Proof of Independence of σ-Algebras: π-Systems, Grouping and Pairs

Throughout, finite products are split and reordered by Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (i), Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (iii) and Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (iv), and intersections of families indexed by $[n]$ directly from the definition; and $[2]=\{1,2\}$ (Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (i), Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (ii)), so a family indexed by $[2]$ is given by its two terms and a list of distinct indices in $[2]$ is $(1)$, $(2)$, $(1,2)$ or $(2,1)$.

Proof of (i). Let $n\in\mathbb{N}$, let $j_{1},\dots,j_{n}\in J'$ be distinct and let $A_{k}\in\mathcal{D}_{j_{k}}$ for $k\in[n]$. Then $j_{1},\dots,j_{n}$ are distinct indices in $J$ and $A_{k}\in\mathcal{C}_{j_{k}}$, so the product formula for $A_{1},\dots,A_{n}$ is that of the independent family $(\mathcal{C}_{j})_{j\in J}$.

Proof of (ii). Step 1: one class at a time. Let $(\mathcal{D}_{j})_{j\in J}$ be an independent family of subsets of $\mathcal{F}$, and let $i\in J$ be such that $\mathcal{D}_{i}$ is a $\pi$-system on $\Omega$. We show that the family $(\mathcal{D}'_{j})_{j\in J}$ with $\mathcal{D}'_{i}=\sigma(\mathcal{D}_{i})$ and $\mathcal{D}'_{j}=\mathcal{D}_{j}$ for $j\neq i$ is independent. Let $\mathcal{I}$ be the set of events $\bigcap_{k=1}^{n}A_{k}$ with $n\in\mathbb{N}$, distinct $j_{1},\dots,j_{n}\in J\setminus\{i\}$ and $A_{k}\in\mathcal{D}_{j_{k}}$, and let

$$\mathcal{L}=\{A\in\mathcal{F}:\ P(A\cap B)=P(A)\,P(B)\text{ for every }B\in\mathcal{I}\}.$$

First, $\mathcal{D}_{i}\subseteq\mathcal{L}$: for $A\in\mathcal{D}_{i}$ and $B=\bigcap_{k=1}^{n}A_{k}\in\mathcal{I}$, independence of $(\mathcal{D}_{j})_{j\in J}$ applied to the distinct indices $i,j_{1},\dots,j_{n}$ and then to $j_{1},\dots,j_{n}$ gives $P(A\cap B)=P(A)\prod_{k=1}^{n}P(A_{k})=P(A)P(B)$.

Next, $\mathcal{L}$ is a $\lambda$-system on $\Omega$. Let $B\in\mathcal{I}$. Clearly $\Omega\in\mathcal{L}$, as $P(\Omega)=1$. If $A\in\mathcal{L}$, then $(\Omega\setminus A)\cap B=B\setminus(A\cap B)$, so by Basic Properties of a Measure (iv) and Basic Properties of a Measure (v)

$$P\bigl((\Omega\setminus A)\cap B\bigr)=P(B)-P(A)P(B)=P(\Omega\setminus A)\,P(B).$$

If $(A_{m})$ is a pairwise disjoint sequence in $\mathcal{L}$, the sets $A_{m}\cap B$ are pairwise disjoint with union $\bigl(\bigcup_{m}A_{m}\bigr)\cap B$, so by countable additivity (Measure, Measure Space, and Probability Measure (i)) and Sums of Sequences in [0,∞]: Real Series, Linearity, Comparison, Rearrangement, Double Series and Monotone Convergence (ix)

$$P\Bigl(\Bigl(\bigcup_{m\in\mathbb{N}}A_{m}\Bigr)\cap B\Bigr)=\sum_{m\in\mathbb{N}}P(A_{m})P(B)=P\Bigl(\bigcup_{m\in\mathbb{N}}A_{m}\Bigr)P(B).$$

By Dynkin's theorem (Dynkin's Pi-Lambda Theorem (ii)), $\sigma(\mathcal{D}_{i})\subseteq\mathcal{L}\subseteq\mathcal{F}$.

