155 L10 : Dynkin’s π-λ Theorem

Video 155.1: Lesson 10: Dynkin’s \pi-\lambda theorem and its use in proving independence of generated \sigma-fields.

In this lesson we will present and then prove Dynkin’s \pi-\lambda theorem, which gives a powerful criterion for showing that a \lambda-system contains the \sigma-field generated by a \pi-system. We then apply this result to show that independent \pi-systems generate independent \sigma-fields.

155.1 Lesson map

Recap: independence of events and independence of classes of events
Definition of independent classes using \Omega to encode lower-order intersections.
Counterexample: independent classes need not generate independent \sigma-fields.
Definition of a \pi-system.
Definition of a \lambda-system.
Equivalent proper-difference condition for \lambda-systems.
A class that is both a \pi-system and a \lambda-system is a \sigma-field.
Intersections of \lambda-systems.
The auxiliary class \mathcal{L}_A=\{B:A\cap B\in\mathcal{L}\}.
Dynkin’s \pi-\lambda theorem.
Independent \pi-systems generate independent \sigma-fields.
Extension to infinite families by checking finite subfamilies.

155.2 Recap: independence for classes of events

Definition 155.1 (Independence for classes of events) Let (\Omega,\mathcal{F},P), be a probability space, with collections (classes) of sets (events) \mathcal{A}_1,\mathcal{A}_2,\ldots,\mathcal{A}_n \subseteq \mathcal{F}

The classes are independent if any choice of events A_i\in\mathcal{A}_i gives independent events A_1,A_2,\ldots,A_n.

For a pair of events, independence means P(A\cap B)=P(A)P(B). For a finite family of classes, the clean version of the definition is:

Definition 155.2 (Alternative definition of independence for classes of events) Let (\Omega,\mathcal{F},P), be a probability space, with collections (classes) of sets (events) from \mathcal{A}_1,\mathcal{A}_2,\ldots,\mathcal{A}_n \subseteq \mathcal{F} are independent if

P(A_1\cap A_2\cap\cdots\cap A_n) = \prod_{i=1}^n P(A_i), \forall A_i \in \mathcal{A}_i. \text{ or } A_i=\Omega.

where each A_i is either chosen from \mathcal{A}_i or is allowed to be \Omega.

Allowing A_i=\Omega folds all lower-order intersections into the same n-fold formula.

For example, with three classes, the pairwise condition for A_1 and A_3 appears by setting A_2=\Omega:

P(A_1\cap \Omega\cap A_3) = P(A_1)P(\Omega)P(A_3) = P(A_1)P(A_3).

155.3 Todays motivating question: Does independence survive generating \sigma-fields?

Our main question today is does independence survive the operation of generating \sigma-fields:

\mathcal{A}_1,\mathcal{A}_2 \text{ independent} \quad \stackrel{???}{\Longrightarrow}\quad \sigma(\mathcal{A}_1),\sigma(\mathcal{A}_2) \text{ independent}.

Hint: The answer is not in general.

It becomes true after adding the key structural assumption: each starting class should be a \pi-system.

155.4 Why the naive theorem fails

events, classes and \sigma-fields

Keeping track of the different levels of structure: events, classes of events, and \sigma-fields is confusing.

The main point jem reiterates is that the sigma-field generation operates on the classes, not on the events!

The key is to remember that a \sigma-field is a special kind of class of events, so we can talk about independence for classes and then ask if it survives when we generate \sigma-fields from those classes.

Since the notation isn’t stellar the example Example 155.1 helps to sort the three out

Example 155.1 (Counterexample: independent coin flips classes yield dependent \sigma-fields) The space for two fair coin coins: \Omega=\{HH,HT,TH,TT\},\quad \mathbb{P}(\{\omega\})=\frac14. Let define events A=\{HH,HT\}, B=\{HH,TT\}, C=\{HT,TT\} and classes \mathcal{A}_1=\{A,B\}, \quad \mathcal{A}_2=\{C\}

The classes \mathcal{A}_1 and \mathcal{A}_2 are independent. e.g. A\cap C = \{HT\}, so P(A\cap C)=\frac14 = \frac12\cdot\frac12 =P(A)P(C).

The original classes pass the independence check.

Note that C = (A\cup B)\setminus(A\cap B) = (A\cup B)\cap (A\cap B)^c \implies C \in \sigma(\mathcal{A}_1). Also C\in\mathcal{A}_2 \implies C\in\sigma(\mathcal{A}_2). The generated \sigma-fields overlap too much: they both contain C.

C isn’t independent of itself! as \mathbb{P}(C\cap C)=\mathbb{P}(C)\neq\mathbb{P}(C)^2.

\sigma(\mathcal{A}_1) \text{ and } \sigma(\mathcal{A}_2) are not independent. \blacksquare

The counterexample works because generating a \sigma-field can create a shared nontrivial event.

155.5 \pi-systems and \lambda-systems

Definition 155.3 (\pi-system) Is a class of sets closed under finite intersections:

A,B\in\mathcal{P} \quad \Longrightarrow \quad A\cap B\in\mathcal{P}.

Every field and every \sigma-field is a \pi-system. However a \pi-system

need not contain \Omega,
need not be closed under complements, and
need not be closed under unions.

This is exactly why it is a weak but useful hypothesis. Billingsley introduces \pi-systems in the uniqueness part of the extension theorem (Billingsley 2012, 43).

A \pi-system remembers only intersection structure.

155.6 Defining \lambda-systems

Definition 155.4 (\lambda-system) is a class \mathcal{L} of subsets of \Omega satisfying:

\Omega\in\mathcal{L}
if A\in\mathcal{L}, then A^c\in\mathcal{L}
if A_1,A_2,\ldots \in \mathcal{L} are pairwise disjoint, then \bigcup_{n=1}^{\infty}A_n\in\mathcal{L}

A \sigma-field is a \lambda-system, but a \lambda-system is weaker because countable unions are required only for disjoint sequences (Billingsley 2012, 43).

155.6.1 Alternative difference axiom for \lambda-systems

Definition 155.5 (\lambda-system) is a class \mathcal{L} of subsets of \Omega satisfying:

\Omega\in\mathcal{L}
if A,B\in\mathcal{L},\ A\subseteq B \quad\Longrightarrow\quad B\setminus A\in\mathcal{L}.
if A_1,A_2,\ldots \in \mathcal{L} are pairwise disjoint, then \bigcup_{n=1}^{\infty}A_n\in\mathcal{L}

Theorem 155.1 (equivalence of \lambda-system definitions) The two definitions of \lambda-systems Definition 155.4 and Definition 155.5 are equivalent.

The second axioms in the definitions are equivalent by themselves. However they are equivalent in the presence of the other \lambda-system axioms.

Proof.

Definition 155.5 (1.) and (2.) \Longrightarrow Definition 155.4 (2.) since we can take B=\Omega in the proper-difference condition to get A^c\in\mathcal{L}.
Definition 155.4 (2.) and (3.) \Longrightarrow Definition 155.5 (2.) because if A\subseteq B, then A is disjoint from B\setminus A, so B\setminus A\in\mathcal{L} by the \lambda-system property.

Conversely, suppose \mathcal{L} is closed under complements and disjoint countable unions. If A,B\in\mathcal{L} and A\subseteq B, then B^c\in\mathcal{L} and A is disjoint from B^c. Therefore A\cup B^c\in\mathcal{L}.

Taking complements, (A\cup B^c)^c = A^c\cap B = B\setminus A \in\mathcal{L}.

Proper differences are often the more convenient closure rule. e.g. using proper differences to avoid reproving complement closure in every auxiliary \lambda-system.

155.6.2 A class that is both a \pi-system and a \lambda-system is a \sigma-field

Theorem 155.2 (A class that is both a \pi-system and a \lambda-system is a \sigma-field) If \mathcal{A} is both a \pi-system and a \lambda-system. Then \mathcal{A} is a \sigma-field c.f. (Billingsley 2012, 43).

