862 CHAPTER 32. QUADRATIC VARIATION

= C2q

∑k=0

E(∥M (τk+1∧ t)∥2

)+E

(∥M (τk ∧ t)∥2

)−2E ((M (τk ∧ t) ,M (τk+1∧ t))) (32.3)

Consider the term E ((M (τk ∧ t) ,M (τk+1∧ t))) . By Doob’s optional sampling theorem formartingales and Corollary 32.1.3 again, this equals

E(E((M (τk ∧ t) ,M (τk+1∧ t)) |Fτk

))= E

((M (τk ∧ t) ,E

(M (τk+1∧ t) |Fτk

)))= E ((M (τk ∧ t) ,M (τk+1∧ t ∧ τk))) = E

(∥M (τk ∧ t)∥2

)It follows 32.3 equals

C2q

∑k=0

E(∥M (τk+1∧ t)∥2

)−E

(∥M (τk ∧ t)∥2

)≤ C2E

(∥M (t)∥2

).

Then from Fatou’s lemma,

E

(∑k≥0

(ξ k,(M (τk+1∧ t)−M (τk ∧ t)))

)2≤

lim infq→∞

E

( q

∑k=0

(ξ k,(M (τk+1∧ t)−M (τk ∧ t)))

)2

≤ C2E(∥M (t)∥2

)■

Now here is an interesting lemma which will be used to prove uniqueness in the mainresult.

32.2 Martingales and Total VariationLemma 32.2.1 Let Ft be a normal filtration and let A(t) ,B(t) be adapted to Ft , con-tinuous, and increasing with A(0) = B(0) = 0 and suppose A(t)−B(t) is a martingale.Then A(t)−B(t) = 0 for all t.

Proof: I shall show A(l) = B(l) where l is arbitrary. Let M (t) be the name of themartingale. Define a stopping time

τ ≡ inf{t > 0 : |M (t)|>C}∧ l∧ inf{t > 0 : A(t)>C}∧ inf{t > 0 : B(t)>C}

where inf( /0)≡ ∞ and denote the stopped martingale Mτ (t)≡M (t ∧ τ) .Then this is also amartingale with respect to the filtration Ft because by Doob’s optional sampling theoremfor martingales. Recall why this is: if s < t,

E (Mτ (t) |Fs)≡ E (M (τ ∧ t) |Fs) = M (τ ∧ t ∧ s) = M (τ ∧ s) = Mτ (s)