Now let $n\in\mathbb{N}$, let $j_{1},\dots,j_{n}\in J$ be distinct and let $A_{k}\in\mathcal{D}'_{j_{k}}$. If no $j_{k}$ equals $i$, the product formula is that of $(\mathcal{D}_{j})_{j\in J}$. Otherwise, reordering, we may assume $j_{1}=i$; for $n=1$ there is nothing to prove, and for $n\ge2$ the event $B=\bigcap_{k=2}^{n}A_{k}$ lies in $\mathcal{I}$ and $A_{1}\in\sigma(\mathcal{D}_{i})\subseteq\mathcal{L}$, so

$$P\Bigl(\bigcap_{k=1}^{n}A_{k}\Bigr)=P(A_{1})\,P(B)=P(A_{1})\prod_{k=2}^{n}P(A_{k})=\prod_{k=1}^{n}P(A_{k}).$$

Step 2: all classes. Let $n\in\mathbb{N}$, let $j_{1},\dots,j_{n}\in J$ be distinct and let $A_{k}\in\sigma(\mathcal{C}_{j_{k}})$. For $r\in\{0,\dots,n\}$ let $(\mathcal{D}^{(r)}_{j})_{j\in J}$ be the family with $\mathcal{D}^{(r)}_{j_{s}}=\sigma(\mathcal{C}_{j_{s}})$ for $s\le r$ and $\mathcal{D}^{(r)}_{j}=\mathcal{C}_{j}$ for the other $j\in J$. We show by induction on $r$ (Induction for Predicative Formulas, and the Order of ℕ below 1 and below n+1 (i)) that $(\mathcal{D}^{(r)}_{j})_{j\in J}$ is independent. For $r=0$ it is $(\mathcal{C}_{j})_{j\in J}$. If $r<n$ and $(\mathcal{D}^{(r)}_{j})_{j\in J}$ is independent, then, as the indices are distinct, its class at $j_{r+1}$ is still the $\pi$-system $\mathcal{C}_{j_{r+1}}$, and Step 1 with $i=j_{r+1}$ turns it into $(\mathcal{D}^{(r+1)}_{j})_{j\in J}$, which is therefore independent. The product formula of $(\mathcal{D}^{(n)}_{j})_{j\in J}$ for $A_{1},\dots,A_{n}$ is the claim.

Proof of (iii). For $L\subseteq J$ let $\mathcal{Q}(L)$ be the set consisting of $\Omega$ and of the events $\bigcap_{l=1}^{p}A_{l}$ with $p\in\mathbb{N}$, distinct $j_{1},\dots,j_{p}\in L$ and $A_{l}\in\mathcal{G}_{j_{l}}$.

Disjoint index sets. Let $L,L'\subseteq J$ be disjoint, $E\in\mathcal{Q}(L)$ and $E'\in\mathcal{Q}(L')$. Then

$$E\cap E'\in\mathcal{Q}(L\cup L')\qquad\text{and}\qquad P(E\cap E')=P(E)\,P(E').\tag{1}$$

If $E=\Omega$ or $E'=\Omega$ this is clear. Otherwise write $E=\bigcap_{l=1}^{p}A_{l}$ and $E'=\bigcap_{l=1}^{p'}A'_{l}$ with indices $j_{1},\dots,j_{p}\in L$ and $j'_{1},\dots,j'_{p'}\in L'$ as above. As $L\cap L'=\emptyset$, the concatenated list $j_{1},\dots,j_{p},j'_{1},\dots,j'_{p'}$ consists of distinct indices, so $E\cap E'$, the intersection of the concatenated list of events, lies in $\mathcal{Q}(L\cup L')$; and independence of $(\mathcal{G}_{j})_{j\in J}$ applied to the concatenated list and to each of the two lists gives $P(E\cap E')=\prod_{l=1}^{p}P(A_{l})\prod_{l=1}^{p'}P(A'_{l})=P(E)P(E')$.

