60.6. SUBMARTINGALE CONVERGENCE THEOREM 1969

60.6.2 Maximal InequalitiesNext I will show that stopping times and the optional sampling theorem, Lemma 60.4.3,can be used to establish maximal inequalities for submartingales very easily.

Lemma 60.6.3 Let {X (k)} be real valued and adapted to the increasing sequence of σ

algebras {Fk} . LetT (ω)≡ inf{k : X (k)≥ λ}

Then T is a stopping time. Similarly,

T (ω)≡ inf{k : X (k)≤ λ}

is a stopping time.

Proof: Is [T ≤ p] ∈Fp for all p?

[T = p] =

∈Fp−1︷ ︸︸ ︷∩p−1

i=1 [X (i)< λ ]∩

∈Fp︷ ︸︸ ︷[X (p)≥ λ ]

Therefore,[T ≤ p] = ∪p

i=1 [T = i] ∈Fp

Theorem 60.6.4 Let {Xk} be a real valued submartingale with respect to the σ algebras{Fk} . Then for λ > 0

λP([

max1≤k≤n

Xk ≥ λ

])≤ E

(X+

n), (60.6.12)

λP([

min1≤k≤n

Xk ≤−λ

])≤ E (|Xn|+ |X1|) , (60.6.13)

λP([

max1≤k≤n

|Xk| ≥ λ

])≤ 2E (|Xn|+ |X1|) . (60.6.14)

Proof: Let T (ω) be the first time Xk (ω) is ≥ λ or if this does not happen for k ≤ n,then T (ω)≡ n. Thus

T (ω)≡min(min{k : Xk (ω)≥ λ} ,n)

Note[T > k] = ∩k

i=1 [Xi < λ ] ∈Fk

and so the complement, [T ≤ k] is also in Fk which shows T is indeed a stopping time.Then 1,T (ω) ,n are stopping times, 1≤ T (ω)≤ n. Therefore, from the optional sam-

pling theorem, Lemma 60.4.3, X1,XT ,Xn is a submartingale. It follows

E (Xn) ≥ E (XT ) =∫[maxk Xk≥λ ]

XT dP+∫[maxk Xk<λ ]

XT dP

=∫[maxk Xk≥λ ]

XT dP+∫[maxk Xk<λ ]

XndP

60.6. SUBMARTINGALE CONVERGENCE THEOREM 196960.6.2 Maximal InequalitiesNext I will show that stopping times and the optional sampling theorem, Lemma 60.4.3,can be used to establish maximal inequalities for submartingales very easily.Lemma 60.6.3 Let {X (k)} be real valued and adapted to the increasing sequence of 0algebras { Fy}. LetT (@) =inf {k: X (k) >A}Then T is a stopping time. Similarly,T (@) =inf {k: X (k) <A}is a stopping time.Proof: Is [T < p] € F, for all p?Therefore,[T<p)=UL,[T=i]eF 1Theorem 60.6.4 Let {X;} be a real valued submartingale with respect to the o algebras{F_}. Then for A >0AP ( max X; > 4]) <E(X,"), (60.6.12)1<k<nAp (| min xs -A]) < E(|Xn|+|Xi|), (60.6.13)1<k<nAP (| mas \X;| >4)) < 2E (|X,|+|X1]). (60.6.14)1<k<nProof: Let T (@) be the first time X;,(@) is > A or if this does not happen for k <n,then T (@) =n. ThusT (@) = min (min {k : X,(@) > A},n)NoteT>kK =k, [Xi < Ale KHand so the complement, [T < k] is also in ¥, which shows T is indeed a stopping time.Then 1,7 (@) ,n are stopping times, 1 < T (@) <n. Therefore, from the optional sam-pling theorem, Lemma 60.4.3, X,,X7,X; is a submartingale. It followsE(X,) 2 E(Xr)= / XrdP+ XrdP[max;, X,>A] [max;, X,<A][ XrdP+ X,dPJ [max X,>A] J [max, X,<A]