1958 CHAPTER 60. CONDITIONAL, MARTINGALES

Proof: I will prove the case where {Xn} is a submartingale and note the other case willonly involve replacing ≥ with =. First recall that from Lemma 60.4.1 FS ⊆FT . Also letm be an upper bound for T . Then it follows from this that

E (|XT |) =m

∑i=1

∫[T=i]|Xi|dP < ∞

with a similar formula holding for E (|XS|). Thus it makes sense to speak of E (XT |FS) .I need to show that if B ∈FS, so that B∩ [S≤ n] ∈Fn for all n, then∫

BXT dP≡

∫B

E (XT |FS)dP≥∫

BXSdP. (60.4.8)

It suffices to do this for B of the special form

B = A∩ [S = i]

because if this is done, then the result follows from summing over all possible values of S.Note that if B = A∩ [S = m] , then XT = XS = Xm and there is nothing to prove in 60.4.8 soit can be assumed i≤ m−1. Then let B be of this form.∫

A∩[S=i]XT dP =

m

∑j=i

∫A∩[S=i]∩[T= j]

XT dP

=m−1

∑j=i

∫A∩[S=i]∩[T= j]

XT dP+∫

A∩[S=i]∩[T≥m]XmdP

And so ∫A∩[S=i]

XT dP =m−1

∑j=i

∫A∩[S=i]∩[T= j]

XT dP+∫

A∩[S=i]∩[T≥m]XmdP (60.4.9)

=m−1

∑j=i

∫A∩[S=i]∩[T= j]

XT dP+∫

A∩[S=i]∩[T≤m−1]CXmdP

≥m−1

∑j=i

∫A∩[S=i]∩[T= j]

XT dP+∫

A∩[S=i]∩[T≤m−1]CXm−1dP

=m−1

∑j=i

∫A∩[S=i]∩[T= j]

XT dP+∫

A∩[S=i]∩[T>m−1]Xm−1dP

provided m−1≥ i because {Xn} is a submartingale and

A∩ [S = i]∩ [T ≤ m−1]C ∈Fm−1

Now combine the top term of the sum with the term on the right to obtain

=m−2

∑j=i

∫A∩[S=i]∩[T= j]

XT dP+∫

A∩[S=i]∩[T≥m−1]Xm−1dP