2246 CHAPTER 65. STOCHASTIC INTEGRATION

Diversion The reason the series converges goes as follows. Estimate

E

∣∣∣∣∣ q

∑k=p

(ψk(t ∧ tn

j+1)−ψk

(t ∧ tn

j))

Φ(tn

j)

gk

∣∣∣∣∣2

H

First consider the mixed terms. Let ∆ψk = ψk

(t ∧ tn

j+1

)−ψk

(t ∧ tn

j

). For l < k,

E((

∆ψkΦ(tn

j)

gk,∆ψ lΦ(tn

j)

gl))

= E(∆ψk∆ψ l

(Φ(tn

j)

gk,Φ(tn

j)

gl))

Now by independence, this equals

E (∆ψk∆ψ l)E((

Φ(tn

j)

gk,Φ(tn

j)

gl))

= E (∆ψk)E (∆ψ l)E((

Φ(tn

j)

gk,Φ(tn

j)

gl))

= 0

Thus you only need to consider the non mixed terms, and the thing you want to estimate isof the form

q

∑k=p

E(∣∣(ψk

(t ∧ tn

j+1)−ψk

(t ∧ tn

j))

Φ(tn

j)

gk∣∣2)

Now by independence again, this equals

q

∑k=p

E((

∆ψkΦ(tn

j)

gk,∆ψkΦ(tn

j)

gk))

=q

∑k=p

E(∆ψ

2k(Φ(tn

j)

gk,Φ(tn

j)

gk))

=q

∑k=p

E(∆ψ

2k)

E(Φ(tn

j)

gk,Φ(tn

j)

gk)

=((

t ∧ tnj+1)−(t ∧ tn

j)) q

∑k=p

E(∣∣Φ(tn

j)

gk∣∣2H

)and this sum is just a part of the convergent infinite sum for∫

Ω

∥∥Φ(tn

j)∥∥2

L2(Q1/2U,H)dP < ∞

Therefore, this converges to 0 as p,q → ∞ and so the sum converges in L2 (Ω,H) asclaimed.

End of diversionThe J and the J−1 cancel in 65.5.12 because J is one to one. It follows that 65.5.11

equalsmn

∑j=0

∞

∑k=1

(ψk(t ∧ tn

j+1)−ψk

(t ∧ tn

j))

Φ(tn

j)

gk+