Advances in Pure Mathematics
Vol.09 No.06(2019), Article ID:93395,44 pages
10.4236/apm.2019.96029

Description of Incomplete Financial Markets for Time Evolution of Risk Assets

Nicholas Simon Gonchar

Bogolyubov Institute for Theoretical Physics of NAS, Kiev, Ukraine

Copyright © 2019 by author(s) and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

Received: June 4, 2019; Accepted: June 27, 2019; Published: June 30, 2019

ABSTRACT

In the paper, a class of discrete evolutions of risk assets having the memory is considered. For such evolutions the description of all martingale measures is presented. It is proved that every martingale measure is an integral on the set of extreme points relative to some measure on it. For such a set of evolutions of risk assets, the contraction of the set of martingale measures on the filtration is described and the representation for it is found. The inequality for the integrals from a nonnegative random value relative to the contraction of the set of martingale measure on the filtration which is dominated by one is obtained. Using these inequalities a new proof of the optional decomposition theorem for super-martingales is presented. The description of all local regular super-martingales relative to the regular set of measures is presented. The applications of the results obtained to mathematical finance are presented. In the case, as evolution of a risk asset is given by the discrete geometric Brownian motion, the financial market is incomplete and a new formula for the fair price of super-hedge is founded.

Keywords:

Random Process, Regular Set of Measures, Optional Doob Decomposition, Local Regular Super-Martingale, Martingale, Discrete Geometric Brownian Motion

1. Introduction

In the paper, the notion of the regular super-martingale relative to the set of equivalent measures is introduced. The necessary and sufficient conditions of the regular super-martingale relative to the set of equivalent measures are found. The notion of the family of equivalent measures consistent with filtration is introduced. Theorem giving the sufficient conditions of the existence of super-martingale and martingale relative to the set of equivalent measures consistent with the filtration is proved. The sufficient conditions of the existence of the set of equivalent measures consistent with the filtration, satisfying the conditions: the mean value of the nonnegative random value relative to these set of measures equal one, are given. Further, we construct the set of equivalent measures consistent with the filtration satisfying the above conditions. First, we give the complete description of the set of equivalent measures satisfying the conditions: the mean value of the nonnegative random value relative to this set of measures equals one. Using the above result we construct the example of the set of equivalent measures consistent with the filtration satisfying the condition: the mean value of the nonnegative random value relative to every measure of this set of measures equals one. The above method we use for the construction of evolution

of risk assets and we describe completely the set of equivalent martingale measures for this evolution. We prove that every martingale measure is an integral on the set of extreme points of the convex set of martingale measures relative to some measure on it. To give a new proof of the optional decomposition for super-martingale we describe the contraction of every martingale measure on the filtration and find the closure of integrals from the integrable random values over all martingale measures. To do this we introduce the notion of the exhaustive decomposition and prove that every separable metric space with the Borel σ-algebra has an exhaustive decomposition. For the integral from the nonnegative random value relative to all martingale measures which is dominated by one, the inequalities for this random value are obtained. This fact gives us the possibility to find a new proof of the optional decomposition for the nonnegative super-martingale. This proof does not use the no-arbitrage arguments and the measurable choice [1] [2] [3] [4] . This paper is a generalization of the results of the paper [5] .

First, the optional decomposition for diffusion processes super-martingale was opened by El Karoui N. and Quenez M. C. [6] . After that, Kramkov D. O. and Follmer H. [1] [2] proved the optional decomposition for the nonnegative bounded super-martingales. Folmer H. and Kabanov Yu. M. [3] [4] proved analogous result for an arbitrary super-martingale. Recently, Bouchard B. and Nutz M. [7] considered a class of discrete models and proved the necessary and sufficient conditions for the validity of the optional decomposition.

The optional decomposition for super-martingales plays the fundamental role for the risk assessment in incomplete markets [1] [2] [6] [8] [9] [10] [11] .

At last, we consider an application of the results obtained to find the new formula for the fair price of super-hedge in the case, as the risk asset evolves by the discrete geometric Brownian motion.

2. Local Regular Super-Martingales Relative to a Set of Equivalent Measures

We assume that on a measurable space { Ω , F } a filtration F m F m + 1 F , m= 0, ¯ , and a set of equivalent measures M on F are given. Further, we

assume that F 0 = { , Ω } and the σ-algebra F = σ ( V n = 1 F n ) is a minimal σ-algebra generated by the algebra V n = 1 F n . A random process ψ = { ψ m } m = 0 is

said to be adapted one relative to the filtration { F m } m = 0 , if ψ m is a F m measurable random value, m = 0 , ¯ .

Definition 1. An adapted random process f = { f m } m = 0 is said to be a super-martingale relative to the filtration F m , m = 0 , ¯ , and the family of equivalent measures M, if E P | f m | < , m = 1 , ¯ , P M , and the inequalities

E P { f m | F k } f k , 0 k m , m = 1 , ¯ , P M , (1)

are valid.

Further, for an adapted process f we use both the denotation { f m , F m } m = 0 and the denotation { f m } m = 0 .

Definition 2. A super-martingale { f m , F m } m = 0 relative to a set of equivalent

measures M is a local regular one, if sup P M E P | f m | < , m = 1 , ¯ , and there exists

an adapted nonnegative increasing random process { g m , F m } m = 0 , g 0 = 0 ,

sup P M E P | g m | < , m = 1 , ¯ , such that { f m + g m , F m } m = 0 is a martingale relative to

every measure from M.

The next elementary Theorem 1 will be very useful later.

Theorem 1. Let a super-martingale { f m , F m } m = 0 , relative to a set of equivalent measures M be such that sup P M E P | f m | < , m = 1 , ¯ . The necessary and sufficient

condition for it to be a local regular one is the existence of an adapted nonnegative

random process { g ¯ m 0 , F m } m = 0 , sup P M E P | g ¯ m 0 | < , m = 1 , ¯ , such that

f m 1 E P { f m | F m 1 } = E P { g ¯ m 0 | F m 1 } , m = 1 , ¯ , P M . (2)

Proof. The necessity. If { f m , F m } m = 0 is a local regular super-martingale, then there exist a martingale { M ¯ m , F m } m = 0 and a non-decreasing nonnegative random process { g m , F m } m = 0 , g 0 = 0 , such that

f m = M ¯ m g m , m = 1 , ¯ . (3)

From here, we obtain the equalities

E P { f m 1 f m | F m 1 } = E P { g m g m 1 | F m 1 } = E P { g ¯ m 0 | F m 1 } , m = 1 , ¯ , P M , (4)

where we introduced the denotation g ¯ m 0 = g m g m 1 0 . It is evident that

E P g ¯ m 0 sup P M E P g m + sup P M E P g m 1 < .

The sufficiency. Suppose that there exists an adapted nonnegative random process g ¯ 0 = { g ¯ m 0 } m = 0 , g ¯ 0 0 = 0 , E P g ¯ m 0 < , m = 1 , ¯ , such that the equalities (2) hold. Let us consider the random process { M ¯ m , F m } m = 0 , where

M ¯ 0 = f 0 , M ¯ m = f m + i = 1 m g ¯ m 0 , m = 1 , ¯ . (5)

It is evident that E P | M ¯ m | < and

E P { M ¯ m 1 M ¯ m | F m 1 } = E P { f m 1 f m g ¯ m 0 | F m 1 } = 0. (6)

Theorem 1 is proved. □

Lemma 1. Any super-martingale { f m , F m } m = 0 relative to a family of measures M for which there hold equalities E P f m = f 0 , m = 1 , ¯ , P M , is a martingale with respect to this family of measures and the filtration F m , m = 1 , ¯ .

Proof. The proof of Lemma 1 see [12] . □

In the next Lemma, we present the formula for calculation of the conditional expectation relative to another measure from M.

Lemma 2. On the measurable space { Ω , F } with the filtration F n on it, let M be a set of equivalent measures and let ξ be an integrable random value. Then, the following formulas

E P 1 { ξ | F n } = E P 2 { ξ φ n P 1 | F n } , n = 1 , ¯ , (7)

are valid, where

φ n P 1 = d P 1 d P 2 [ E P 2 { d P 1 d P 2 | F n } ] 1 , P 1 , P 2 M . (8)

Proof. The proof of Lemma 2 is evident. □

3. Local Regular Super-Martingales Relative to a Set of Equivalent Measures Consistent with the Filtration

Definition 3. On a measurable space { Ω , F } with a filtration F n on it, a set of equivalent measures M we call consistent with the filtration F n , if for every pair of measures ( Q 1 , Q 2 ) M 2 the set of measures

R s k ( A ) = A E Q 1 { d Q 2 d Q 1 | F k } E Q 1 { d Q 2 d Q 1 | F s } d Q 1 , A F , k s n , n = 0 , ¯ , (9)

belongs to the set M, where M 2 is a direct product of the set M by itself.

Lemma 3. On the measurable space { Ω , F } with the filtration F n on it, the set of measures

M = { Q , Q ( A ) = A α ( ω ) d P , A F , Q ( Ω ) = 1 } (10)

is a consistent one with the filtration F n , if P is a measure on { Ω , F } and a random value α ( ω ) runs over all nonnegative random values, satisfying the condition P ( { ω , α ( ω ) > 0 } ) = 1 .

Proof. Suppose that ( Q 1 , Q 2 ) belongs to M 2 . Then, d Q 2 d Q 1 = α 2 ( ω ) α 1 ( ω ) and P ( { ω , d Q 2 d Q 1 > 0 } ) = 1 , since the equalities P ( { ω , 0 < α 1 ( ω ) < } ) = 1 , P ( { ω , 0 < α 2 ( ω ) < } ) = 1 are true. It is evident that

R s k ( A ) = A E Q 1 { d Q 2 d Q 1 | F k } E Q 1 { d Q 2 d Q 1 | F s } d Q 1 = A E Q 1 { d Q 2 d Q 1 | F k } E Q 1 { d Q 2 d Q 1 | F s } α 1 ( ω ) d P , A F , k s n , n = 0 , ¯ . (11)

It is easy to see that

P ( { ω , E Q 1 { d Q 2 d Q 1 | F k } E Q 1 { d Q 2 d Q 1 | F s } α 1 ( ω ) > 0 } ) = 1 , k s , (12)

since

P ( { ω , E Q 1 { d Q 2 d Q 1 | F k } > 0 } ) = 1 , k s , (13)

P ( { ω , 0 < E Q 1 { d Q 2 d Q 1 | F s } < } ) = 1 , s n , n = 0 , ¯ . (14)

The last equality follows from the equivalence of the measures Q 1 , Q 2 and P. Altogether, it means that the set of measures R s k , k s n , n = 0 , ¯ , belongs to the set M. The same is true for the pair ( Q 2 , Q 1 ) M 2 . Lemma 3 is proved. □

Theorem 2. On the measurable space { Ω , F } with the filtration F n on it, let the set of equivalent measures M be consistent with the filtration F n . Then,

for every nonnegative random value ξ such that sup P M E P ξ < , the random process { f n , F n } n = 0 is a super-martingale relative to the set of measures M, where f n = ess sup P M E P { ξ | F n } , n = 0 , ¯ .

