2277

= E((Yj,∆ jM (t))H E

((Yk,∆kM (t))H |Ftk

))= E

((Yj,∆ jM (t))H

(Yk,E

(∆kM (t) |Ftk

))H

)= E

((Yj,∆ jM (t))H (Yk,0)H

)= 0

Now

mn−1

∑k=0

E((Yk,M (t ∧ tk+1)−M (t ∧ tk))

2H

)=

mn−1

∑k=0

E((

Yk,(Mtk+1 −Mtk

)(t))2

H

)It follows from 66.0.4

E

(mn−1

∑k=0

(Yk,M (t ∧ tk+1)−M (t ∧ tk))H

)2=

mn−1

∑k=0

E((

Yk,(Mtk+1 −Mtk

)(t))2

H

)

=mn−1

∑k=0

E([(

Yk,(Mtk+1 −Mtk

)(t))]

+Nk (t))

where Nk is a martingale equal to 0 for t ≤ tk. Then this equals

mn−1

∑k=0

E([(

Yk,(Mtk+1 −Mtk

)(t))])

From Lemma 66.0.5

≤ E

(mn−1

∑k=0∥Yk∥2

H([M]tk+1 (t)− [M]tk (t)

))

= E

(mn−1

∑k=0∥Yk∥2

H([M](tnk+1∧ t

)− [M] (tn

k ∧ t)))

(66.0.5)

= E(∫ t

0∥Y∥2

H d [Mτ p ]

)= E

(∫ t

0∥Y∥2

H d [M]τ p

)Note that everything makes sense because it is assumed that ∥Yk∥M∗ ∈ L2 (Ω). This provesthe following lemma.

Lemma 66.0.6 Let ∥Y (t)∥(Mτ p)∗ ∈ L2 (Ω) for each t, where Y is an elementary functionand let τ p be a stopping time for which Mτ p is a L2 martingale. Then

E

(∣∣∣∣∫ t

0(Y,dMτ p)

∣∣∣∣2)≤ E

(∫ t

0∥Y∥2

H d [M]τ p

)The condition that ∥Y (t)∥(Mτ p)∗ ∈ L2 (Ω) ensures that

E((Yk,Mτ p (t ∧ tk+1)−Mτ p (t ∧ tk+1))

2H

)always is finite.