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Cuchulainn
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QuoteDiscrete 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$Not sure I understand/agree with this. Can you elaborate? It looks like an abuse of notation.I assume when you write $\omega_1$ you mean a (measurable) set?If we take a similar problem, i.e.. define the delta function as a measure:m(A) = 1 if 0 is an element of A (A is a subset of the real line)m(A) = 0 otherwiseThen this delta measure does not have a Radon-Nikodym derivative and is not a standard Riemann or Lebesgue integral. It is a special case of a Schwarz distribution which does have a rigorous definition.
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emac
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I am simply looking at the probability space $(\Omega, F, \mu)$ where $\Omega=\{\omega_1, \omega_2\}$, $F$ is the power set of $\Omega$ and $\mu$ is defined as I did it. The OP asked for the simplest possible example. A probability space with two possible outcomes is about as simple as it gets.I guess I should really write $\mu (\{\omega_1\})$ etc. since ,as you say, I am evaluating $\mu$ on the set $\{\omega_1\}$As for your examples, I don't get what you are saying. The measure you describe doesn't have a density w.r.t. Lebesque measure as they do not have the same null sets.

emac
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Maybe I should be clearer: the Radon-Nikodym theorem is about the existence of a density between two absolutely continuous measures. In finance, the measures could be, for example, the risk-neutral measures corresponding to different numeraires (e.g. foreign and domestic bonds), or the physical measure and the risk-neutral measure.What is the density? it is basically a measure of how distorted probabilities of certain events become by moving from one measure to another. In finance, it is often given by a stochastic exponential. Girsanov's theorem tells you how dynamics of processes change when you change measure by a Radon-Nikodym density which is a stochastic exponential.In finance, you work on probability spaces which are complicated - if you work on the sample path space (Wiener space) then it is infinite-dimensional etc. etc. so obviously there are technicalities in proving the Radon-Nikodym theorem and Girsanov's theorem etc. On the other hand, if you want to intuitively understand what a Radon-Nikodym derivative/density is, it is easy. Think of the simplest (non-trivial) probability space and simplest measure you can. For me, this is tossing a coin. There are two outcomes: H or T. Now pick two probability measures on this space, which don't assign either outcome 0. In the example below, I picked a fair coin and a biased coin. Then to calculate the Radon-Nikodym derivative between the two measures, it is just the ratio of the respective probabilities. (This is also known as the likelihood ratio!)It is completely clear what the Radon-Nikodym derivative is and what, intuitively, it means.As with everything else in maths, once you understand it is easy. Thinking about R-N theorem for more complicated probability spaces and measures is more technical, but the idea is the same.

Orbit
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QuoteOriginally posted by: CactusManWho here is into this and can talk about it?Oh I'm into it alright. I think one of the clearest and most intuitive explanations can be found in Salih Neftci's intro book. Rather than type it out here, I would suggest check out his book.

Cuchulainn
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One example in fixed income:LetD(t, s) == discount factor at s as seen from t (present value of one unit of currency)P(t, s) == price at time t of pure discount bond on at s-maturity pure discount bondQ == risk-neutral measureQ_T == forward measureThe Radon-Nikodym in this case is:dQ_T/dQ = D(0,T)/P(0,T)
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Cuchulainn
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QuoteAs with everything else in maths, once you understand it is easy. Thinking about R-N theorem for more complicated probability spaces and measures is more technical, but the idea is the same. Why is the integral form of RN not referred to? It more intuitive and rigorous. Shreve II has RN as well. He states it in terms of equivalent measures (i.e. they agrees on what sets of measure/probability zero are).Take the normal distribution. I claim n(x) (it is a a.s. r.v with Expectation 1) is the RN derivative. Agree/disagree?
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emac
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QuoteWhy is the integral form of RN not referred to? It more intuitive and rigorous. More intuitive and rigorous than what? We were asked for the simplest Radon-Nikodym derivative we can think of. I picked one which is a function on a set with two elements. This is extremely simple.QuoteTake the normal distribution. I claim n(x) (it is a a.s. r.v with Expectation 1) is the RN derivative. What is n(x)? And what does "a.s. r.v." mean? Of course if $\nu$ is the law of a Normal random variable and $\mu$ is Lebesgue measure then the Radon-Nikodym derivative $\frac{d\nu}{d\mu}$ is the "usual" density of a normal random variable.