Proof. Let Q M , then, due to Lemma 2, for every P M

E P { ξ | F n } = E Q { ξ | d P d Q E Q { d P d Q | F n } | F n } . (15)

If to put instead of the measure P the measure R s k , k s n , for the pair of measures ( Q , P ) we obtain

E R s k { ξ | F n } = E Q { ξ | d R s k d Q E Q { d R s k d Q | F n } | F n } = E Q { ξ E Q { d P d Q | F k } E Q { d P d Q | F s } | F n } , (16)

where we took into account the equality

E Q { d R s k d Q | F n } = E Q { E Q { d P d Q | F k } E Q { d P d Q | F s } | F n } = 1 , k s n . (17)

From the formula (16), it follows the equality

ess sup P M E P { ξ | F n } = ess sup T R n E P { ξ T | F n } , (18)

where R n is a set of martingales T = { T m } m = 0 relative to the measure Q such

that T m = 1 , m n , T m = E Q { d P d Q | F m } E Q { d P d Q | F s } , m s n , P M . The definition of

ess sup for the uncountable set of random values see [13] . It is evident that T n T n 1 . Let us consider

E Q { ess sup P M E P { ξ | F n } | F n 1 } = E Q { ess sup T R n E P { ξ T | F n } | F n 1 } = E Q { sup i 1 E P { ξ T i | F n } | F n 1 } = E Q { lim k max 1 i k E P { ξ T i | F n } | F n 1 } = lim k E Q { max 1 i k E P { ξ T i | F n } | F n 1 } = lim k E P { ξ T τ k | F n 1 } ess sup T R n E Q { ξ T | F n 1 } ess sup T R n 1 E Q { ξ T | F n 1 } = ess sup P M E Q { ξ | F n 1 } , (19)

where

τ 1 = 1 , (20)

τ i = { τ i 1 , E P { ξ T τ i 1 | F n } > E P { ξ T i | F n } , i , E P { ξ T τ i 1 | F n } E P { ξ T i | F n } , i = 2 , k ¯ . (21)

Lemma 2 is proved. □

Theorem 3. On the measurable space { Ω , F } , F = σ ( V i = 1 F i ) , let M be a set

of equivalent measures being consistent with the filtration F n . If there exists a nonnegative random value ξ 1 such that E P ξ = 1 , P M , then E P { ξ | F n } , P M , is a local regular martingale.

Proof. Due to Lemma 2, the random process { f n , F n } n = 0 , where

f n = ess sup P M E P { ξ | F n } , n = 0 , ¯ , is a super-martingale relative to the set of

measures M, that is,

E Q { ess sup P M E P { ξ | F n } | F n 1 } ess sup P M E P { ξ | F n 1 } , Q M , n = 0, ¯ . (22)

From the inequality (22), it follows the inequality

E Q ess sup P M E P { ξ | F n } 1, n = 0, ¯ . (23)

Since E Q ess sup P M E P { ξ | F n } E Q E Q { ξ | F n } = 1 , we have

E Q ess sup P M E P { ξ | F n } = 1, Q M , n = 0, ¯ . (24)

The inequalities (22) and the equalities (24) give the equalities

E Q { ess sup P M E P { ξ | F n } | F n 1 } = ess sup P M E P { ξ | F n 1 } , Q M , n = 1, ¯ , (25)

which are true with the probability 1. The last means that { f n , F n } n = 0 is a martingale relative to the set of measures M, where f n = ess sup P M E P { ξ | F n } , n = 0, ¯ . With the probability 1, lim n ess sup P M E P { ξ | F n } = f , where the random value

f is F measurable one. From the inequality (23) and Fatou Lemma [13] [14] , we obtain

E P f 1, P M . (26)

Prove that f = ξ . Going to the limit in the inequality

ess sup P M E P { ξ | F n } E P 1 { ξ | F n } , (27)

as n , we obtain the inequality

f ξ . (28)

From the inequality (26) and the inequality (28), we obtain the inequalities 1 E P f E P ξ = 1 . Or, E P f = 1 . The equalities E P f = 1 , E P ξ = 1 and the inequality (28) give the equality f = ξ with the probability 1. Lemma 3 is proved. □

Lemma 4. On the measurable space { Ω , F } with the filtration F n on it, let there exist k equivalent measures P 1 , , P k , k > 1 , and a nonnegative random value ξ 0 1 be such that

E P i { ξ 0 | F n } = E P 1 { ξ 0 | F n } , E P i ξ 0 = 1 , i = 2 , k ¯ , n = 0 , ¯ . (29)

Then, there exists the set of equivalent measures M consistent with the filtration F n , satisfying the condition E P ξ 0 = 1 , P M .

Proof. Let us consider the set of equivalent measures M, satisfying the condition

E P { ξ 0 | F n } = E P 1 { ξ 0 | F n } , n = 0 , ¯ , P M . (30)

Such a set of measures is a nonempty one. Suppose that Q 1 , Q 2 M , then

E Q 1 { ξ 0 | F n } = E Q 2 { ξ 0 | F n } , n = 0 , ¯ . (31)

Let us prove that the formula

E Q 1 { ξ 0 E Q 1 { d Q 2 d Q 1 | F k } E Q 1 { d Q 2 d Q 1 | F s } | F n } = E Q 1 { ξ 0 | F n } , n s k , n = 0 , ¯ , (32)

is valid. Let s n . Then, from the equalities (31), we have

E Q 1 { E Q 2 { ξ 0 | F s } | F n } = E Q 1 { ξ 0 | F n } .

Let k s . Then,

E Q 1 { E Q 2 { ξ 0 | F s } | F n } = E Q 1 { E Q 2 { E Q 2 { ξ 0 | F k } | F s } | F n } = E Q 1 { E Q 1 { E Q 2 { ξ 0 | F k } d Q 2 d Q 1 E Q 1 { d Q 2 d Q 1 | F s } | F s } | F n } = E Q 1 { E Q 2 { ξ 0 | F k } d Q 2 d Q 1 E Q 1 { d Q 2 d Q 1 | F s } | F n }

= E Q 1 { E Q 1 { ξ 0 | F k } d Q 2 d Q 1 E Q 1 { d Q 2 d Q 1 | F s } | F n } = E Q 1 { E Q 1 { ξ 0 | F k } E Q 1 { d Q 2 d Q 1 | F k } E Q 1 { d Q 2 d Q 1 | F s } | F n } = E Q 1 { ξ 0 | E Q 1 { d Q 2 d Q 1 | F k } E Q 1 { d Q 2 d Q 1 | F s } | F n } .

This proves the formula (32). To finish the proof of Lemma 4, it needs to prove that the set of measures

R s k ( A ) = A E Q 1 { d Q 2 d Q 1 | F k } E Q 1 { d Q 2 d Q 1 | F s } d Q 1 , A F , k s n , n = 0 , ¯ , (33)

belongs to the set M. Really,

E R s k { ξ 0 | F n } = E Q 1 { ξ 0 | d R s k d Q 1 E Q 1 { d R s k d Q 1 | F n } | F n } = E Q 1 { ξ 0 E Q 1 { d Q 2 d Q 1 | F k } E Q 1 { d Q 2 d Q 1 | F s } | F n } = E Q 1 { ξ 0 | F n } , (34)

where we took into account the equality

E Q 1 { d R s k d Q 1 | F n } = E Q 1 { E Q 1 { d Q 2 d Q 1 | F k } E Q 1 { d Q 2 d Q 1 | F s } | F n } = 1 , k s n . (35)

From this, it follows that the set of measures R s k M . This proves the consistence with the filtration of the set of measures M. Lemma 4 is proved. □

The next Lemma 5 is a key statement in the construction of the set of measures satisfying the conditions of Lemma 4.

On a probability space { Ω , F , P } , let ξ be a random value, satisfying the conditions

0 < P ( { ω , ξ > 0 } ) < 1 , 0 < P ( { ω , ξ < 0 } ) . (36)

Denote Ω + = { ω , ξ ( ω ) > 0 } , Ω = { ω , ξ ( ω ) 0 } and let F , F + be the restrictions of the σ-algebra F on the sets Ω and Ω + , correspondingly. Suppose that P and P + are the contractions of the measure P on the σ-algebras F , F + , correspondingly. Consider the measurable space with measure { Ω × Ω + , F × F + , μ } , which is a direct product of the measurable spaces with measures { Ω , F , P } and { Ω + , F + , P + } , where μ = P × P + . Introduce the denotations

ξ + ( ω ) = { ξ ( ω ) , ω { ξ ( ω ) > 0 } , 0 , ω { ξ ( ω ) 0 } , (37)

ξ ( ω ) = { ξ ( ω ) , ω { ξ ( ω ) 0 } , 0 , ω { ξ ( ω ) > 0 } . (38)

Then, ξ ( ω ) = ξ + ( ω ) ξ ( ω ) .

On the measurable space { Ω × Ω + , F × F + , P × P + } , we assume that the set of nonnegative measurable functions α ( ω 1 , ω 2 ) , satisfying the conditions

μ ( { ( ω 1 , ω 2 ) Ω × Ω + , α ( ω 1 , ω 2 ) > 0 } ) = P ( Ω + ) P ( Ω ) , (39)

Ω Ω + α ( ω 1 , ω 2 ) ξ ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) < , (40)

Ω Ω + α ( ω 1 , ω 2 ) d μ ( ω 1 , ω 2 ) = 1 , (41)

is a nonempty set. Such assumptions are true for the nonempty set of bounded random values α ( ω 1 , ω 2 ) , for example, if the random value ξ is an integrable one relative to the measure P.

Lemma 5. On the probability space { Ω , F , P } , let a random value ξ satisfy the conditions (36) and let a measure Q be equivalent to the measure P and such that E Q ξ = 0 . Then, for the measure Q the following representation

Q ( A ) = Ω Ω + χ A ( ω 1 ) α ( ω 1 , ω 2 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) + Ω Ω + χ A ( ω 2 ) α ( ω 1 , ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) , A F , (42)

is valid for those random value α ( ω 1 , ω 2 ) that satisfy the conditions (39)-(41).