The \pi-system supplies the finite intersections; the \lambda-system supplies the disjoint countable union.

Proof. The first two \sigma-field properties are immediate:

\Omega\in\mathcal{A} because \mathcal{A} is a \lambda-system, and

A\in\mathcal{A} \quad\Longrightarrow\quad A^c\in\mathcal{A} again by the \lambda-system property.

We will be disjointizing a countable union to prove \sigma-field closure.

Now take any sequence A_1,A_2,\ldots\in\mathcal{A}. These sets need not be disjoint, so define disjoint pieces:

B_1=A_1 \quad B_2=A_2\cap A_1^c \quad B_3=A_3\cap A_2^c\cap A_1^c,

and in general B_k = A_k\cap\bigcap_{j<k}A_j^c.

The complements A_j^c are in \mathcal{A} because \mathcal{A} is a \lambda-system. The finite intersections are in \mathcal{A} because \mathcal{A} is a \pi-system. Thus each B_k\in\mathcal{A}. The B_k are disjoint, so the \lambda-system property gives \bigcup_{k=1}^{\infty}B_k\in\mathcal{A}.

But \bigcup_{k=1}^{\infty}B_k = \bigcup_{k=1}^{\infty}A_k.

Therefore \mathcal{A} is closed under countable unions, and hence is a \sigma-field.

Turn arbitrary countable unions into disjoint unions by peeling off the earlier sets.

155.7 Two auxiliary \lambda-system lemmas

155.7.1 Intersections of \lambda-systems

Theorem 155.3 (Intersections of \lambda-systems) If \mathcal{L}_1 and \mathcal{L}_2 are \lambda-systems on the same underlying set \Omega, then \mathcal{L}_1\cap\mathcal{L}_2 is also a \lambda-system.

Here the intersection is an intersection of classes of sets. A set belongs to \mathcal{L}_1\cap\mathcal{L}_2 when the entire set belongs to both classes. We are not taking a set from \mathcal{L}_1 and intersecting it with a set from \mathcal{L}_2.

Proof. The proof mirrors the analogous fact for \sigma-fields:

\Omega lies in both \mathcal{L}_1 and \mathcal{L}_2.
If A lies in both, then A^c lies in both.
If A_1,A_2,\ldots are disjoint and each lies in both, then \bigcup_n A_n lies in both.

155.7.2 The smallest \lambda-system containing a class

This also justifies defining the smallest \lambda-system containing a class as the intersection of all \lambda-systems containing it, exactly as generated \sigma-fields are defined by intersections of \sigma-fields (Billingsley 2012, 21).

Intersections let us speak of the smallest \lambda-system containing a given class.

Theorem 155.4 (The section lemma: \mathcal{L}_A) Let \mathcal{L} be a \lambda-system and fix A\subseteq\Omega. Define \mathcal{L}_A = \{B\subseteq\Omega: A\cap B\in\mathcal{L}\}

If A\in\mathcal{L}, then \mathcal{L}_A is a \lambda-system.

Blackboard slide defining L_A as the class of subsets B of Omega such that A intersection B belongs to L. It states that if A belongs to the lambda-system L, then L_A is itself a lambda-system. The first check is Omega, since A intersection Omega equals A. — Definition of the auxiliary class L sub A.

Blackboard slide proving that L_A is closed under complements. From B in L_A, it gets A intersection B in L. Since A intersection B is contained in A, the proper difference A minus A intersection B is in L. The algebra simplifies this set to A intersection B complement, so B complement is in L_A. — Complement closure for the auxiliary class L sub A.

Blackboard slide proving that L_A is closed under countable disjoint unions. If B_1, B_2, and so on are disjoint sets in L_A, then A intersection B_i are disjoint sets in L. Their union equals A intersection the union of the B_i, so the union of the B_i belongs to L_A. — Countable disjoint union closure for L sub A.

Proof. First, A\cap\Omega=A\in\mathcal{L}, so \Omega\in\mathcal{L}_A.

For complements, suppose B\in\mathcal{L}_A. Then

A\cap B\in\mathcal{L}, \qquad A\cap B\subseteq A, \qquad A\in\mathcal{L}.

By the proper-difference version of the \lambda-system axiom,

\begin{aligned} \mathcal{L} &\ni A\setminus(A\cap B) \\ &= A\cap(A\cap B)^c \\ &= A\cap(A^c\cup B^c) \\ &= A\cap B^c \\ &\implies B^c\in\mathcal{L}_A \end{aligned}

Finally, let B_1,B_2,\ldots be disjoint sets in \mathcal{L}_A. Then A\cap B_1,A\cap B_2,\ldots are disjoint sets in \mathcal{L}. Since \mathcal{L} is a \lambda-system, \bigcup_{n=1}^{\infty}(A\cap B_n) \in\mathcal{L}.

Distributing intersection over union gives

\bigcup_{n=1}^{\infty}(A\cap B_n) = A\cap\left(\bigcup_{n=1}^{\infty}B_n\right).

So \bigcup_{n=1}^{\infty}B_n \in\mathcal{L}_A

155.8 Dynkin’s \pi-\lambda theorem

Theorem 155.5 (Dynkin’s \pi-\lambda theorem) Let \mathcal{P} be a \pi-system and let \mathcal{L} be a \lambda-system. If

\mathcal{P}\subseteq\mathcal{L},

then

\sigma(\mathcal{P})\subseteq\mathcal{L}.

Billingsley states this as Theorem 3.2 and uses it immediately for uniqueness of probability measures on generated \sigma-fields (Billingsley 2012, 43–45).

Dynkin’s theorem is a bootstrap: prove a \lambda-system contains a \pi-system, then get the generated \sigma-field for free.

Slide with a portrait of Eugene Dynkin and the statement of Dynkin's pi-lambda theorem: if P is a pi-system, L is a lambda-system, and P is contained in L, then sigma of P is contained in L. — Dynkin’s pi-lambda theorem.

155.8.1 Proof idea

Let \mathcal{L}_0 be the smallest \lambda-system containing \mathcal{P}. Since \mathcal{L} is a \lambda-system containing \mathcal{P},

\mathcal{P} \subseteq \mathcal{L}_0 \subseteq \mathcal{L}.

If we can prove that \mathcal{L}_0 is also a \pi-system, then \mathcal{L}_0 is both a \pi-system and a \lambda-system, hence a \sigma-field. Then minimality of \sigma(\mathcal{P}) gives

\sigma(\mathcal{P}) \subseteq \mathcal{L}_0 \subseteq \mathcal{L}.

So the whole proof reduces to showing:

A,B\in\mathcal{L}_0 \quad\Longrightarrow\quad A\cap B\in\mathcal{L}_0.

Blackboard slide defining L_0 as the smallest lambda-system containing P. It shows P contained in L_0 contained in L. The slide explains that if L_0 is a sigma-field, then sigma of P is contained in L_0 and hence in L. — Minimal lambda-system containing the pi-system.

Blackboard slide saying that since L_0 is already a lambda-system, it is enough to show L_0 is a pi-system. The target is: if A and B are in L_0, then A intersection B is in L_0. — Reducing the proof to showing L_0 is a pi-system.

155.8.2 Claim 1: intersections with sets in \mathcal{P} preserve \mathcal{P}

Take C\in\mathcal{P}. Since \mathcal{P}\subseteq\mathcal{L}_0, we also have C\in\mathcal{L}_0. Define

\mathcal{L}_C = \{D\subseteq\Omega:C\cap D\in\mathcal{L}_0\}.

By the section lemma, \mathcal{L}_C is a \lambda-system.

If D\in\mathcal{P}, then because \mathcal{P} is a \pi-system,

C\cap D\in\mathcal{P} \subseteq \mathcal{L}_0.

