Math 7244 week 6
We saw that progressive implies adapted and measurable. In the other direction, however, adapted and measurable does not imply progressive. To see an example, let η be a uniform [0, 1] random variable and define a process X by
Xt=
(
1 if t = η 0 otherwise. Then, writing N for the negligible sets, define a filtration Ft by
Ft =
(
σ(N ) if t < 1 σ(η, N ) if t ≥ 1,
with F∞= σ(∪tFt). Then for fixed t, Xt takes only two values, 0 and 1, and
{ω : Xt = 1} = {ω : Xt= 0}c
is a negligible set, and so is Ft-measurable. This means X is adapted. It is also measurable
since t − η is F∞× B([0, ∞))-measurable, and Xt(ω) = 1{t−η=0}. However it cannot be
progressive, since if it were, the map (ω, s) 7→ Xs(ω) on the interval [0, 1/2] would be
F1/2× B([0, 1/2])-measurable. This would mean that
{(ω, s) : Xs(ω) = 1 and s ≤ 1/2} = {(ω, s) : η = s ≤ 1/2}
would be F1/2× B([0, 1/2])-measurable, and so its projection onto the first coordinate {ω :
η ≤ 1/2} would be F1/2-measurable, which it is not.
In line with the previous example, although adapted measurable processes are not pro-gressive, the previous example has a progressive modification (just consider the zero process). This is true generally.
Theorem 0.1 (Chung-Doob). Every adapted measurable process has a progressive modifi-cation.
Moving on, we see that a adapted right-continuous (or left-continuous) process is pro-gressive.
Proposition 0.2. Let (Xt)t≥0 be a random process with values in a metric space (E, d).
Suppose that X is adapted and that its sample paths are right-continuous (or left-continuous). Then X is progressive.
Proof. We assume right continuity, as left continuity is similar. We will approximate the process by others which are obviously progressive. For any t ≥ 0 fixed, and for any n ≥ 1 and s ∈ [0, t], define a random variable Xn
s by
and Xn
t = Xt. (This is a “right-endpoint” approximation.) The right-continuity of sample
paths implies that for all ω ∈ Ω and s ∈ [0, t], one has Xs(ω) = lim
n→∞X n s(ω).
Indeed, this is clear for s = t. Otherwise, if > 0, choose δ > 0 such that if s0 satisfies s ≤ s0 < s + δ, then d(Xs0, Xs) < . Then for n large enough, one can find k such that
s ∈ [(k − 1)t/n, kt/n) and t/n < δ. Then
d(Xsn, Xs) = d(Xkt/n, Xs) < .
Now, Xn
t is progressive because if A is a Borel subset of E, then
{(ω, s) ∈ Ω × [0, t] : Xn s(ω) ∈ A} = {Xt∈ A} × {t} [ n [ k=1 {Xkt/n∈ A} × (k − 1)t n , kt n ! .
Both sets on the right are in Ft× B([0, t]). But now since Xs(ω) (viewed as a function of
(ω, s) for s ∈ [0, t]) is a pointwise limit of Xsn(ω), the limiting function is also Ft× B([0,
t])-measurable. In other words, X is progressive.
Given these discussions, we define the progressive sigma-field. It allows us to phrase the notion of progressive as a measurability relative to one sigma-field.
Definition 0.3. The collection P of all sets A ∈ F ×B([0, ∞)) such that the process Xt(ω) =
1A(ω, t) is progressive is a sigma-field on Ω × [0, ∞), the progressive sigma-field.
One can show:
• A ⊂ Ω × [0, ∞) is in P if and only if for every t ≥ 0, the set A ∩ (Ω × [0, t]) belongs to Ft× B([0, t]).
• A process X is progressive if and only if the mapping (ω, t) 7→ Xt(ω) is measurable
relative to P.
0.1
Stopping times and their sigma-fields
Recall the notion of a stopping time and its sigma-field.
