Kenneth Kuttler

26.4. INDEPENDENT EVENTS AND σ ALGEBRAS 723

Lemma 26.4.8 If {Xk}rk=1 are independent random variables having values in Z a

separable metric space, and if gk is a Borel measurable function, then {gk (Xk)}rk=1 is

also independent.

Proof: First consider the claim about {gk (Xk)}rk=1. Letting O be an open set in Z,

(gk ◦Xk)−1 (O) =X−1

(g−1

k (O))=X−1

k (Borel set) ∈ σ (Xk) .

It follows (gk ◦Xk)−1 (E) is in σ (Xk) whenever E is Borel because the sets whose inverse

images are measurable includes the Borel sets. Thus σ (gk ◦Xk)⊆ σ (Xk) . ■

Theorem 26.4.9 Suppose {Xi}mi=1 are independent random variables with values

in X a Banach space, then ∏mi=1 ∥Xi∥

∫Ω ∏

mi=1 ∥Xi∥dP = ∏

mi=1∫

Ω∥Xi∥dP.

Proof: The real valued random variables ∥Xi∥ are respectively measurable in σ (Xi) andso, from Corollary 26.4.7 and the independence of ∥Xi∥ ,∫ m

∏i=1∥Xi∥dµ =

∫∞

0· · ·∫

∞

0µ (∩m

i=1 [∥Xi∥i > ti])dt1 · · ·dtm

=∫

∞

0· · ·∫

∞

∏i=1

depends only on tiµ ([∥Xi∥i > ti])dt1 · · ·dtm

∏i=1

∫∞

0µ ([∥Xi∥i > ti])dti =

∏i=1

∫Ω

∥Xi∥dP ■

Note that if each Xi ∈ L1 and these are independent, then their product is also in L1.Maybe this would be a good place to put a really interesting result known as the Doob

Dynkin lemma. This amazing result is illustrated with the following diagram in whichX = (X1, · · · ,Xm). By Proposition 26.1.6 σ (X) = σ (X1, · · · ,Xn) , the expression on theright being the smallest σ algebra such that each Xi is measurable. The following diagramsummarizes this result.

(Ω,σ (X))Y→ F

↘ ◦g↗

(∏mi=1 Ei,B (∏m

i=1 Ei))

You start with Y and can write it as the composition g◦X provided Y is σ (X) measurable.

Lemma 26.4.10 Let (Ω,F ) be a measure space and let Xi : Ω→ Ei where Ei is aseparable Banach space. Suppose also that Y : Ω→ F where F is a separable Banachspace. Then Y is σ (X1, · · · ,Xm) measurable if and only if there exists a Borel measurablefunction g : ∏

mi=1 Ei→ F such that Y = g(X1, · · · ,Xm).

Proof: First suppose Y (ω) = f XW (ω) where f ∈ F and W ∈ σ (X1, · · · ,Xm) . Thenby Proposition 26.1.6, W is of the form (X1, · · · ,Xm)

−1 (B)≡X−1 (B) where B is Borel in∏

mi=1 Ei. Therefore,

Y (ω) = f XX−1(B) (ω) = f XB (X (ω)) .

26.4. INDEPENDENT EVENTS AND o ALGEBRAS 723Lemma 26.4.8 If {X;};_, are independent random variables having values in Z aseparable metric space, and if g, is a Borel measurable function, then {gx (Xx) },— isalso independent.Proof: First consider the claim about {g; (X;)};—,. Letting O be an open set in Z,(g,0 Xx) | (0) = x;! (a (O)) = x;! (Borel set) € o (Xx).It follows (g,0X,) | (E) is in o (X,) whenever E is Borel because the sets whose inverseimages are measurable includes the Borel sets. Thus o (g,0 X,) Co(X,).Theorem 26.4.9 Suppose {X;}"", are independent random variables with valuesin X a Banach space, then TY", ||Xi|| fo M44 ||Xil| dP = TTL, So ||Xill a.Proof: The real valued random variables ||X;|| are respectively measurable in o (X;) andso, from Corollary 26.4.7 and the independence of ||X;|| ,/Tixiliaei=li vf (My (Xl; > ti]) dt, -++dtm0 0co co m_ depends only on ¢;[ of [ [u (Xi; > t))dti--- dtm0 0 j=]IT exile > alan =T] fixie mi=170 j=1 72Note that if each X; € L! and these are independent, then their product is also in L!,Maybe this would be a good place to put a really interesting result known as the DoobDynkin lemma. This amazing result is illustrated with the following diagram in whichX = (Xj,-::,Xm). By Proposition 26.1.6 o (X) = o (Xj,--- ,X,), the expression on theright being the smallest o algebra such that each X; is measurable. The following diagramsummarizes this result.(Q,0(X)) * Fx 8~ 0 /(iti Ei, A UT Ei)You start with Y and can write it as the composition go X provided Y is o (X ) measurable.Lemma 26.4.10 Let (Q,.) be a measure space and let X; : Q — E; where E; is aseparable Banach space. Suppose also that Y :Q— F where F is a separable Banachspace. Then Y is 0 (X1,--- ,Xm) measurable if and only if there exists a Borel measurablefunction g : JL, E; > F such that Y = g(X,--- ,Xm).Proof: First suppose Y (@) = f Zw (@) where f € F and W € o(X,:-- ,Xm). Thenby Proposition 26.1.6, W is of the form (X1,--- ,Xm) | (B) = _X~!(B) where B is Borel in;_1 Ei. Therefore,¥ (@) = ¥x-1(p (0) = F%o(X (0).