59.3. INDEPENDENCE 1867

independent. Furthermore, if the random variables have values in R, and they are allbounded, then

E

(r

∏i=1

Xi

)=

r

∏i=1

E (Xi) .

More generally, the above formula holds if it is only known that each Xi ∈ L1 (Ω;R) and

r

∏i=1

Xi ∈ L1 (Ω;R) .

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

(gk ◦Xk)−1 (O) = X−1

k

(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) and this provesthe first part of the lemma.

Let X1 = ∑mi=1 ciXEi ,X2 = ∑

mj=1 d jXFj where P(EiFj) = P(Ei)P(Fj). Then

∫X1X2dP = ∑

i, jd jciP(Ei)P(Fj) =

(∫X1dP

)(∫X2dP

)In general for X1,X2 independent, there exist sequences of bounded simple functions

{sn} ,{tn}

measurable with respect to σ (X1) and σ (X2) respectively such that sn→ X1 pointwise andtn→ X2 pointwise. Then from the above and the dominated convergence theorem,∫

X1X2dP = limn→∞

∫sntndP = lim

n→∞

(∫sndP

)(∫tndP

)=

(∫X1dP

)(∫X2dP

)Next suppose there are m of these independent bounded random variables. Then ∏

mi=2 Xi ∈

σ (X2, · · · ,Xm) and by Lemma 59.3.4 the two random variables X1 and ∏mi=2 Xi are inde-

pendent. Hence from the above and induction,∫ m

∏i=1

XidP =∫

X1

m

∏i=2

XidP =∫

X1dP∫ m

∏i=2

XidP =m

∏i=1

∫XidP

Now consider the last claim. Replace each Xi with Xni where this is just a truncation of

the form

Xni ≡

 Xi if |Xi| ≤ nn if Xi > n−n if Xi < n