Every measure Q, given by the formula (42), with the random value α ( ω 1 , ω 2 ) , satisfying the conditions (39)-(41) is equivalent to the measure P and is such that E Q ξ = 0 . For the measure Q, the canonical representation

Q ( A ) = Ω Ω + χ A ( ω 1 ) α 1 ( ω 1 , ω 2 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) + Ω Ω + χ A ( ω 2 ) α 1 ( ω 1 , ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) , A F , (43)

is valid, where

α 1 ( ω 1 , ω 2 ) = ψ 1 ( ω 1 ) ψ 2 ( ω 2 ) [ ξ ( ω 1 ) + ξ + ( ω 2 ) ] d , ( ω 1 , ω 2 ) Ω × Ω + , (44)

ψ 1 ( ω 1 ) = Ω + α ( ω 1 , ω 2 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d P ( ω 2 ) , ω 1 Ω , (45)

ψ 2 ( ω 2 ) = Ω α ( ω 1 , ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d P ( ω 1 ) , ω 2 Ω + , (46)

d = Ω ξ ( ω 1 ) ψ 1 ( ω 1 ) d P ( ω 1 ) = Ω + ξ + ( ω 2 ) ψ 2 ( ω 2 ) d P ( ω 2 ) . (47)

Proof. From the Lemma 5 conditions,

Q ( A ) = A ψ ( ω ) d P , P ( { ω , ψ ( ω ) > 0 } ) = 1 , (48)

Ω ψ ( ω ) ξ ( ω ) d P ( ω ) = 0. (49)

The condition (49) means

Ω + ψ 2 ( ω 2 ) ξ + ( ω 2 ) d P ( ω 2 ) = Ω ψ 1 ( ω 1 ) ξ ( ω 1 ) d P ( ω 1 ) = d > 0 , (50)

where

ψ 1 ( ω ) = { ψ ( ω ) , ω Ω , 0 , ω Ω + , (51)

ψ 2 ( ω ) = { ψ ( ω ) , ω Ω + , 0 , ω Ω . (52)

Let us put

α ( ω 1 , ω 2 ) = ψ 1 ( ω 1 ) ψ 2 ( ω 2 ) [ ξ ( ω 1 ) + ξ + ( ω 2 ) ] d , ( ω 1 , ω 2 ) Ω × Ω + . (53)

Then, for such α ( ω 1 , ω 2 ) the equality (39) is true. Moreover,

Ω Ω + α ( ω 1 , ω 2 ) ξ ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) = d 2 < , (54)

Ω Ω + α ( ω 1 , ω 2 ) d μ ( ω 1 , ω 2 ) = Ω ψ 1 ( ω 1 ) d P ( ω 1 ) + Ω + ψ 2 ( ω 2 ) d P ( ω 2 ) = 1 , (55)

E Q ξ = Ω Ω + α ( ω 1 , ω 2 ) ξ ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) + Ω Ω + α ( ω 1 , ω 2 ) ξ ( ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) = 0 , (56)

since ξ ( ω 1 ) = ξ ( ω 1 ) , ω 1 Ω , ξ ( ω 2 ) = ξ + ( ω 2 ) , ω 2 Ω + .

Let us prove the last statement of Lemma 5. Suppose that the representation (42) for the measure Q, satisfying the conditions (39)-(41), is valid. Taking into account the denotations (45)-(47), we obtain

Q ( A ) = Ω χ A ( ω 1 ) ψ 1 ( ω 1 ) d P ( ω 1 ) + Ω + χ A ( ω 2 ) ψ 2 ( ω 2 ) d P ( ω 2 ) , (57)

0 = E Q ξ = Ω ξ ( ω 1 ) ψ 1 ( ω 1 ) d P ( ω 1 ) + Ω + ξ ( ω 2 ) ψ 2 ( ω 2 ) d P ( ω 2 ) = Ω ξ ( ω 1 ) ψ 1 ( ω 1 ) d P ( ω 1 ) + Ω + ξ + ( ω 2 ) ψ 2 ( ω 2 ) d P ( ω 2 ) . (58)

If to introduce the denotation

ψ ( ω ) = { ψ 1 ( ω ) , ω Ω , ψ 2 ( ω ) , ω Ω + , (59)

then we obtain the representation

Q ( A ) = A ψ ( ω ) d P ( ω ) , (60)

where P ( ψ 1 ( ω ) > 0 ) = P ( Ω ) , P ( ψ 2 ( ω ) > 0 ) = P ( Ω + ) .

The last formula proves the equivalence of the measures Q and P. At last, to prove the canonical representation (43) it is sufficient to substitute the expression (44) for α 1 ( ω 1 , ω 2 ) into the expression (43) for Q ( A ) . We obtain the expression (57) for Q ( A ) . Then, if to substitute the expressions (45), (46) for ψ 1 ( ω 1 ) , ψ 2 ( ω 2 ) into the expression (57) for Q ( A ) , we obtain that the canonical representation for Q ( A ) is true. This proves Lemma 5. □

For further investigations, the next Theorem 4 is very important [5] .

Theorem 4. The necessary and sufficient conditions of the local regularity of the nonnegative super-martingale { f m , F m } m = 0 relative to a set of equivalent measures M are the existence of F m -measurable random values ξ m 0 A 0 , m = 1 , ¯ , such that

f m f m 1 ξ m 0 , E P { ξ m 0 | F m 1 } = 1 , P M , m = 1 , ¯ . (61)

Proof. The necessity. Without loss of generality, we assume that f m a for a certain real number a > 0 . Really, if it is not so, then we can come to the consideration of the super-martingale { f m + a , F m } m = 0 . Thus, let { f m , F m } m = 0 be a nonnegative local regular super-martingale. Then, there exists a nonnegative

adapted random process { g m } m = 0 , g 0 = 0 , such that sup P M E P g m < ,

f m 1 E P { f m | F m 1 } = E P { g m | F m 1 } , P M , m = 1 , ¯ . (62)

Let us put ξ m 0 = f m + g m f m 1 , m = 1 , ¯ . Then, ξ m 0 A 0 and from the equalities

(62) we obtain E P { ξ m 0 | F m 1 } = 1 , P M , m = 1 , ¯ . It is evident that the inequalities (61) are valid.

The sufficiency. Suppose that the conditions of Theorem 4 are valid. Then, f m f m 1 + f m 1 ( ξ m 0 1 ) . Introduce the denotation g m = f m + f m 1 ξ m 0 . Then,

g m 0 , sup P M E P g m sup P M E P f m + sup P M E P f m 1 < , m = 1 , ¯ . The last equality and

the inequalities give

f m = f 0 + i = 1 m f i 1 ( ξ i 0 1 ) i = 1 m g i , m = 1 , ¯ . (63)

Let us consider the random process { M m , F m } m = 0 , where

M m = f 0 + i = 1 m f i 1 ( ξ i 0 1 ) . Then, E P { M m | F m 1 } = M m 1 , P M , m = 1 , ¯ . Theorem 4 is proved. □

4. Construction of the Regular Set of Measures

In the next two Lemmas, we investigate the closure of a convex set of equivalent measures presented in Lemma 5 by the formula (42). First, we consider the countable case.

Suppose that Ω 1 contains the countable set of elementary events and let F 1 be a σ-algebra of all subsets of the set Ω 1 . Let P 1 be a measure on the σ-algebra F 1 . We assume that P 1 ( ω i ) = p i > 0 , i = 1 , ¯ . On the probability space { Ω 1 , F 1 , P 1 } , let us consider a nonnegative random value ξ 1 , satisfying the conditions

0 < P 1 ( { ω Ω 1 , η 1 ( ω ) < 0 } ) < 1 , 0 < P 1 ( { ω Ω 1 , η 1 ( ω ) > 0 } ) ,

E P 1 | η 1 ( ω ) | < , (64)

where we introduced the denotation η 1 ( ω ) = ξ 1 ( ω ) 1 . On the measurable space { Ω 1 , F 1 } , let us consider the set of measures M 1 , which are equivalent to the measure P 1 and are given by the formula

Q ( A ) = ω 1 Ω 1 ω 2 Ω 1 + χ A ( ω 1 ) α ( ω 1 , ω 2 ) η 1 + ( ω 2 ) η 1 ( ω 1 ) + η 1 + ( ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) + ω 1 Ω 1 ω 2 Ω 1 + χ A ( ω 2 ) α ( ω 1 , ω 2 ) η 1 ( ω 1 ) η 1 ( ω 1 ) + η 1 + ( ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) , A F 1 , (65)

where η 1 ( ω ) = η 1 + ( ω ) η 1 ( ω ) , Ω 1 + = { ω , η 1 ( ω ) > 0 } , Ω 1 = { ω , η 1 ( ω ) 0 } . Introduce the denotations F 1 + = Ω 1 + F 1 , F 1 = Ω 1 F 1 . Let P 1 be a contraction of the measure P 1 on the σ-algebra F 1 and let P 1 + be a contraction of the measure P 1 on the σ-algebra F 1 + . On the probability space { Ω 1 × Ω 1 + , F 1 × F 1 + , P 1 × P 1 + } , the set of random value α ( ω 1 , ω 2 ) satisfies the conditions

P 1 × P 1 + ( { ( ω 1 , ω 2 ) Ω 1 × Ω 1 + , α ( ω 1 , ω 2 ) > 0 } ) = P 1 ( Ω 1 + ) P 1 ( Ω 1 ) , (66)

ω 1 Ω ω 2 Ω + α ( ω 1 , ω 2 ) η 1 ( ω 1 ) η 1 + ( ω 2 ) η 1 ( ω 1 ) + η 1 + ( ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) < , (67)

ω 1 Ω 1 ω 2 Ω 1 + α ( ω 1 , ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) = 1. (68)

On the probability space { Ω 1 × Ω 1 + , F 1 × F 1 + , P 1 × P 1 + } , all the bounded strictly positive random values α ( ω 1 , ω 2 ) the above conditions satisfy. Introduce into the set of all measures on { Ω 1 , F 1 } the metrics

ρ ( Q 1 , Q 2 ) = i = 1 | Q 1 ( ω i ) Q 2 ( ω i ) | . (69)

Lemma 6. The closure of the set of measures M 1 in metrics (69) contains the set of measures

ν ω 1 , ω 2 ( A ) = χ A ( ω 1 ) η 1 + ( ω 2 ) η 1 ( ω 1 ) + η 1 + ( ω 2 ) + χ A ( ω 2 ) η 1 ( ω 1 ) η 1 ( ω 1 ) + η 1 + ( ω 2 ) (70)

for ω 1 Ω 1 , ω 2 Ω 1 + , A F 1 . For every bounded random value f ( ω ) , the closure of the set of points E Q f , Q M 1 , in metrics ρ ( x , y ) = | x y | , x , y R 1 , contains the points E ν ω 1 , ω 2 f , ( ω 1 , ω 2 ) Ω 1 × Ω 1 + .