Therefore D\in\mathcal{L}_C, and hence

\mathcal{P}\subseteq\mathcal{L}_C.

Blackboard slide for Claim 1. For C in P, it defines L_C as all D such that C intersection D is in L_0. Since C is in L_0, L_C is a lambda-system. If D is in P, then C intersection D is in P because P is a pi-system, so D belongs to L_C. Thus P is contained in L_C. — Claim 1 in the proof of Dynkin’s theorem.

155.8.3 Claim 2: one set from \mathcal{P} and one from \mathcal{L}_0

For C\in\mathcal{P}, Claim 1 gives \mathcal{P}\subseteq\mathcal{L}_C. Since \mathcal{L}_C is a \lambda-system containing \mathcal{P}, minimality of \mathcal{L}_0 gives

\mathcal{L}_0\subseteq\mathcal{L}_C.

Thus if D\in\mathcal{L}_0, then D\in\mathcal{L}_C, which means

C\cap D\in\mathcal{L}_0.

C\in\mathcal{P},\ D\in\mathcal{L}_0 \quad\Longrightarrow\quad C\cap D\in\mathcal{L}_0.

Blackboard slide for Claim 2. If C is in P and D is in L_0, then C intersection D is in L_0. The proof uses that L_C is a lambda-system containing P, so by minimality L_0 is contained in L_C. — Claim 2 in the proof of Dynkin’s theorem.

155.8.4 Claim 3: reverse the roles

Now take C\in\mathcal{L}_0. We want to show

\mathcal{P}\subseteq\mathcal{L}_C.

Let D\in\mathcal{P}. By Claim 2, with the \mathcal{P}-set D and the \mathcal{L}_0-set C,

D\cap C \in \mathcal{L}_0.

Since D\cap C=C\cap D, this means D\in\mathcal{L}_C. Hence

\mathcal{P}\subseteq\mathcal{L}_C.

Blackboard slide for Claim 3. If C is in L_0, then P is contained in L_C. For D in P, Claim 2 gives C intersection D in L_0, so D is in L_C. — Claim 3 in the proof of Dynkin’s theorem.

155.8.5 Claim 4: two arbitrary sets from \mathcal{L}_0

Finally take A\in\mathcal{L}_0. By the section lemma, \mathcal{L}_A is a \lambda-system. Claim 3 gives

\mathcal{P}\subseteq\mathcal{L}_A.

Therefore, by minimality,

\mathcal{L}_0\subseteq\mathcal{L}_A.

If B\in\mathcal{L}_0, then B\in\mathcal{L}_A, so

A\cap B\in\mathcal{L}_0.

Thus \mathcal{L}_0 is a \pi-system. Since it is already a \lambda-system, it is a \sigma-field. This proves Dynkin’s \pi-\lambda theorem.

Blackboard slide for Claim 4. If A and B are in L_0, then A intersection B is in L_0. The proof defines L_A, uses Claim 3 to get P contained in L_A, uses minimality to get L_0 contained in L_A, and concludes B in L_A. — Claim 4 completes the proof of Dynkin’s theorem.

155.9 Independent \pi-systems generate independent \sigma-fields

Now return to the motivation of the lesson.

Theorem 155.6 (Independence theorem for generated \sigma-fields) Let

\mathcal{A}_1,\ldots,\mathcal{A}_n

be independent \pi-systems in a probability space (\Omega,\mathcal{F},P). Then

\sigma(\mathcal{A}_1),\ldots,\sigma(\mathcal{A}_n)

are independent.

This is the promised corrected form of the false naive theorem. The extra \pi-system assumption prevents the counterexample pathology.

Define

\mathcal{B}_i = \mathcal{A}_i\cup\{\Omega\}.

Adding \Omega does not break the \pi-system property, because intersecting with \Omega changes nothing. Also,

\sigma(\mathcal{B}_i)=\sigma(\mathcal{A}_i),

since \Omega belongs to every \sigma-field anyway.

Blackboard slide stating that if A_1 through A_n are independent pi-systems, then the generated sigma-fields sigma of A_1 through sigma of A_n are mutually independent. It begins the proof by defining B_i as A_i union {Omega}, noting that B_i is still a pi-system and generates the same sigma-field as A_i. — Independent pi-systems generate independent sigma-fields.

The independence of the original classes is equivalent to saying that for all B_i\in\mathcal{B}_i,

P(B_1\cap\cdots\cap B_n) = \prod_{i=1}^nP(B_i).

Fix arbitrary sets

B_2\in\mathcal{B}_2, \ldots, B_n\in\mathcal{B}_n.

Define

\mathcal{L} = \left\{ A\in\mathcal{F}: P(A\cap B_2\cap\cdots\cap B_n) = P(A)\prod_{i=2}^nP(B_i) \right\}.

The goal is to show that \mathcal{L} is a \lambda-system containing \mathcal{B}_1. Then Dynkin’s theorem gives

\sigma(\mathcal{B}_1) \subseteq \mathcal{L},

which is the first step in replacing \mathcal{B}_1 by \sigma(\mathcal{A}_1).

To put a \sigma on one coordinate, freeze every other coordinate and make the unfrozen coordinate a \lambda-system.

Blackboard slide recalling the independence formula for the B_i classes and then fixing B_2 through B_n. It defines L as the class of A in F such that P(A intersection B_2 intersection through B_n) factors as P(A) times the probabilities of B_2 through B_n. — Constructing a lambda-system to upgrade one independent pi-system.

155.9.1 Showing \mathcal{L} is a \lambda-system

First, \Omega\in\mathcal{L} because

P(\Omega\cap B_2\cap\cdots\cap B_n) = P(B_2\cap\cdots\cap B_n) = \prod_{i=2}^nP(B_i) = P(\Omega)\prod_{i=2}^nP(B_i).

Blackboard slide showing that Omega belongs to the constructed class L. Plugging Omega into the definition reduces the left side to the probability of B_2 intersection through B_n, which factors by independence and equals P(Omega) times the product of the B_i probabilities. — Omega belongs to the constructed lambda-system.

For complements, let A\in\mathcal{L} and abbreviate

B=B_2\cap\cdots\cap B_n.

Then

B=(A\cap B)\cup(A^c\cap B),

and the two pieces are disjoint. By finite additivity,

P(B)=P(A\cap B)+P(A^c\cap B).

Because A\in\mathcal{L},

P(A\cap B) = P(A)\prod_{i=2}^nP(B_i).

Also, by independence of B_2,\ldots,B_n,

P(B)=\prod_{i=2}^nP(B_i).

Therefore

P(A^c\cap B) = \left(1-P(A)\right)\prod_{i=2}^nP(B_i) = P(A^c)\prod_{i=2}^nP(B_i).

Thus A^c\in\mathcal{L}.

Blackboard slide proving complement closure for the constructed class L. It writes B as B_2 intersection through B_n, decomposes B into A intersection B and A complement intersection B, uses finite additivity, and solves for the probability of A complement intersection B to show it factors correctly. — Complement closure for the constructed lambda-system.

For countable disjoint unions, let A_1,A_2,\ldots\in\mathcal{L} be disjoint. Then the sets

A_m\cap B_2\cap\cdots\cap B_n

are also disjoint. Countable additivity gives

\begin{aligned} P\left(\left(\bigcup_{m=1}^{\infty}A_m\right)\cap B_2\cap\cdots\cap B_n\right) &= P\left(\bigcup_{m=1}^{\infty}(A_m\cap B_2\cap\cdots\cap B_n)\right)\\ &= \sum_{m=1}^{\infty}P(A_m\cap B_2\cap\cdots\cap B_n)\\ &= \sum_{m=1}^{\infty}P(A_m)\prod_{i=2}^nP(B_i)\\ &= \left(\sum_{m=1}^{\infty}P(A_m)\right)\prod_{i=2}^nP(B_i)\\ &= P\left(\bigcup_{m=1}^{\infty}A_m\right)\prod_{i=2}^nP(B_i). \end{aligned}

\bigcup_{m=1}^{\infty}A_m\in\mathcal{L}.

