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There are three ways to define the exchangeable $\sigma$-algebra of a stochastic process: one via symmetric functions, another via $n$-symmetric functions and a third via exchangeable events. Why are these three formulations equivalent?

Denote by $S(n)$ ($n\in\mathbb{N}_1$) (with $\mathbb{N}_1=\left\{1,2,3,\dots\right\}$ the set of positive integers) the set consisting of all bijections $\rho:\mathbb{N}_1\rightarrow\mathbb{N}_1$ that leave all numbers $>n$ unchanged.

If $D$, $E$ are non-empty sets, $x=\left(x_1, x_2, \dots\right)\in E^{\mathbb{N}_1}$ and $\rho:\mathbb{N}_1\rightarrow\mathbb{N}_1$, $x^\rho$ will denote $\left(x_{\rho(1)},x_{\rho(2)},\dots\right)$. If $f:D\rightarrow E^{\mathbb{N}_1}$ and $\rho:\mathbb{N}_1\rightarrow\mathbb{N}_1$, $f^\rho:D\rightarrow E^{\mathbb{N}_1}$ will denote the function $f^\rho(x):=[f(x)]^\rho$. $f:E^{\mathbb{N}_1}\rightarrow D$ will be called $n$-symmetric (for a fixed $n\in\mathbb{N}_1$) iff $f(x)=f(x^\rho)$ for all $x\in E$ and $\rho\in S(n)$. $f$ will be called symmetric iff it is $n$-symmetric for all $n\in\mathbb{N}_1$.

Let $X=\left(X_n\right)_{n\in\mathbb{N}_1}$ be a sequence of random objects defined over the measurable space $\left(\Omega,\mathcal{A}\right)$, each taking values in the measurable space $\left(E,\mathcal{B}\right)$.

If $B\in\otimes_{n=1}^\infty\mathcal{B}$ (the product $\sigma$-algebra), we will define $B^\rho:=\left\{X^\rho\in B\right\}$ for $\rho:\mathbb{N}_1\rightarrow\mathbb{N}_1$. $B$ will be called range-side $n$-symmetric (for a fixed $n\in\mathbb{N}_1$) iff $B^\rho=X^{-1}(B)$ for every $\rho\in S(n)$. $B$ will be called range-side symmetric iff it is range-side $n$-symmetric for all $n\in\mathbb{N}_1$. It is easy to verify that the collection of range-side $n$-symmetric events is a sub-$\sigma$-algebra of $\otimes_{n=1}^\infty\mathcal{B}$. It will be denoted by $\mathcal{D}_n$ and referred to as the range-side $n$-exchangeable $\sigma$-algebra of $X$. It is likewise easy to see that $\bigcap_{n=1}^\infty\mathcal{D}_n$ is the collection of range-side symmetric events, which is therefore a $\sigma$-algebra too. It will be denoted by $\mathcal{D}$ and referred to as the range-side exchangeable $\sigma$-algebra of $X$.

With $\mathfrak{B}$ denoting the Borel field on the real line, define

$$\begin{array}{lcl} \mathcal{E}_n & := & \sigma\left(\bigcup\left\{\left.\left(f\circ X\right)^{-1}\left(\mathfrak{B}\right):\right|\space f:E^{\mathbb{N}_1}\rightarrow\mathbb{R}\space\mathrm{is}\otimes_{n=1}^\infty\mathcal{B}\space/\space\mathfrak{B}\space\mathrm{-measurable}\space\mathrm{and}\space n\mathrm{-symmetric}\right\}\right) \\ \mathcal{E} & := & \sigma\left(\bigcup\left\{\left.\left(f\circ X\right)^{-1} \left(\mathfrak{B}\right):\right|\space f:E^{\mathbb{N}_1}\rightarrow\mathbb{R}\space\mathrm{is}\space\otimes_{n=1}^\infty\mathcal{B}\space/\space\mathfrak{B}\space\mathrm{-measurable}\space\mathrm{and}\space\mathrm{symmetric}\right\}\right) \\ \mathcal{F}_n & := & X^{-1}\left(\mathcal{D}_n\right) \\ \mathcal{F} & := & X^{-1}\left(\mathcal{D}\right) \end{array}$$

