5.a Uniform integrability, last elements, closure. Alast elementfor the partially ordered index set T is an element ∞ ∈ T satisfyingt≤ ∞, for allt∈ T. Such an element is uniquely determined, if it exists. A submartingale X = (Xt,Ft)t∈T is calledclosed, if the index setT has a last element. In this case the random variable X∞is called the last element of Xand satisfies
Xt≤E(X∞|Ft), ∀t∈ T.
Closed supermartingales and closed martingales are defined similarly. The last element X∞ of a closed martingaleX satisfiesXt=E(X∞|Ft), for allt∈ T. The existence of last elements has strong consequences:
5.a.0. (i) If the submartingale X = (Xt)is closed, then the family{Xt+|t∈ T }is uniformly integrable.
(ii) If the martingaleX = (Xt)is closed, then X itself is uniformly integrable.
Proof. (i) Let X = (Xt) be a submartingale with a last element. Then so is the process (Xt+), according to 3.a.0. Thus 0 ≤ Xt+ ≤ E(X∞+|Ft), t ∈ T, and the uniform integrability of the family {Xt+ | t ∈ T } now follows from the uniform integrability of the family {EFt(X∞+)| t∈ T } (2.b.13). (ii) follows directly from 2.b.13.
Consider now a submartingale X = (Xt,Ft)t∈T, where the index set T does not have a last element. Choosing∞ ∈ T and decreeing thatt≤ ∞, for allt∈ T, T can be enlarged to a partially ordered index setT ∪ {∞}with last element ∞. The filtration (Ft) can also be extended by setting
F∞=σ
t∈TFt
.
The question is now if the submartingale X can be extended also, that is, if there exists a random variable X∞ such that the process (Xt,Ft)t∈T ∪{∞} is still a submartingale, that is, such that X∞ is F∞-measurable, E(X∞+) < ∞ and Xt ≤ E(X∞|Ft), for all t ∈ T. A random variable X∞ having these proper- ties will be called a last element f or the submartingale X. The submartingale X will be called closeable if there exists a last element X∞ for X. In this case (Xt,Ft)t∈T ∪{∞}is a closed submartingale extending X.
A last element for the supermartingaleX is not uniquely determined: IfX∞is a last element for X andZ ≥0 isF∞-measurable with E(Z) finite, thenX∞+Z is also a last element forX.
Closeable supermartingales and martingales and last elements for these are defined similarly. Note thatX∞is a last element for the martingaleX if and only ifX∞∈L1(P) isF∞-measurable andXt=E(X∞|Ft), for allt∈ T. Equivalently X∞ is a last element for the martingaleX if and only if it is a last element forX both as a submartingale and as a supermartingale.
Note for example that each nonnegative supermartingale X = (Xt) has a last element, namelyX∞= 0. However, ifX happens to be a martingale, thenX∞= 0 will be a last element for the martingaleX only ifXt= 0, for allt∈ T.
In the case of martingale sequences, that is, T = N with the usual order, F∞=σ
n≥1Fn
and the convergence theorems yield the following:
5.a.1. (i) The submartingale sequence X = (Xn,Fn)is closeable, if and only if the sequence(Xn+)is uniformly integrable. In this caseX∞= limnXn existsP-as. and is a last element for X.
(ii) The martingale sequence (Xn,Fn) is closeable, if and only if it is uniformly integrable. In this case the last element X∞ is uniquely determined as X∞ = limnXn,P-as.
Proof. (i) From 5.a.0 it follows that the sequence (Xn+) is uniformly integrable, if X is closeable. Conversely, if the sequence (Xn+) is uniformly integrable (and hence L1-bounded), then, according to the Submartingale Convergence Theorem 4.a.3.(i),(iii)X∞= lim supnXn (defined everywhere and F∞-measurable) is a last element for the submartingaleX.
(ii) This follows similarly from 5.a.0 and 4.a.5. Only the uniqueness of the last elementX∞requires additional argument. Indeed, letZ be any last element forX, that is Z ∈L1(P)F∞-measurable andXn =E(Z|Fn), for alln≥1. Then, from Levi’s theorem, Xn→Z,P-as., that is,Z= limnXn,P-as.
Remark. The uniform integrability of the sequence (Xn+) follows, for example, if Xn≤Z,n≥1, for some integrable random variableZ.