Blackboard slide proving countable disjoint union closure for L. It expands the probability of the union of A_m intersected with B_2 through B_n, uses countable additivity, uses the defining factorization for each A_m, factors out the constant product of probabilities of the B_i, and then recombines the sum as the probability of the union of the A_m. — Countable disjoint union closure for the constructed lambda-system.

155.9.2 Apply Dynkin’s theorem once

Because the original classes are independent, every B_1\in\mathcal{B}_1 satisfies the defining factorization for \mathcal{L}. Hence

\mathcal{B}_1\subseteq\mathcal{L}.

Since \mathcal{B}_1 is a \pi-system and \mathcal{L} is a \lambda-system, Dynkin’s theorem gives

\sigma(\mathcal{B}_1) \subseteq\mathcal{L}.

But \sigma(\mathcal{B}_1)=\sigma(\mathcal{A}_1), so for any

A_1\in\sigma(\mathcal{A}_1),

and any fixed B_2\in\mathcal{B}_2,\ldots,B_n\in\mathcal{B}_n,

P(A_1\cap B_2\cap\cdots\cap B_n) = P(A_1)\prod_{i=2}^nP(B_i).

Thus

\sigma(\mathcal{A}_1),\mathcal{B}_2,\ldots,\mathcal{B}_n

are independent.

Blackboard slide concluding the first upgrade step. Since L is a lambda-system and B_1 is a pi-system contained in L, Dynkin's theorem gives sigma of B_1 contained in L. Since sigma of B_1 equals sigma of A_1, any A_1 from sigma of A_1 satisfies the desired factorization with B_2 through B_n. — Dynkin theorem upgrades the first pi-system to its generated sigma-field.

155.9.3 Iterate the argument

Now fix

A_1\in\sigma(\mathcal{A}_1), \qquad B_3\in\mathcal{B}_3, \ldots, B_n\in\mathcal{B}_n,

and define a new class

\mathcal{L} = \left\{ A\in\mathcal{F}: P(A_1\cap A\cap B_3\cap\cdots\cap B_n) = P(A_1)P(A)\prod_{i=3}^nP(B_i) \right\}.

The same proof shows that this new \mathcal{L} is a \lambda-system containing \mathcal{B}_2. Dynkin’s theorem gives

\sigma(\mathcal{B}_2) \subseteq \mathcal{L},

so \mathcal{B}_2 can be upgraded to \sigma(\mathcal{A}_2).

Repeating the argument for \mathcal{A}_3,\ldots,\mathcal{A}_n yields

\sigma(\mathcal{A}_1), \sigma(\mathcal{A}_2), \ldots, \sigma(\mathcal{A}_n)

mutually independent.

Blackboard slide explaining the iteration step. After upgrading A_1 to sigma of A_1, fix A_1 from that generated sigma-field and fix B_3 through B_n. Define a new lambda-system in the second coordinate and apply Dynkin's theorem to upgrade B_2 to sigma of A_2. Continue until all classes have been upgraded. — Iterating the pi-lambda upgrade over the remaining classes.

For an infinite collection of independent \pi-systems, the result follows by applying the finite theorem to every finite subcollection. This works because independence of an infinite family is defined by independence of all finite subfamilies.

The infinite-family version follows because independence is checked on finite subfamilies.

155.10 Takeaway

Dynkin’s \pi-\lambda theorem is a minimality machine. It says that once a \lambda-system contains a \pi-system, it must contain everything generated from that \pi-system as a \sigma-field.

The lesson uses that machine twice:

Abstractly, to prove the theorem itself by showing the minimal \lambda-system over a \pi-system is also a \pi-system.
Probabilistically, to prove that independent \pi-systems remain independent after generating \sigma-fields.

This is also why \lambda-systems are tuned to probability: their closure condition is countable additivity over disjoint unions, which is the central operation for probability measures (Billingsley 2012, 45).

