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[MATH] under [MATH] are all distinct because they differ in the first entry. As a consequence, we get that the number of orbits is [MATH] |
To each orbit there corresponds one vector in [MATH] , namely the sum over its elements. This follows directly from the structure of a permutation module, where to each cycle of the permutation there corresponds one eigenvector with eigenvalue [MATH] |
The algebra [MATH] also has a natural [MATH] -grading, arising from the eigenspace decomposition corresponding to the action of [MATH] . Indeed, for every [MATH] , define [MATH] . We call this subspace the [MATH] -homogeneous component of weight [MATH] . Because of the definition of [MATH] , a Lie bracket [MATH] in [MA... |
Since the homogeneous components [MATH] are [MATH] -invariant, they are, in turn, [MATH] -graded by this basic fact of linear algebra: |
Lemma 5 Let [MATH] be a field and [MATH] [MATH] -vector space. Let [MATH] be a linear map and [MATH] [MATH] -invariant subspace. Assume that an element [MATH] of [MATH] can be written in the form |
[MATH] , where [MATH] are eigenvectors of [MATH] corresponding to distinct eigenvalues [MATH] . Then [MATH] for all [MATH] Hence, we can write |
[EQUATION] The subspaces on the right-hand side are permuted by [MATH] , according to the formula [MATH] . Hence, for every prime [MATH] and positive integer [MATH] , the subspaces [MATH] and |
[MATH] are [MATH] -modules. The basis of [MATH] consists of eigenvectors for the action of [MATH] which are permuted by [MATH] in orbits of length [MATH] . Each of these orbits spans a [MATH] -module of dimension [MATH] Hence, we have that |
[EQUATION] Using Proposition and the [MATH] -invariance of both subspaces, we get the corresponding formula for [MATH] 3. The ideal [MATH] |
Let [MATH] denote the ideal [MATH] . We have already mentioned that [MATH] is degree-homogeneous and [MATH] -invariant. Hence, the same properties hold for [MATH] . In this section, for every natural number [MATH] , we bound the dimension of the subspaces [MATH] |
Since [MATH] and [MATH] are eigenvectors of [MATH] with corresponding eigenvalues [MATH] , a basis for the subspace [MATH] is given by those Lie brackets [MATH] belonging to [MATH] for which there exists a permutation [MATH] in the symmetric group [MATH] such that [MATH] , for some [MATH] In the next proposition, we bo... |
Proposition 6 For every prime number [MATH] and natural number [MATH] , let [MATH] denote the dimension of the homogeneous subspace [MATH] . Then |
[EQUATION] Proof. First, we recall that [MATH] is the subspace of homogeneous elements of [MATH] of degree [MATH] and it is [MATH] . Since none of the [MATH] is centralized by [MATH] , we clearly have [MATH] . More in general, we have |
[EQUATION] Hence, we obtain [MATH] The subspace [MATH] has basis given by the Lie brackets [MATH] in [MATH] for which [MATH] . This implies that the last index [MATH] is uniquely determined by the previous ones and hence, [MATH] . Notice that this bound is only sharp for [MATH] , because the set [MATH] is a basis for [... |
The subspace [MATH] is [MATH] , where [MATH] ranges over a basis of [MATH] and [MATH] . Hence, we have [MATH] . Once again, we point out that the equality trivially holds for [MATH] , and for [MATH] . The inequality is instead strict for larger values of [MATH] . These bounds together give the recursive formula |
[EQUATION] from which the conclusion follows by induction. This proposition concludes this section and the study of the ideal [MATH] . Indeed, the bound on the dimension of its homogeneous components [MATH] is all we need in the proof of Proposition |
4. The ideal [MATH] Let [MATH] denote the smallest [MATH] -invariant ideal containing the derived subalgebra [MATH] . Since [MATH] is not [MATH] -invariant, then [MATH] is strictly bigger than the ideal generated simply by [MATH] . Indeed, in this section we will prove that [MATH] is generated, as ideal, by the [MATH] ... |
We start determining the generators of the subalgebra [MATH] . As already pointed out, the centralizer [MATH] is degree-homogeneous and a basis for the subspaces |
[MATH] is given in Proposition . The subalgebra [MATH] is [MATH] , where [MATH] and [MATH] lie in the basis of [MATH] . Those Lie brackets are trivial unless one of the entries is [MATH] . Hence, the derived subalgebra of [MATH] is [MATH] , where [MATH] lies in the basis of [MATH] |