Definition 0.4. A random variable T : Ω → [0, ∞] is a stopping time of the filtration (Ft)
if for every t ≥ 0, one has {T ≤ t} ∈ Ft. The sigma-field of the past before T is
Recall that if T is a stopping time, also {T < t} ∈ Ft for all t ≥ 0, and moreover
{T = ∞} = (∪∞n=1{T ≤ n})c∈ F∞.
In the previous section we introduced the right-continuous version of a filtration as (Ft+).
Proposition 0.5. Write Gt= Ft+.
1. A random variable T : Ω → [0, ∞] is a stopping time of (Gt) if and only if {T < t} ∈ Ft
for every t > 0. Equivalently, T ∧ t is Ft-measurable for every t > 0.
2. Let T be a stopping time of (Gt). Then
GT = {A ∈ F∞: for all t > 0, A ∩ {T < t} ∈ Ft}.
We then define FT + = GT.
Proof. If T is a stopping time of (Gt), then for t > 0,
{T < t} = [
q∈Q:0≤q<t
{T ≤ q}.
Each set on the right is in Gq, and for q < t, this is in Ft. Conversely, if {T < t} ∈ Ft for
every t > 0, then for t ≥ 0 and s > t,
{T ≤ t} = \
q∈Q:t<q<s
{T < q} ∈ Fs.
This is true for all s > t, so {T ≤ t} ∈ Ft+ = Gt.
For the second claim of item 1, we first note that T ∧ t being Ft-measurable is equivalent
to the statement that for every s < t, {T ≤ s} ∈ Ft. Indeed, assuming that T ∧ t is Ft
-measurable, if s < t, then {T ≤ s} = {T ∧ t ≤ s} ∈ Ft. Conversely, if for every s < t,
{T ≤ s} ∈ Ft, then for any s ∈ R,
{T ∧ t ≤ s} = (
{T ≤ s} if s ≤ t Ω if s > t , which are both sets in Ft, so {T ∧ t} is Ft-measurable.
Now suppose that T ∧ t is Ft-measurable for every t > 0, and write {T < t} =
∪s∈Q:0≤s<t{T ≤ s}. By the above equivalence, this intersection is in Ft, so by the first
part of item 1, T is a stopping time of (Gt). Conversely, if T is a stopping time of (Gt), then
{T ≤ s} ∈ Gs ⊂ Ft whenever s < t, and so T ∧ t is Ft-measurable.
Moving to item 2, if A ∈ GT, then A ∩ {T ≤ t} ∈ Gt for every t ≥ 0, so for t > 0,
A ∩ {T < t} = [
q∈Q:0≤q<t
Because A ∩ {T ≤ q} ∈ Gq ⊂ Ft for every q < t, the above is in Ft. Conversely, if
A ∩ {T < t} ∈ Ft for every t > 0, then for t ≥ 0 and s > t, one has
A ∩ {T ≤ t} = \
q∈Q:t<q<s
(A ∩ {T < q}) ∈ Fs.
Therefore A ∩ {T ≤ t} ∈ Gt= Ft+, and so A ∈ GT.
The following hold for stopping times and their sigma-fields.
• For every stopping time T , we have FT ⊂ FT +. If the filtration (Ftis right-continuous,
then FT + = FT.
Proof. This follows directly from the definition. For instance, if A ∈ FT, then for
t ≥ 0, A ∩ {T ≤ t} ∈ Ft ⊂ Ft+, so A ∈ FT +.
• If T = t is a constant stopping time, then FT = Ft and FT + = Ft+.
Proof. Both are similar, so we show the first. If A ∈ Ft, then for s ≥ 0,
A ∩ {T ≤ s} = (
A if s ≥ t ∅ if s < t,
and both are in Fs. Conversely, if A ∈ FT, then one has A = A ∩ {T ≤ t} ∈ Ft.
• Let T be a stopping time. Then T is FT-measurable.
Proof. We already proved this.