--- title: "L10 : Dynkin's π-λ Theorem" subtitle: "Measure Theoretic Probability - Jem Corcoran" date: 2026-04-17 keywords: [measure theory, probability, sigma algebra, outer measure, lemmas] --- ::: {#vid-C99-l10 .column-margin} {{< video https://youtu.be/d83SdJvFt1M >}} Lesson 10: Dynkin's $\pi$-$\lambda$ theorem and its use in proving independence of generated $\sigma$-fields. ::: In this lesson we will present and then prove [Dynkin's $\pi$-$\lambda$ theorem](#sec-dynkin-theorem), which gives a powerful criterion for showing that a $\lambda$-system contains the $\sigma$-field generated by a $\pi$-system. We then apply this result to show that independent $\pi$-systems generate independent $\sigma$-fields. ## Lesson map {#sec-lesson-map} - [Recap: independence of events and independence of classes of events](#sec-independent-classes) - [Definition of independent classes using $\Omega$ to encode lower-order intersections](#sec-independent-classes). - [Counterexample: independent classes need not generate independent $\sigma$-fields](#sec-independent-classes). - [Definition of a $\pi$-system](#sec-pi-systems). - [Definition of a $\lambda$-system](#sec-lambda-systems). - [Equivalent proper-difference condition for $\lambda$-systems](#sec-lambda-systems). - [A class that is both a $\pi$-system and a $\lambda$-system is a $\sigma$-field](#sec-pi-lambda-systems). - [Intersections of $\lambda$-systems](#sec-lambda-systems). - [The auxiliary class $\mathcal{L}_A=\{B:A\cap B\in\mathcal{L}\}$](#sec-lambda-systems). - [Dynkin's $\pi$-$\lambda$ theorem](#sec-dynkin-theorem). - [Independent $\pi$-systems generate independent $\sigma$-fields](#sec-independent-pi-systems). - [Extension to infinite families by checking finite subfamilies](#sec-infinite-families). ## Recap: independence for classes of events {#sec-independent-classes} ::: {#def-independent-classes} ## Independence for classes of events Let $(\Omega,\mathcal{F},P),$ be a probability space, with collections (classes) of sets (events) $\mathcal{A}_1,\mathcal{A}_2,\ldots,\mathcal{A}_n \subseteq \mathcal{F}$ The classes are independent if any choice of events $A_i\in\mathcal{A}_i$ gives independent events $A_1,A_2,\ldots,A_n$. ::: For a pair of events, **independence** means $P(A\cap B)=P(A)P(B).$ For a finite family of classes, the clean version of the definition is: ::: {#def-independent-classes-alt} ## Alternative definition of independence for classes of events Let $(\Omega,\mathcal{F},P),$ be a probability space, with collections (classes) of sets (events) from $\mathcal{A}_1,\mathcal{A}_2,\ldots,\mathcal{A}_n \subseteq \mathcal{F}$ are independent if $$ P(A_1\cap A_2\cap\cdots\cap A_n) = \prod_{i=1}^n P(A_i), \forall A_i \in \mathcal{A}_i. \text{ or } A_i=\Omega. $$ ::: where each $A_i$ is either chosen from $\mathcal{A}_i$ or is allowed to be $\Omega$. [Allowing $A_i=\Omega$ folds all lower-order intersections into the same $n$-fold formula.]{.mark} For example, with three classes, the pairwise condition for $A_1$ and $A_3$ appears by setting $A_2=\Omega$: $$ P(A_1\cap \Omega\cap A_3) = P(A_1)P(\Omega)P(A_3) = P(A_1)P(A_3). $$ ## Todays motivating question: Does independence survive generating $\sigma$-fields? {#sec-motivating-question} Our main question today is [does independence survive the operation of generating $\sigma$-fields:]{.mark} $$ \mathcal{A}_1,\mathcal{A}_2 \text{ independent} \quad \stackrel{???}{\Longrightarrow}\quad \sigma(\mathcal{A}_1),\sigma(\mathcal{A}_2) \text{ independent}. $$ Hint: The answer is **not in general**. It becomes true after adding the key structural assumption: [each starting class should be a $\pi$-system.]{.mark} ## Why the naive theorem fails {#sec-naive-failure} ::: {.callout-warning} ## events, classes and $\sigma$-fields Keeping track of the different levels of structure: events, classes of events, and $\sigma$-fields is confusing. The main point jem reiterates is that the sigma-field generation operates on the classes, not on the events! The key is to remember that a $\sigma$-field is a special kind of class of events, so we can talk about independence for classes and then ask if it survives when we generate $\sigma$-fields from those classes. Since the notation isn't stellar the example @exm-counterexample-independent-classes helps to sort the three out ::: ::: {#exm-counterexample-independent-classes} ## Counterexample: independent coin flips classes yield dependent $\sigma$-fields The space for two fair coin coins: $\Omega=\{HH,HT,TH,TT\},\quad \mathbb{P}(\{\omega\})=\frac14.$ Let define events $$A=\{HH,HT\}, B=\{HH,TT\}, C=\{HT,TT\} $$ and classes $$\mathcal{A}_1=\{A,B\}, \quad \mathcal{A}_2=\{C\} $$ The classes $\mathcal{A}_1$ and $\mathcal{A}_2$ are independent. e.g. $A\cap C = \{HT\}$, so $P(A\cap C)=\frac14 = \frac12\cdot\frac12 =P(A)P(C).$ [The original classes pass the independence check.]{.mark} Note that $C = (A\cup B)\setminus(A\cap B) = (A\cup B)\cap (A\cap B)^c \implies C \in \sigma(\mathcal{A}_1)$. Also $C\in\mathcal{A}_2 \implies C\in\sigma(\mathcal{A}_2).$ [The generated $\sigma$-fields overlap too much: they both contain $C$.]{.mark} $C$ isn't independent of itself! as $\mathbb{P}(C\cap C)=\mathbb{P}(C)\neq\mathbb{P}(C)^2.$ $\sigma(\mathcal{A}_1) \text{ and } \sigma(\mathcal{A}_2)$ are not independent. $\blacksquare$ ::: [The counterexample works because generating a $\sigma$-field can create a shared nontrivial event.]{.mark} ## $\pi$-systems and $\lambda$-systems {#sec-pi-lambda-systems} ::: {#def-pi-system} ## $\pi$-system Is a class of sets closed under finite intersections: $$ A,B\in\mathcal{P} \quad \Longrightarrow \quad A\cap B\in\mathcal{P}. $$ ::: Every field and every $\sigma$-field is a $\pi$-system. However a $\pi$-system - need not contain $\Omega$, - need not be closed under complements, and - need not be closed under unions. This is exactly why it is a *weak but useful hypothesis*. Billingsley introduces $\pi$-systems in the uniqueness part of the extension theorem [@billingsley2012probability p.43]. [A $\pi$-system remembers only intersection structure.]{.mark} ## Defining $\lambda$-systems {#sec-lambda-systems} ::: {#def-lambda-system} ## $\lambda$-system is a class $\mathcal{L}$ of subsets of $\Omega$ satisfying: 1. $\Omega\in\mathcal{L}$ 2. if $A\in\mathcal{L}$, then $A^c\in\mathcal{L}$ 3. if $A_1,A_2,\ldots \in \mathcal{L}$ are pairwise disjoint, then $$ \bigcup_{n=1}^{\infty}A_n\in\mathcal{L} $$ ::: A $\sigma$-field is a $\lambda$-system, but a $\lambda$-system is weaker because countable unions are required only for **disjoint** sequences [@billingsley2012probability p.43]. ### Alternative difference axiom for $\lambda$-systems ::: {#def-lambda-system-alt} ## $\lambda$-system is a class $\mathcal{L}$ of subsets of $\Omega$ satisfying: 1. $\Omega\in\mathcal{L}$ 2. if $A,B\in\mathcal{L},\ A\subseteq B \quad\Longrightarrow\quad B\setminus A\in\mathcal{L}.$ 3. if $A_1,A_2,\ldots \in \mathcal{L}$ are pairwise disjoint, then $$ \bigcup_{n=1}^{\infty}A_n\in\mathcal{L} $$ ::: ::: {#thm-equivalent-of-lambda-definitions} ## equivalence of $\lambda$-system definitions The two definitions of $\lambda$-systems @def-lambda-system and @def-lambda-system-alt are equivalent. ::: [The second axioms in the definitions are equivalent by themselves. However they are equivalent in the presence of the other $\lambda$-system axioms.]