74.6. THE ITO FORMULA 2517

converges in probability to a ≥ 0. If you take the expectation of the square of the otherfactor, it is no larger than

∥B∥E

(qk−1

∑j=1

∥∥( I−Pn)∆M (t j)∥∥2

W

)

= ∥B∥E

(qk−1

∑j=1

∥∥( I−Pn)(M(t j+1

)−M (t j)

)∥∥2W

)

= ∥B∥qk−1

∑j=1

E(∥∥( I−Pn)

(M(t j+1

)−M (t j)

)∥∥2W

)Then∥∥( I−Pn)

(M(t j+1∧ t

)−M (t j ∧ t)

)∥∥2W =

[(1−Pn)Mt j+1 − (1−Pn)Mt j

](t)+N (t)

= [(1−Pn)M]t j+1 (t)− [(1−Pn)M]t j (t)+N (t)

for N (t) a martingale. In particular, taking t = tqk , the above reduces to

∥B∥qk−1

∑j=1

E(∥∥( I−Pn)

(M(t j+1

)−M (t j)

)∥∥2W

)= ∥B∥

qk−1

∑j=1

E([(1−Pn)M]

(t j+1

)− [(1−Pn)M] (t j)

)= ∥B∥E

([(1−Pn)M]

(tqk

))= ∥B∥E

(∥∥(1−Pn)M(tqk

)∥∥2W

)From maximal theorems, Theorem 62.9.4,

∥B∥E

(suptqk

∥∥(1−Pn)M(tqk

)∥∥2W

)≤ 2∥B∥E

(∥(1−Pn)M (T )∥2

W

)and this on the right converges to zero as n→ ∞ by assumption that M (t) is in L2 and thedominated convergence theorem. In particular, this shows that(

qk−1

∑j=1

∣∣⟨B( I−Pn)∆M (t j) ,( I−Pn)∆M (t j)⟩∣∣2)1/2

converges to 0 in L2 (Ω) independent of k as n→ ∞.Thus the expression in 74.6.26 is of the form fkgnk where fk converges in probability to

a1/2 as k→ ∞ and gnk converges in probability to 0 as n→ ∞ independent 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(

∣∣∣ fk−a1/2∣∣∣+2δ

∣∣∣a1/2∣∣∣> ε

)+P(2Cδ |gnk|> ε)

74.6. THE ITO FORMULA 2517converges in probability to a > 0. If you take the expectation of the square of the otherfactor, it is no larger thana=!iaie (E e—epam eyd=!= jae ($F Yr—Aa Orn) ekK! 5= (81) & (IF) (4 (1) ~My)j=ThenI|( 7 — Pr) (M (t41 At) —M(t;At)) ||, = [1 — Pa) M*! — (1 — P,) M4] (1) +N (0)= [(1— Pa) MJ7*" (1) —[(1 = Pa) MIP (1) +N (0)for N (t) a martingale. In particular, taking t = tg, , the above reduces toKa) 2(3 2 2 (||P) (M (nn) —M))|Iv)j=dal||| » E (((1— Pa) M] (ti41) — [1 — Pa) M] (27)= [BIE ((1—Py)M] (tq) = BILE (|| — Pa) (0a) li)From maximal theorems, Theorem 62.9.4,2 2Be (wel P44) <2|[BI|E (||P) M (7) I)qkand this on the right converges to zero as n —> co by assumption that M (t) is in L* and thedominated convergence theorem. In particular, this shows thatK-1 1/2(5 Kar nya) (1A aM)converges to 0 in L? (Q) independent of k as n > ~.Thus the expression in 74.6.26 is of the form f,g,x, where f, converges in probability toa'/? as k > co and gn converges in probability to 0 as n — © independent of k. Now thisimplies f; 274 converges in probability to 0. Here is why.P(\|feSnk| > €)) < P(26| fel > €) +P (2C5 |8nt| > €)< P(25|fe—a'?| +26 |a!/?| > ©) + P(2Cs legntl > €)