In the next proposition we produce a set of generators of the ideal [MATH] Proposition 7 The smallest ideal containing [MATH] and invariant under the action of the group [MATH] is generated, as an ideal, by the [MATH] -homogeneous components of the generators of [MATH] |
Proof. Consider [MATH] , where [MATH] belongs to the basis of [MATH] given in Proposition . We can write [MATH] as a linear combination of eigenvectors of [MATH] , namely [MATH] , where [MATH] is in the [MATH] -homogeneous component of weight [MATH] . Since [MATH] is in [MATH] , we necessarily have that [MATH] , for al... |
[MATH] , where the sum in parenthesis represents its [MATH] -homogeneous component of weight [MATH] . Applying [MATH] to it, we get |
[EQUATION] where this should be interpreted as a sum over [MATH] , where [MATH] varies from [MATH] to [MATH] This is exactly the homogeneous component of weight [MATH] |
We have just shown that the set of the [MATH] -homogeneous components of the generators of [MATH] is permuted by [MATH] . This implies that the ideal [MATH] -invariant. Moreover, this ideal is also [MATH] -invariant because, by construction, its generators are [MATH] -homogeneous. It contains the subalgebra [MATH] beca... |
To prove the opposite inclusion, it is sufficient to show that all [MATH] -homogeneous components of the generators of [MATH] are in [MATH] . This is true by Lemma |
We can now bound the dimension of the degree-homogeneous components of [MATH] Proposition 8 For every prime number [MATH] and natural number [MATH] , let [MATH] denote the dimension of the subspace [MATH] , the homogeneous component of [MATH] of degree [MATH] . Then |
[EQUATION] Proof. First, observe that the ideal [MATH] is contained in the derived subalgebra of [MATH] . This implies trivial intersection with the subspace [MATH] of homogeneous elements of degree [MATH] and hence [MATH] |
More generally, if [MATH] ranges in the set of generators of [MATH] and [MATH] denotes its [MATH] -homogeneous component of weight [MATH] , by the previous result we can write |
[EQUATION] Hence, we have [MATH] The subalgebra [MATH] , which is non-trivial only when [MATH] , is [MATH] , where [MATH] is an element of the basis of [MATH] and [MATH] . Indeed, the Lie brackets [MATH] are trivial by construction. Hence, we get |
[MATH] As previously remarked, the subalgebra [MATH] is [MATH] , where [MATH] is in the basis of [MATH] . Hence, its dimension is not greater than the dimension of [MATH] , namely [MATH] , by Proposition . Using the expression for [MATH] computed in Proposition , we get the following recursive formula |
[EQUATION] We now prove the desired bound for [MATH] by induction on [MATH] . Observe that the subspace [MATH] is one-dimensional, [MATH] . According to Proposition , its [MATH] -homogeneous components generate [MATH] . Hence, the bound holds for [MATH] . Suppose now that the bound is true for [MATH] . Using the above ... |
[EQUATION] This concludes the proof. This also concludes the section about the ideal [MATH] . As in the case of the ideal [MATH] , all we will need later on is the bound for the dimension of its homogeneous components. |
5. Proofs of Proposition and Theorem In this last section, we will prove Proposition by explicit construction of the Lie algebras [MATH] belonging to the family [MATH] . We will then apply Lazard correspondence to prove Theorem , where we will need to impose some conditions on the underlying field of each [MATH] in [MA... |
Proof of Proposition For every prime number [MATH] , define [MATH] to be the quotient algebra [MATH] . Since the two ideals [MATH] and [MATH] are graded, the same is true for their sum. In particular, every homogeneous component is given by the sum of the corresponding ones: [MATH] . This implies that [MATH] inherits f... |
[EQUATION] Substituting in this formula the expression for [MATH] from Proposition , and the upper bounds for [MATH] and [MATH] , respectively from Proposition and , we get |
[EQUATION] We observe that the right-hand side is a polynomial function of the prime number [MATH] , whose leading term is [MATH] and comes from the formula for [MATH] |
The algebra [MATH] admits the same Frobenius group [MATH] , as a group of automorphisms. Indeed, since [MATH] and [MATH] are [MATH] -invariant, the generator [MATH] of [MATH] and the generator [MATH] of [MATH] induce automorphisms of the quotient algebra, defined in the canonical way. The orders of [MATH] and [MATH] re... |