{.mark} ::: {.proof} 1. @def-lambda-system-alt (1.) and (2.) $\Longrightarrow$ @def-lambda-system (2.) since we can take $B=\Omega$ in the proper-difference condition to get $A^c\in\mathcal{L}$. 2. @def-lambda-system (2.) and (3.) $\Longrightarrow$ @def-lambda-system-alt (2.) because if $A\subseteq B$, then $A$ is disjoint from $B\setminus A$, so $B\setminus A\in\mathcal{L}$ by the $\lambda$-system property. Conversely, suppose $\mathcal{L}$ is closed under complements and disjoint countable unions. If $A,B\in\mathcal{L}$ and $A\subseteq B$, then $B^c\in\mathcal{L}$ and $A$ is disjoint from $B^c$. Therefore $A\cup B^c\in\mathcal{L}.$ Taking complements, $(A\cup B^c)^c = A^c\cap B = B\setminus A \in\mathcal{L}.$ ::: [Proper differences are often the more convenient closure rule. e.g. using proper differences to avoid reproving complement closure in every auxiliary $\lambda$-system.]{.mark} ### A class that is both a $\pi$-system and a $\lambda$-system is a $\sigma$-field ::: {#thm-pi-lambda-systems-are-sigma-fields} ## A class that is both a $\pi$-system and a $\lambda$-system is a $\sigma$-field If $\mathcal{A}$ is both a $\pi$-system and a $\lambda$-system. Then $\mathcal{A}$ is a $\sigma$-field c.f. [@billingsley2012probability p.43]. ::: The $\pi$-system supplies the finite intersections; the $\lambda$-system supplies the disjoint countable union. ::: {.proof} The first two $\sigma$-field properties are immediate: $\Omega\in\mathcal{A}$ because $\mathcal{A}$ is a $\lambda$-system, and $A\in\mathcal{A} \quad\Longrightarrow\quad A^c\in\mathcal{A}$ again by the $\lambda$-system property. [We will be disjointizing a countable union to prove $\sigma$-field closure.]{.mark} Now take any sequence $A_1,A_2,\ldots\in\mathcal{A}$. These sets need not be disjoint, so define disjoint pieces: $B_1=A_1 \quad B_2=A_2\cap A_1^c \quad B_3=A_3\cap A_2^c\cap A_1^c$, and in general $B_k = A_k\cap\bigcap_{j<k}A_j^c.$ The complements $A_j^c$ are in $\mathcal{A}$ because $\mathcal{A}$ is a $\lambda$-system. The finite intersections are in $\mathcal{A}$ because $\mathcal{A}$ is a $\pi$-system. Thus each $B_k\in\mathcal{A}$. The $B_k$ are disjoint, so the $\lambda$-system property gives $\bigcup_{k=1}^{\infty}B_k\in\mathcal{A}.$ But $\bigcup_{k=1}^{\infty}B_k = \bigcup_{k=1}^{\infty}A_k.$ Therefore $\mathcal{A}$ is closed under countable unions, and hence is a $\sigma$-field. ::: [Turn arbitrary countable unions into disjoint unions by peeling off the earlier sets.]{.mark} ## Two auxiliary $\lambda$-system lemmas {#sec-aux-lambda-lemmas} ### Intersections of $\lambda$-systems {#sec-lambda-systems} ::: {#thm-intersection-of-lambda-systems} ### Intersections of $\lambda$-systems If $\mathcal{L}_1$ and $\mathcal{L}_2$ are $\lambda$-systems on the same underlying set $\Omega$, then $\mathcal{L}_1\cap\mathcal{L}_2$ is also a $\lambda$-system. ::: Here the intersection is an intersection of **classes of sets**. A set belongs to $\mathcal{L}_1\cap\mathcal{L}_2$ when the entire set belongs to both classes. We are not taking a set from $\mathcal{L}_1$ and intersecting it with a set from $\mathcal{L}_2$. ::: {.proof} The proof mirrors the analogous fact for $\sigma$-fields: - $\Omega$ lies in both $\mathcal{L}_1$ and $\mathcal{L}_2$. - If $A$ lies in both, then $A^c$ lies in both. - If $A_1,A_2,\ldots$ are disjoint and each lies in both, then $\bigcup_n A_n$ lies in both. ::: ### The smallest $\lambda$-system containing a class {#sec-lambda-systems} This also justifies defining the **smallest $\lambda$-system containing a class** as the intersection of all $\lambda$-systems containing it, exactly as generated $\sigma$-fields are defined by intersections of $\sigma$-fields [@billingsley2012probability p.21]. [Intersections let us speak of the smallest $\lambda$-system containing a given class.]{.mark} ::: {#thm-section-lemma} ## The section lemma: $\mathcal{L}_A$ Let $\mathcal{L}$ be a $\lambda$-system and fix $A\subseteq\Omega$. Define $\mathcal{L}_A = \{B\subseteq\Omega: A\cap B\in\mathcal{L}\}$ If $A\in\mathcal{L}$, then $\mathcal{L}_A$ is a $\lambda$-system. ::: :::{.column-margin #fig-l10-slide-11} ![Definition of the auxiliary class L sub A.](images/l10-s11.png){fig-alt="Blackboard slide defining L_A as the class of subsets B of Omega such that A intersection B belongs to L. It states that if A belongs to the lambda-system L, then L_A is itself a lambda-system. The first check is Omega, since A intersection Omega equals A." group="slides"} The class $\mathcal{L}_A$ collects all sets that keep us inside $\mathcal{L}$ after intersecting with $A$. ::: :::{.column-margin #fig-l10-slide-12} ![Complement closure for the auxiliary class L sub A.](images/l10-s12.png){fig-alt="Blackboard slide proving that L_A is closed under complements. From B in L_A, it gets A intersection B in L. Since A intersection B is contained in A, the proper difference A minus A intersection B is in L. The algebra simplifies this set to A intersection B complement, so B complement is in L_A." group="slides"} The proper-difference trick is exactly why the alternative axiom was useful. ::: :::{.column-margin #fig-l10-slide-13} ![Countable disjoint union closure for L sub A.](images/l10-s13.png){fig-alt="Blackboard slide proving that L_A is closed under countable disjoint unions. If B_1, B_2, and so on are disjoint sets in L_A, then A intersection B_i are disjoint sets in L. Their union equals A intersection the union of the B_i, so the union of the B_i belongs to L_A." group="slides"} Intersecting a disjoint union with $A$ preserves disjointness. ::: ::: {.proof} First, $A\cap\Omega=A\in\mathcal{L},$ so $\Omega\in\mathcal{L}_A.$ For complements, suppose $B\in\mathcal{L}_A$. Then $$ A\cap B\in\mathcal{L}, \qquad A\cap B\subseteq A, \qquad A\in\mathcal{L}. $$ By the proper-difference version of the $\lambda$-system axiom, $$ \begin{aligned} \mathcal{L} &\ni A\setminus(A\cap B) \\ &= A\cap(A\cap B)^c \\ &= A\cap(A^c\cup B^c) \\ &= A\cap B^c \\ &\implies B^c\in\mathcal{L}_A \end{aligned} $$ Finally, let $B_1,B_2,\ldots$ be disjoint sets in $\mathcal{L}_A$. Then $A\cap B_1,A\cap B_2,\ldots$ are disjoint sets in $\mathcal{L}$. Since $\mathcal{L}$ is a $\lambda$-system, $\bigcup_{n=1}^{\infty}(A\cap B_n) \in\mathcal{L}.$ Distributing intersection over union gives $$ \bigcup_{n=1}^{\infty}(A\cap B_n) = A\cap\left(\bigcup_{n=1}^{\infty}B_n\right). $$ So $\bigcup_{n=1}^{\infty}B_n \in\mathcal{L}_A$ ::: ## Dynkin's $\pi$-$\lambda$ theorem {#sec-dynkin-theorem} ::: {#thm-dynkin-pi-lambda} ## Dynkin's $\pi$-$\lambda$ theorem Let $\mathcal{P}$ be a $\pi$-system and let $\mathcal{L}$ be a $\lambda$-system. If $$ \mathcal{P}\subseteq\mathcal{L}, $$ then $$ \sigma(\mathcal{P})\subseteq\mathcal{L}. $$ ::: Billingsley states this as Theorem 3.2 and uses it immediately for uniqueness of probability measures on generated $\sigma$-fields [@billingsley2012probability pp.43-45]. [Dynkin's theorem is a bootstrap: prove a $\lambda$-system contains a $\pi$-system, then get the generated $\sigma$-field for free.]