A martingale inequality D.H.Fremlin University of Essex, - PDF document

filename n14513.tex Version of 28.5.14 A martingale inequality D.H.Fremlin University of Essex, Colchester, England Theorem Suppose that (Ω , Σ , µ ) is a probability space, Σ 0 ⊆ . . . ⊆ Σ n are σ -subalgebras of Σ, ( Y 0 , . . . , Y n ) is a martingale adapted to (Σ 0 , . . . , Σ n ), and X i : Ω → [ − 1 , 1] is Σ i -measurable for each i < n . Set Z = � n − 1 i =0 X i × ( Y i +1 − Y i ). 1 Then Pr( | Z | ≥ M ) ≤ M 2 / 3 (1 + E ( | Y n | )) for every M > 0. 1

2 Theorem If ( Y 0 , . . . , Y n ) is a martingale adapted to (Σ 0 , . . . , Σ n ), X i : Ω → [ − 1 , 1] is Σ i -measurable for i < n , and Z = � n − 1 i =0 X i × ( Y i +1 − Y i ), 1 then Pr( | Z | ≥ M ) ≤ M 2 / 3 (1 + E ( | Y n | )) for every M > 0. Doob’s maximal inequality If ( Y 0 , . . . , Y n ) is a martingale, and Z = max i ≤ n | Y i | , then Pr( | Z | ≥ M ) ≤ 1 M E ( | Y n | ) for every M > 0. Measure Theory

3 Theorem If ( Y 0 , . . . , Y n ) is a martingale adapted to (Σ 0 , . . . , Σ n ), X i : Ω → [ − 1 , 1] is Σ i -measurable for i < n , and Z = � n − 1 i =0 X i × ( Y i +1 − Y i ), 1 then Pr( | Z | ≥ M ) ≤ M 2 / 3 (1 + E ( | Y n | )) for every M > 0. A fractionally sharper theorem If ( Y 0 , . . . , Y n ) is a martingale adapted to (Σ 0 , . . . , Σ n ), X i : Ω → [ − 1 , 1] is Σ i -measurable for i < n , and Z = � n − 1 i =0 X i × ( Y i +1 − Y i ), then Pr( | Z | ≥ M ) ≤ K 2 M 2 + 1 K E ( | Y n | ) for all K , M > 0. (Set K = M 2 / 3 to get the original version.) Case 1 Suppose that | Y n | ≤ a.e. K . Then Pr( | Z | ≥ M ) ≤ K 2 M 2 . D.H.Fremlin

4 Case 1 Suppose that | Y n | ≤ a.e. K . Then Pr( | Z | ≥ M ) ≤ K 2 M 2 . proof We have n − 1 n − 1 � � E ( Z 2 ) = E ( X i × X j × ( Y i +1 − Y i ) × ( Y j +1 − Y j )) i =0 j =0 n − 1 � E ( X 2 i × ( Y i +1 − Y i ) 2 ) = i =0 (because if i < j , X i × X j × ( Y i +1 − Y i ) is Σ j -measurable, while 0 is a conditional expectation of Y j +1 − Y j on Σ j ) n − 1 � E (( Y i +1 − Y i ) 2 ) ≤ i =0 n − 1 � E ( Y 2 i +1 − Y 2 = i ) − 2 E ( Y i × ( Y i +1 − Y i )) i =0 n − 1 � E ( Y 2 i +1 − Y 2 i ) = E ( Y 2 n ) − E ( Y 2 0 ) ≤ K 2 = i =0 and the result follows at once. Measure Theory

5 Case 2 Suppose that whenever i < n and max j ≤ i | Y j ( ω ) | < K ≤ | Y i +1 ( ω ) | then | Y i +1 ( ω ) | = K . Then Pr( | Z | ≥ M ) ≤ K 2 M 2 + 1 K E ( | Y n | ). proof Set Y ′ i ( ω ) = 0 if | Y 0 ( ω ) | ≥ K, = Y i ( ω ) if | Y j ( ω ) | < K for every j ≤ i, = Y k ( ω ) if 0 < k ≤ i, | Y j ( ω ) | < K for every j < k, | Y k ( ω ) | = K and set Z ′ = � n − 1 i =0 X i × ( Y ′ i +1 − Y ′ i ). By Case 1, Pr( | Z ′ | ≥ M ) ≤ K 2 M 2 , so Pr( | Z | ≥ M ) ≤ Pr( | Z ′ | ≥ M ) + Pr( Z ′ � = Z ) ≤ K 2 M 2 + Pr( ∃ i, Y ′ i � = Y i ) ≤ K 2 M 2 + Pr( ∃ i, | Y i | ≥ K ) ≤ K 2 M 2 + 1 K E ( | Y n | ) by Doob’s inequality. D.H.Fremlin