The fixed-point subspace [MATH] is trivial. Indeed, in case of a coprime action, the fixed points in the quotient algebra are covered by the fixed points in the original Lie algebra (see for example Khu93 , Theorem 1.6.2] ). For the same reason, if we require that, for every prime [MATH] , the characteristic of the und... |
It only remains to show that for every integer [MATH] , there exists a Lie algebra in [MATH] with nilpotency class at least [MATH] . In other words, this algebra must have a non-trivial Lie bracket of degree [MATH] . Because the lower bound for [MATH] has positive leading coefficient, for every fixed [MATH] we can alwa... |
Proof of Theorem Every [MATH] is a metabelian [MATH] -graded Lie algebras with [MATH] . According to a theorem of Shumyatsky Shu05 , Theorem 3.3] , its nilpotency class is at most [MATH] . We can therefore produce a similar result for groups under the extra condition that, for every algebra [MATH] (and hence [MATH] ), ... |
Increasing [MATH] with [MATH] in our construction was only required for an application of the Lazard correspondence. Hence, it is not unnatural to ask the following |
Question We wonder if it is possible to construct a family of [MATH] -groups, of unbounded nilpotency class, where [MATH] is a fixed prime number, whose elements [MATH] satisfy: |
(1) [MATH] admits a metacyclic Frobenius group of automorphisms; (2) the centralizer of the Frobenius kernel in [MATH] is trivial; |
(3) the centralizer of the Frobenius complement in [MATH] is abelian. We conclude this section by proving that the fixed points of [MATH] in [MATH] are covered by the fixed points of the same group in [MATH] even when the action is not coprime. |
Lemma 9 For every prime number [MATH] let [MATH] [MATH] [MATH] and [MATH] defined as above. Let [MATH] denote the ideal [MATH] , so that [MATH] . Then we have |
[EQUATION] Proof. We can restrict ourselves to the degree-homogeneous case and, denoting the subspace [MATH] by [MATH] , show that |
[EQUATION] Recall that [MATH] and [MATH] Because [MATH] and [MATH] are [MATH] -invariant we have [EQUATION] whence [EQUATION] A set of generators for [MATH] is given by the generators of [MATH] together with the generators of [MATH] . In other words, since the generators of [MATH] , described in Propositions and , are ... |
It is not hard to show that the generators of [MATH] are permuted by [MATH] in orbits of length [MATH] , whose elements are all in different [MATH] -homogeneous components (one for each [MATH] , with [MATH] ). The same holds for the generators of [MATH] . The key point here is that each simple Lie bracket involved in o... |
[EQUATION] Now consider a basis [MATH] for the subspace [MATH] . From the above equation it follows that [MATH] has basis [EQUATION] |
If we complete the basis of [MATH] to a basis of [MATH] , by adding the vectors [MATH] , then a basis of [MATH] is [EQUATION] where, with a little abuse of notation, the cosets are named by their representative. The desired conclusion follows. |
We remark that the result of the previous lemma strongly relies on the definitions of the algebras involved. Indeed, a crucial step in the proof is the fact that both [MATH] and [MATH] are free [MATH] -modules. |
# Source: arxiv 1806.05557 # Title: Martingales and Super-martingales Relative to a Convex Set of Equivalent Measures # Sections: all # Downloaded: 2026-03-03T04:46:55.257450+00:00 |
Martingales and Super-martingales Relative to a Convex Set of Equivalent Measures Nicholas S. Gonchar Bogolyubov Institute for Theoretical Physics of NAS, Kiev, Ukraine |
Email: mhonchar@i.ua Abstract In the paper, the martingales and super-martingales relative to a convex set of equivalent measures are systematically studied. The notion of local regular super-martingale relative to a convex set of equivalent measures is introduced and the necessary and sufficient conditions of the loca... |
Keywords Random process; Convex set of equivalent measures; Optional Doob decomposition; Local regular super-martingale; martingale; Fair price of contingent claim. |
Introduction In the paper, a new method of investigation of martingales and super-martingales relative to a convex set of equivalent measures is developed. A new proof that the essential supremum over the set of regular martingales, nonnegative random value and a convex set of equivalent measures, is a super-martingale... |