Proof. Let us choose the set of equivalent measures Q ε defined by α ε ( ω 1 , ω 2 ) , 0 < ε < 1 , and given by the law:

α ε ( ω 1 0 , ω 2 0 ) = 1 ε P 1 ( ω 1 0 ) P 1 ( ω 2 0 ) , ω 1 0 Ω 1 , ω 2 0 Ω 1 + ,

α ε ( ω 1 , ω 2 ) = ε α 0 ε ( ω 1 , ω 2 ) , α 0 ε ( ω 1 , ω 2 ) = 1 ω 1 ω 1 0 ω ω 2 0 P ( ω 1 ) P ( ω 2 ) , ( ω 1 , ω 2 ) ( ω 1 0 , ω 2 0 ) ,

ω 1 Ω 1 , ω 2 Ω 1 + .

It is evident that α ε ( ω 1 , ω 2 ) > 0 , ( ω 1 , ω 2 ) Ω 1 × Ω 1 + , for every 1 > ε > 0 , and satisfy the equality

( ω 1 , ω 2 ) Ω 1 × Ω 1 + α ε ( ω 1 , ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) = 1. (71)

Then,

Q ε ( ω 1 0 ) = ω 2 Ω 1 + α ε ( ω 1 0 , ω 2 ) η 1 + ( ω 2 ) η 1 ( ω 1 0 ) + η 1 + ( ω 2 ) P 1 ( ω 1 0 ) P 1 ( ω 2 ) , (72)

Q ε ( ω 2 0 ) = ω 1 Ω 1 α ε ( ω 1 , ω 2 0 ) η 1 ( ω 1 ) η 1 ( ω 1 ) + η 1 + ( ω 2 0 ) P 1 ( ω 1 ) P 1 ( ω 2 0 ) . (73)

Q ε ( ω 1 0 ) = ( 1 ε ) η 1 + ( ω 2 0 ) η 1 ( ω 1 0 ) + η 1 + ( ω 2 0 ) + ε ω 2 Ω 1 + , ω 2 ω 2 0 α 0 ε ( ω 1 0 , ω 2 ) η 1 + ( ω 2 ) η 1 ( ω 1 0 ) + η 1 + ( ω 2 ) P 1 ( ω 1 0 ) P 1 ( ω 2 ) , (74)

Q ε ( ω 2 0 ) = ( 1 ε ) η 1 ( ω 1 0 ) η 1 ( ω 1 0 ) + η 1 + ( ω 2 0 ) + ε ω 1 Ω 1 , ω 1 ω 1 0 α 0 ε ( ω 1 , ω 2 0 ) η 1 ( ω 1 ) η 1 ( ω 1 ) + η 1 + ( ω 2 0 ) P 1 ( ω 1 ) P 1 ( ω 2 0 ) . (75)

If ω 1 ω 1 0 , ω 2 ω 2 0 , then

Q ε ( ω 1 ) = ε ω 2 Ω 1 + α 0 ε ( ω 1 , ω 2 ) η 1 + ( ω 2 ) η 1 ( ω 1 0 ) + η 1 + ( ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) , (76)

Q ε ( ω 2 ) = ε ω 1 Ω 1 α 0 ε ( ω 1 , ω 2 ) η 1 ( ω 1 ) η 1 ( ω 1 ) + η 1 + ( ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) . (77)

The distance between the measures Q ε and ν ω 1 0 , ω 2 0 is given by the formula

ρ ( Q ε , ν ω 1 0 , ω 2 0 ) = ε + ε ω 2 Ω 1 + , ω 2 ω 2 0 α 0 ε ( ω 1 0 , ω 2 ) η 1 + ( ω 2 ) η 1 ( ω 1 0 ) + η 1 + ( ω 2 ) P 1 ( ω 1 0 ) P 1 ( ω 2 ) + ε ω 1 Ω 1 , ω 1 ω 1 0 α 0 ε ( ω 1 , ω 2 0 ) η 1 ( ω 1 ) η 1 ( ω 1 ) + η 1 + ( ω 2 0 ) P 1 ( ω 1 ) P 1 ( ω 2 0 ) + ε ω 1 Ω 1 , ω 1 ω 1 0 ω 2 Ω 1 + α 0 ε ( ω 1 , ω 2 ) η 1 + ( ω 2 ) η 1 ( ω 1 0 ) + η 1 + ( ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) + ε ω 2 Ω 1 + , ω 2 ω 2 0 ω 1 Ω 1 α 0 ε ( ω 1 , ω 2 ) η 1 ( ω 1 ) η 1 ( ω 1 ) + η 1 + ( ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) . (78)

Since

ω 2 Ω 1 + , ω 2 ω 2 0 α 0 ε ( ω 1 0 , ω 2 ) η 1 + ( ω 2 ) η 1 ( ω 1 0 ) + η 1 + ( ω 2 ) P 1 ( ω 1 0 ) P 1 ( ω 2 ) + ω 1 Ω 1 , ω 1 ω 1 0 α 0 ε ( ω 1 , ω 2 0 ) η 1 ( ω 1 ) η 1 ( ω 1 ) + η 1 + ( ω 2 0 ) P 1 ( ω 1 ) P 1 ( ω 2 0 ) 1 ,

ω 1 Ω 1 , ω 1 ω 1 0 ω 2 Ω 1 + α 0 ε ( ω 1 , ω 2 ) η 1 + ( ω 2 ) η 1 ( ω 1 0 ) + η 1 + ( ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) 1 ,

ω 2 Ω 1 + , ω 2 ω 2 0 ω 1 Ω 1 α 0 ε ( ω 1 , ω 2 ) η 1 ( ω 1 ) η 1 ( ω 1 ) + η 1 + ( ω 2 ) P 1 ( ω 1 ) P 1 ( ω 2 ) 1 ,

we obtain

ρ ( Q ε , ν ω 1 0 , ω 2 0 ) 4 ε .

Let us prove the second part of Lemma 6. It is evident that the inequality

| E Q ε f E ν ω 1 0 , ω 2 0 f | 4 ε sup ω Ω 1 | f ( ω ) | (79)

is true. Due to arbitrariness of the small ε , Lemma 6 is proved. □

Definition 4. Let { Ω 1 , F 1 } be a measurable space. The decomposition A n , k , n , k = 1 , ¯ , of the space Ω 1 we call exhaustive one if the following conditions are valid:

1) A n , k F 1 , A n , k A n , s = , k s , k = 1 A n , k = Ω 1 , n = 1 , ¯ ;

2) the ( n + 1 ) -th decomposition is a sub-decomposition of the n-th one, that is, for every j, A n + 1, j A n , k for a certain k = k ( j ) ;

3) the minimal σ-algebra containing all A n , k , n , k = 1 , ¯ , coincides with F 1 .

The next Remark 1 is important for the construction of the filtration having the exhaustive decomposition.

Remark 1. Suppose that the measurable spaces { Ω 1 , F 1 } and { Ω 2 , F 2 } have the exhaustive decompositions A n , k 1 , n , k = 1 , ¯ and A m , s 2 , m , s = 1 , ¯ , respectively, then the measurable space { Ω 1 × Ω 2 , F 1 × F 2 } also have the exhaustive decomposition B n , k s , n = 1 , ¯ , k , s = 1 , ¯ , B n , k s = A n , k 1 × A n , s 2 , k , s = 1 , ¯ , n = 1 , ¯ . Really,

1) A n , k 1 × A n , s 2 F 1 × F 2 , A n , k 1 × A n , s 2 A n , t 1 × A n , r 2 = , ( k , s ) ( t , r ) ,

k , s = 1 B n , k s = Ω 1 × Ω 2 , n = 1 , ¯ ;

2) the ( n + 1 ) -th decomposition is a sub-decomposition of the n-th one, that is, for every k , s B n + 1, k s B n , i j for a certain i = i ( k ) , j = j ( s ) ;

3) the minimal σ-algebra containing all B n , k s , n , k , s = 1 , ¯ , coincides with F 1 × F 2 .

In the next Lemma we give the sufficient condition of the existence of exhaustive decomposition.

Lemma 7. Let { Ω 1 , F 1 } be a measurable space with a complete separable metric space Ω 1 and Borel σ-algebra F 1 on it. Then { Ω 1 , F 1 } has an exhaustive decomposition.

Proof. If { ω 1 , , ω n , } is a countable dense set in Ω 1 , then we denote

B ( ω n , ε m ) = { ω Ω 1 , ρ ( ω , ω n ) < ε m } , n , m = 1 , ¯ , (80)

the countable set of open balls as ε m runs all positive rational numbers, where ρ ( ω 1 , ω 2 ) , ω 1 , ω 2 Ω 1 is a metric in Ω 1 . Prove that F 1 = σ ( B ( ω n , ε m ) , n , m = 1 , ¯ ) , where σ ( B ( ω n , ε m ) , n , m = 1 , ¯ ) is a minimal σ-algebra generated by the sets (80). For this purpose let us prove that for every open set A Ω 1 the representation

A = n k N 1 , m s Q + 1 B ( ω n k , ε m s ) (81)

is true, where N 1 is a subset of positive integers, and Q + 1 is a subset of positive rational numbers. Let us denote { ω 1 A , , ω n A , } = A { ω 1 , , ω n , } . Suppose that ω 0 A , then d = inf ω A ¯ \ A ρ ( ω 0 , ω ) > 0 , where A ¯ is a closure of the set A.