$\mathcal{Q}(L)$ is a $\pi$-system with $\sigma(\mathcal{Q}(L))=\sigma\bigl(\bigcup_{j\in L}\mathcal{G}_{j}\bigr)$. For $E\in\mathcal{Q}(L)$, $j\in L$ and $A'\in\mathcal{G}_{j}$ we have $E\cap A'\in\mathcal{Q}(L)$: if $E=\Omega$, then $E\cap A'=A'$; otherwise $E=\bigcap_{l=1}^{p}A_{l}$ as above, and if $j=j_{l}$ for some (necessarily unique) $l$, replacing $A_{l}$ by $A_{l}\cap A'\in\mathcal{G}_{j}$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix)) writes $E\cap A'$ in the required form, while if $j$ is none of the $j_{l}$, appending $j$ and $A'$ to the list does. For $E'=\bigcap_{l=1}^{p'}A'_{l}\in\mathcal{Q}(L)$, induction on $p'$ with $E\cap\bigcap_{l=1}^{p'+1}A'_{l}=\bigl(E\cap\bigcap_{l=1}^{p'}A'_{l}\bigr)\cap A'_{p'+1}$ then gives $E\cap E'\in\mathcal{Q}(L)$, and $E\cap\Omega=E$. Taking $p=1$ gives $\bigcup_{j\in L}\mathcal{G}_{j}\subseteq\mathcal{Q}(L)$; and $\mathcal{Q}(L)\subseteq\sigma\bigl(\bigcup_{j\in L}\mathcal{G}_{j}\bigr)$, a $\sigma$-algebra containing every $\mathcal{G}_{j}$ with $j\in L$ and closed under intersections of two sets (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii), Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix)), by induction on $p$. The equality follows by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (v) and Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iv); in particular $\sigma(\mathcal{Q}(J_{k}))=\mathcal{H}_{k}$ for $k\in K$.

$(\mathcal{Q}(J_{k}))_{k\in K}$ is independent. Let $n\in\mathbb{N}$, let $k_{1},\dots,k_{n}\in K$ be distinct and let $H_{s}\in\mathcal{Q}(J_{k_{s}})$ for $s\in[n]$. For $r\in[n]$ let $L_{r}=\bigcup_{s\in[r]}J_{k_{s}}$. We show by induction on $r$ that

$$\bigcap_{s=1}^{r}H_{s}\in\mathcal{Q}(L_{r})\qquad\text{and}\qquad P\Bigl(\bigcap_{s=1}^{r}H_{s}\Bigr)=\prod_{s=1}^{r}P(H_{s}).$$

For $r=1$ this is clear. If it holds for some $r<n$, then $J_{k_{r+1}}$ is disjoint from $L_{r}$, as the sets $J_{k}$ are pairwise disjoint and $k_{r+1}$ differs from $k_{1},\dots,k_{r}$, so (1) with $E=\bigcap_{s=1}^{r}H_{s}$ and $E'=H_{r+1}$ gives the claim for $r+1$. The case $r=n$ is the product formula.

By (ii), $(\sigma(\mathcal{Q}(J_{k})))_{k\in K}=(\mathcal{H}_{k})_{k\in K}$ is independent.

Proof of (iv). We first show, for subsets $\mathcal{C}$ and $\mathcal{D}$ of $\mathcal{F}$, that the family $(\mathcal{C}_{k})_{k\in[2]}$ with $\mathcal{C}_{1}=\mathcal{C}$ and $\mathcal{C}_{2}=\mathcal{D}$ is independent if and only if

$$P(C\cap D)=P(C)\,P(D)\qquad(C\in\mathcal{C},\ D\in\mathcal{D}).\tag{2}$$

If the family is independent, its product formula for the list $(1,2)$ is (2). Conversely, assume (2). For the lists $(1)$ and $(2)$ both sides of the product formula are $P(A_{1})$; for $(1,2)$ the product formula is (2) with $C=A_{1}$ and $D=A_{2}$, and for $(2,1)$ it is (2) with $C=A_{2}$ and $D=A_{1}$, as $A_{1}\cap A_{2}=A_{2}\cap A_{1}$.

For $\sigma$-algebras $\mathcal{G}$ and $\mathcal{H}$, (2) is the defining identity of the independence of $\mathcal{G}$ and $\mathcal{H}$ (Independence of Classes of Events and of σ-Algebras (iii)), which proves the equivalence.

Proof of (v). By the case of subsets of $\mathcal{F}$ in the proof of the equivalence ((iv)), the family $(\mathcal{C}_{k})_{k\in[2]}$ with $\mathcal{C}_{1}=\mathcal{C}$ and $\mathcal{C}_{2}=\mathcal{D}$ is independent. By (ii) the family $(\sigma(\mathcal{C}_{k}))_{k\in[2]}$ is independent, so $\sigma(\mathcal{C})$ and $\sigma(\mathcal{D})$ are independent by the equivalence ((iv)).