A notion of local regular super-martingale is introduced and the necessary and sufficient conditions are found under that the above defined super-martingales are local regular ones. The last fact allowed us to describe the local regular super-martingales. It is proved that the existence of a nontrivial martingale relat... |
An important notion of the complete convex set of equivalent measures is introduced. It is proved that any super-martingale relative to the complete convex set of equivalent measures on a measurable space with the finite set of elementary events is a local regular one. The notion of the complete convex set of equivalen... |
The definition of the fair price of contingent claim is introduced. The sufficient conditions of the existence of the fair price of contingent claim are presented. The conditions that the introduced notion coincides with classical one are given. |
All these notions are used in the case as the convex set of equivalent measures is a set of equivalent martingale measures for the evolution of both risk and non risk assets. The formula for the fair price of Standard Contract with Option of European type in an incomplete market is found. |
The notion of the complete convex set of equivalent measures permits us to give a new proof of the optional decomposition for a nonnegative super-martingale. This proof does not use the no-arbitrage arguments and the measurable choice |
First, the optional decomposition for diffusion processes super-martingale was opened by by El Karoui N. and Quenez M. C. . After that, Kramkov D. O. and Follmer H. |
proved the optional decomposition for the nonnegative bounded super-martingales. Folmer H. and Kabanov Yu. M. proved analogous result for an arbitrary super-martingale. Recently, Bouchard B. and Nutz M. |
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 Considered in the paper problem is a generalization of the corresponding one that appeared in mathematical finance about the optional decomposition for a super-martingale and which is related with t... |
Our statement of the problem unlike the above-mentioned one and it is more general: a super-martingale relative to a convex set of equivalent measures is given and it is necessary to find the conditions for the super-martingale and the set of measures under that the optional decomposition exists. |
The generality of our statement of the problem is that we do not require that the considered set of measures was and that is important for the proof of the optional decomposition in these papers. |
Local regular super-martingales relative to a convex set of equivalent measures. We assume that on a measurable space [MATH] a filtration [MATH] and a family of convex set of equivalent measures [MATH] on [MATH] are given. Further, we assume that [MATH] and the [MATH] -algebra [MATH] is a minimal [MATH] -algebra [MATH] |
A random process [MATH] is said to be adapted one relative to the filtration [MATH] if [MATH] is a [MATH] measurable random value, [MATH] |
Definition 1 An adapted random process [MATH] is said to be a super-martingale relative to the filtration [MATH] and the convex family of equivalent measures [MATH] if [MATH] and the inequalities |
[EQUATION] are valid. Further, for an adapted process [MATH] we use both the denotation [MATH] and the denotation [MATH] Definition 2 |
A super-martingale [MATH] relative to a convex set of equivalent measures M is a local regular one if [MATH] and there exists an adapted nonnegative increasing random process [MATH] |
[MATH] such that [MATH] is a martingale relative to every measure from [MATH] The next elementary Theorem will be very useful later. |
Theorem 1 Let a super-martingale [MATH] relative to a convex set of equivalent measures M be such that [MATH] The necessary and sufficient condition for it to be a local regular one is the existence of an adapted nonnegative random process [MATH] |
[MATH] such that [EQUATION] Proof. Necessity. If [MATH] is a local regular super-martingale, then there exist a martingale [MATH] and a non-decreasing nonnegative random process [MATH] |
[MATH] such that [EQUATION] From here we obtain the equalities [EQUATION] [EQUATION] where we introduced the denotation [MATH] It is evident that [MATH] |
Sufficiency. Suppose that there exists an adapted nonnegative random process [MATH] [MATH] such that the equalities ( ) hold. Let us consider the random process [MATH] where |