{.mark} :::{.column-margin #fig-l10-slide-14} ![Dynkin's pi-lambda theorem.](images/l10-s14.png){fig-alt="Slide with a portrait of Eugene Dynkin and the statement of Dynkin's pi-lambda theorem: if P is a pi-system, L is a lambda-system, and P is contained in L, then sigma of P is contained in L." group="slides"} This theorem is the main engine of the lesson. ::: ### Proof idea Let $\mathcal{L}_0$ be the smallest $\lambda$-system containing $\mathcal{P}$. Since $\mathcal{L}$ is a $\lambda$-system containing $\mathcal{P}$, $$ \mathcal{P} \subseteq \mathcal{L}_0 \subseteq \mathcal{L}. $$ If we can prove that $\mathcal{L}_0$ is also a $\pi$-system, then $\mathcal{L}_0$ is both a $\pi$-system and a $\lambda$-system, hence a $\sigma$-field. Then minimality of $\sigma(\mathcal{P})$ gives $$ \sigma(\mathcal{P}) \subseteq \mathcal{L}_0 \subseteq \mathcal{L}. $$ So the whole proof reduces to showing: $$ A,B\in\mathcal{L}_0 \quad\Longrightarrow\quad A\cap B\in\mathcal{L}_0. $$ :::{.column-margin #fig-l10-slide-15} ![Minimal lambda-system containing the pi-system.](images/l10-s15.png){fig-alt="Blackboard slide defining L_0 as the smallest lambda-system containing P. It shows P contained in L_0 contained in L. The slide explains that if L_0 is a sigma-field, then sigma of P is contained in L_0 and hence in L." group="slides"} The proof narrows the goal to showing $\mathcal{L}_0$ is closed under finite intersections. ::: :::{.column-margin #fig-l10-slide-16} ![Reducing the proof to showing L_0 is a pi-system.](images/l10-s16.png){fig-alt="Blackboard slide saying that since L_0 is already a lambda-system, it is enough to show L_0 is a pi-system. The target is: if A and B are in L_0, then A intersection B is in L_0." group="slides"} Once $\mathcal{L}_0$ is a $\pi$-system, the earlier theorem turns it into a $\sigma$-field. ::: ### Claim 1: intersections with sets in $\mathcal{P}$ preserve $\mathcal{P}$ Take $C\in\mathcal{P}$. Since $\mathcal{P}\subseteq\mathcal{L}_0$, we also have $C\in\mathcal{L}_0$. Define $$ \mathcal{L}_C = \{D\subseteq\Omega:C\cap D\in\mathcal{L}_0\}. $$ By the section lemma, $\mathcal{L}_C$ is a $\lambda$-system. If $D\in\mathcal{P}$, then because $\mathcal{P}$ is a $\pi$-system, $$ C\cap D\in\mathcal{P} \subseteq \mathcal{L}_0. $$ Therefore $D\in\mathcal{L}_C$, and hence $$ \mathcal{P}\subseteq\mathcal{L}_C. $$ :::{.column-margin #fig-l10-slide-17} ![Claim 1 in the proof of Dynkin's theorem.](images/l10-s17.png){fig-alt="Blackboard slide for Claim 1. For C in P, it defines L_C as all D such that C intersection D is in L_0. Since C is in L_0, L_C is a lambda-system. If D is in P, then C intersection D is in P because P is a pi-system, so D belongs to L_C. Thus P is contained in L_C." group="slides"} The first claim uses only the $\pi$-system closure of $\mathcal{P}$. ::: ### Claim 2: one set from $\mathcal{P}$ and one from $\mathcal{L}_0$ For $C\in\mathcal{P}$, Claim 1 gives $\mathcal{P}\subseteq\mathcal{L}_C$. Since $\mathcal{L}_C$ is a $\lambda$-system containing $\mathcal{P}$, minimality of $\mathcal{L}_0$ gives $$ \mathcal{L}_0\subseteq\mathcal{L}_C. $$ Thus if $D\in\mathcal{L}_0$, then $D\in\mathcal{L}_C$, which means $$ C\cap D\in\mathcal{L}_0. $$ So $$ C\in\mathcal{P},\ D\in\mathcal{L}_0 \quad\Longrightarrow\quad C\cap D\in\mathcal{L}_0. $$ :::{.column-margin #fig-l10-slide-18} ![Claim 2 in the proof of Dynkin's theorem.](images/l10-s18.png){fig-alt="Blackboard slide for Claim 2. If C is in P and D is in L_0, then C intersection D is in L_0. The proof uses that L_C is a lambda-system containing P, so by minimality L_0 is contained in L_C." group="slides"} Minimality of $\mathcal{L}_0$ upgrades $\mathcal{P}\subseteq\mathcal{L}_C$ to $\mathcal{L}_0\subseteq\mathcal{L}_C$. ::: ### Claim 3: reverse the roles Now take $C\in\mathcal{L}_0$. We want to show $$ \mathcal{P}\subseteq\mathcal{L}_C. $$ Let $D\in\mathcal{P}$. By Claim 2, with the $\mathcal{P}$-set $D$ and the $\mathcal{L}_0$-set $C$, $$ D\cap C \in \mathcal{L}_0. $$ Since $D\cap C=C\cap D$, this means $D\in\mathcal{L}_C$. Hence $$ \mathcal{P}\subseteq\mathcal{L}_C. $$ :::{.column-margin #fig-l10-slide-19} ![Claim 3 in the proof of Dynkin's theorem.](images/l10-s19.png){fig-alt="Blackboard slide for Claim 3. If C is in L_0, then P is contained in L_C. For D in P, Claim 2 gives C intersection D in L_0, so D is in L_C." group="slides"} The symmetry $C\cap D=D\cap C$ lets Claim 2 be reused. ::: ### Claim 4: two arbitrary sets from $\mathcal{L}_0$ Finally take $A\in\mathcal{L}_0$. By the section lemma, $\mathcal{L}_A$ is a $\lambda$-system. Claim 3 gives $$ \mathcal{P}\subseteq\mathcal{L}_A. $$ Therefore, by minimality, $$ \mathcal{L}_0\subseteq\mathcal{L}_A. $$ If $B\in\mathcal{L}_0$, then $B\in\mathcal{L}_A$, so $$ A\cap B\in\mathcal{L}_0. $$ Thus $\mathcal{L}_0$ is a $\pi$-system. Since it is already a $\lambda$-system, it is a $\sigma$-field. This proves Dynkin's $\pi$-$\lambda$ theorem. :::{.column-margin #fig-l10-slide-20} ![Claim 4 completes the proof of Dynkin's theorem.](images/l10-s20.png){fig-alt="Blackboard slide for Claim 4. If A and B are in L_0, then A intersection B is in L_0. The proof defines L_A, uses Claim 3 to get P contained in L_A, uses minimality to get L_0 contained in L_A, and concludes B in L_A." group="slides"} Claim 4 proves the missing finite-intersection closure. ::: ## Independent $\pi$-systems generate independent $\sigma$-fields {#sec-independent-pi-systems} Now return to the motivation of the lesson. ::: {#thm-independent-generated-sigma-fields} ## Independence theorem for generated $\sigma$-fields Let $$ \mathcal{A}_1,\ldots,\mathcal{A}_n $$ be independent $\pi$-systems in a probability space $(\Omega,\mathcal{F},P)$. Then $$ \sigma(\mathcal{A}_1),\ldots,\sigma(\mathcal{A}_n) $$ are independent. ::: This is the promised corrected form of the false naive theorem. The extra $\pi$-system assumption prevents the counterexample pathology. Define $$ \mathcal{B}_i = \mathcal{A}_i\cup\{\Omega\}. $$ Adding $\Omega$ does not break the $\pi$-system property, because intersecting with $\Omega$ changes nothing. Also, $$ \sigma(\mathcal{B}_i)=\sigma(\mathcal{A}_i), $$ since $\Omega$ belongs to every $\sigma$-field anyway. :::{.column-margin #fig-l10-slide-21} ![Independent pi-systems generate independent sigma-fields.](images/l10-s21.png){fig-alt="Blackboard slide stating that if A_1 through A_n are independent pi-systems, then the generated sigma-fields sigma of A_1 through sigma of A_n are mutually independent. It begins the proof by defining B_i as A_i union {Omega}, noting that B_i is still a pi-system and generates the same sigma-field as A_i." group="slides"} Adding $\Omega$ makes the independence formula uniform across all subfamilies. ::: The independence of the original classes is equivalent to saying that for all $B_i\in\mathcal{B}_i$, $$ P(B_1\cap\cdots\cap B_n) = \prod_{i=1}^nP(B_i). $$ Fix arbitrary sets $$ B_2\in\mathcal{B}_2, \ldots, B_n\in\mathcal{B}_n. $$ Define $$ \mathcal{L} = \left\{ A\in\mathcal{F}: P(A\cap B_2\cap\cdots\cap B_n) = P(A)\prod_{i=2}^nP(B_i) \right\}. $$ The goal is to show that $\mathcal{L}$ is a $\lambda$-system containing $\mathcal{B}_1$. Then Dynkin's theorem gives $$ \sigma(\mathcal{B}_1) \subseteq \mathcal{L}, $$ which is the first step in replacing $\mathcal{B}_1$ by $\sigma(\mathcal{A}_1)$. [To put a $\sigma$ on one coordinate, freeze every other coordinate and make the unfrozen coordinate a $\lambda$-system.]