Talk:Radon–Nikodym theorem

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This article incorporates material from PlanetMath, which is licensed under the GFDL.

So presumably not just a function in the statement, but integrable? Charles Matthews 11:16, 11 Oct 2003 (UTC)

Yes, I'll make this explicit in the article. Pete 12:39, 11 Oct 2003 (UTC)

Another question is: f is not unique (actually, it is unique almost sure), as far as I know (but I may be perfectly wrong). If this is the case I think it is worth mentioning it (something like the equality holds for f and g whenever f=g a.s.). Pfortuny 12:48, 11 Mar 2004 (UTC)

The second LaTeX formula bothers me. First of all, what is A supposed to be? I would guess a random variable rather than a set (like above). Also, shouldn't it rather be

$E_Q(X) = E_P\left( \frac{dQ}{dP} X \right)$

(E_P and E_Q switched?). Not sure though, else I would change it myself. DrZ 13:50, 16 Mar 2004 (UTC)

I changed the last bit, presumably A is supposed to be same arbitrary measurable set as from the first formula? Pete/Pcb21 (talk) 14:10, 16 Mar 2004 (UTC)

But what is the expectation operator applied to a set supposed to mean? DrZ 14:50, 16 Mar 2004 (UTC)

That was wrong. I think now it is right. There is no expectation operator applied to a set, so you were right in your concern. Pfortuny 15:15, 16 Mar 2004 (UTC)

1 proof
2 measures on an algebra (not sigma)
3 Proof for signed measure
4 "Financial mathematics uses the theorem extensively" ?
5 as to complex measures...

[edit] proof

heh, article claims theorem is from functinal analysis yet doesn't give a (more elegant) functional analytic proof. Mct mht 11:49, 10 June 2006 (UTC)

[edit] measures on an algebra (not sigma)

definition: ν <<< μ : for all e>0 exists d>0 for all A in algebra (μ(A)<d -> ν(A)<e)

lemma: (ν <<< μ) -> (ν << μ) lemma: for sigma additive ν, μ on a sigma-algebra (ν << μ) -> (ν <<< μ)

definition: simple-function : finite linear combination of indicator functions of measurable sets.

there's also a theorem that states: let ν <<< μ positive meaures on an algebra, let e>0 then exists a simple-function f s.t. ||ν-fdμ||<e.

with completeness of L1(μ), the regular Radon-Nikodym theorem follows. --itaj 00:22, 15 June 2006 (UTC)

I don't understand what you are trying to say. But either way I don't think this belongs in this article, rather a new article containing this should be created. Oleg Alexandrov (talk) 03:07, 15 June 2006 (UTC)

it looks like itaj is saying the following: <<< is a continuity condition that's equivalent to the usual absolte continuity <<, for countably additive measures. the notation ||.|| looks like it means the total mass of a measure. a suitable f is then obtained by approximation using simple functions. question is: when the assumption that $L 1$ being complete is invoked, aren't you implicitly extending to a σ-algebra? Mct mht 03:25, 15 June 2006 (UTC)

what i meant was to discuss that theorem i described. to see if it's to be added here or maybe make another article. and the last comment says that this theorem together with the theorem that

L 1

is complete can prove the regular radon-nikodym theorem. --itaj 20:59, 15 June 2006 (UTC)

well, that's the question. when you use the assumption that $L 1$ is complete, implicitly you're extending to a sigma algebra anyway, no? in other words, the support of functions in $L 1$ might as well be addded to the algebra to make it a sigma algebra. so maybe it's not so much of an improvement. Mct mht 21:06, 15 June 2006 (UTC)

what i was saying is this: the theorem i described is a version of radon-nikodym for an additive measure over an algebra (not sigma) and is weaker. and this one does not follow from regular radon nikodym (as far as i know, not in a canonical way). but the regular radon nikodym (for sigma-additive measures over sigma-algebra) does follow from this one (using this theorem and also using the theorem that L1 of sigma-additive measure over sigma-algebra is complete). --itaj 17:00, 14 February 2007 (UTC)

[edit] Proof for signed measure

The article says, "If ν is a signed measure, then it can be Hahn–Jordan decomposed as ν = ν+−ν− where one of the measures is finite". Yes, this is true. However, how can we be sure that the non-fininite measure is σ-finite? Is it always true? If yes, then any nonnegative measure is σ-finite? Jackzhp 02:34, 10 December 2006 (UTC)

You are right, one should have mentioned that ν was σ-finite. I fixed that now. And no, not every measure is sigma-finite. Oleg Alexandrov (talk) 03:13, 15 February 2007 (UTC)

In the "Radom-Nikodym derivative" section the article states

   [...] if μ is a nonnegative σ-finite measure, and ν is a finite-valued signed
   or complex measure such that , there is μ-integrable real- or
   complex-valued function g on X

It is however sufficient for the signed measure to be σ-finite. In this case, g will not be μ-integrable in general, because it is not necessarily finite. Maybe the following would be more accurate:

   [...] if μ is a nonnegative σ-finite measure, and ν is a σ-finite signed or
   complex measure such that , there is a μ-(quasi-)integrable real- or
   complex-valued function g on X

By quasi-integrable I mean that the integral exists, but it is not necessarily finite.--Drizzd 17:39, 23 February 2007 (UTC)

[edit] "Financial mathematics uses the theorem extensively" ?

The article doesn't explain why this should be.

Does it become particularly important when treating stochastic processes to work with tools defined for probability measures on arbitrary sets, rather than probability densities on real numbers? Jheald 12:34, 4 March 2007 (UTC)

I suspect so. The Wiener process is used in financial mathematics, and certainly that's not a distribution on the real line. Michael Hardy 01:31, 5 March 2007 (UTC)

[edit] as to complex measures...

In the section on the Radon-Nikodym-derivative, it is stated that the theorem also holds in case

ν is a finite-valued signed or complex measure such that $|\nu| \ll \mu$

However, I just caught the theorem (as noted by user:Drizzd above) stating that $ν$ also needs to be $σ$ -finite (which is, as far as I know, not the same as finite-valued...).

I also noted somewhere else that (at least in this setting), $|\nu|\ll\mu\iff\nu\ll\mu$ (please correct me if I'm wrong with this!). Wouldn't therefore, replacing the total variation measure $| ν |$ by just $ν$ make this entry easier to use?