According to Klenke (Definition 12.6 and Remark 12.7),

  1. $\mathcal{E}_n=\mathcal{F}_n$ for all $n\in\mathbb{N}_1$ and
  2. $\mathcal{E}=\bigcap_{n=1}^\infty\mathcal{E}_n=\mathcal{F}$

The collection in 1 is the domain-side $n$-exchangeable $\sigma$-algebra of $X$; its members are the domain-side $n$-symmetric events of $X$. The collection in 2 is the domain-side exchangeable $\sigma$-algebra of $X$, or simply the exchangeable $\sigma$-algebra of $X$; its members are the domain-side symmetric events of $X$, or simply the symmetric events of $X$.

I see why $\mathcal{E}_n\subseteq\mathcal{F}_n$ and why $\mathcal{E}\subseteq\bigcap_{n=1}^\infty\mathcal{E}_n\subseteq\bigcap_{n=1}^\infty\mathcal{F}_n=\mathcal{F}$. Any help with proving the converse containments required to establish Klenke's claims, namely $\mathcal{F}_n\subseteq\mathcal{E}_n$ and $\mathcal{F}\subseteq\mathcal{E}$, will be appreciated.


Comments

There are a couple possible alternative definitions of an exchangeable $\sigma$-field. After defining them, i will explore their relation to the $\sigma$-algebras defined above.

The first alternative definition is due to David Freedman (p. 39). An event $B\in\otimes_{n=1}^\infty\mathcal{B}$ will be called range-side $n$-symmetric of the 2nd kind iff $x\in B\implies x^\rho\in B$ for all $x\in E^{\mathbb{N}_1}$ and all $\rho\in S(n)$. It will be called range-side symmetric of the 2nd kind iff it is $n$-symmetric of the 2nd kind for all $n\in\mathbb{N}_1$. The collection of range-side $n$-symmetric events of the 2nd kind (for a fixed $n\in\mathbb{N}_1$) constitutes a $\sigma$-algebra that will be denoted $\mathcal{D}_n^{(2)}$ and called the range-side $n$-exchangeable $\sigma$-algebra of the 2nd kind. The intersection $\bigcap_{n=1}^\infty\mathcal{D}_n^{(2)}$ is the collection of range-side symmetric events of the 2nd kind, which is therefore a $\sigma$-algebra that will be denoted $\mathcal{D}^{(2)}$ and called the range-side exchangeable $\sigma$-algebra of the 2nd kind. $X^{-1}\left(\mathcal{D}_n^{(2)}\right)$ will be denoted $\mathcal{F}_n^{(2)}$ and called the domain-side $n$-exchangeable $\sigma$-algebra of the 2nd kind of $X$, its members - the domain-side $n$-symmetric events of the 2nd kind of $X$. Similarly, $X^{-1}\left(\mathcal{D}^{(2)}\right)$ will be denoted $\mathcal{F}^{(2)}$ and called the domain-side exchangeable $\sigma$-algebra of the 2nd kind of $X$ (or simply the exchangeable $\sigma$-algebra of the 2nd kind of $X$), its members - the domain-side symmetric events of the 2nd kind of $X$ (or simply the symmetric events of the 2nd kind of $X$).

