pobazee
Senior Member
Posts: 246
Joined: Sep 2003
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Wed Sep 24, 03 05:08 PM
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N (Newton)
Here is a section from my work-in-progress, sorry, the notation is messy and I didn't not have time to correct them when I pasted it over.
3.2 Space of Information Representation and EMH
A not too stringent approach in measure theory that has an appeal here is to take as given a class of complete metric spaces. To facilitate the discussion, we provide some definitions and results leading up to our characterization of the space of information representation as Souslin space.
3.2.1 Souslin Measurable Space
Let (L, 't) be a topological space, and (L, 't) is called a Polish space if it is separable and there exist a metric on L for which L and 't is complete. Some examples of Polish space are; (i) Rd with it usual topology, (ii) separable Banach space with it norm topology, (iii) compact Hausdorff space with countable basis, and (iv) the space [0, 1]N consisting of all sequences of zeros and ones. We recall that the subset of topological space L is called Go if it is the intersection of sequence of open subsets of L, and Fa if it is the union of sequence of closed subsets of L. If L is a Polish space, then a subspace of L is Polish if and only if it is a Go in L. Every Polish space is Homeomorphic to Go in [0,1]N. If L is a Polish space, then each open and closed subsets of L is analytic. If L is Polish space, then each Borel subset of L is analytic. Suppose that L is a Polish space, the universally measurable subsets of L are those that are universally measurable with respect to (L, ~L)). Every analytic subset of L (Polish space) is universally measurable.
Let L be a metrizable topological space (m.t.s). L is called a Lusin space if it is homeomorphic to Borel subset of compact metrizable space. Every Polish space is Lusin, and every Lusin space is Souslin. Let L be m.t.s, and L is said to be Souslin if it is homeomorphic to analytic subset of compact metrizable space. A measurable space (L, fij is Souslin if it is measurably isomorphic to a measurable space (.o.,9f where .0. is Souslin metrizable space and .9is the Borel a-field of .0.. Every Souslin space is separable. If 't , 't are two comparable topologies in Souslin space, then the Borel a-algebra of 't and 't coincide. The Borel a-algebra of Souslin is countably generated. Let L be a Souslin space, then ~L) is countably separated. Let L be a Souslin space, then (L, ~L)) is analytic measurable space. Let (iJ be the separated and countably generated sub-algebra of {d, and let (L, fij be analytic measurable space, then (iJ = [rl, Let (.0., Sf be analytic measurable space it is a Blackwell ~ if (L, fij is a Souslin space.
(3.2.1) Fact: (P.-A. Meyer) Let (L,fij be Souslin space. Every Souslin measurable space (L, fij is a Radon space.
Proof See Schwartz [1973, p.123 and additional insights in pp.125-126].
Let L be H.t.s., then L is called a Radon space if every finite Borel measure on L is a Radon measure. Suppose that Z is H.t.s, and let L be Radon topological. subspace of Z Then L is universally measurable. The countable unions and intersections of Radon spaces are Radon. Note that Kcro is the c°u.ntable intersection of countable unions of compact spaces (= ("'I.u K.), and every continuous image of Kcro is called K-analytic. Suppose that 1 is a continuous onto map from L to X, and let 1 be the associated map for finite Radon measures. If it is K-analytic, then 1 is surjective. Given two comparable topologies in K-analytic set L, the spaces of finite Radon measures on them coincide, and the identity maps on them are universally Lusin measurable. Moreover, suppose 1 is a map from topological space L equipped with Radon measure U to Souslin space X. If U is Borel measurable, then U is Lusin measurable.
The results here are of particular interest to modeling consideration in dynamic security market. We offer them to draw immediate attention to the work of Huang (1985a&b) on continuous information structure. A function 1 on R with values in topological space is called regulated (or regular) if and only if it has a left and a right limits at every points. Suppose that X is a complete metric space, and let D be a countable dense subset of ~. The set of mappings of D to X that are the restrictions of regular, right continuous maps of ~ into X is Borel in the XID (= X restricted to set D).
3.2.2 Disintegration 01 Measure and EMH
As noted earlier, the existence of a disintegration of measure is fundamental to understanding informational efficiency of the market model. Roughly speaking, the existence of a disintegration of measure means that agents actions (beliefs) depend measurably on their information or that they put a full probability measure on the distinguishable events (if the Borel a-algebra is countably separated or generated).
To give a more intuitive interpretation, X be measurable space equipped with a finite positive Radon measure P that has a total mass of 1 defined on X., and the space X is a probability space. Interpret X as the state of the world, and X as the set of distinguishable events X is assumed to be drawn according to X. After which agents form their posterior probability for ro (denoted by P) by conditioning on their information. Thus, the mapping X characterizes the agents information (Bayesian updating). In this view, the existence of a disintegration could be interpreted to mean that agents use limiting posterior probability assessments for every event.