5.a.2 Example. Here we construct a nonzero, nonnegative martingaleX = (Xn) which converges to zero almost surely. The nonnegativity implies that Xn1 = E(Xn) =E(X1) > 0, for all n ≥ 1. Thus the martingale X is L1-bounded and does not converge to zero in norm. It follows that X is not uniformly integrable.
ConsequentlyX does not admit a last element, although it converges almost surely.
To obtain suchXlet Ω be [0,1[ equipped with the Borel sets and Lebesgue mea- sure. Inductively define a sequence (Pn) of partitions of Ω into intervals as follows:
P0:={[0,1]}and the intervals inPn+1arise from those inPnby bisection. In short, Pn consists of the dyadic intervalsIkn:= [k/2n,(k+ 1)/2n[,k= 0,1, . . . ,2n−1.
Now let Fn := σ(Pn) and set Xn := 2n1I0n = 2n1[0,1/2n[. Clearly Xn ≥ 0 and Xn →0 at all points ω= 0. We have to show that (Xn,Fn) is a martingale.
Since I0n ∈ Fn, Xn isFn-measurable. It remains to be shown thatE(Xn+11A) = E(Xn1A), for all setsA∈ Fn. Since each setA∈ Fn is a (finite) union of intervals Ikn,k= 0,1, . . . ,2n−1, it will suffice to show thatE(Xn+11Ikn) =E(Xn1Ikn), for all k= 0,1, . . . ,2n−1. Indeed, both integrals equal one, ifk= 0, and both are equal to zero, for all otherk.
44 5.b Sampling of closed submartingale sequences.
5.b Sampling of closed submartingale sequences. Let (Fn) be a filtration on (Ω,F, P), F∞ =
Fn =σ(
nFn) and set N∞ := N∪ {∞}. Assume that X = (Xn,Fn)1≤n<∞ is a submartingale sequence. Recall that alast element Z for the submartingale X is an F∞-measurable random variable Z such that E(Z+) <∞
and Xn≤E(Z|Fn), 1≤n <∞. (1)
Thus, if we set X∞=Z, then (Xn)1≤n≤∞ is again a submartingale. Inequality (1) is reversed in the case of a last element of a supermartingale and replaced with an equality in the case of a martingale.
We have seen in 5.a.1 that a submartingale sequence has a last element if and only if the sequence (Xn+) is uniformly integrable. In this caseX∞= limnXnexists P-as. and is a last element forX. More precisely, sinceF∞is not assumed to contain theP-null sets,Z = lim supnXn, which is defined everywhere,F∞-measurable and equals the limit limnXn almost surely, is a last element forX.
A last element Z provides a random variableX∞:=Z which closes the sub- martingale (Xn)1≤n<∞, that is, the extended sequence (Xn)1≤n≤∞ is still a sub- martingale (and hence a closed submartingale). Moreover the random variable XT
is now defined for each optional timeT : Ω→N∞. Here we setXT =X∞=Z on the set [T =∞].
We will now show that a closed submartingale X = (Xn)1≤n≤∞ satisfies the Optional Sampling Theorem 3.b.1 for all optional times S, T : Ω → N∞. The general case will be put together from the following two special cases:
(i) X has the formXn=E(Z|Fn) (5.b.0).
(ii) X is anonnegativesupermartingale and thus a supermartingale with last ele- mentX∞= 0 (5.b.1).
5.b.0. Let Z ∈ E(P) with E(Z+) < ∞ and set Xn = E(Z|Fn) for 1 ≤ n ≤ ∞. Then for any two optional times S, T: Ω→N∞ we have
(a) E XS+
<∞ andXS =E(Z|FS).
(b) S≤T implies XS =E(XT|FS).
Remark. IfZ∈L1(P), then (Xn) is a martingale with last elementZ. Proof. (a) We have Xn+≤EFn(Z+) (2.b.12) and soE
1AXn+
≤E 1AZ+
, for all sets A∈ Fn. It follows that
E XS+
=
1≤n≤∞
E
Xn+; [S =n]
≤
1≤n≤∞
E
Z+; [S =n]
=E(Z+)<∞. Thus XS ∈ E(P). As XS is FS-measurable (3.b.0.(c)), it will now suffice to show that E
1AXS
=E 1AZ
, for all setsA∈ FS. If A∈ FS, then A∩[S=n]∈ Fn, for alln∈N∞, and it follows that
E 1AXS
=
1≤n≤∞E
XS;A∩[S=n]
=
1≤n≤∞E
Xn;A∩[S=n]
=
1≤n≤∞E
Z;A∩[S=n]
=E 1AZ
.