Let the point ω k 0 A belong to the ball C ( ω 0 , d 8 ) = { ω Ω 1 , ρ ( ω 0 , ω ) < d 8 } and

let us consider the ball

C ( ω k 0 , d 8 + ρ ( ω 0 , ω k 0 A ) ) = { ω Ω 1 , ρ ( ω k 0 , ω ) < d 8 + ρ ( ω 0 , ω k 0 A ) } . The point ω 0 belongs to this ball and for every ω C ( ω k 0 , d 8 + ρ ( ω 0 , ω k 0 A ) ) the inequality

ρ ( ω 0 , ω ) ρ ( ω 0 , ω k 0 A ) + ρ ( ω k 0 A , ω ) < d 8 + 2 ρ ( ω 0 , ω k 0 A ) < 3 d 8 (82)

is true. Therefore C ( ω k 0 , d 8 + ρ ( ω 0 , ω k 0 A ) ) C ( ω 0 , 3 d 8 ) . Let the rational number ε k 0 satisfies the inequalities

d 8 + 2 ρ ( ω 0 , ω k 0 A ) < ε k 0 < 3 d 8 , (83)

then C ( ω k 0 A , ε k 0 ) C ( ω 0 , d 2 ) , since for every ω C ( ω k 0 A , ε k 0 ) , ρ ( ω 0 , ω ) ρ ( ω 0 , ω k 0 ) + ρ ( ω k 0 , ω ) < d 8 + ε k 0 < d 2 . So, for ω 0 A we found ω k 0 { ω 1 , , ω n , } and the rational number ε k 0 such that ω 0 C ( ω k 0 A , ε k 0 ) C ( ω 0 , d 2 ) A . The last prove the needed statement. To complete the proof of Lemma 7 let us construct the exhaustive decomposition. Let us renumber the sets B ( ω n , ε m ) putting by D 1 = B ( ω 1 , ε 1 ) , D 2 = B ( ω 1 , ε 2 ) , D 3 = B ( ω 2 , ε 1 ) , and so on. We put that { A 1 k } k = 1 consists of two sets D 1 and D ¯ 1 = Ω 1 \ D 1 . If the set { A n k } k = 1 is constructed, then the set { A n + 1 k } k = 1 we construct from the various set of the kind A n k D n + 1 , A n k D ¯ n + 1 . By construction the minimal σ-algebra σ { A n k , n , k = 1 , ¯ } = σ { B ( ω n , ε m ) , n , m = 1 , ¯ } . Taking into account the previous part of the proof we have σ { A n k , n , k = 1 , ¯ } = F 1 . Lemma 7 is proved. □

Lemma 8. Let a measurable space { Ω , F } have an exhaustive decomposition and let ξ be an integrable random value relative to the measure P, satisfying the conditions (36). Then, the closure of the set of measure Q, given by the formula (42), relative to the pointwise convergence of measures contains the set of measures

ν ω 1 , ω 2 ( A ) = χ A ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) + χ A ( ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) , A F , ( ω 1 , ω 2 ) Ω × Ω + , (84)

for those ( ω 1 , ω 2 ) Ω × Ω + which have the full measure μ = P × P + For every integrable finite valued random value f ( ω ) relative to all measures Q, the closure in metrics ρ ( x 1 , x 2 ) = | x 1 x 2 | , x 1 , x 2 R 1 , of the set of real numbers

E Q f = Ω Ω + f ( ω 1 ) α ( ω 1 , ω 2 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) + Ω Ω + f ( ω 2 ) α ( ω 1 , ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) , (85)

when α ( ω 1 , ω 2 ) runs over all random values satisfying the conditions (39), (41), contains the set of numbers

f ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) + f ( ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) , ( ω 1 , ω 2 ) Ω × Ω + . (86)

Proof. On a probability space { Ω , F , P } , let ξ be an integrable random value, satisfying the conditions (36). As before, let Ω + = { ω , ξ ( ω ) > 0 } , Ω = { ω , ξ ( ω ) 0 } and let F , F + be the restrictions of the σ-algebra F on the sets Ω and Ω + , correspondingly. Suppose that P and P + are the contractions of the measure P on the σ-algebras F , F + , correspondingly. Consider the probability space { Ω × Ω + , F × F + , P × P + } which is a direct product of the probability spaces { Ω , F , P } and { Ω + , F + , P + } . Due to Lemma 8 conditions and Remark 1, the measurable space { Ω × Ω + , F × F + } has the exhaustive decomposition B n , k s , k , s = 1 , ¯ , n = 1 , ¯ . Denote F n the minimal σ-algebra generated by decomposition B n , k s , k , s = 1 , ¯ . It is evident

that F n F n + 1 . Moreover, σ ( V n = 1 F n ) = F × F + . On the probability space

{ Ω × Ω + , F × F + , P × P + } , for every integrable finite valued random value f ( ω 1 , ω 2 ) the sequence E μ { f ( ω 1 , ω 2 ) | F n } converges to f ( ω 1 , ω 2 ) with probability one, as n , since it is a regular martingale. It is evident that for those B n , k s for which μ ( B n , k s ) 0

E μ { f ( ω 1 , ω 2 ) | F n } = B n , k s f ( ω 1 , ω 2 ) d μ μ ( B n , k s ) , ( ω 1 , ω 2 ) B n , k s . (87)

Denote D 0 = n , k , s , μ ( B n , k s ) = 0 B n , k s . It is evident that μ ( D 0 ) = 0 . For every

( ω 1 , ω 2 ) Ω × Ω + \ D 0 , the formula (87) is well defined and is finite. Let D 1 be the subset of the set Ω × Ω + \ D 0 , where the limit of the left hand side of the formula (87) does not exist. Then, μ ( D 1 ) = 0 . For every ( ω 1 , ω 2 ) Ω × Ω + \ ( D 0 D 1 ) , the right hand side of the formula (87) converges to f ( ω 1 , ω 2 ) . For ( ω 1 , ω 2 ) Ω × Ω + \ ( D 0 D 1 ) , denote A n = A n ( ω 1 , ω 2 ) those set B n , k s for which ( ω 1 , ω 2 ) B n , k s for a certain k , s . Then, for every integrable finite valued f ( ω 1 , ω 2 )

lim n A n f ( ω 1 , ω 2 ) d μ μ ( A n ) = f ( ω 1 , ω 2 ) . (88)

Let us consider the sequence

α n ε n ( ω 1 , ω 2 ) = ( 1 ε n ) χ A n ( ω 1 , ω 2 ) μ ( A n ) + ε n χ Ω × Ω + \ A n ( ω 1 , ω 2 ) μ ( Ω × Ω + \ A n ) , (89)

where 0 < ε n < 1 , lim n ε n = 0 . Such a sequence α n ε n ( ω 1 , ω 2 ) satisfies the conditions (39)-(41) and

Q n ε n ( A ) = Ω Ω + χ A ( ω 1 ) α n ε n ( ω 1 , ω 2 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) + Ω Ω + χ A ( ω 2 ) α n ε n ( ω 1 , ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) = ( 1 ε n ) A n χ A ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) μ ( A n ) + ( 1 ε n ) A n χ A ( ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) μ (An)

+ ε n Ω × Ω + \ A n χ A ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) μ ( Ω × Ω + \ A n ) + ε n Ω × Ω + \ A n χ A ( ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) μ ( Ω × Ω + \ A n ) . (90)

From the formula (90), we obtain

lim n Q n ε n ( A ) = χ A ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) + χ A ( ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) , A F , ( ω 1 , ω 2 ) Ω × Ω + \ ( D 0 D 1 ) . (91)

Further,

E Q n ε n f ( ω ) = Ω Ω + f ( ω 1 ) α n ε n ( ω 1 , ω 2 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) + Ω Ω + f ( ω 2 ) α n ε n ( ω 1 , ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) = ( 1 ε n ) A n f ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) μ ( A n ) + ( 1 ε n ) A n f ( ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) μ (An)

+ ε n Ω × Ω + \ A n f ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) μ ( Ω × Ω + \ A n ) + ε n Ω × Ω + \ A n f ( ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) d μ ( ω 1 , ω 2 ) μ ( Ω × Ω + \ A n ) . (92)

From the formula (92), we obtain

lim n E Q n ε n f ( ω ) = f ( ω 1 ) ξ + ( ω 2 ) ξ ( ω 1 ) + ξ + ( ω 2 ) + f ( ω 2 ) ξ ( ω 1 ) ξ ( ω 1 ) + ξ + ( ω 2 ) , ( ω 1 , ω 2 ) Ω × Ω + \ ( D 0 D 1 ) . (93)

Lemma 8 is proved. □

The next Theorem 5 is a consequence of Lemma 5.

Theorem 5. On the probability space { Ω , F , P } , for the nonnegative random value ξ 1 the set of measures M 0 on the measurable space { Ω , F } , being equivalent to the measure P, satisfies the condition

E Q ξ = 1 , Q M 0 , (94)

if and only if as for Q M 0 the representation

Q ( A ) = Ω Ω + χ A ( ω 1 ) α ( ω 1 , ω 2 ) ( ξ 1 ) + ( ω 2 ) ( ξ 1 ) ( ω 1 ) + ( ξ 1 ) + ( ω 2 ) d μ ( ω 1 , ω 2 ) + Ω Ω + χ A ( ω 2 ) α ( ω 1 , ω 2 ) ( ξ 1 ) ( ω 1 ) ( ξ 1 ) ( ω 1 ) + ( ξ 1 ) + ( ω 2 ) d μ ( ω 1 , ω 2 ) , A F , (95)

is true, where on the measurable space { Ω × Ω + , F × F + , P × P + } , the random value α ( ω 1 , ω 2 ) satisfies the conditions

μ ( { ( ω 1 , ω 2 ) Ω × Ω + , α ( ω 1 , ω 2 ) > 0 } ) = P ( Ω + ) P ( Ω ) , (96)

Ω Ω + α ( ω 1 , ω 2 ) ( ξ 1 ) ( ω 1 ) ( ξ 1 ) + ( ω 2 ) ( ξ 1 ) ( ω 1 ) + ( ξ 1 ) + ( ω 2 ) d μ ( ω 1 , ω 2 ) < , (97)

Ω Ω + α ( ω 1 , ω 2 ) d μ ( ω 1 , ω 2 ) = 1. (98)

We introduced above the following denotations: μ = P × P + , P is a contraction of the measure P on the set Ω = { ω Ω , ξ 1 0 } , P + is a contraction of the measure P on the set Ω + = { ω Ω , ξ 1 > 0 } , F = Ω F , F + = Ω + F .

It is evident that the set of measure M 0 is a nonempty one, since it contains those measures Q, for which the random value α ( ω 1 , ω 2 ) is bounded, since E Q | ξ 1 | < . The set of measure M 0 is consistent with the filtration { F 0 , F } on the measurable space { Ω , F } , where F 0 = { , Ω } .

Theorem 6. On the probability space { Ω , F , P } with the filtration F n on it, the set of measures M 0 , given by the formula (95), is consistent with the filtration F n , if and only if, as E Q { ξ | F n } , Q M 0 , is a local regular martingale.

Proof. The necessity. Let the set of measures M 0 be consistent with the filtration. Then, due to Theorem 3, E Q { ξ | F n } , Q M 0 , is a local regular martingale.