• Let T be a stopping time and A ∈ F∞. Set
TA(ω) = (
T (ω) if ω ∈ A ∞ if ω /∈ A. Then A ∈ FT if and only if TA is a stopping time.
Proof. For every t ∈ [0, ∞),
{TA≤ t} = A ∩ {T ≤ t}.
The right side is in Ft for every t if and only if A ∈ FT. The left side is in Ftfor every
t if and only if TA is a stopping time.
Proof. We need only show that FS ⊂ FT because if this holds for an arbitrary filtration
(Ft), we just apply it to (Ft+) to get the other result. To show this, if A ∈ FS, then
A ∩ {T ≤ t} = (A ∩ {S ≤ t}) ∩ {T ≤ t} ∈ Ft,
and so A ∈ FT.
• Let S, T be stopping times. Then S ∧ T and S ∨ T are stopping times and FS∧T =
FS∩ FT. Also, {S ≤ T } ∈ FS∧T and {S = T } ∈ FS∧T.
Proof. We have
{S ∧ T ≤ t} = {S ≤ t} ∪ {T ≤ t} ∈ Ft
{S ∨ T ≤ t} = {S ≤ t} ∩ {T ≤ t} ∈ Ft,
so that S ∧ T and S ∨ T are stopping times. Since S ∧ T ≤ S and ≤ T , we obtain FS∧T ⊂ (FS∩ FT). Conversely, if A ∈ FS∩ FT, then
A{S ∧ T ≤ t} = (A ∩ {S ≤ t}) ∪ (A ∩ {T ≤ t}) ∈ Ft,
and so A ∈ FS∧T.
Last, for every t ≥ 0,
{S ≤ T } ∩ {T ≤ t} = {S ≤ t} ∩ {T ≤ t} ∩ {S ∧ t ≤ T ∧ t} ∈ Ft,
and
{S ≤ T } ∩ {S ≤ t} = {S ∧ t ≤ T ∧ t} ∩ {S ≤ t} ∈ Ft,
because S ∧ t and T ∧ t are both Ft-measurable (as they are measurable relative to
FS∧t ⊂ Ft and FT ∧t ⊂ FT). Therefore {S ≤ T } ∈ FS∩ FT = FS∧T. For the final
statement, {S = T } = {S ≤ T } ∩ {T ≤ S}.
• If (Sn) is a monotone increasing sequence of stopping times, then S = limn→∞Sn is
also a stopping time. Proof. For every t ≥ 0,
{S ≤ t} = ∩n{Sn≤ t} ∈ Ft.
• If (Sn) is a monotone decreasing sequence of stopping times, then S = limn→∞Sn is a
stopping time of the filtration (Ft+), and
Proof. For t > 0,
{S < t} = ∪n{Sn< t} ∈ Ft,
so by our previous proposition, S is a stopping time of (Ft+). Furthermore, FS+ ⊂ FSn+
for every n since Sn ≥ S. Conversely, if A ∈ ∩nFSn+, then
A ∩ {S < t} = ∪n(A ∩ {Sn< t}) ∈ Ft,
and so A ∈ FS+.
• If (Sn) is a monotone decreasing sequence of stopping times which is also stationary
(that is, for every ω, there is an integer N (ω) such that Sn(ω) = S(ω) for every
n ≥ N (ω)), then S = limn→∞Sn is also a stopping time and
FS = ∩nFSn.
Proof. For every t ≥ 0,
{S ≤ t} = ∪n{Sn≤ t} ∈ Ft,
so S is a stopping time. If A ∈ ∩nFSn, then
A ∩ {S ≤ t} = ∪n(A ∩ {Sn≤ t}) ∈ Ft,
giving A ∈ FS.
• Let T be a stopping time. A function ω 7→ Y (ω) defined on the set {T < ∞} and taking values in the measurable set (E, E ) is FT-measurable if and only if for every
t ≥ 0, the restriction of Y to the set {T ≤ t} is Ft-measurable.
