2144 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALE

It suffices to verify 63.5.20. Let { fn} and {gn} be sequences of elementary functionsconverging to f and g in GM ∩GN . By Lemma 63.5.2,

E((∫ t

0fndM,

∫ t

0gndN

)U

)=∫

∫ t

0fngnd [M,N]

Then by the Holder inequality and the above definition,

limn→∞

E((∫ t

0fndM,

∫ t

0gndN

)U

)= E

((∫ t

0f dM,

∫ t

0gdN

)U

)Consider the right side which equals

14

∫Ω

∫ t

0fngnd [M+N]dP− 1

4

∫Ω

∫ t

0fngnd [M−N]dP

Now from Lemma 63.5.4,∣∣∣∣∫Ω

∫ t

0fngnd [M+N]dP−

∫Ω

∫ t

0f gd [M+N]dP

∣∣∣∣=

∣∣∣∣∫Ω

∫ t

0fngndνM+NdP−

∫Ω

∫ t

0f gdνM+NdP

∣∣∣∣≤ 2

(∫Ω

∫ t

0| fngn− f g|dνMdP+

∫Ω

∫ t

0| fngn− f g|dνNdP

)and by the choice of the fn and gn, these both converge to 0. Similar considerations applyto ∣∣∣∣∫

∫ t

0fngnd [M−N]dP−

∫Ω

∫ t

0f gd [M−N]dP

∣∣∣∣and show

limn→∞

∫Ω

∫ t

0fngnd [M,N] =

∫Ω

∫ t

0f gd [M,N]

63.6 Another Limit For Quadratic VariationThe problem to consider first is to define an integral∫ t

0f dM

where f has values in H ′ and M is a continuous martingale having values in H. For thesake of simplicity assume M (0) = 0. The process of definition is the same as before. Firstconsider an elementary function

f (t)≡m−1

∑k=0

fkX(tk,tk+1] (t) (63.6.22)

where fk is measurable into H ′ with respect to Ftk . Then define∫ t

0f dM ≡

m−1

∑k=0

fk (M (t ∧ tk+1)−M (t ∧ tk)) ∈ R (63.6.23)

2144 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALEIt suffices to verify 63.5.20. Let {f,} and {g,} be sequences of elementary functionsconverging to f and g in Gy NG. By Lemma 63.5.2,E (( [ toam, | “end ) ) = [FF fuse.)Then by the Holder inequality and the above definition,tine (( same fear) )—e((f vom foe), )Consider the right side which equalszl f sisnaimemar—> [| fugnd iM jarNow from Lemma 63.5.4,Lf tosoaim rman ff realm enjar= LL | SnsndtvorsvaP — I [ teavusvaP|<2 ( Lf \fosn—folavmar+ ff Jnén — feld¥vaPand by the choice of the f,, and g,, these both converge to 0. Similar considerations applyt: Lf toeutt—mjar ff peal njaP|tim, [| [ fugnainl =f) [reat63.6 Another Limit For Quadratic VariationThe problem to consider first is to define an integral[samwhere f has values in H’ and M is a continuous martingale having values in H. For thesake of simplicity assume M (0) = 0. The process of definition is the same as before. Firstconsider an elementary functionand showm—1fO=Y fe Mate O (63.6.22)k=0where fj is measurable into H’ with respect to F;,. Then definet m—1[ fau= Y fe(M(t\tes1) —M(tm)) ER (63.6.23)k=0