The sufficiency. Suppose that E Q { ξ | F n } , Q M 0 , is a local regular martingale. Let us prove that, if Q 1 , Q 2 M 0 , then the set of measures

R s k ( A ) = A E Q 2 { d Q 1 d Q 2 | F k } E Q 2 { d Q 1 d Q 2 | F s } d Q 2 , A F , k s n , n = 0 , ¯ , (99)

belongs to the set M 0 . For this, it is to prove that E R s k ( ξ 1 ) = 0 , or E R s k ξ = 1 . Really, if E Q 1 ξ = 1 , E Q 2 ξ = 1 , then

E R s k ξ = E Q 2 ξ E Q 2 { d Q 1 d Q 2 | F k } E Q 2 { d Q 1 d Q 2 | F s } = E Q 2 E Q 2 { ξ | F k } d Q 1 d Q 2 E Q 2 { d Q 1 d Q 2 | F s } = E Q 2 E Q 2 { E Q 2 { ξ | F k } d Q 1 d Q 2 E Q 2 { d Q 1 d Q 2 | F s } | F s } = E Q 2 E Q 1 { E Q 2 { ξ | F k } | F s } = E Q 2 E Q 1 { E Q 1 { ξ | F k } | F s } = E Q 2 E Q 1 { ξ | F s } = E Q 2 E Q 2 { ξ | F s } = E Q 2 ξ = 1. (100)

Theorem 6 is proved. □

Theorem 7. On the probability space { Ω , F , P } with the filtration F n on it, the set of measures M 0 , given by the formula (95), is consistent with the filtration F n , if and only if there exists not depending on ( ω 1 , ω 2 ) Ω × Ω + the random process { m n , F n } n = 0 such that

E ν ω 1 , ω 2 { ξ | F n } = m n , n = 1 , ¯ , (101)

for those ( ω 1 , ω 2 ) Ω × Ω + that have the full measure μ = P × P + , where

ν ω 1 , ω 2 ( A ) = χ A ( ω 1 ) ( ξ 1 ) + ( ω 2 ) ( ξ 1 ) ( ω 1 ) + ( ξ 1 ) + ( ω 2 ) + χ A ( ω 2 ) ( ξ 1 ) ( ω 1 ) ( ξ 1 ) ( ω 1 ) + ( ξ 1 ) + ( ω 2 ) , A F , ( ω 1 , ω 2 ) Ω × Ω + . (102)

Proof. The necessity. Suppose that the set of measures M 0 , given by the formula (95), is consistent with the filtration F n . Due to Theorem 6, E Q { ξ | F n } , Q M 0 , is a local regular martingale. Then, E Q { ξ | F n } = m n . Using Lemma 8, we obtain E ν ω 1 , ω 2 { ξ | F n } = m n for those ( ω 1 , ω 2 ) Ω × Ω + that have the full measure μ .

The sufficiency. If the formula (101) is true, then E Q { ξ | F n } = m n , Q M 0 . From this, it follows that E Q { ξ | F n } , Q M 0 , is a local regular martingale. Theorem 7 is proved. □

Definition 5. On the probability space { Ω , F , P } with the filtration F n on it, the consistent with the filtration F n subset of the measures M of the set of the measures M 0 generating by the nonnegative random value ξ 1 , E Q ξ = 1 , Q M 0 , we call the regular set of measures.

Let { Ω , F , P } be a probability space. On the measurable space { Ω , F } with the filtration F n on it, let M M 0 be a set of regular measures, where the set M 0 is generated by the nonnegative random value ξ 1 . Denote by { m n , F n } n = 0 the regular martingale, where m n = E Q { ξ | F n } , Q M , n = 1 , ¯ . Assume that M n is a contraction of the set of regular measures M onto the σ-algebra F n . Every Q n M n is equivalent to P n , where P n is a contraction of the measure P on the σ-algebra F n . For every Q n M n , we have E Q n [ m n m n 1 ] = 0 . Therefore, for the measure Q n M n the representation

Q n ( A ) = Ω n × Ω n + χ A ( ω 1 ) α n ( ω 1 , ω 2 ) [ m n m n 1 ] + ( ω 2 ) [ m n m n 1 ] ( ω 1 ) + [ m n m n 1 ] + ( ω 2 ) d μ n ( ω 1 , ω 2 ) + Ω n × Ω n + χ A ( ω 2 ) α n ( ω 1 , ω 2 ) [ m n m n 1 ] ( ω 1 ) [ m n m n 1 ] ( ω 1 ) + [ m n m n 1 ] + ( ω 2 ) d μ n ( ω 1 , ω 2 ) , A F n , (103)

Ω n = { ω 1 Ω , [ m n m n 1 ] ( ω 1 ) 0 } ,

Ω n + = { ω 2 Ω , [ m n m n 1 ] ( ω 2 ) > 0 } ,

is true, where, on the measurable space { Ω n × Ω n + , F n × F n + } , the random value α n ( ω 1 , ω 2 ) satisfies the conditions

μ n ( { ( ω 1 , ω 2 ) Ω n × Ω n + , α n ( ω 1 , ω 2 ) > 0 } ) = P n ( Ω + ) P n ( Ω ) , (104)

Ω n Ω n + α n ( ω 1 , ω 2 ) [ m n m n 1 ] ( ω 1 ) [ m n m n 1 ] + ( ω 2 ) [ m n m n 1 ] ( ω 1 ) + [ m n m n 1 ] + ( ω 2 ) d μ n ( ω 1 , ω 2 ) < , (105)

Ω n Ω n + α n ( ω 1 , ω 2 ) d μ n ( ω 1 , ω 2 ) = 1. (106)

Here, the measure μ n = P n × P n + is given on the measurable space { Ω n × Ω n + , F n × F n + } and it is a direct product of the measures P n and P n + , where the measure P n + is a contraction of the measure P n on the σ-algebra F n + = Ω n + F n and P n is a contraction of the measure P n on the σ-algebra F n = Ω n F n . It is evident that the regular set of measures M is a convex set of measure.

Definition 6. On the probability space { Ω , F , P } with the filtration F n on it, denote by A 0 the set of all nonnegative integrable random values ζ relative to the set of regular measures M, satisfying the conditions:

E P ζ = 1 , P M . (107)

Due to Theorem 3, { E P { ζ | F n } , F n } n = 0 is a regular martingale relative to the set of measures M.

It is evident that the set A 0 is a nonempty one, since it contains the random value ζ = 1 . The more interesting case is as A 0 contains more than one element. So, further we consider the regular set of measure M with the set A 0 , containing more than one element.

In the next Lemma 9, using Lemma 5, we construct a set of measures consistent with the filtration. On the probability space { Ω 1 0 , F 1 0 , P 1 0 } , let us consider a nonnegative random value ξ 1 , satisfying the conditions

0 < P 1 0 ( { ω 1 Ω 1 0 , η 1 ( ω 1 ) < 0 } ) < 1 ,

0 < P 1 0 ( { ω 1 Ω 1 0 , η 1 ( ω 1 ) > 0 } ) , (108)

where we introduced the denotation η 1 ( ω 1 ) = ξ 1 ( ω 1 ) 1 . Described in Lemma 5 the set of equivalent measures to the measure P 1 0 and such that E Q η 1 ( ω 1 ) = 0 , we denote by M 1 . Let us construct the infinite direct product of the measurable

spaces { Ω i 0 , F i 0 } , i = 1 , ¯ , where Ω i 0 = Ω 1 0 , F i 0 = F 1 0 . Denote Ω = i = 1 Ω i 0 . On

the space Ω , under the σ-algebra F we understand the minimal σ-algebra,

generated by the sets i = 1 G i , G i F i 0 , where in the last product only the finite set

of G i do not equal Ω i 0 . On the measurable space { Ω , F } , under the filtration

F n we understand the minimal σ-algebra generated by the sets i = 1 G i , G i F i 0 ,

where G i = Ω i 0 for i > n . We consider the probability space { Ω , F , P } , where

P = i = 1 P i 0 , P i 0 = P 1 0 , i = 1 , ¯ .

On the measurable space { Ω , F } , we introduce into consideration the set of

measures M, where Q belongs to M, if Q = i = 1 Q i , Q i M 1 . We denote by M Q 0 a subset of the set M of those measures Q = i = 1 Q i , Q i M 1 , for which only the

finite set of Q i does not coincide with the measure Q 0 M 1 .

Lemma 9. On the measurable space { Ω , F } with the filtration F n on it, there exists consistent with the filtration F n the set of measures M 0 and the nonnegative random variable ξ 0 such that E Q ξ 0 = 1 , Q M 0 , if the random value ξ 1 , satisfying the conditions (108), is bounded.

Proof. To prove Lemma 9, we need to construct a nonnegative bounded random value ξ 0 on the measurable space { Ω , F } and a set of equivalent measures M 0 on it, such that E Q ξ 0 = 1 , Q M 0 , and to prove that the set of measures M 0 is consistent with the filtration F n . From the Lemma 9 conditions, the random value η 1 ( ω 1 ) = ξ 1 ( ω 1 ) 1 is also bounded. Let us put

ξ 0 = i = 1 [ 1 + a i ( ω 1 , , ω i 1 ) η i ( ω i ) ] , (109)

where the random values a i ( ω 1 , , ω i 1 ) are F i 1 -measurable, i = 1 , ¯ , they satisfy the conditions 0 < a i ( ω 1 , , ω i 1 ) b i < 1 . The constants b i are such

that i = 1 b i < , the random value η i ( ω i ) is given on { Ω i 0 , F i 0 , P i 0 } and is

distributed as η 1 ( ω 1 ) on { Ω 1 0 , F 1 0 , P 1 0 } . From this, it follows that the random

value ξ 0 is bounded by the constant i = 1 [ 1 + C b i ] , where C > 0 and it is such

that | η i ( ω i ) | < C , i = 1 , ¯ . It is evident that E Q ξ 0 = 1 , Q M Q 0 . Really,

E Q i = 1 n [ 1 + a i ( ω 1 , , ω i 1 ) η i ( ω i ) ] = E Q 0 n 1 i = 1 n 1 [ 1 + a i ( ω 1 , , ω i 1 ) η i ( ω i ) ] × E Q n [ 1 + a n 1 ( ω 1 , , ω i 1 ) η n ( ω n ) ] , (110)

where Q = i = 1 Q i , Q 0 n 1 = i = 1 n 1 Q i ,

E Q n [ 1 + a n 1 ( ω 1 , , ω n 1 ) η i ( ω n ) ] = [ 1 + a n 1 ( ω 1 , , ω n 1 ) E Q n η n ( ω n ) ] = 1. (111)

From the last equality, we have

E Q i = 1 n [ 1 + a i ( ω 1 , , ω i 1 ) η i ( ω i ) ] = 1. (112)

Since ξ 0 = lim n i = 1 n [ 1 + a i ( ω 1 , , ω i 1 ) η i ( ω i ) ] , from the equality (112) and the

possibility to go to the limit under the mathematical expectation, we prove the needed statement. Let us prove the existence of the set of measures M 0 consistent with the filtration F n . If Q M Q 0 , then

E Q { ξ 0 | F n } = i = 1 n [ 1 + a i ( ω 1 , , ω i 1 ) η i ( ω i ) ] , Q M Q 0 . (113)

Due to Lemma 4, there exists a set of measures M 0 such that it is consistent with the filtration and M 0 M Q 0 , E Q ξ 0 = 1 , Q M 0 . The set M 0 is a linear convex span of the set M Q 0 . It means that the set of measures M 0 is consistent with the filtration. Lemma 9 is proved. □

Remark 2 The boundedness of the random value ξ 1 is not essential. For the applications, the case, as a i ( ω 1 , , ω i 1 ) = 0 , i n + 1 , is very important (see Section 8). In this case, Lemma 9 is true as the random value η 1 is an integrable one. The random value ξ 0 is also integrable one relative to every measures from the set M 0 and it is F n -measurable one.