The second alternative set of definitions derives from Klenke's definition (12.1) of an exchangeable sequence of random objects over a probability space. Suppose, then, that a measure $\mu$ is defined on $\left(\Omega,\mathcal{A}\right)$ and denote by $\mu_X$ its distribution, i.e. the measure induced on $\left(E^{\mathbb{N}_1},\otimes_{n=1}^\infty\mathcal{B}\right)$ by $X$. For each $\rho\in\mathbb{N}_1\rightarrow\mathbb{N}_1$ write $\mu^\rho:=\mu_{X^\rho}$, i.e. the measure induced on $\left(E^{\mathbb{N}_1},\otimes_{n=1}^\infty\mathcal{B}\right)$ by $X^\rho$. Then $B\in\otimes_{n=1}^\infty\mathcal{B}$ will be called a range-side $n$-symmetric event of the 3rd kind iff $\mu_X(B)=\mu^\rho(B)$ for all $\rho\in S(n)$ (for a fixed $n\in\mathbb{N}_1$). It will be called a range-side symmetric event of the 3rd kind iff it is an $n$-symmetric event of the 3rd kind for all $n\in\mathbb{N}_1$. The collection of all range-side $n$-symmetric events of the 3rd kind (for a fixed $n\in\mathbb{N}_1$) is a $\sigma$-algebra that will be denoted $\mathcal{D}_n^{(3)}$ and called the range-side $n$-exchangeable $\sigma$-algebra of the 3rd kind. $\bigcap_{n=1}^\infty\mathcal{D}_n^{(3)}$ is the collection of range-side exchangeable events of the 3rd kind, which is therefore a $\sigma$-algebra that will be denoted $\mathcal{D}^{(3)}$ and called the range-side exchangeable $\sigma$-algebra of the 3rd kind. A "domain-side ($n$-)exchangeable $\sigma$-algebra of the 3rd kind" will not be defined.

It is interesting to note that the definition of $\mathcal{D}^{(2)}$ makes no use of either $X$ or $\mu$, the definition of $\mathcal{D}$ relies on $X$ but not on $\mu$ and the definition of $\mathcal{D}^{(3)}$ relies on both $X$ and $\mu$.

It is evident that $\mathcal{D}_n^{(2)}\subseteq\mathcal{D}_n\subseteq\mathcal{D}_n^{(3)}$ and that $\mathcal{D}^{(2)}\subseteq\mathcal{D}\subseteq\mathcal{D}^{(3)}$. Similarly, $\mathcal{F}_n^{(2)}\subseteq\mathcal{F}_n$ and $\mathcal{F}^{(2)}\subseteq\mathcal{F}$.

It is also easy to check that

$$\begin{array}{lcl} \mathcal{D}_n^{(2)} & = & \bigcup\left\{\left.f^{-1}\left(\mathfrak{B}\right):\right|\space f:E^{\mathbb{N}_1}\rightarrow\mathbb{R}\space\mathrm{is}\otimes_{n=1}^\infty\mathcal{B}\space/\space\mathfrak{B}\space\mathrm{-measurable}\space\mathrm{and}\space n\mathrm{-symmetric}\right\} \\ \mathcal{D}^{(2)} & = & \bigcup\left\{\left.f^{-1} \left(\mathfrak{B}\right):\right|\space f:E^{\mathbb{N}_1}\rightarrow\mathbb{R}\space\mathrm{is}\space\otimes_{n=1}^\infty\mathcal{B}\space/\space\mathfrak{B}\space\mathrm{-measurable}\space\mathrm{and}\space\mathrm{symmetric}\right\} \end{array}$$

Hence $\mathcal{E}_n=\mathcal{F}_n^{(2)}$ and $\mathcal{E}=\mathcal{F}^{(2)}$



References

  1. Klenke, Achim. "Probability Theory: A Comprehensive Course", 2008
  2. Freedman, David. "Markov Chains", 1971
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    $\begingroup$ May I suggest that you consider editing the question offline until it has reached a stable state? 20 edits in the span of 7 hours might be viewed as excessive by some users. Cheers. $\endgroup$
    – cardinal
    Mar 17, 2013 at 17:57

1 Answer 1

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Fix an $n\in\mathbb{N}_1$. I will show that $\mathcal{F}_n\subseteq\mathcal{E}_n$. The proposition $\mathcal{F}\subseteq\mathcal{E}$ can be shown analogously. Since $\mathcal{F}_n^{(2)}=\mathcal{E}_n$, it suffices to show that $\mathcal{F}_n\subseteq\mathcal{F}_n^{(2)}$. In light of the equalities $\mathcal{F}_n=X^{-1} \left(\mathcal{D}_n\right)$ and $\mathcal{F}_n^{(2)}=X^{-1} \left(\mathcal{D}_n^{(2)}\right)$, this amounts to showing that to every $B\in\mathcal{D}_n$ there's some $B'\in\mathcal{D}_n^{(2)}$ such that $X^{-1}(B)=X^{-1}(B')$.