6 Lemma Suppose that (Ω , Σ , µ ) is a probability space, Σ 0 ⊆ . . . ⊆ Σ n are σ -subalgebras of Σ, ( Y 0 , . . . , Y n ) is a martingale adapted to (Σ 0 , . . . , Σ n ), and X i : Ω → [ − 1 , 1] is Σ i -measurable for each i < n . Then there are a probability space (Ω ′ , Σ ′ , µ ′ ), σ -subalgebras Σ ′ 0 ⊆ . . . ⊆ Σ ′ 2 n of Σ ′ , a martingale ( Y ′ 0 , . . . , Y ′ 2 n ) adapted to (Σ ′ 0 , . . . , Σ ′ 2 n ), and a Σ ′ 2 i -measurable random variable X ′ 2 i for each i < n , such that (i) whenever i < n and | Y ′ 2 i ( ω ′ ) | < K then either | Y ′ 2 i +1 ( ω ′ ) | = K or | Y ′ 2 i +1 ( ω ′ ) | < K and | Y ′ 2 i +2 ( ω ′ ) | < K , (ii) Y ′ 0 , Y ′ 2 , . . . , Y ′ 2 n , X ′ 0 , X ′ 2 , . . . , X ′ 2 n − 2 have the same joint distribution as Y 0 , Y 1 , . . . , Y n , X 0 , X 1 , . . . , X n − 1 . proof of theorem Set X ′ 2 i +1 = X ′ 2 i for i < n , Z ′ = � 2 n − 1 j ) = � n − 1 j =0 X ′ j × ( Y ′ j +1 − Y ′ i =0 X ′ 2 i × ( Y ′ 2 i +2 − Y ′ 2 i ). Then Z and Z ′ have the same distribution so Pr( | Z | ≥ M ) = Pr( | Z ′ | ≥ M ) ≤ K 2 M 2 + 1 K E ( | Y ′ 2 n | ) (by Case 2) = K 2 M 2 + 1 K E ( | Y n | ) . Measure Theory

7 Lemma Suppose that ( Y 0 , . . . , Y n ) is a martingale adapted to (Σ 0 , . . . , Σ n ), and X i : Ω → [ − 1 , 1] is Σ i -measurable for each i < n . Then there are a probability space (Ω ′ , Σ ′ , µ ′ ), a martingale ( Y ′ 0 , . . . , Y ′ 2 n ) adapted to (Σ ′ 0 , . . . , Σ ′ 2 n ), and a Σ ′ 2 i -measurable random variable X ′ 2 i for each i < n , such that (i) whenever i < n and | Y ′ 2 i ( ω ′ ) | < K then either | Y ′ 2 i +1 ( ω ′ ) | = K or | Y ′ 2 i +1 ( ω ′ ) | < K and | Y ′ 2 i +2 ( ω ′ ) | < K , (ii) Y ′ 0 , Y ′ 2 , . . . , Y ′ 2 n , X ′ 0 , X ′ 2 , . . . , X ′ 2 n − 2 have the same joint distribution as Y 0 , . . . , Y n , X 0 , . . . , X n − 1 . Proving the lemma: basic case Take n = 1, Σ 1 = Σ, Σ 0 = {∅ , Ω } , Y 0 = γ = E ( Y 1 ) where | γ | < K . Set Ω ′ = Ω × [0 , 1] with product measure µ ′ , domain Σ ′ 2 ; set Σ ′ 0 = {∅ , Ω ′ } , Y ′ 0 ( ω, t ) = Y 0 ( ω ) = γ , X ′ 0 ( ω, t ) = X 0 ( ω ), 2 ( ω, t ) = Y 1 ( ω ). Seek a partition ( G + , G − , H ) of Ω ′ such that Y ′ 2 dµ ′ = Kµ ′ G + , 2 dµ ′ = − Kµ ′ G − � � G + Y ′ G − Y ′ and H ⊆ F × [0 , 1] where F = { ω : | Y 1 ( ω ) | < K } . Then we can take Σ ′ 1 to have atoms G + , G − and H . D.H.Fremlin