Below, we describe one class of evolutions of risk assets satisfying no arbitrage condition [15] - [20] and give the complete description of the set of equivalent martingale measures.

On the introduced measurable space { Ω , F , P } we consider the evolution of the risk asset given by the law

S n = S n 1 ( 1 + a n ( ω 1 , , ω n 1 ) ) η n ( ω n ) , n = 1 , N ¯ , (114)

where the random values a i ( ω 1 , , ω i 1 ) are F i 1 -measurable, i = 1 , N ¯ , they satisfy the conditions 0 < a i ( ω 1 , , ω i 1 ) 1 , η 1 ( ω 1 ) = ξ 1 ( ω 1 ) 1 , the random value η i ( ω i ) is given on { Ω i 0 , F i 0 , P i 0 } and is distributed as η 1 ( ω 1 ) on { Ω 1 0 , F 1 0 , P 1 0 } . The main aim is to describe the set of martingale measures for the evolution of risk asset given by the formula (114). This problem we solve in Theorem 8.

Below, we describe completely the regular set of measures in the case as

ξ 0 = i = 1 N [ 1 + a i ( ω 1 , , ω i 1 ) η i ( ω i ) ] , N < , 0 < a i ( ω 1 , , ω i 1 ) 1 , i = 1 , N ¯ , and

the random value ξ 1 is an integrable one relative to the measure P 1 0 . For this purpose, we introduce the denotations: Ω 1 = { ω 1 Ω 1 0 , η 1 ( ω 1 ) 0 } , Ω 1 + = { ω 1 Ω 1 0 , η 1 ( ω 1 ) > 0 } , P 1 is a contraction of the measure P 1 0 on the σ-algebra F 1 , P 1 + is a contraction of the measure P 1 0 on the σ-algebra F 1 + , F 1 = Ω 1 F 1 0 , F 1 + = Ω 1 + F 1 0 .

Denote U 1 = Ω 1 × Ω 1 + and introduce the measure μ 1 = P 1 × P 1 + on the σ-algebra G 1 = F 1 × F 1 + . Let us introduce the measurable space { V , L , μ } ,

where V = i = 1 N U i , U i = U 1 , i = 1 , N ¯ , is a direct product of the spaces U i = Ω i × Ω i + , Ω i = Ω 1 , Ω i + = Ω 1 + , L = i = 1 N G i is a direct product of the σ-algebras G i = G 1 , i = 1 , N ¯ . At last, let μ = i = 1 N μ i be a direct product of the measures μ i = μ 1 , i = 1 , N ¯ , and let ν v = i = 1 N ν ω i 1 , ω i 2 , v = { ( ω 1 1 , ω 1 2 ) , , ( ω N 1 , ω N 2 ) } ,

be a direct product of the measures ν ω i 1 , ω i 2 , i = 1 , N ¯ , which is a countable additive function on the σ-algebra F N for every v V , where

ν ω i 1 , ω i 2 ( A i ) = χ A i ( ω i 1 ) η i + ( ω i 2 ) η i ( ω i 1 ) + η i + ( ω i 2 ) + χ A i ( ω i 2 ) η i ( ω i 1 ) η i ( ω i 1 ) + η i + ( ω i 2 ) (115)

for ω i 1 Ω i , ω i 2 Ω i + , A i F i 0 .

In the next Theorem 8, we assume that the random value η 1 ( ω 1 ) is an integrable one.

Theorem 8. On the measurable space { Ω , F } with the filtration F n on it, every measure Q of the regular set of measures M for the random value

ξ 0 = i = 1 N [ 1 + a i ( ω 1 , , ω i 1 ) η i ( ω i ) ] , N < , 0 < a i ( ω 1 , , ω i 1 ) 1 , i = 1 , N ¯ , has

the representation

Q ( A ) = V α ( v ) ν v ( A ) d μ ( v ) , (116)

where the random value α ( v ) satisfies the conditions

μ ( { v V , α ( v ) > 0 } ) = [ P 1 0 ( Ω 1 ) P 1 0 ( Ω 1 + ) ] N , (117)

V α ( v ) i = 1 N η i ( ω i 1 ) η i + ( ω i 2 ) η i ( ω i 1 ) + η i + ( ω i 2 ) d μ ( v ) < , (118)

V α ( v ) d μ ( v ) = 1. (119)

Proof. To prove Theorem, it needs to prove that the countable additive measure ν v ( A ) at every fixed v V is a measurable map from the measurable space { V , L } into the measurable space { [ 0,1 ] , B ( [ 0,1 ] ) } for every fixed

A F N . For A = i = 1 N A i , A i F i 0 , ν v ( A ) is a measurable map from the measurable space { V , L } into the measurable space { [ 0,1 ] , B ( [ 0,1 ] ) } . The family of sets of the kind i I E i , E i = s = 1 N A s i , A s i F s 0 , where E i E j = , the set I is an

arbitrary finite set, forms the algebra of the sets that we denote by U 0 . From the

countable additivity of ν v ( A ) , ν v ( i I E i ) = i I ν v ( E i ) is a measurable map

from the measurable space { V , L } into the measurable space { [ 0,1 ] , B ( [ 0,1 ] ) } . Let T be a class of the sets from the minimal σ-algebra Σ generated by U 0 for every subset E of that ν v ( E ) is a measurable map from the measurable space

{ V , L } into the measurable space { [ 0,1 ] , B ( [ 0,1 ] ) } . Let us prove that T is a monotonic class. Suppose that E i E i + 1 , i = 1 , ¯ , E i T . Then,

ν v ( E i ) ν v ( E i + 1 ) . From this, it follows that lim i ν v ( E i ) is a measurable map from the measurable space { V , L } into the measurable space { [ 0,1 ] , B ( [ 0,1 ] ) } . But, ν v ( E i + 1 \ E i ) = ν v ( E i + 1 ) ν v ( E i ) is a measurable map from { V , L } into { [ 0,1 ] , B ( [ 0,1 ] ) } . From this equality, it follows that the set E i + 1 \ E i belongs to

the class T. Since i = 1 E i = E 1 i = 1 [ E i + 1 \ E i ] , we have

lim n ν v ( E n ) = ν v ( E 1 ) + lim n i = 1 n ν v ( E i + 1 \ E i ) = ν v ( E 1 ) + i = 1 ν v ( E i + 1 \ E i ) = ν v ( E 1 i = 1 [ E i + 1 \ E i ] ) = ν v ( i = 1 E i ) . (120)

The equalities (120) mean that i = 1 E i belongs to T, since ν v ( i = 1 E i ) is a

measurable map of { V , L } into { [ 0,1 ] , B ( [ 0,1 ] ) } . Suppose that

E i E i + 1 , E i T , i = 1 , ¯ . Then, this case is reduced to the previous one by the

note that the sequence E ¯ i = i = 1 N Ω i 0 \ E i , i = 1 , ¯ is monotonically increasing. From this, it follows that E ¯ = i = 1 E ¯ i T . Therefore, i = 1 E i = i = 1 N Ω i 0 \ i = 1 E ¯ i T .

Thus, T is a monotone class. But, U 0 T . Hence, T contains the minimal monotone class generated by the algebra U 0 , that is, m ( U 0 ) = Σ , therefore, Σ T . Thus, ν v ( E ) is a measurable map of { V , L } into { [ 0,1 ] , B ( [ 0,1 ] ) } for A Σ . The fact that the random value α ( v ) satisfies the conditions (117)-(119) means that Q, given by the formula (116), is a countable additive function of sets and E Q ξ 0 < . Moreover, E Q ξ 0 = 1 . It is evident that

E Q { ξ 0 | F n } = i = 1 n [ 1 + a i ( ω 1 , , ω i 1 ) η i ( ω i ) ] , Q M . Due to Lemma 4, this proves that the set M is a regular set of measure. Theorem 8 is proved. □

Remark 3. The representation (116) for the regular set of measures M means that M is a convex set of equivalent measures. Since the random value α ( v ) runs all bounded random values, satisfying the conditions (117 - 119), it is easy to show that the set of measures ν v ( A ) , v V , A F N , is the set of extreme points for the set M.

Let us introduce the denotations

m n ( { ω } ) = i = 1 n [ 1 + a i ( ω 1 , , ω i 1 ) η i ( ω i ) ] , 1 n N < , (121)

0 < a i ( ω 1 , , ω i 1 ) 1 , i = 1 , N ¯ .

{ ω } = { ω 1 , , ω N } , { ω } i = { ω 1 i , , ω N i } , i = 1 , 2 , Ω n = i = 1 n Ω i 0 , n = 1 , N ¯ ,

{ ω } n 1 = { ω 1 1 , , ω n 1 } , { ω } n i = { ω 1 i , , ω n i } , i = 1 , 2 ,

Ω N n = i = n + 1 N Ω i 0 , n = 1 , N 1 ¯ ,

Ω ¯ n = { { ω } n Ω n , [ m n m n 1 ] ( { ω } n ) 0 } = Ω n 1 × { ω n Ω n 0 , η n ( ω n ) 0 } ,

Ω ¯ n + = { { ω } n Ω n , [ m n m n 1 ] ( { ω } n ) > 0 } = Ω n 1 × { ω n Ω n 0 , η n ( ω n ) > 0 } ,

Ω n = { { ω } Ω N , [ m n m n 1 ] ( { ω } ) 0 } = Ω n 1 × { ω n Ω n 0 , η n ( ω n ) 0 } × Ω N n ,

Ω n + = { { ω } Ω N , [ m n m n 1 ] ( { ω } ) > 0 } = Ω n 1 × { ω n Ω n 0 , η n ( ω n ) > 0 } × Ω N n . (122)

Note that the σ-algebra F n is generated by sets of the kind G = i = 1 N G i , where G i F i 0 , i = 1 , n ¯ , G i = Ω i 0 , i = n + 1 , N ¯ . Denote P n = i = 1 n P i 0 the contraction of the measure P N = i = 1 N P i 0 onto the σ-algebra F n . Further we use the denotations

P n and P n + which are the contractions the measure P n onto the σ-algebras F n Ω n and F n Ω n + , correspondingly. If the measure Q belongs to the set of martingale measures (116), then E Q { m n | F n 1 } = m n 1 , or E Q [ m n m n 1 ] = 0 . From this, for the measure Q the representation