Proof. First assume that for every t ≥ 0, the restriction of Y to {T ≤ t} is Ft
-measurable. Then for every measurable subset A of E, {Y ∈ A} ∩ {T ≤ t} ∈ Ft⊂ F∞.
Let t → ∞ to obtain {Y ∈ A} ∈ F∞, and so {Y ∈ A} is a set in F∞ such that for
every t ≥ 0, its intersection with {T ≤ t} is in Ft, meaning it is in FT, and so Y is
FT-measurable.
Conversely, if Y is FT-measurable, then {Y ∈ A} ∈ FT, and so {Y ∈ A} ∩ {T ≤ t} ∈
Ft, and this means the restriction of Y to {T ≤ t} is Ft-measurable.
We will often need to stop a process at a stopping time, and hope that the stopped process is FT-measurable. We showed this before with BM, but it is true for progressive
Theorem 0.6. Let (Xt)t≥0 be a progressive process with values in a measurable space (E, E ),
and let T be a stopping time. Then the function
ω 7→ XT(ω) := XT (ω)(ω),
which is defined on {T < ∞}, is FT-measurable.
Proof. To use the last stopping time property above, let t ≥ 0. Then the restriction of the function ω 7→ XT(ω) to the event {T ≤ t} is ω 7→ XT ∧t(ω), which is the composition
Φ2(Φ1(ω)), where Φ1 : {T ≤ t} → Ω × [0, t] is given by
Φ1(ω) = (ω, T (ω) ∧ t)
and Φ2 : Ω × [0, t] → E is given by
Φ2(ω, s) = Xs(ω).
The second is measurable from Ft× B([0, t]) to E since X is progressive. The first is
mea-surable from Ft to Ft× B([0, t]) because T ∧ t is Ft-measurable. Taking the composition,
the restriction of ω 7→ XT(ω) to {T ≤ t} is Ft-measurable, and this completes the proof by
the last item above.
We often need to approximate stopping times by discrete ones, and as we have seen before, we should approximate them from above.
Proposition 0.7. Let T be a stopping time and let S be and FT-measurable random variable
with values in [0, ∞] such that S ≥ T . Then S is also a stopping time. In particular, if T is a stopping time, Tn= ∞ X k=0 k + 1 2n 1{k2−n<T ≤(k+1)2−n}+ ∞ · 1{T =∞}
for n = 0, 1, . . . defines a sequence of stopping times that decreases to T .
Proof. Intuitively speaking, to check if S has occurred by time t, we can check if T has occurred, and at that time T , we know the value of S, so we can then check if S ≤ t. Rigorously, for t ≥ 0, since {S ≤ t} is FT-measurable,
{S ≤ t} = {S ≤ t} ∩ {T ≤ t} ∈ Ft,
and so S is a stopping time. The second fact follows because Tn ↓ T by construction, and
Tn ≥ T and is a function of T , so is FT-measurable (as T is).
Last we give two main examples of stopping times.
1. If X has right-continuous paths and O is an open set in E, then TO= inf{t ≥ 0 : Xt∈ O}
is a stopping time of (Ft+).
2. If X has continuous paths and F is a closed set in E, then TF = inf{t ≥ 0 : Xt ∈ F }
is a stopping time of (Ft).
Proof. For the first item, for t > 0,
{TO< t} = ∪s∈Q:0≤s<t{Xs ∈ O} ∈ Ft.
For the second item, for t ≥ 0, {TF ≤ t} = inf 0≤s≤td(Xs, F ) = 0 = inf s∈Q:0≤s≤td(Xs, F ) = 0 ∈ Ft.
0.2
Continuous-time martingales
The idea of martingale carries over from discrete to continuous time as a sort of fair game. We will take a filtered space (Ω, F , (Ft), P).
