2446 CHAPTER 72. THE HARD ITO FORMULA

Proof: Consider the formula in Lemma 72.3.3.

|X (t)|2 = |X (s)|2 +2∫ t

s⟨Y (u) ,X (t)⟩du+2(X (s) ,M (t)−M (s))

+ |M (t)−M (s)|2−|X (t)−X (s)− (M (t)−M (s))|2 (72.4.6)

Now let t j denote a point of Pk from Lemma 72.2.1. Then for t j > 0,X (tk) is just the valueof X at tk but when t = 0, the definition of X (0) in this step function is X (0)≡ 0. Thus

|X (tm)|2−|X0|2 =m−1

∑j=1

∣∣X (t j+1)∣∣2− ∣∣X (t j)

∣∣2 + |X (t1)|2−|X0|2

Using the formula in Lemma 72.3.3, for t = tm this yields

|X (tm)|2−|X0|2 = 2m−1

∑j=1

∫ t j+1

t j

⟨Y (u) ,X rk (u)⟩du+

+2m−1

∑j=1

(∫ t j+1

t j

Z (u)dW,X (t j)

)H+

m−1

∑j=1

∣∣M (t j+1)−M (t j)

∣∣2−

m−1

∑j=1

∣∣X (t j+1)−X (t j)−

(M(t j+1

)−M (t j)

)∣∣2+2∫ t1

0⟨Y (u) ,X (t1)⟩du+2

(X0,∫ t1

0Z (u)dW

)+ |M (t1)|2

−|X (t1)−X0−M (t1)|2 (72.4.7)

Of course

2∫ t1

0⟨Y (u) ,X (t1)⟩du+2

(X0,∫ t1

0Z (u)dW

)+ |M (t1)|2

converges to 0 for a.e. ω as k→∞ because the norms of the partitions converge to 0 and thestochastic integral is continuous off a set of measure zero. Actually this is not completelyclear for the first of the above terms. This term is dominated by(∫ t1

0∥Y (u)∥p′ du

)1/p(∫ T

0∥X r

k (u)∥p du

)1/p

≤ C (ω)

(∫ t1

0∥Y (u)∥p′ du

)1/p

Hence this converges to 0 for a.e. ω . At this time, not much is known about the last term in72.4.7, but it is negative and is about to be neglected anyway.The Ito isometry implies theother two terms converge to 0 in L1 (Ω) also, in addition to converging for a.e. ω . At thistime, not much is known about the last term in 72.4.7, but it is negative and is about to beneglected anyway.