Who here is into this and can talk about it?

- Cuchulainn
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QuoteOriginally posted by: CactusManWho here is into this and can talk about it?I once generalized it as undergrad (AFAIR to the complex case), prompted by Cornelis Lanczos, DIAS. It's a part of measure theory. It's a fancy derivative. It is used in finance to change the measure from foreign risk probability measure to domestic risk probability measure (Brigo/Mercurio). That's about it IMO.Why do you want to know?// I found measure theory awful boring at university.

Last edited by Cuchulainn on November 20th, 2014, 11:00 pm, edited 1 time in total.

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- Cuchulainn
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QuoteOriginally posted by: CactusManWho here is into this and can talk about it?I see this is not the first time you have asked

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- MiloRambaldi
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QuoteOriginally posted by: CactusManWho here is into this and can talk about it?I am not into this anymore, because it is a part of pure math which I don't do anymore for various reasons.I did use it in my thesis to prove some esoteric results.The theorem states that given two measures [$]\mu[$] an [$]\nu[$] on [$]X[$] satisfying certain conditions, there exists a measurable function [$]f:X\to[0,\infty)[$] such that $$\nu(A)=\int_A f\, \mathrm d\mu$$ for every measurable [$]A\subseteq X[$]. [$]f[$] is written [$]\frac{\mathrm d\nu}{\mathrm d\mu}[$] and is called the Radon-Nikodym derivative. It holds, for example, when [$]X=\mathbb R[$] is the real line, [$]\mu[$] is the Lebesgue measure and [$]\nu[$] measures all of the intervals and satisfies [$]\nu(A)=0[$] whenever [$]\mu(A)=0[$] (i.e. [$]\nu\ll\mu[$]). Since the proof (at least the usual one) is non-constructive (i.e. does not say how to construct the RN-derivative), I'm doubtful about applications.Girsanov's theorem is often used in finance for change of measure (I guess that's what Cuchulainn example is), which involves a RN-derivative. However, the RN-derivative is defined in the statement of Girsanov's theorem as the stochastic exponential of the given stochastic process. This means that the Radon-Nikodym theorem is not invoked to prove Girsanov's theorem. The RN-theorem is used to prove the existence of conditional expectations, which are needed to do any stochastic calculus. So it is used indirectly in finance in that sense.

Last edited by MiloRambaldi on November 27th, 2014, 11:00 pm, edited 1 time in total.

- ThinkDifferent
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is this thread some kind of a joke? if not, why is it in the Off Topic?

What is the very simplest example of the Radon-Nikodym derivative you can think of?

- MiloRambaldi
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QuoteOriginally posted by: CactusManWhat is the very simplest example of the Radon-Nikodym derivative you can think of?Perhaps you want to rephrase that as "the simplest non-trivial example"?Otherwise, the RN-derivative of [$]\mu[$] with respect to itself is [$]1[$], i.e. [$]\frac{\mathrm d\mu}{\mathrm d\mu}[$] is almost everywhere equal to 1 (almost everywhere refers to the fact that two RN-derivatives that only differ on a set of measure zero are equivalent).

- Cuchulainn
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QuoteOriginally posted by: MiloRambaldiQuoteOriginally posted by: CactusManWhat is the very simplest example of the Radon-Nikodym derivative you can think of?Perhaps you want to rephrase that as "the simplest non-trivial example"?Otherwise, the RN-derivative of [$]\mu[$] with respect to itself is [$]1[$], i.e. [$]\frac{\mathrm d\mu}{\mathrm d\mu}[$] is almost everywhere equal to 1 (almost everywhere refers to the fact that two RN-derivatives that only differ on a set of measure zero are equivalent).Indeed; a "RN for the uninitiated/challenged" is just not on the cards; the underlying theory used to be the stuff of advanced senior sophomore pure maths degrees(I heard it's graduate level these days). There is indeed no simple example; you have to know a lot of measure theory. Period.Anyhoo, a special case/analogy when measure degenerates into the Riemann integral case is the Fundamental Theorem of CalculusGiven this example, the next step is measure...In this case eq. (2) is the analogy of RN and (3) that of RN derivative.Application?? not much but it seems the RN derivative satisfies a Kolmogorov PDE That should be computable, equation (21).

Last edited by Cuchulainn on January 3rd, 2015, 11:00 pm, edited 1 time in total.

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- Cuchulainn
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QuoteOriginally posted by: ThinkDifferentis this thread some kind of a joke? if not, why is it in the Off Topic?I have requested Admin to move this to Student.

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QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: ThinkDifferentis this thread some kind of a joke? if not, why is it in the Off Topic?I have requested Admin to move this to Student.inappropriate.C_ct_sM_n is not a studentMaybe "General" or "Technical"

QuoteOriginally posted by: ppauperQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: ThinkDifferentis this thread some kind of a joke? if not, why is it in the Off Topic?I have requested Admin to move this to Student.inappropriate.C_ct_sM_n is not a studentMaybe "General" or "Technical"He is a beginner in Radon?Nikodym hence appropriate.

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Discrete probability space with two outcomes [$]\omega_1[$] and [$]\omega_2 [$]. Let the probability measure [$]\mu[$] assign them each [$]\mu(\omega _1)=\mu(\omega_2)=1/2[$]. Define a new probability measure $\nu$ which assigns them [$]\nu(\omega_1)=1/4[$] and [$]\nu(\omega_2)=3/4[$]. Then, the Radon-Nikodym derivative [$]\frac{d\mu}{d\nu} [$] is the random variable[$]\frac{d\mu}{d\nu}(\omega_1) = (1/2)/(1/4) = 2[$][$]\frac{d\mu}{d\nu}(\omega_2) = (1/2) / (3/4) = 2/3 [$]

- Cuchulainn
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QuoteOriginally posted by: ppauperQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: ThinkDifferentis this thread some kind of a joke? if not, why is it in the Off Topic?I have requested Admin to move this to Student.inappropriate.C_ct_sM_n is not a studentMaybe "General" or "Technical""Student" is a role, i.e. someone who is learning.Age and marital status etc. is irrelevant.So it is not the person asking the question, it is the level of the topic, in this case Student.That was my rationale.

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