63.2. THE QUADRATIC VARIATION 2123

This last sum equals Pn (t) defined as

2 ∑k≥0

(M (τn

k) ,(M(t ∧ τ

nk+1)−M (t ∧ τ

nk)))≡ Pn (t) (63.2.6)

This is because in the kth term, if t ≥ τnk , then it reduces to(

M (τnk) ,(M(t ∧ τ

nk+1)−M (t ∧ τ

nk)))

while if t < τnk , then the term reduces to 0 which is also the same as(

M (τnk) ,(M(t ∧ τ

nk+1)−M (t ∧ τ

nk)))

.

This is a finite sum because eventually, for large enough k, τnk = T . However the number

of nonzero terms depends on ω . This is not a good thing. However, a little more can besaid. In fact the sum also converges in L2 (Ω). Say ||M (t,ω)|| ≤C.

E

( q

∑k≥p

(M (τn

k) ,(M(t ∧ τ

nk+1)−M (t ∧ τ

nk))))2



=q

∑k≥p

E((

M (τnk) ,(M(t ∧ τ

nk+1)−M (t ∧ τ

nk)))2)+ mixed terms (63.2.7)

Consider one of these mixed terms for j < k.

E

M

(τ

nj),

∆ j︷ ︸︸ ︷

M(t ∧ τ

nj+1)−M

(t ∧ τ

nj) ·

M (τnk) ,

∆k︷ ︸︸ ︷

M(t ∧ τ

nk+1)−M (t ∧ τ

nk)



Then it equals

E(E((

M(τ

nj),∆ j)(

M(τ

nj),∆k)|Fτk

))= E

((M(τ

nj),∆ j)

E((

M(τ

nj),∆k)|Fτk

))= E

((M(τ

nj),∆ j)(

M(τ

nj),E(∆k|Fτk

)))= 0

Now since the mixed terms equal 0, it follows from 63.2.7, that expression is dominated by

C2q

∑k≥p

E(∣∣∣∣M (t ∧ τ

nk+1)−M (t ∧ τ

nk)∣∣∣∣2)