2454 CHAPTER 72. THE HARD ITO FORMULA

−qk−1

∑j=1

(X(t j+1

)−X (t j)−

(M(t j+1

)−M (t j)

),( I−Pn)

(M(t j+1

)−M (t j)

))(72.6.15)

The sum in 72.6.15 is dominated by(qk−1

∑j=1

∣∣X (t j+1)−X (t j)−

(M(t j+1

)−M (t j)

)∣∣2)1/2

·

(qk−1

∑j=1

∣∣( I−Pn)(M(t j+1

)−M (t j)

)∣∣2)1/2

(72.6.16)

Now it is known that ∑qk−1j=1

∣∣X (t j+1)−X (t j)−

(M(t j+1

)−M (t j)

)∣∣2 converges in proba-bility to a. If you take the expectation of the other factor it is

E

(qk−1

∑j=1

∣∣∣∣( I−Pn)∫ t j+1

t j

Z (s)dW (s)∣∣∣∣2)

=qk−1

∑j=1

E

(∣∣∣∣∫ t j+1

t j

( I−Pn)Z (s)dW (s)∣∣∣∣2)

=qk−1

∑j=1

E(∫ t j+1

t j

||( I−Pn)Z (s)||2L2(Q1/2U,H)

)ds

≤ E(∫ T

0||( I−Pn)Z (s)||2

L2(Q1/2U,H) ds)

=∫

Ω

∫ T

0

∞

∑i=n+1

(Z (s) ,ei)2 dsdP

The integrand converges to 0 as n→∞ and is dominated by ∑∞i=1 (Z (s) ,ei)

2 which is givento be in L1 ([0,T ]×Ω). Therefore, it converges to 0.

Thus the expression in 72.6.16 is of the form fkgnk where fk converges in probabilityto a as k→ ∞ and gnk converges in probability to 0 as n→ ∞ independently of k. Now thisimplies fkgnk converges in probability to 0. Here is why.

P([| fkgnk|> ε]) ≤ P(2δ | fk|> ε)+P(2Cδ |gnk|> ε)

≤ P(2δ | fk−a|+2δ |a|> ε)+P(2Cδ |gnk|> ε)

where δ | fk|+Cδ |gkn|> | fkgnk| and limδ→0 Cδ = ∞. Pick δ small enough that ε−2δ |a|>ε/2. Then this is dominated by

≤ P(2δ | fk−a|> ε/2)+P(2Cδ |gnk|> ε)

Fix n large enough that the second term is less than η . Now taking k large enough, theabove is less than η . It follows the expression in 72.6.16 and consequently in 72.6.15converges to 0 in probability.