[EQUATION] It is evident that [MATH] and [EQUATION] Theorem is proved. Lemma 1 Any super-martingale [MATH] relative to a family of measures [MATH] for which there hold equalities [MATH] |
[MATH] is a martingale with respect to this family of measures and the filtration [MATH] Proof. The proof of Lemma see .∎ Description of local regular super-martingales relative to a convex set of equivalent measures |
Below, we describe the local regular super-martingales relative to a convex set of equivalent measures [MATH] Lemma 2 On a measurable space [MATH] with filtration [MATH] on it, let [MATH] be a sub [MATH] -algebra of the [MATH] -algebra [MATH] and let [MATH] be a finite family of nonnegative bounded random values. Then ... |
[EQUATION] Proof. We have the inequalities [EQUATION] Therefore, [EQUATION] The last implies [EQUATION] In the next Lemma, we present the formula for calculation of the conditional expectation relative to another measure from [MATH] |
Lemma 3 On a measurable space [MATH] with a filtration [MATH] on it, let [MATH] be a convex set of equivalent measures and let [MATH] be a bounded random value. Then the following formulas |
[EQUATION] are valid, where [EQUATION] Proof. The proof of Lemma is evident.∎ Let [MATH] be a family of equivalent measures on a measurable space [MATH] and let us introduce the denotation [MATH] for a convex set of equivalent measures |
[EQUATION] Lemma 4 If [MATH] is an integrable random value relative to the set of equivalent measures [MATH] , then the formula [EQUATION] |
is valid almost everywhere relative to the measure [MATH] Proof. The definition of [MATH] for non countable family of random variables see |
. Using the formula [EQUATION] where [MATH] [MATH] we obtain the inequality [EQUATION] or, [EQUATION] On the other side [EQUATION] |
Therefore, [EQUATION] Lemma is proved. Lemma 5 On a measurable space [MATH] with a filtration [MATH] on it, let [MATH] be a nonnegative bounded random value. If [MATH] are [MATH] measurable and [MATH] |
then the inequalities [EQUATION] are valid. Proof. From Lemma and Lemma conditions relative to the density of one measure with respect to another, we have |
[EQUATION] From the equality ( 21 ) we obtain the inequality [EQUATION] Lemma is proved. In this section, we assume that the conditions of Lemma relative to the density of one measure with respect to another are true. |
Lemma 6 On a measurable space [MATH] with a filtration [MATH] on it, let [MATH] be a nonnegative random value which is integrable relative to the set of equivalent measures [MATH] Then the inequalities |
[EQUATION] are valid. Proof. Using Lemma inequalities for the nonnegative bounded [MATH] and the formula [EQUATION] where [MATH] |
we prove Lemma inequalities. Let us consider the case, as [MATH] Let [MATH] be a sequence of bounded random values converging to [MATH] monotonuosly. Then |
[EQUATION] Due to the monotony convergence of [MATH] to [MATH] as [MATH] we can pass to the limit under the conditional expectations on the left and right sides in the inequalities ( 25 ) that proves Lemma |
Lemma 7 On a measurable space [MATH] with filtration [MATH] on it, for every nonnegative integrable random value [MATH] relative to a set of equivalent measures [MATH] the inequalities |
[EQUATION] are valid. Lemma is a consequence of Lemma Lemma 8 On a measurable space [MATH] with a filtration [MATH] on it, let [MATH] be a nonnegative integrable random value with respect to a set of equivalent measures [MATH] and such that |
[EQUATION] then the random process [MATH] is a martingale relative to a convex set of equivalent measures [MATH] Proof. Due to Lemma , a random process |
[MATH] is a super-martingale, that is, [EQUATION] Or, [MATH] From the other side, [EQUATION] The above inequalities imply [MATH] |
The last equalities lead to the equalities [MATH] The fact that [MATH] is a super-martingale relative to the set of measures [MATH] and the above equalities prove Lemma , since the Lemma conditions are valid. |
In the next Theorem we denote [MATH] the minimal [MATH] -algebra [MATH] Theorem 2 Let [MATH] be a measurable space with a filtration [MATH] on it and let [MATH] be a nonnegative integrable random value with respect to a set of equivalent measures [MATH] |
The necessary and sufficient conditions of the local regularity of the super-martingale [MATH] where [EQUATION] is its uniform integrability relative to the set of measure [MATH] and the fulfillment of the equalities |
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