Let (formula) be a complete measure space. Let e be a Banach space over R. A function f on .0. with values in e is said to strongly measurable if there is a sequence X of step functions, X such that for X almost everywhere X in e as X. A function X is U-integrable if f is strongly measurable and if (formula). Where X assign to each X, the norm of X in e. Let (formula) be a measure space and g- be a sub a-algebra. Let (formula) be U-integrable. A function f is called the conditional expectation (c,e) of X if X is U-integrable and for each (formula) The integral X is the expectation of the indicator function of A with respect to X. The interpretation is that a trader with prior X hold the belief X about the expected value of f given that the true state is X.
Let X be a sub a-algebra of X and let X be a family of positive measure on X for X. The family X is called disintegration of v with respect to G if tile following properties are satisfied; (i) for (formula) and (ii) for (formula) is (formula). We note that if A = X then by property (ii), and for all v-almost X, is a probability measure. Notice that for a function f on X, the function X is the X.
Examples: Let F that is complete (formula), and let (formula) The measure (formula) on X is the conditional probability of v given B. Similarly, [I ydv/v(B)] is the c.e. of f given B.
Case (i) Now, suppose that .0. = U ER such that for all (formula), and (formula). Hence (B) is the X partition of X. We now show that there is a X and disintegration for this case. First take I to be any subset of X, so that X form a a-algebra as I varies over X. Let X denote this a-algebra of (formula). Let f be v-integrable function with value in e, and let f be given by X.
Thus, f satisfy the definition of the X of X; and X is then the (formula): Likewise for the family X roe X of measures on f let X = Y
Then (v:>roe.o. is by definition the disintegration of v w.r.t. .!):
Case (ii) Let .9':= (0, ~ rJ. .0.) be a a-algebra. Let f be v-integrable function on .0. with value in e. Let
(formula)
Then rY is e-valued, rY E .9'and v-integrable. Therefore, we have
(formula)
and
(formula)
Hence, Y is the X of f . Now, for the family (formula) measures on F let
(formula)
We conclude that this family satisfies the definition of a disintegration of measure V w.r.t. & This end the examples.
We provide two theorems that addresses the existence and uniqueness of disintegration of measure. The proof these theorems is intimately related to that of c.e.
(3.2.2.1) Fact. Let X be Souslin measurable space, and let v be positive Radon measure on .0.. Let .9'be a-algebra contained in X. Then there exist a disintegration of measure of V.
Proof See Ramaswamy/Schwartz (1975) theorem (III, §5, 43). The proof actually follow from theorem (III, §3, 38). .
(3.2.2.2) Fact. Let (.0.,.9; be Souslin measurable space, and let v be positive Radon measure on .9: Let .9'be a-algebra contained in X and X are two disintegration of X then (formula)for all v-almost X.
Proof See Ramaswamy/Schwartz (1975) theorems (ill, §3, 36) and (ill, §5, 42). .
The following results give the necessary and sufficient conditions for X to be to disintegration of X.
(3.2.2.3) Proposition. Let X be a family of positive measure. For the necessary and sufficient conditions for X to be disintegration of X it is enough to verify the following: (i) The function X taking X to V: is in (formula) (iii) for all X and for all v-almost <0, v.9' is carried by B or by Bc according to whether X or X
Proof Necessity: Let X be the disintegration of X Then conditions (i) and (ii) follows from the definition of disintegration. For condition (iii); let X and let X be the X. Whence, (formula) and (formula). Thus, for all v-almost (formula) and (formula). Therefore, for all v-almost X is carried by (formula) and is carried by (formula).
Sufficiency: Let X be the c.e. of IB w.r.t .9: We need to show that, for A e !land {;rle .9; SA IBdv = SA V1B)dv.1n equivalent form, we need to show that v(A n B) = S A v ~B)dv. That is (formula)). Notice that for all v-almost CO, and CO e A; v: is carried by A. Hence, for all v-almost CO, and CO e A; v co(B n A ) = 0, and
(formula). Therefore,
(formula). Recall that
(formula)
Since for all v-almost CO, V; is carried by A c if co e A c, then (formula).
Thus, (formula)
We observe that the existence of disintegration of measure is not a mere mathematical artifice rather it is fundamental to almost all results in finance that appeal to conditional expectation formula. The reason is that the existence of disintegration of measure Jl with respect to sub a-algebra X immediately implies the existence of conditional expectation with respect to X for nonnegative functions in X. Indeed, one is tempted in a reasonably way to say that the existence and uniqueness of a disintegration of measure in the market model implies informational efficiency under reasonable assumptions about the measurable space (X) We claim that a primitive assumption on market model that implies informational efficiency is that:
(3.2.2.*) Assumption: The analytic measurable space (X) is a Blackwell space.
In addition, for economically meaningful spaces, saying that X is a Blackwell measurable space is equivalent to it being Souslin space.
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"The Bourbaki were Puritans, and Puritans are strongly opposed to pictorial representations of truths of their faith."
--- Pierre Cartier, June 18, 1997
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