63.6. ANOTHER LIMIT FOR QUADRATIC VARIATION 2145

Lemma 63.6.1 The kth term in the above sum is a martingale and the integral is also amartingale.

Proof: Let σ be a stopping time with two values. Then

E ( fk (M (σ ∧ tk+1)−M (σ ∧ tk)))

= E(E(

fk (M (σ ∧ tk+1)−M (σ ∧ tk)) |Ftk

))= E

(fkE((M (σ ∧ tk+1)−M (σ ∧ tk)) |Ftk

))= 0

and it works the same with σ replaced with t. Hence by the lemma about recognizingmartingales, Lemma 63.1.1, each term is a martingale and so it follows that the integral∫ t

0 f dM is also a martingale.Note also that, since M is continuous, this is a continuous martingale.As before, it is important to estimate this.

E

(∣∣∣∣∫ t

0f dM

∣∣∣∣2)≤?

Consider a mixed term. For j < k, it follows from measurability considerations that

E(( fk (M (t ∧ tk+1)−M (t ∧ tk)))

(f j(M(t ∧ t j+1

)−M (t ∧ t j)

)))= E

(E[( fk (M (t ∧ tk+1)−M (t ∧ tk)))

(f j(M(t ∧ t j+1

)−M (t ∧ t j)

))|Ftk

])= E

((f j(M(t ∧ t j+1

)−M (t ∧ t j)

))fkE[(M (t ∧ tk+1)−M (t ∧ tk)) |Ftk

])= 0

Therefore,

E

(∣∣∣∣∫ t

0f dM

∣∣∣∣2)

= E

(m−1

∑k=0| fk (M (t ∧ tk+1)−M (t ∧ tk))|2

)

≤ E

(m−1

∑k=0∥ fk∥2 |M (t ∧ tk+1)−M (t ∧ tk)|2

)

= E

(m−1

∑k=0∥ fk∥2 ([Mtk+1 −Mtk

](t)+Nk (t)

))

= E

(m−1

∑k=0∥ fk∥2 ([Mtk+1

](t)−

[Mtk](t)+Nk (t)

))

= E

(m−1

∑k=0∥ fk∥2 ([M] (t ∧ tk+1)− [M] (t ∧ tk)+Nk (t))

)

= E

(m−1

∑k=0∥ fk∥2 ([M] (t ∧ tk+1)− [M] (t ∧ tk))

)