794 CHAPTER 29. MARTINGALES

≤ 1λ

n

∑k=1

∫Ak

E (Xn|Fk)dP≤ 1λ

n

∑k=1

∫Ak

E (Xn|Fk)+ dP

≤ 1λ

n

∑k=1

∫Ak

E(X+

n |Fk)

dP =1λ

n

∑k=1

∫Ak

X+n dP≤ 1

λ

∫Ω

X[max1≤k≤n Xk>λ ]X+n dP. ■

Now suppose Xk is a martingale with values in W a Banach space. Then from thetheorem on conditional expectation, E (∥Xk+1∥|Fk) ≥ ∥E (Xk+1|Fk)∥ = ∥Xk∥ . Thus k→∥Xk∥ is a sub-martingale and so one gets the following interesting corollary.

Corollary 29.3.15 Let Xn be a martingale with values in a Banach space W. Then forλ > 0,

P([

max1≤k≤n

∥Xk∥> λ

])≤ 1

λ

∫Ω

X[max1≤k≤n∥Xk∥>λ ] ∥Xn∥dP≤ 1λ

∫Ω

∥Xn∥dP

Now suppose Xk is a martingale with values in W a Banach space. For p> 1,k→∥Xk∥p

is a sub-martingale because

E (∥Xk+1∥p |Fk)≥ (E (∥Xk+1∥|Fk))p ≥ ∥E (Xk+1|Fk)∥p = ∥Xk∥p

Therefore, from the definition of the Lebesgue integral of a positive function,∫Ω

(max

1≤k≤n∥Xk∥

)p

dP =∫

max1≤k≤n

∥Xk∥p dP =∫

0P([

max1≤k≤n

∥Xk∥p > λ

])dλ

Change variables λ = µ p and using the Doob estimate, Theorem 29.3.14,

=∫

0P([

max1≤k≤n

∥Xk∥> λ1/p])

dλ = p∫

0P([

max1≤k≤n

∥Xk∥> µ

])µ

p−1dµ

To save on notation, let X∗n ≡max1≤k≤n ∥Xk∥ . Then using Lemma 29.3.13 as needed,

≤∫

0

pµ p−1

µ

∫Ω

X[X∗n >µ] ∥Xn∥dPdµ =∫

∥Xn (ω)∥∫

0

pµ p−1

µX[X∗n >µ]dµdP

Then p−1 = p/q where 1/p+1/q = 1,

≤∫

∫ X∗n

0pµ

p−2 ∥Xn∥dµdP =p

p−1

∫Ω

(X∗n )p/q ∥Xn∥dP

≤ pp−1

(∫Ω

(X∗n )p)1/q(∫

∥Xn∥p dP)1/p

This proves the following version of the above Doob estimate.

Theorem 29.3.16 Let n→ Xn be a martingale with values in a Banach space Wand Xn ∈ Lp (Ω;W ) , p > 1, then for X∗n ≡maxk≤n {∥Xk∥} , then∫

(X∗n )p dP≤ p

p−1

(∫Ω

(X∗n )p)1/q(∫

∥Xn∥p dP)1/p

In fact, (∫Ω

(X∗n )p dP

)1/p

≤ pp−1

(∫Ω

∥Xn∥p dP)1/p

794 CHAPTER 29. MARTINGALESly + ! + +< re aac \A)aP=>y | x, dP< x | Zlomsrcrcnxioa]% dP./Now suppose X; is a martingale with values in W a Banach space. Then from thetheorem on conditional expectation, E (||Xx41|||-F) > ||E (Xia41|-Fe)|| = ||Xx||. Thus & >||X;|| is a sub-martingale and so one gets the following interesting corollary.Corollary 29.3.15 Let X, be a martingale with values in a Banach space W. Then forA>0,1 1(| mas aul >a]) <7 ff Pjpaneacastoa] Malla? < xf [XelaNow suppose X; is a martingale with values in W a Banach space. For p > 1,k — ||X;||?is a sub-martingale becauseE (||Xevill? | Fe) = (E (Xerill | Fe)? 2 WE Xe] Fell? = [Xell?Therefore, from the definition of the Lebesgue integral of a positive function,P oo[ (max 1%) dP= | max [xr ap= [ P (| max \|X,|? >a] )aaQ \isk<n Q1<k<n 0 1<k<nChange variables A = yl” and using the Doob estimate, Theorem 29.3.14,_[* 1/p _ ° p-l=| (| max xl] > A |)a=e/ P (| ama xl > a) duTo save on notation, let X* = max) <g<» ||X,||. Then using Lemma 29.3.13 as needed,= pur! [ [ —_S Rixesyj||Xnl|dPdu = | ||\Xn(@ o Qiye. ydudP< | a I Ziseon InlldPae = fell [PinguThen p — 1 = p/q where 1/p+1/q=1,Xn 2 P /< ff" pur xallduae =P f(x?!" [xp\laPaJo p—1Je<i ( hes") (fear) ”This proves the following version of the above Doob estimate.Theorem 29.3.16 Let n Xn be a martingale with values in a Banach space Wand X, € L? (Q;W),p > 1, then for X* = maxz<y {||X¢||}, then[xnrars 2 ([m”) Vs (isirar) up(/, (xj)PaP) " < oI (/, Xd?) vpIn fact,