{.mark} :::{.column-margin #fig-l10-slide-22} ![Constructing a lambda-system to upgrade one independent pi-system.](images/l10-s22.png){fig-alt="Blackboard slide recalling the independence formula for the B_i classes and then fixing B_2 through B_n. It defines L as the class of A in F such that P(A intersection B_2 intersection through B_n) factors as P(A) times the probabilities of B_2 through B_n." group="slides"} The constructed $\mathcal{L}$ is tailored to the one factorization we want. ::: ### Showing $\mathcal{L}$ is a $\lambda$-system First, $\Omega\in\mathcal{L}$ because $$ P(\Omega\cap B_2\cap\cdots\cap B_n) = P(B_2\cap\cdots\cap B_n) = \prod_{i=2}^nP(B_i) = P(\Omega)\prod_{i=2}^nP(B_i). $$ :::{.column-margin #fig-l10-slide-23} ![Omega belongs to the constructed lambda-system.](images/l10-s23.png){fig-alt="Blackboard slide showing that Omega belongs to the constructed class L. Plugging Omega into the definition reduces the left side to the probability of B_2 intersection through B_n, which factors by independence and equals P(Omega) times the product of the B_i probabilities." group="slides"} The $\Omega$ case uses independence of $\mathcal{B}_2,\ldots,\mathcal{B}_n$. ::: For complements, let $A\in\mathcal{L}$ and abbreviate $$ B=B_2\cap\cdots\cap B_n. $$ Then $$ B=(A\cap B)\cup(A^c\cap B), $$ and the two pieces are disjoint. By finite additivity, $$ P(B)=P(A\cap B)+P(A^c\cap B). $$ Because $A\in\mathcal{L}$, $$ P(A\cap B) = P(A)\prod_{i=2}^nP(B_i). $$ Also, by independence of $B_2,\ldots,B_n$, $$ P(B)=\prod_{i=2}^nP(B_i). $$ Therefore $$ P(A^c\cap B) = \left(1-P(A)\right)\prod_{i=2}^nP(B_i) = P(A^c)\prod_{i=2}^nP(B_i). $$ Thus $A^c\in\mathcal{L}$. :::{.column-margin #fig-l10-slide-24} ![Complement closure for the constructed lambda-system.](images/l10-s24.png){fig-alt="Blackboard slide proving complement closure for the constructed class L. It writes B as B_2 intersection through B_n, decomposes B into A intersection B and A complement intersection B, uses finite additivity, and solves for the probability of A complement intersection B to show it factors correctly." group="slides"} Complement closure comes from decomposing the frozen intersection $B$ into the $A$ and $A^c$ parts. ::: For countable disjoint unions, let $A_1,A_2,\ldots\in\mathcal{L}$ be disjoint. Then the sets $$ A_m\cap B_2\cap\cdots\cap B_n $$ are also disjoint. Countable additivity gives $$ \begin{aligned} P\left(\left(\bigcup_{m=1}^{\infty}A_m\right)\cap B_2\cap\cdots\cap B_n\right) &= P\left(\bigcup_{m=1}^{\infty}(A_m\cap B_2\cap\cdots\cap B_n)\right)\\ &= \sum_{m=1}^{\infty}P(A_m\cap B_2\cap\cdots\cap B_n)\\ &= \sum_{m=1}^{\infty}P(A_m)\prod_{i=2}^nP(B_i)\\ &= \left(\sum_{m=1}^{\infty}P(A_m)\right)\prod_{i=2}^nP(B_i)\\ &= P\left(\bigcup_{m=1}^{\infty}A_m\right)\prod_{i=2}^nP(B_i). \end{aligned} $$ So $$ \bigcup_{m=1}^{\infty}A_m\in\mathcal{L}. $$ :::{.column-margin #fig-l10-slide-25} ![Countable disjoint union closure for the constructed lambda-system.](images/l10-s25.png){fig-alt="Blackboard slide proving countable disjoint union closure for L. It expands the probability of the union of A_m intersected with B_2 through B_n, uses countable additivity, uses the defining factorization for each A_m, factors out the constant product of probabilities of the B_i, and then recombines the sum as the probability of the union of the A_m." group="slides"} Countable additivity is exactly why a $\lambda$-system, not an arbitrary class, appears here. ::: ### Apply Dynkin's theorem once Because the original classes are independent, every $B_1\in\mathcal{B}_1$ satisfies the defining factorization for $\mathcal{L}$. Hence $$ \mathcal{B}_1\subseteq\mathcal{L}. $$ Since $\mathcal{B}_1$ is a $\pi$-system and $\mathcal{L}$ is a $\lambda$-system, Dynkin's theorem gives $$ \sigma(\mathcal{B}_1) \subseteq\mathcal{L}. $$ But $\sigma(\mathcal{B}_1)=\sigma(\mathcal{A}_1)$, so for any $$ A_1\in\sigma(\mathcal{A}_1), $$ and any fixed $B_2\in\mathcal{B}_2,\ldots,B_n\in\mathcal{B}_n$, $$ P(A_1\cap B_2\cap\cdots\cap B_n) = P(A_1)\prod_{i=2}^nP(B_i). $$ Thus $$ \sigma(\mathcal{A}_1),\mathcal{B}_2,\ldots,\mathcal{B}_n $$ are independent. :::{.column-margin #fig-l10-slide-26} ![Dynkin theorem upgrades the first pi-system to its generated sigma-field.](images/l10-s26.png){fig-alt="Blackboard slide concluding the first upgrade step. Since L is a lambda-system and B_1 is a pi-system contained in L, Dynkin's theorem gives sigma of B_1 contained in L. Since sigma of B_1 equals sigma of A_1, any A_1 from sigma of A_1 satisfies the desired factorization with B_2 through B_n." group="slides"} One application of Dynkin's theorem replaces one $\pi$-system by its generated $\sigma$-field. ::: ### Iterate the argument Now fix $$ A_1\in\sigma(\mathcal{A}_1), \qquad B_3\in\mathcal{B}_3, \ldots, B_n\in\mathcal{B}_n, $$ and define a new class $$ \mathcal{L} = \left\{ A\in\mathcal{F}: P(A_1\cap A\cap B_3\cap\cdots\cap B_n) = P(A_1)P(A)\prod_{i=3}^nP(B_i) \right\}. $$ The same proof shows that this new $\mathcal{L}$ is a $\lambda$-system containing $\mathcal{B}_2$. Dynkin's theorem gives $$ \sigma(\mathcal{B}_2) \subseteq \mathcal{L}, $$ so $\mathcal{B}_2$ can be upgraded to $\sigma(\mathcal{A}_2)$. Repeating the argument for $\mathcal{A}_3,\ldots,\mathcal{A}_n$ yields $$ \sigma(\mathcal{A}_1), \sigma(\mathcal{A}_2), \ldots, \sigma(\mathcal{A}_n) $$ mutually independent. :::{.column-margin #fig-l10-slide-27} ![Iterating the pi-lambda upgrade over the remaining classes.](images/l10-s27.png){fig-alt="Blackboard slide explaining the iteration step. After upgrading A_1 to sigma of A_1, fix A_1 from that generated sigma-field and fix B_3 through B_n. Define a new lambda-system in the second coordinate and apply Dynkin's theorem to upgrade B_2 to sigma of A_2. Continue until all classes have been upgraded." group="slides"} The same $\lambda$-system construction is recycled one coordinate at a time. ::: For an infinite collection of independent $\pi$-systems, the result follows by applying the finite theorem to every finite subcollection. This works because independence of an infinite family is defined by independence of all finite subfamilies. [The infinite-family version follows because independence is checked on finite subfamilies.]{.mark} ## Takeaway {#sec-takeaway} Dynkin's $\pi$-$\lambda$ theorem is a minimality machine. It says that once a $\lambda$-system contains a $\pi$-system, it must contain everything generated from that $\pi$-system as a $\sigma$-field. The lesson uses that machine twice: 1. **Abstractly**, to prove the theorem itself by showing the minimal $\lambda$-system over a $\pi$-system is also a $\pi$-system. 2. **Probabilistically**, to prove that independent $\pi$-systems remain independent after generating $\sigma$-fields. This is also why $\lambda$-systems are tuned to probability: their closure condition is countable additivity over disjoint unions, which is the central operation for probability measures [@billingsley2012probability p.45].