728 CHAPTER 26. INDEPENDENCE

The Kolmogorov zero one law follows next. It has to do with something called a tailevent.

Definition 26.7.2 Let {Fn} be a sequence of σ algebras. Then Tn ≡ σ(∪∞

k=nFk)

where this means the smallest σ algebra which contains each Fk for k ≥ n. Then a tailevent is a set which is in the σ algebra, T ≡ ∩∞

n=1Tn.

As usual, (Ω,F ,P) is the underlying probability space such that all σ algebras arecontained in F .

Lemma 26.7.3 Suppose {Fn}∞

n=1 are independent σ algebras and suppose A is a tailevent and Aki ∈Fki , i = 1, · · · ,m are given sets. Then

P(Ak1 ∩·· ·∩Akm ∩A

)= P

(Ak1 ∩·· ·∩Akm

)P(A)

Proof: Let K be the π system consisting of finite intersections of the form

Bm1 ∩Bm2 ∩·· ·∩Bm j

where Bmi ∈Fki for ki > max{k1, · · · ,km} ≡ N. Thus σ (K ) = σ(∪∞

i=N+1Fi). Now let

G ≡{

B ∈ σ (K ) : P(Ak1 ∩·· ·∩Akm ∩B

)= P

(Ak1 ∩·· ·∩Akm

)P(B)

}Then clearly K ⊆ G . It is also true that G is closed with respect to complements andcountable disjoint unions. By the lemma on π systems, G = σ (K ) = σ

(∪∞

i=N+1Fi).

Since A is in σ(∪∞

i=N+1Fi)

due to the assumption that it is a tail event, it follows that

P(Ak1 ∩·· ·∩Akm ∩A

)= P

(Ak1 ∩·· ·∩Akm

)P(A) ■

Theorem 26.7.4 Suppose the σ algebras, {Fn}∞

n=1 are independent and supposeA is a tail event. Then P(A) either equals 0 or 1.

Proof: Let A ∈T ≡ ∩∞n=1Tn ≡ ∩∞

n=1σ(∪∞

k=nFk). I want to show that P(A) = P(A)2.

Since A is in T , it is in each σ(∪∞

k=nFk). Let K denote sets of the form Ak1 ∩ ·· · ∩Akm

for some m, Ak j ∈Fk j where each k j > n. Thus K is a π system and

σ (K ) = σ(∪∞

k=n+1Fk)≡Tn+1

LetG ≡

{B ∈Tn+1 ≡ σ

(∪∞

k=n+1Fk)

: P(A∩B) = P(A)P(B)}

Thus K ⊆ G because

P(Ak1 ∩·· ·∩Akm ∩A

)= P

(Ak1 ∩·· ·∩Akm

)P(A)

by Lemma 26.7.3. However, it is routine that G is closed with respect to countable disjointunions and complements. Therefore by the Lemma on π systems Lemma 9.3.2 on Page243, it follows G = σ (K ) = σ

(∪∞

k=n+1Fk).

Thus for any B ∈ σ(∪∞

k=n+1Fk)= Tn+1,P(A∩B) = P(A)P(B). However, A is in all

of these Tn+1 and so P(A∩A) = P(A) = P(A)2 so P(A) equals either 0 or 1. ■What sorts of things are tail events of independent σ algebras?

728 CHAPTER 26. INDEPENDENCEThe Kolmogorov zero one law follows next. It has to do with something called a tailevent.Definition 26.7.2 Le: { Fn} be a sequence of 6 algebras. Then Jy = 6 (Up_»F xk)where this means the smallest o algebra which contains each ¥, for k >n. Then a tailevent is a set which is in the o algebra, 7 = V1 T,.As usual, (Q,.¥,P) is the underlying probability space such that all o algebras arecontained in ¥.Lemma 26.7.3 Suppose {Fn}, are independent o algebras and suppose A is a tailevent and Ax, € Fx,, i= 1,--+ ,mare given sets. ThenP (Ag, A+++ Ag, AA) = P (Ag, +++ MAg,,) P(A)Proof: Let .% be the z system consisting of finite intersections of the formBin, VB 1-+* Bm;where Bm, € Fx, for ki > max {ki ,-++ ,km} =N. Thus 6 (.#) = 6 (Uy, Fi) . Now letG={BEo(L):P (Ag, ++: MAR, OB) = P (Ag, O-+ Ag, ) P(B)}Then clearly “ CY. It is also true that Y is closed with respect to complements andcountable disjoint unions. By the lemma on 7 systems, Y = 0 (.#) = 0 (UZ.y41Fi) -Since A is ino (UZ 41Fi) due to the assumption that it is a tail event, it follows thatP (Ag, A+++ Ag, OA) = P (Ag, O--+OAg,,) P(A)Theorem 26.7.4 suppose the o algebras, {Fn}; _, are independent and supposeA is a tail event. Then P(A) either equals 0 or 1.Proof: Let A ¢ J =N_,.% =MP_,0 (UL, Fx). L want to show that P(A) = P(A)’.Since A is in 7, it is in each o (U_,, Fx). Let % denote sets of the form Ag, N---NAk,for some m, Ax, € Fx; where each kj > n. Thus .% is a 7 system andO(H) = 6 (Unit Fe) = FavLetG={BE Fu, = 6 (UR Fx) : P(ANB) = P(A) P(B)}Thus .#” C Y becauseP (Ag, A+++ Ag, AA) = P (Ag, +++ MAg,,) P(A)by Lemma 26.7.3. However, it is routine that Y is closed with respect to countable disjointunions and complements. Therefore by the Lemma on 7 systems Lemma 9.3.2 on Page243, it follows Y = 6 (%) = 6 (UC 4) Fx):Thus for any B € 6 (Ug_,4; Fk) = Tnz1,P (ANB) = P(A) P(B). However, A is in allof these 7%, and so P(ANA) = P(A) = P(A)’ so P(A) equals either 0 or 1.What sorts of things are tail events of independent o algebras?