Independence extended from generating pi-systems Created 2026-09-24 Updated 2026-09-24
If two pi-systems satisfy the probability factorization pairwise, their generated sigma-algebras are independent. For one event in the first family, the events in the second sigma-algebra satisfying the factorization form a Dynkin system; apply Dynkin lemma, then repeat with the roles reversed.
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 4 25J b Solution Created 2026-09-24 Updated 2026-10-05
If is measure-preserving, equality holds on every measurable set and hence on the specified -system. Conversely, let . Both finite measures have equal total mass. Therefore , complements of members remain in by subtracting from the finite total mass, and disjoint countable unions remain in by countable additivity. Thus is a Dynkin system.
It contains the generating pi-system , so Dynkin lemma gives . ConsequentlyThis is exactly the measure-preserving property.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 1 27K a Solution Created 2026-09-24 Updated 2026-09-25
Dynkin lemma, also called the pi-lambda theorem, states that if a Dynkin system contains a pi-system , then
Let be the smallest Dynkin system containing . It is enough to prove that is a sigma-algebra, because then
First fix and defineThis is a Dynkin system. It contains because ; closure under relative complements follows fromand closure under disjoint unions follows by distributing over the union. Since is closed under intersections, . Minimality therefore givesThus whenever and .
Now fix and defineThe same argument shows that is a Dynkin system, and the preceding paragraph shows that it contains . Hence . We have proved that is itself a pi-system.
A Dynkin system that is also a pi-system is a sigma-algebra: it is closed under arbitrary finite intersections, hence finite unions by complements, and any countable union can be disjointified before using closure under disjoint unions. Therefore , proving the lemma.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 1 27K b Solution Created 2026-09-24 Updated 2026-09-25
Fix and letThis is a Dynkin system. It contains . If , thenso it is closed under complements. Countable additivity gives closure under countable disjoint unions.
The hypothesis says . Since is a pi-system, Dynkin lemma yieldsThus the factorization holds for every and every .
Now fix such a and defineThe identical calculation makes a Dynkin system. The first step gives , so a second application of Dynkin lemma givesTherefore the probability factorization holds for everywhich is precisely independence of the two sigma-algebras. This is independence extended from generating pi-systems.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 201 1 d Solution Created 2026-09-24 Updated 2026-09-25
Let . The limit from part (c) is -measurable because each is. If , then for some , and for every ,The convergence lets us pass to the limit and obtain .
The sets on which this identity holds form a Dynkin system, and is a generating pi-system. The Dynkin lemma therefore extends the identity to every . Consequently has both defining properties of , established here without appealing to the general existence theorem.