Cuchulainn
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QuoteOriginally posted by: emacQuoteWhy is the integral form of RN not referred to? It more intuitive and rigorous. More intuitive and rigorous than what? We were asked for the simplest Radon-Nikodym derivative we can think of. I picked one which is a function on a set with two elements. This is extremely simple.QuoteTake the normal distribution. I claim n(x) (it is a a.s. r.v with Expectation 1) is the RN derivative. What is n(x)? And what does "a.s. r.v." mean? Of course if $\nu$ is the law of a Normal random variable and $\mu$ is Lebesgue measure then the Radon-Nikodym derivative $\frac{d\nu}{d\mu}$ is the "usual" density of a normal random variable.More rigorous than 'notation' dQ_T/dQ. That's what worries about the _original_ example you posted.Don't n, a.s., r.v.? this is worrying..a.s. used to be p.p. (Presque partout).
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Cuchulainn
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QuoteOf course if ν is the law of a Normal random variable and μ is Lebesgue measure then the Radon-Nikodym derivative dν/dμ is the "usual" density of a normal random variable.That's a good example IMO.
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emac
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QuoteMore rigorous than 'notation' dQ_T/dQ. That's what worries about the _original_ example you posted.Don't n, a.s., r.v.? this is worrying..a..s used to be p.p. (Presque partout). dQ_T/dQ is perfectly rigorous. The R-N states the existence of a measurable function, which most people call dQ_T/dQ or d(\mu)/d(\nu) or whatever.I know what a.s. means - it doesn't make sense in the way you used it. A normal random variable is not 'almost surely' a random variable. This does not make sense.

Cuchulainn
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QuoteOriginally posted by: emacQuoteMore rigorous than 'notation' dQ_T/dQ. That's what worries about the _original_ example you posted.Don't n, a.s., r.v.? this is worrying..a..s used to be p.p. (Presque partout). dQ_T/dQ is perfectly rigorous. The R-N states the existence of a measurable function, which most people call dQ_T/dQ or d(\mu)/d(\nu) or whatever.I know what a.s. means - it doesn't make sense in the way you used it. A normal random variable is not 'almost surely' a random variable. This does not make sense.It's how Shreve formulates it. A definition of a.s. equality is hereI see the formulation that the r.v. is unique up to a set of probability (~ measure..) zero. It's an equivalence class, yes?// If you want to solve RN numerically, which formulation is mote intuitive??
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emac
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I know what almost sure means. You did not say two random variables were equal almost surely, you saidQuoteit is a a.s. r.v What is an almost sure random variable?

Cuchulainn
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QuoteOriginally posted by: emacI know what almost sure means. You did not say two random variables were equal almost surely, you saidQuoteit is a a.s. r.v What is an almost surely random variable?True, I did not say that, did I. BTW I was quoting Shreve which I think is a good explanation of RN.Maybe cross-wired. See my previous post. But you are probably correct. How would to formulate it to make it unambiguous? Maybe Rudin's definition is better where RN derivative is in the Banach space L_1 (mu) and we are done because they are equivalence classes of measurable functions.Then a.s. and r.v. just get subsumed as special cases. QuotedQ_T/dQ is perfectly rigorous. The R-N states the existence of a measurable function, which most people call dQ_T/dQ or d(\mu)/d(\nu) or whatever.I agree. IMO it is the accepted notation for that function in the statement of the RN theorem.
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Cuchulainn
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Quote I'm doubtful about applications.Maybe it has escaped everyone's attention... but ..RN derivative _is_ the likelihood function in parameter estimation in SDEs.Or am I missing something here?
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Cuchulainn
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QuoteOriginally posted by: OrbitQuoteOriginally posted by: CactusManWho here is into this and can talk about it?Oh I'm into it alright. I think one of the clearest and most intuitive explanations can be found in Salih Neftci's intro book. Rather than type it out here, I would suggest check out his book.Believe it or not, I once attended a course by the late prof Salih on "advanced maths for derivatives".. He had a knack for explaining things.His formulation of RN is an ordinary differential equation which makes it tangible.But discussing measure theory form an economics background is not optimal IMO. The bespoke example by Brigo and Mercurio is good IMO.
Last edited by Cuchulainn on January 27th, 2015, 11:00 pm, edited 1 time in total.
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