Definition 0.9. An adapted real-valued process (Xt)t≥0 with E|Xt| < ∞ for every t is called
1. a martingale if for every 0 ≤ s < t, E[Xt | Fs] = Xs,
2. a supermartingale if for every 0 ≤ s < t, E[Xt | Fs] ≤ Xs,
3. a submartingale if for every 0 ≤ s < t, E[Xt| Fs] ≥ Xs.
As in the discrete case,
• If X is a submartingale, then −X is a supermartingale.
• If X is a martingale (or submartingale or supermartingale) then for 0 ≤ s ≤ t, EXs=
EXt (or ≥ or ≤).
• The easiest example of a martingale is similar to the discrete case: take a random variable Z with E|Z| < ∞ and set
• What Le Gall calls “an important example” is as follows. We say that (Zt)t≥0 with
values in R or Rd has independent increments relative to (F
t) if it is adapted and if for
every 0 ≤ s < t, the variable Zt− Zs is independent of Fs. This is true for BM with
the canonical filtration. For such processes Z, then
1. if E|Zt| < ∞ for all t, then ˜Zt= Zt− EZt is a martingale,
2. if EZt2 < ∞ for all t, then Yt= ( ˜Zt)2− E( ˜Zt)2 is a martingale, and
3. if for some θ ∈ R, one has EeθZt < ∞ for all t, then
Xt=
eθZt
EeθZt
is a martingale.
Proof. This follows from straightforward calculations. All the variables are in L1 and
the processes are clearly adapted. To check the martingale property is direct. For the second, if 0 ≤ s < t, then E[(Z˜t)2 | Fs] = E[( ˜Zs+ ˜Zt− ˜Zs)2 | Fs] = ( ˜Zs)2+ 2 ˜ZsE[Z˜t− ˜Zs | Fs] + E[( ˜Zt− ˜Zs)2 | Fs] = ( ˜Zs)2+ E( ˜Zt− ˜Zs)2 = ( ˜Zs)2+ E( ˜Zt)2− 2E ˜ZsZ˜t+ E( ˜Zs)2 = ( ˜Zs)2+ E( ˜Zt)2− E( ˜Zs)2.
Here we have used
E[Z˜sZ˜t] = Eh ˜ZsE[Z˜t| Fs]
i
= E( ˜Zs)2.
This implies item 2. For item 3,
E[Xt| Fs] = eθZs E[eθ(Zt−Zs) | Fs] EeθZs Eeθ(Zt−Zs) = eθZs EeθZs = Xs.
Sometimes we want to consider a BM relative to a different filtration.
Definition 0.10. A real-valued process B = (Bt)t≥0 is an (Ft)-BM if B is a BM and if B
is adapted and has independent increments relative to (Ft). A similar definition is made for
d-dimensional (Ft)-BM.
One can show in particular that if B is a d-dimensional BM, and (FB
t ) is the canonical
filtration (possibly also completed), then B is a d-dimensional (FtB)-BM. From the above, the following are then martingales with continuous paths:
Bt, Bt2− t, e
The latter process is called an exponential martingale of BM.
If Z = N is a Poisson process with parameter λ and (Ft) is the canonical filtration of N ,
then one can show that Z has independent increments, and so Nt− λt, (Nt− λt)2− λt, exp θNt− λt(eθ− 1)
are martingales, but they do not have modifications with continuous paths. In other words, not all continuous-time martingales can be modified to have continuous paths.
Proposition 0.11. Let (Xt)t≥0 be an adapted process and f : R → [0, ∞) be a convex
function with Ef (Xt) < ∞ for every t ≥ 0.
1. If (Xt) is a martingale, then (f (Xt)) is a submartingale.
2. If (Xt) is a submartingale, and if f is nondecreasing, then (f (Xt)) is a submartingale.
Proof. These are analogues of discrete-time statements, and their proofs are identical. By Jensen,
E[f (Xt) | Fs] ≥ f (E[Xt| Fs]) ≥ f (Xs).
The last inequality holds as equality when (Xt) is a martingale. If it is only a submartingale,