(b) IfS ≤T, then FS ⊆ FT and so, using (a), E(XT|FS) =E(E(Z|FT)| FS) = E(Z|FS) =XS.
5.b.1. Assume that (Xn)1≤n<∞ is a nonnegative supermartingale. Then X∞ = 0 is a last element for X. IfS, T : Ω→N∞are any optional times with S≤T, then XS, XT ∈L1(P) and we haveXS≥E(XT|FS).
Remark. Here we setXS =X∞= 0 on the set [S=∞] and the same forXT. Proof. Let us first show that XS is integrable. For n ≥ 1 apply the Optional Sampling Theorem 3.b.1 to the bounded optional times 1, S∧n ≥1 to conclude that E
XS∧n
≤ E(X1). Since Xn ≥ 0 = X∞, we have XS ≤ lim infnXS∧n and Fatou’s Lemma now shows that E(XS) ≤ lim infnE(XS∧n) ≤ E(X1) < ∞. Combined with the nonnegativity ofXS this shows that XS ∈L1(P).
Assume now that S ≤T. To see that XS ≥E(XT|FS) we must show that E
1AXS
≥ E 1AXT
, for all sets A ∈ FS. Let A ∈ FS and n ≥ 1. Then A∩[S≤n]∈ FS∧n (3.b.0.(g)). Thus the Optional Sampling Theorem 3.b.1 applied to the bounded optional timesS∧n≤T∧nyields
E
XS∧n;A∩[S ≤n]
≥E
XT∧n;A∩[S≤n]
≥E
XT∧n;A∩[T ≤n]
, the second inequality following from A∩[S ≤n] ⊇ A∩[T ≤ n] and XT∧n ≥ 0.
Since XS∧n = XS on the set [S ≤ n] (and a similar fact for XT∧n), this can be rewritten asE
XS;A∩[S ≤n]
≥E
XT;A∩[T ≤n]
. Letting n↑ ∞and using increasing convergence it follows that
E
XS;A∩[S <∞]
≥E
XT;A∩[T <∞] .
As XS =X∞= 0 on the set [S =∞] and XT = 0 on the set [T =∞], this can be rewritten as E
1AXS
≥E 1AXT
, as desired.
5.b.2 Discrete Optional Sampling Theorem. Let X = (Xn)1≤n≤∞ be a closed submartingale sequence. Then we have E
XT+
<∞andXS ≤E(XT|FS), for all optional times S, T : Ω→N∞ withS≤T.
Proof. LetZ =X∞. ThenE(Z+)<∞and Xn ≤E(Z|Fn). SetAn =E(Z|Fn).
Then An and−Xn are supermartingales and so Bn =An−Xn is a nonnegative supermartingale satisfying Xn =An−Bn. Defining A∞ =E(Z|F∞) =Z =X∞ and B∞ = 0 we have X∞ =A∞−B∞. Here 5.b.0 and 5.b.1 apply to A and B respectively: IfS, T : Ω→N∞are optional times withS≤T, thenBT ∈L1(P) and XS =AS−BS, XT =AT −BT. MoreoverBS ≥E(BT|FS) andAS =E(AT|FS) and thus
XS =AS−BS ≤E(AT|FS)−E(BT|FS) =E(XT|FS).
It remains to be shown only that E(XT+) < ∞. Since BT ≥ 0 we have XT+ ≤ A+T = 1[AT≥0]AT. Moreover, according to 5.b.0, AT = E(Z|FT) and integrating this equality over the set [AT ≥0]∈ FT yields
E(XT+)≤E(A+T) =E
1[AT≥0]AT
=E
1[AT≥0]Z
≤E(Z+)<∞.
46 5.b Sampling of closed submartingale sequences.
Remark. XS is FS-measurable (3.b.0.(c)). According to 2.b.4.(i) the inequality XS ≤E(XT|FS) is thus equivalent withE
1AXS
≤E 1AXT
, for all setsA∈ FS, and implies in particular that E(XS)≤E(XT). The corresponding fact for closed supermartingales (all inequalities reversed) follows easily from this. IfX is a closed martingale, then it is both a closed supermartingale and a closed submartingale with the same last element. The conclusion of 5.b.2 then becomesXT ∈L1(P) and XS =E(XT|FS), for all optional timesS, T : Ω→N∞ withS≤T.