8 Construction when n = 1, Σ 0 = {∅ , Ω } , Ω ′ = Ω × [0 , 1], | E ( Y 1 ) | < K , 2 ( ω, t ) = Y 1 ( ω ). Seek a partition ( G + , G − , H ) of Ω ′ such that Y ′ 2 dµ ′ = Kµ ′ G + , 2 dµ ′ = − Kµ ′ G + � � G + Y ′ G − Y ′ and H ⊆ F × [0 , 1] where F = { ω : | Y 1 ( ω ) | < K } . � ( α α α ) If E Y 1 ≥ KµE where E = Ω \ F , try G α = ( E × [0 , 1]) ∪ ( F × [0 , α ]) for α ∈ [0 , 1]. 2 dµ ′ ≥ Kµ ′ G α ; if α = 1, 2 dµ ′ < Kµ ′ G α ; so for a G α Y ′ G α Y ′ � � If α = 0, suitable α can take G + = G α , G − = ∅ . H G + α E F ( β β ) Similarly if � E Y 1 ≤ − KµE . β Measure Theory

9 Seek a partition ( G + , G − , H ) of Ω ′ such that 2 dµ ′ = Kµ ′ G + , 2 dµ ′ = − Kµ ′ G + � � G + Y ′ G − Y ′ and H ⊆ F × [0 , 1] where F = { ω : | Y 1 ( ω ) | < K } . γ ) If − KµE < � ( γ γ E Y 1 < KµE ; set E + = { ω : Y 1 ( ω ) ≥ K } , E − = { ω : Y 1 ( ω ) ≤ − K } , V α = ( E + × [0 , 1]) ∪ ( E − × [0 , α ]) for α ∈ [0 , 1]. 2 dµ ′ = Kµ ′ V α . Now set V α Y ′ � Then there is an α such that W β = ( E + × [ β, 1]) ∪ ( E − × [ αβ, 1]) for β ∈ [0 , 1]. W β Y ′ 2 = − Kµ ′ W β . Take � Then there is a β such that G + = ( E + × [0 , β [) ∪ ( E − × [0 , αβ [). G − = W β , β G - H α V G + α αβ E + E - F D.H.Fremlin

10 Vector-valued extensions? Lemma Let U be a Banach space. Sup- pose that (Ω , Σ , µ ) is a probability space, Σ 0 ⊆ . . . ⊆ Σ n are σ -subalgebras of Σ, ( Y 0 , . . . , Y n ) is a martingale of Bochner integrable U -valued functions adapted to (Σ 0 , . . . , Σ n ), and X i : Ω → [ − 1 , 1] is Σ i -measurable for each i < n . Then there are a probability space (Ω ′ , Σ ′ , µ ′ ), σ -subalgebras Σ ′ 0 ⊆ . . . ⊆ Σ ′ 2 n of Σ ′ , a U -valued Bochner martingale ( Y ′ 0 , . . . , Y ′ 2 n ) adapted to (Σ ′ 0 , . . . , Σ ′ 2 n ), and a Σ ′ 2 i -measurable random variable X ′ 2 i for each i < n , such that (i) whenever i < n and � Y ′ 2 i ( ω ′ ) � < K then either � Y ′ 2 i +1 ( ω ′ ) � = K or � Y ′ 2 i +1 ( ω ′ ) � < K and � Y ′ 2 i +2 ( ω ′ ) � < K , (ii) Y ′ 0 , Y ′ 2 , . . . , Y ′ 2 n , X ′ 0 , X ′ 2 , . . . , X ′ 2 n − 2 have the same joint distribution as Y 0 , Y 1 , . . . , Y n , X 0 , X 1 , . . . , X n − 1 . Theorem Let U be a Hilbert space. Suppose that ( Y 0 , . . . , Y n ) is a U - valued Bochner martingale adapted to (Σ 0 , . . . , Σ n ), and X i : Ω → [ − 1 , 1] is Σ i -measurable for each i < n . Set Z = � n − 1 i =0 X i × ( Y i +1 − Y i ). 1 Then Pr( � Z � ≥ M ) ≤ M 2 / 3 (1 + E ( � Y n � )) for every M > 0. Measure Theory