomath's favourite tool in creating monsters) and the Continuum hypothesis, then the answer is yes: either X is countable or has the cardinality of the continuum; in the first case, it's trivial, in the second, well-order both X and [0,1], sample uniformly from [0,1] and get back to X using the bijection (probably there's no need to use the continuum hypothesis to create this monster). So, by counting, there are 2c real-valued random variables. However, since every characteristic function is continuous, there are only c characteristic functions. Unless I made some big mistake, I think I proved that not all real-valued random variables have a characteristic function (in fact, I guess I showed that most of them don't - the problem with pathologies is that usually pathologies are the rule). Darth Albmont (talk) 16:33, 19 November 2008 (UTC)
replies: text: The characteristic function is just the Fourier transform of the distribution. (Thus it is really about distribution rather than random variable itself.) It is well-defined always. At least if we treat a distribution as a probability measure on the Borel sigma-field (be it completed or not). No place for pathologies here. Boris Tsirelson (talk) 20:17, 23 March 2009 (UTC)
replies: text: Characteristic function is an integral \phi_X(t) = \mathrm{E}\,e^{itX} = \int e^{itx}dF_X(x). Since expectation operator exists for any random variable, and the corresponding integral is absolutely integrable for real values of t: \left| \int e^{itx}dF_X(x) \right| \leq \int |e^{itx}|dF_X(x) = \int dF_X(x) = 1, then characteristic function is well-defined for any random variable, even a "pathological one". Stpasha (talk) 21:20, 16 June 2009 (UTC)
thread_title:  Pathological examples 

message: The article says that, if phi is a characteristic function, then so is Im(phi). But one property all characteristic functions phi have is phi(0) = 1, so Im(phi)(o) = Im(phi (0)) = Im(1) = 0, which is not 1. So, surely, if phi is a characteristic function, then Im(phi) never is? I tried correcting this by removing Im(phi) from the list, but I never saved my edit. I previewed it, only to find out I'd messed up the code on all the other symbols for Re(phi) etc. Could someone who knows what they're doing with the code please remove Im(phi)? (Unless it actually does belong there, but for the life of me I cannot see how!) 90.206.183.244 (talk) 15:38, 6 July 2009 (UTC)
replies: text: Thank you. You are right. I did. Boris Tsirelson (talk) 18:36, 6 July 2009 (UTC)
thread_title:  This can't be right, surely? 

message: Under "properties" the article states "the characteristic function of a symmetric random variable is real-valued and even".  A similar remark is made in the caption for graph of the characteristic function of a uniform distribution.  But is this really generally true, or true only if the random variable is symetric around zero.  For example, a normal or Cauchy distribution with a non-zero mean is symetric, but it appears that the characteristic functions would retain complex values.  But I admit that I am weak on complex math, so maybe those i's that seem to be there drop out. Rlendog (talk) 13:54, 21 August 2009 (UTC)
replies: text: You are right. Thanks. I just added "around the origin". Boris Tsirelson (talk) 15:07, 21 August 2009 (UTC)
thread_title: Properties - symmetry

message: What's T suppose to denote? How different is this from t?  —Preceding unsigned comment added by 161.53.64.70 (talk) 08:53, 7 October 2009 (UTC)
replies: text: It was supposed to be small t — the k×p matrix argument of the c.f.  … stpasha »  19:58, 7 October 2009 (UTC)
thread_title:  Definition for k×p-dimensional random matrix 

message: This text appears in the Definition Section.
.. I may be blind, but I don't see any parentheses in that section..
OldMacDonalds (talk) 03:45, 19 February 2014 (UTC)
thread_title:  "and the last formula in parentheses is valid" 

message: This [1] looks like a very useful statement, relating independence of X,Y to the characteristic functions \varphi_{(X,Y)} and \varphi_{X},\varphi_{Y}. Is there a reason for the absence? (If not, I can add it.)
Theorem (Kac's theorem, from [1]). Let X,Y \in L^1 be \mathbb{R}^d-valued random variables. Then the following statements are equivalent.
  1. X,Y are independent
  2. \forall \eta,\xi \in \mathbb{R}^d,  \mathbb{E}e^{i \langle(X,Y), (\xi,\eta)\rangle} = \mathbb{E}e^{i \, X \cdot \xi} \cdot \mathbb{E}e^{i \, Y \cdot \eta}
[1] http://math.stackexchange.com/questions/287138/moment-generating-functions-characteristic-functions-of-x-y-factor-implies-x/287321#287321  — Preceding unsigned comment added by Ceacy (talk • contribs) 11:01, 30 May 2016 (UTC)
thread_title:  Relation to independence/ Kac's Theorem 

message: Theorem. If a is (possibly) an atom of X (in the univariate case this means a point of discontinuity of FX ) then
* If X is scalar:
:: F_X(a) - F_X(a-0) = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{+T}e^{-ita}\varphi_X(t)\,dt
But F_X(a-0) = F_X(a) so this quantity is trivially zero. Is this supposed to reflect some kind of limit?
Manybytes (talk) 03:48, 24 September 2019 (UTC)
replies: text: Sure, this is (a usual, convenient notation for) a limit: F_X(a-0) = \lim_{b\to a,b<a} F_X(b). See one-sided limit. Well, F_X(a-) is better; I guess, F_X(a-0) is out of fashion. Boris Tsirelson (talk) 04:25, 24 September 2019 (UTC)
thread_title:  This can't be right: F(A) - F(A - 0)? 

message: The introduction contains this sentence:
"If a random variable admits a probability density function, then the characteristic function is the Fourier transform of the probability density function."
But among the various definitions of the Fourier transform, all of them involve an integral involving exp(-ixt) or exp(-2πixt), with a negative sign inside the exponential.
Doesn't this mean that the characteristic function is more like the inverse Fourier transform of the density function (if any)? 2601:200:C000:1A0:BC00:5039:DB55:E9EC (talk) 18:51, 29 July 2022 (UTC)
replies: text: I agree that the inverse Fourier transform is closer than the forward Fourier transform.  But there is also the factor of 2 to worry about.  The inverse Fourier transform is something like \int F(\nu) e^{2 \pi \nu t} d\nu.  With the substitution \omega = 2 \pi \nu that becomes \frac{1}{2\pi} \int F(\omega/2\pi) e^{\omega t} d\omega.  Either way, there is a factor of 2 to be reckoned with.  What if we say that the characteristic function is closely related to the inverse Fourier transform?  —Quantling (talk | contribs) 19:02, 29 July 2022 (UTC)
thread_title:  Is it really the Fourier transform, or only *almost* the Fourier transform?