2132 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALE

whereτ ≡ inf{t ∈ [0,T ] : ||M (t)||> λ}

Thus N is a martingale and N (0)= 0. In fact N (t)= 0 as long as t ≤ τ . As usual inf( /0)≡∞.Note

[τ < ∞] = [M∗ > λ ]⊇ [N∗ > 0] .

This is because to say τ < ∞ is to say there exists t < T such that ||M (t)||> λ which is thesame as saying M∗ > λ . Thus the first two sets are equal. If τ = ∞, then from the formulafor N (t) above, N (t) = 0 for all t ∈ [0,T ] and so it can’t happen that N∗ > 0. Thus the thirdset is contained in [τ < ∞] as claimed.

Let β > 2 and let δ ∈ (0,1) . Then

β −1 > 1 > δ > 0

Consider the following which is set up to use the good lambda inequality.

Sr ≡ [M∗ > βλ ]∩[([M] (T ))1/2 ≤ rδλ

]where 0 < r < 1.It is shown that Sr corresponds to hitting “this before that” and there is anestimate for this which involves P([N∗ > 0]) which is bounded above by P([M∗ > λ ]) asdiscussed above. This will satisfy the hypotheses of the good lambda inequality.

Claim: For ω ∈ Sr, N (t) hits λ2(

1−δ2).

Proof of claim: For ω ∈ Sr, there exists a t < T such that ||M (t)||> βλ and so usingCorollary 63.3.3,

N (t) ≥ |||M (t)||− ||Mτ (t)|||2− [M−Mτ ] (t)≥ |βλ −λ |2− [M] (t)

≥ (β −1)2λ

2−δ2λ

2

which shows that N (t) hits (β −1)2λ

2 − δ2λ

2 for ω ∈ Sr. By the intermediate valuetheorem, it also hits λ

2(

1−δ2)

. This proves the claim.

Claim: N (t)(ω) never hits −δ2λ

2 for ω ∈ Sr.

Proof of claim: Suppose t is the first time N (t) reaches −δ2λ

2. Then t > τ and so

N (t) = −δ2λ

2 ≥ |||M (t)||−λ |2− [M] (t)+ [Mτ ] (t)

≥ −r2λ

2,

a contradiction since r < 1. This proves the claim.Therefore, for all ω ∈ Sr, N (t)(ω) reaches λ

2(

1−δ2)

before it reaches −δ2λ

2. Itfollows

P(Sr)≤ P(

N (t) reaches λ2(

1−δ2)

before −δ2λ

2)

and because of Theorem 62.11.3 this is no larger than

P([N∗ > 0])δ

2

λ2(

1−δ2)−(−δ

2) = P([N∗ > 0])δ

2 ≤ δ2P([M∗ > λ ]) .

2132 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALEwhereT = inf {t € [0,7]: ||M(a)|| >A}Thus N is a martingale and N (0) = 0. In fact N (t) = 0as long ast < t. As usual inf (0) =o.Note[tT <0] = [M* >A] D [N* > 0].This is because to say T < 0 is to say there exists t < T such that ||M (t)|| > A which is thesame as saying M* > A. Thus the first two sets are equal. If tT =o, then from the formulafor N (t) above, N (t) =0 for all t € [0, 7] and so it can’t happen that N* > 0. Thus the thirdset is contained in [t < co] as claimed.Let B > 2 and let 6 € (0,1). ThenB-1>1>6>0Consider the following which is set up to use the good lambda inequality.S, =(M* > BA}o [cm (rT) < r5A|where 0 <r < 1.It is shown that S, corresponds to hitting “this before that” and there is anestimate for this which involves P ([N* > 0]) which is bounded above by P([M* > A]) asdiscussed above. This will satisfy the hypotheses of the good lambda inequality.Claim: For @ € S,, N (t) hits A? (1 — 5°) .Proof of claim: For @ € S,, there exists a t < T such that ||M(t)|| > BA and so usingCorollary 63.3.3,[I|M()|| — [MF (||? — [MM] (1) > [BA — AP? — [MI] (0)(B —1)2a2— 82a?N(t) >=which shows that N(t) hits (8 —1)?4?— 6A? for @ € S,. By the intermediate valuetheorem, it also hits A7 (1 — 5°) . This proves the claim.Claim: N (t) (@) never hits —6°A” for @ € S,.Proof of claim: Suppose t is the first time N (t) reaches —68°A*. Then t > t and so242 2Nit) = —6A°> |||M(1)||—Al —[M] (1) + [M"] (0)> —P 27s’,a contradiction since r < 1. This proves the claim.Therefore, for all @ € S,, N(t)(@) reaches A” (1 — 5°) before it reaches —5°A”. ItfollowsP(S;) <P (v (t) reaches A? (1 — 5°) before — 52°)and because of Theorem 62.11.3 this is no larger thanaanw (1 - 5°) - GuaP([N* > OJ) = P([N* > 0]) 6° < &°P([M* > A)}).