I will proceed as follows. To every set $A\in E^{\mathbb{N}_1}$ (not necessarily measurable) i will define a set $\overline{A}\in E^{\mathbb{N}_1}$ called the $n$-symmetric closure of $A$, or, since $n$ is fixed, simply the symmetric closure of $A$. I will then show that $B':=\overline{B}$ satisfies the requirement by showing that

  1. If $A\in\otimes_{n=1}^\infty\mathcal{B}$, $\overline{A}\in\mathcal{D}_n^{(2)}$, i.e.

    a) For all $x\in\overline{A}$, $x^\rho\in\overline{A}$ for all $\rho\in S(n)$

    b) $\overline{A}\in\otimes_{n=1}^\infty\mathcal{B}$

  2. If $A\in\mathcal{D}_n$, $X^{-1}(A)=X^{-1}\left(\overline{A}\right)$

I will start with an auxiliary definition. A set $A\in E^{\mathbb{N}_1}$ will be called $n$-symmetric (or simply symmetric, since $n$ is fixed) iff $\forall x\in E^{\mathbb{N}_1}\forall\rho\in S(n), x\in A\implies x^\rho\in A$.

It is easy to verify that if $\mathcal{C}$ is a collection of symmetric sets, $\bigcap\mathcal{C}$ is symmetric too. Given $A\in E^{\mathbb{N}_1}$, the $n$-symmetric closure of $A$, to be denoted $\mathrm{SymClo}_n{A}$, is defined as the smallest $n$-symmetric set containing $A$: $\mathrm{SymClo}_n{A}:=\bigcap\left\{B\in E^{\mathbb{N}_1}\left|:\space A\subseteq B,\space B\space\mathrm{is}\space n\mathrm{-symmetric}\right.\right\}$. Since $n$ is fixed, we will simplify the notation and write $\overline{A}:=\mathrm{SymClo}_n(A)$. Note that $A\subseteq\overline{A}$ and $\overline{A}$ is symmetric. So property 1a is verified.

It is easy to check the following characterization of $\overline{A}$: $x\in\overline{A}\iff\exists\rho\in S(n),\space x^{\rho}\in A$. From this, in light of the fact that

$$\pi_\rho:E^{\mathbb{N}_1}\rightarrow E^{\mathbb{N}_1},\space\space\pi_\rho(x):=x^\rho$$

is $\otimes_{n=1}^\infty\mathcal{B}\space/\space\otimes_{n=1}^\infty\mathcal{B}$-measurable for all $\rho\in S(n)$, as can be easily verified, we get that if $A\in\otimes_{n=1}^\infty\mathcal{B}$, $\overline{A}=\bigcup_{\rho\in S(n)}\pi_\rho^{-1}(A)\in\otimes_{n=1}^\infty\mathcal{B}$. So property 1b is verified.

Finally, from the same characterization follows property 2. Indeed, let $A\in\mathcal{D}_n$ and suppose to the contrary that $X^{-1}(A)\subsetneq X^{-1}\left(\overline{A}\right)$. Let $\omega\in X^{-1}\left(\overline{A}\right)\backslash X^{-1}(A)$. Let $y\in\overline{A}\backslash A$ such that $y=X(\omega)$. From the characterization property, there's some $\rho\in S(n)$ such that $y':=y^\rho\in A$. Then $X^\rho\left(\omega\right)=y'$. But then, by the definition of $\mathcal{D}_n$, $\omega\in X^{-1}(A)$, a contradiction.

It is interesting to note that even though $\mathcal{D}_n^{(2)}\subseteq\mathcal{D}_n$ and $X^{-1}\left(D_n\right)=X^{-1}\left(\mathcal{D}_n^{(2)}\right)$, in general $\mathcal{D}_n^{(2)}\neq\mathcal{D}_n$. For example, if $X$ is constant, $X=c$, than any $\otimes_{n=1}^\infty\mathcal{B}$-measurable set that contains $c$ belongs to $\mathcal{D}_n$, but only the symmetric ones belong to $\mathcal{D}_n^{(2)}$.

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