Q ( A 1 ) = Ω n × Ω n + χ A 1 ( { ω } 1 ) α ( { ω } 1 ; { ω } 2 ) [ m n m n 1 ] + ( { ω } 2 ) [ m n m n 1 ] ( { ω } 1 ) + [ m n m n 1 ] + ( { ω } 2 ) d [ P N × P N + ] + Ω n × Ω n + χ A 1 ( { ω } 2 ) α ( { ω } 1 ; { ω } 2 ) [ m n m n 1 ] ( { ω } 1 ) [ m n m n 1 ] ( { ω } 1 ) + [ m n m n 1 ] + ( { ω } 2 ) d [ P N × P N + ] , A 1 F n , (123)

is true if the random value α ( { ω } 1 ; { ω } 2 ) > 0 satisfies the condition

Ω n × Ω n + α ( { ω } 1 ; { ω } 2 ) d [ P N × P N + ] = 1. (124)

Since for the set A 1 the representation A 1 = A × i = n + 1 N Ω i 0 , is true, where A F ¯ n = i = 1 n F i 0 , then for the contraction Q n of the measure Q onto the σ-algebra F ¯ n the representation

Q n ( A ) = Ω ¯ n × Ω ¯ n + χ A ( { ω } n 1 ) α n 1 ( { ω } n 1 ; { ω } n 2 ) [ m n m n 1 ] + ( { ω } n 2 ) [ m n m n 1 ] ( { ω } n 1 ) + [ m n m n 1 ] + ( { ω } n 2 ) d [ P ¯ n × P ¯ n + ] + Ω ¯ n × Ω ¯ n + χ A ( { ω } n 2 ) α n 1 ( { ω } n 1 ; { ω } n 2 ) [ m n m n 1 ] ( { ω } n 1 ) [ m n m n 1 ] ( { ω } n 1 ) + [ m n m n 1 ] + ( { ω } n 2 ) d [ P ¯ n × P ¯ n + ] , A F ¯ n , (125)

is true, where we introduced the denotations P ¯ n and P ¯ n + which are the contractions of the measure P n onto the σ-algebras F ¯ n Ω ¯ n and F ¯ n Ω ¯ n + , correspondingly,

α n 1 ( { ω } n 1 ; { ω } n 2 ) = Ω N n × Ω N n α ( { ω } 1 ; { ω } 2 ) d [ P N n × P N n ] ,

P N n = i = n + 1 N P i 0 , Ω ¯ n × Ω ¯ n + α n 1 ( { ω } n 1 ; { ω } n 2 ) d [ P ¯ n × P ¯ n + ] = 1. (126)

In the set Ω ¯ n × Ω ¯ n + let us introduce the transformation

T n ( { ω } n 1 ; { ω } n 2 ) = ( T n 1 ( { ω } n 1 ) ; T n 2 ( { ω } n 2 ) ) ,

T n 1 ( { ω } n 1 ) = { { ω } n 1 2 , ω n 1 } , T n 2 ( { ω } n 2 ) = { { ω } n 1 1 , ω n 2 } , n = 1 , N ¯ . (127)

By the definition we put that for n = 1 the transformation T 1 is identical one. Introduce the denotations

ν n 1 ( { ω } n 1 ; { ω } n 2 ) = [ m n m n 1 ] + ( { ω } n 2 ) φ n ( { ω } n 1 ; { ω } n 2 ) , (128)

ν n 2 ( { ω } n 1 ; { ω } n 2 ) = [ m n m n 1 ] ( { ω } n 1 ) φ n ( { ω } n 1 ; { ω } n 2 ) . (129)

φ n 1 ( { ω } n 1 ; { ω } n 2 ) = [ m n m n 1 ] ( { ω } n 1 ) + [ m n m n 1 ] + ( { ω } n 2 ) , (130)

φ n ( { ω } n 1 ; { ω } n 2 ) = φ n 1 ( { ω } n 1 ; { ω } n 2 ) + φ n 1 ( T n ( { ω } n 1 ; { ω } n 2 ) ) . (131)

Theorem 9. Let Ω 1 0 be a complete separable metric space and F 1 0 be a Borel σ-algebra on it. If the condition

Ω n f ( { ω } n ) d P n < , (132)

is true for F n -measurable nonnegative random value f ( { ω } n ) , then the closure of the set of points E Q n f ( { ω } n ) , Q n M n , in metrics ρ ( x , y ) = | x y | on the real line contains the set of points

f ( { ω } n 1 ) ν n 1 ( { ω } n 1 ; { ω } n 2 ) + f ( { ω } n 2 ) ν n 2 ( { ω } n 1 ; { ω } n 2 ) + f ( T n 1 ( { ω } n 1 ) ) ν n 1 ( T n ( { ω } n 1 ; { ω } n 2 ) ) + f ( T n 2 ( { ω } n 2 ) ) ν n 2 ( T n ( { ω } n 1 ; { ω } n 2 ) ) , n = 1 , N ¯ . (133)

Proof. Let us find the conditions for the measurable functions α n 1 ( { ω } n 1 ; { ω } n 2 ) under which E Q n { m n | F ¯ n 1 } = m n 1 . Introduce the denotation

α n 0 ( { ω } n 1 ; { ω } n 2 ) = α n 1 ( { ω } n 1 ; { ω } n 2 ) [ m n m n 1 ] ( { ω } n 1 ) + [ m n m n 1 ] + ( { ω } n 2 ) . (134)

Let the set B belongs to F ¯ n 1 , then

E Q n χ B ( { ω } n 1 ) [ m n m n 1 ] ( { ω } n ) = Ω ¯ n × Ω ¯ n + χ B ( { ω } n 1 1 ) α n 0 ( { ω } n 1 ; { ω } n 2 ) [ m n m n 1 ] ( { ω } n 1 ) × [ m n m n 1 ] + ( { ω } n 2 ) d [ P ¯ n × P ¯ n + ] + Ω ¯ n × Ω ¯ n + χ B ( { ω } n 1 2 ) α n 0 ( { ω } n 1 ; { ω } n 2 ) [ m n m n 1 ] ( { ω } n 2 ) × [ m n m n 1 ] ( { ω } n 1 ) d [ P ¯ n × P ¯ n + ] . (135)

If to take into account the relations

[ m n m n 1 ] ( { ω } n ) = m n 1 ( { ω } n 1 ) a n ( { ω } n 1 ) η n ( ω n ) ,

[ m n m n 1 ] + ( { ω } n ) = m n 1 ( { ω } n 1 ) a n ( { ω } n 1 ) η n + ( ω n ) ,

[ m n m n 1 ] ( { ω } n ) = m n 1 ( { ω } n 1 ) a n ( { ω } n 1 ) η n ( ω n ) , (136)

and introduce the denotations

θ 1 ( { ω } n 1 ; { ω } n 2 ) = m n 1 ( { ω } n 1 1 ) a n ( { ω } n 1 1 ) × m n 1 ( { ω } n 1 2 ) a n ( { ω } n 1 2 ) α n 0 ( { ω } n 1 ; { ω } n 2 ) , (137)

θ 2 ( { ω } n 1 ; { ω } n 2 ) = m n 1 ( { ω } n 1 1 ) a n ( { ω } n 1 1 ) × m n 1 ( { ω } n 1 2 ) a n ( { ω } n 1 2 ) α n 0 ( { ω } n 1 ; { ω } n 2 ) , (138)

we obtain

E Q n χ B ( { ω } n 1 ) [ m n m n 1 ] ( { ω } n ) = Ω ¯ n × Ω ¯ n + χ B ( { ω } n 1 1 ) θ 1 ( { ω } n 1 ; { ω } n 2 ) η n ( ω n 1 ) η n + ( ω n 2 ) d [ P ¯ n × P ¯ n + ] + Ω ¯ n × Ω ¯ n + χ B ( { ω } n 1 2 ) θ 2 ( { ω } n 1 ; { ω } n 2 ) η n ( ω n 1 ) η n + ( ω n 2 ) d [ P ¯ n × P ¯ n + ]

= { η n ( ω n 1 ) 0 } × { η n ( ω n 2 ) > 0 } d [ P 1 ( ω n 1 ) × P 1 ( ω n 2 ) ] η n ( ω n 1 ) η n + ( ω n 2 ) × Ω n 1 × Ω n 1 χ B ( { ω } n 1 1 ) [ θ 1 ( { ω } n 1 ; { ω } n 2 ) θ 1 ( { ω } n 1 2 , ω n 1 ; { ω } n 1 1 , ω n 2 ) ] × d [ P n 1 ( { ω } n 1 1 ) × P n 1 ( { ω } n 1 2 ) ] . (139)

It is evident that the expression (139) equals zero for every B F ¯ n 1 if and only if as

θ 1 ( { ω } n 1 ; { ω } n 2 ) θ 1 ( { ω } n 1 2 , ω n 1 ; { ω } n 1 1 , ω n 2 ) = 0. (140)

The last equality (140) is valid if the equality

α n 0 ( { ω } n 1 1 , ω n 1 ; { ω } n 1 2 , ω n 2 ) = α n 0 ( { ω } n 1 2 , ω n 1 ; { ω } n 1 1 , ω n 2 ) (141)

is true.

Now if for α n 2 ( { ω } n 1 ; { ω } n 2 ) > 0 satisfying the condition

Ω ¯ n × Ω ¯ n + α n 2 ( { ω } n 1 ; { ω } n 2 ) d [ P ¯ n × P ¯ n + ] = 1 (142)

to put

α n 0 ( { ω } n 1 ; { ω } n 2 ) = α n 2 ( { ω } n 1 ; { ω } n 2 ) + α n 2 ( T n ( { ω } n 1 ; { ω } n 2 ) ) φ n ( { ω } n 1 ; { ω } n 2 ) , (143)

then

Q n ( A ) = Ω ¯ n × Ω ¯ n + χ A ( { ω } n 1 ) α n 0 ( { ω } n 1 ; { ω } n 2 ) [ m n m n 1 ] + ( { ω } n 2 ) d [ P ¯ n × P ¯ n + ] + Ω ¯ n × Ω ¯ n + χ A ( { ω } n 2 ) α n 0 ( { ω } n 1 ; { ω } n 2 ) [ m n m n 1 ] ( { ω } n 1 ) d [ P ¯ n × P ¯ n + ] (144)

is a probability measure on the σ-algebra F ¯ n .

Taking into account the denotation (134) and the formula (143) we obtain that the measure

Q n ( A ) = Ω ¯ n × Ω ¯ n + χ A ( { ω } n 1 ) α n 1 ( { ω } n 1 ; { ω } n 2 ) [ m n m n 1 ] + ( { ω } n 2 )