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In my paper "On the inverse best approximation property of systems of subspaces of a Hilbert space" I introduced the Inverse marginal property (IMP) for a collection of $\sigma$-algebras.

Let $(\Omega,\mathcal{F},\mu)$ be a probability space and $\mathcal{F}_1,...,\mathcal{F}_n$ be sub-$\sigma$-algebras of $\mathcal{F}$. We will say that the collection $\mathcal{F}_1,...,\mathcal{F}_n$ possesses the inverse marginal property (IMP) if for arbitrary random variables $\xi_1,...,\xi_n$ such that

(1) $\xi_k$ is $\mathcal{F}_k$-measurable, $k=1,2,...,n$;

(2) $E|\xi_k|^2<\infty$, $k=1,2,...,n$;

(3) $E\xi_1=E\xi_2=...=E\xi_n$,

there exists a random variable $\xi$ such that $E|\xi|^2<\infty$ and $E(\xi|\mathcal{F}_k)=\xi_k$ for all $k=1,2,...,n$.

The simplest example of a collection of sub-$\sigma$-algebras which possesses the IMP is a system of pairwise independent sub-$\sigma$-algebras. In this case a needed random variable $\xi$ can be defined by $\xi:=\xi_1+...+\xi_n-(n-1)a$, where $a:=E\xi_1=E\xi_2=...=E\xi_n$.

Question: is the IMP a new notion or it is well-known? Have you seen this property or something similar in the literature?

I will be very grateful for any comments on the IMP.

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  • $\begingroup$ If one restrict to $\xi$ such that $\mathbb{E}(\xi)=0$. Is your IMP equivalent to $$ L^2(\Omega,\mathcal{F}_1,\mu)\oplus L^2(\Omega,\mathcal{F}_2,\mu)\oplus\cdots \oplus L^2(\Omega,\mathcal{F}_n,\mu)$$? $\endgroup$
    – RaphaelB4
    Apr 30, 2020 at 15:38
  • $\begingroup$ Unfortunately, I do not understand your question. Please specify the question. $\endgroup$ Apr 30, 2020 at 18:48
  • $\begingroup$ Just that the vector space generated by the variable $\mathcal{F}_i$ measurable are in direct sum : $\forall \xi_1,\cdots,\xi_n\in L^2(\Omega,\mathcal{F_1},\mu)\times...\times L^2(\Omega,\mathcal{F_n},\mu)$. $\xi_1+\cdots +\xi_n=0\Rightarrow \xi_1=0,\cdots,\xi_n=0$. $\endgroup$
    – RaphaelB4
    Apr 30, 2020 at 20:12
  • $\begingroup$ No, IMP is not equivalent to the linear independence of the marginal subspaces $L^2_0(\mathcal{F}_1),...,L^2_0(\mathcal{F}_n)$. One can show that a collection of $\sigma$-algebras $\mathcal{F}_1,...,\mathcal{F}_n$ possesses the IMP if and only if the marginal subspaces $L^2_0(\mathcal{F}_1),...,L^2_0(\mathcal{F}_n)$ are linearly independent and their sum is closed in $L^2(\mathcal{F})$. For details see my paper "On the inverse best approximation property of systems of subspaces of a Hilbert space" (available on ArXiv). $\endgroup$ Apr 30, 2020 at 21:15
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    $\begingroup$ Please choose between "article" and "paper" once and for all, and stop making these needless changes. $\endgroup$
    – S. Carnahan
    May 3, 2020 at 4:34

1 Answer 1

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I've never heard the term 'inverse marginal property', but the notion is somewhat familiar. Take any square integrable martingale $\xi_i$, $i=1,2,...$, and let $\{\mathcal{F}_n \}$ be it's natural filtration. Then I suspect for any $N < \infty$, the initial sequence $\{ \mathcal{F}_n, n \leq N \}$ has this property.

The difference is that $\xi_n$ are not arbitrary $\mathcal{F}_n$ measurable random variables - they're rather special. I expect that given our filtration, it is easy to construct a sequence $\{ \zeta_n \}$ adapted to $\mathcal{F}_n$ that break your IMP.

In fact, modifying your example slightly, $\xi_{k} = a + \sum_{i\leq k}(\xi_i - a)$, $k = 1, \cdots, n$ is a martingale w.r.t $\{ \mathcal{F_k} \}$. It is not an arbitrary sequence.

Could you construct a sequence of $\sigma$-algebras that satisfy IMP for any adapted sequence of random variables?

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  • $\begingroup$ In the definition of the IMP the needed $\xi$ must exist for arbitrary random variables $\xi_1,...,\xi_n$ that satisfy (1), (2), (3) in the Question. $\endgroup$ Apr 27, 2020 at 18:57
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    $\begingroup$ If $\mathcal{F_1}\subset\mathcal{F}_2\subset...\subset\mathcal{F}_n$, then the collection $\mathcal{F}_1,...,\mathcal{F}_n$ does not possess the IMP (if the probability space is not trivial). In fact, if $\xi_1,...,\xi_n$ are random variables that satisfy conditions (1),(2),(3) in the Question, then the needed $\xi$ exists if and only if $\xi_1,...,\xi_n$ is a martingale with respect to $\{\mathcal{F}_1,...,\mathcal{F}_n\}$. If this is the case, then one can take $\xi=\xi_n$. $\endgroup$ Apr 27, 2020 at 19:02
  • $\begingroup$ For examples of collections of $\sigma$-algebras that possess the IMP see Section 5 of my paper "On the inverse best approximation property of systems of subspaces of a Hilbert space" (it is available on ArXiv). $\endgroup$ Apr 27, 2020 at 19:06
  • $\begingroup$ If $\mathcal{F}_1\subset...\subset\mathcal{F}_n$, then the collection $\mathcal{F}_1,...,\mathcal{F}_n$ does not possess the IMP (if the $\sigma$-algebras are not trivial). $\endgroup$ Apr 27, 2020 at 19:22

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