will be a continuous submartingale and a < b. Let
X
(t)
≡
(Y (t) − a)
_{+} + a so X
(0)
≥ a. Then X is also a submartingale. It is an
increasing convex function of one. If Y
(t)
has an upcrossing of
[a,b]
, then X
(t)
starts off
at a and ends up at least as large as b. If X
(t)
has an upcrossing of
[a,b]
then it must
start off at a since it cannot be smaller and it ends up at least as large as b.
Thus we can count the upcrossings of Y
(t)
by considering the upcrossings of
X
(t)
.
The next task is to consider an upcrossing estimate as was done before for discrete
submartingales.
τ0 ≡ min (i(nf{{t > 0 : X (t) = a} ,M ), } )
τ1 ≡ min i(nf{t > 0 : (X (t ∨τ0)− X (τ0))+ = b− a},M ),
τ2 ≡ min inf t > 0 : (X (τ1)− X (t∨ τ1))+ = b− a ,M ,
( { } )
τ3 ≡ min i(nf{t > 0 : (X (t ∨τ2)− X (τ2))+ = b− a},M ),
τ4 ≡ min inf t > 0 : (X (τ3)− X (t∨ τ3))+ = b− a ,M ,
..
.
If X
(t)
is never a, then τ_{0} = M and there are no upcrossings. It is obvious τ_{1}≥ τ_{0} since
otherwise, the inequality could not hold. Thus the evens have X
(τ2k)
= a and
X
(τ2k+1)
= b.
Lemma 61.10.1The above τ_{i}are stopping times for t ∈
[0,M ]
.
Proof:It is obvious that τ_{0} is a stopping time because it is the minimum of M and
the first hitting time of a closed set by a continuous adapted process. Consider a stopping
time η ≤ M and let
{ }
σ ≡ inf t > 0 : (X (t∨η) − X (η))+ = b− a
I claim that t → X
(t∨η)
− X
(η)
is adapted to ℱ_{t}. Suppose α ≥ 0 and consider
[ ]
(X (t∨ η)− X (η))+ > α (61.10.36)
(61.10.36)
The above set equals
([(X (t∨η )− X (η))+ > α]∩ [η ≤ t])∩ ([(X (t∨ η)− X (η))+ > α] ∩[η > t])
Consider the second of the above two sets. Since α ≥ 0, this set is ∅. This is because for
η > t, X
(t ∨η)
− X
(η)
= 0. Now consider the first. It equals
[(X (t∨η) − X (η)) > α]∩ [η ∨t ≤ t],
+
a set of ℱ_{t∨η} intersected with
[η ∨ t ≤ t]
and so it is in ℱ_{t} from properties of stopping
times.
If α < 0, then 61.10.36 reduces to Ω, also in ℱ_{t}. Therefore, by Proposition 61.7.2, σ is
a stopping time because it is the first hitting time of a closed set of a continuous adapted
process. It follows that σ ∧M is also a stopping time. Similarly t → X
(η)
−X
(t∨η )
is
adapted and
{ }
σ ≡ inf t > 0 : (X (η)− X (t∨η))+ = b− a
is also a stopping time from the same reasoning. It follows that the τ_{i} defined above are
all stopping times. ■
Note that in the above, if η = M, then σ = M also. Thus in the definition of the
τ_{i}, if any τ_{i} = M, it follows that also τ_{i+1} = M and so there is no change in
the stopping times. Also note that these stopping times τ_{i} are increasing as i
increases.
Let
nM ∑n --X-(τ2k+1)−-X-(τ2k)--
U[a,b] ≡ εli→m0 ε+ X (τ2k+1) − X (τ2k)
k=0
Note that if an upcrossing occurs after τ_{2k} on
[0,M ]
, then τ_{2k+1}> τ_{2k} because there
exists t such that
(X (t∨ τ2k)− X (τ2k)) = b − a
+
However, you could have τ_{2k+1}> τ_{2k} without an upcrossing occuring. This happens
when τ_{2k}< M and τ_{2k+1} = M which may mean that X
(t)
never again climbs to b. You
break the sum into those terms where X
(τ2k+1)
−X
(τ2k)
= b−a and those where this is
less than b − a. Suppose for a fixed ω, the terms where the difference is b − a are for
k ≤ m. Then there might be a last term for which X
(τ2k+1)
−X
(τ2k)
< b−a because it
fails to complete the up crossing. There is only one of these at k = m + 1. Then the above
sum is
∑m
≤ --1-- X (τ2k+1) − X (τ2k)+ --X-(M-)−-a--
b − ak=0 ε+ X (M )− a
1 ∑n X (M )− a
≤ b-−-a X (τ2k+1) − X (τ2k)+ ε+-X-(M-)−-a-
k=0
--1--∑n
≤ b − a X (τ2k+1) − X (τ2k)+ 1
k=0
Then U_{}
[a,b]
^{nM} is clearly a random variable which is at least as large as the number of
upcrossings occurring for t ≤ M using only 2n + 1 of the stopping times. From the
optional sampling theorem,
∫
E (X (τ2k))− E (X (τ2k−1)) = X (τ2k) − X (τ2k−1)dP
∫Ω ( )
= E X (τ2k)|ℱτ2k−1 − X (τ2k−1)dP
∫Ω
≥ X (τ2k− 1) − X (τ2k−1)dP = 0
Ω
Note that, X
(τ2k)
= a while X
(τ2k−1)
= b so the above may seem surprising.
However, the two stopping times can both equal M so this is actually possible. For
example, it could happen that X
(t)
= a for all t ∈
[0,M ]
.
Next, take the expectation of both sides,
( ) n
E U nM ≤ -1--∑ E (X (τ2k+1))− E (X (τ2k))+ 1
[a,b] b− ak=0
n n
--1--∑ --1--∑
≤ b− a E (X (τ2k+1)) − E (X (τ2k))+ b − a E (X (τ2k))− E (X (τ2k−1)) +1
k=0 k=1
1 1 ∑n
= b−-a-(E (X (τ1)) − E(X (τ0)))+ b-−-a E (X (τ2k+1))− E (X (τ2k−1)) +1
k=1
≤ --1--(E(X (τ2n+1))− E (X (τ0)))+ 1
b− a
≤ --1--(E(X (M ))− a)+ 1
b− a
which does not depend on n. The last inequality follows because 0 ≤ τ_{2n+1}≤ M and
X
(t)
is a submartingale. Let n →∞ to obtain
( M ) -1---
E U[a,b] ≤ b− a (E (X (M )) − a)+ 1
where U_{}
[a,b]
^{M} is an upper bound to the number of upcrossings of
{X (t)}
on
[0,M ]
.
This proves the following interesting upcrossing estimate.
Lemma 61.10.2Let
{Y (t)}
be a continuous submartingale adapted to a normalfiltration ℱ_{t}for t ∈
[0,M ]
. Then if U_{[a,b]
}^{M}is defined as the above upper bound to thenumber of upcrossings of
{Y (t)}
for t ∈
[0,M ]
, then this is a random variable and
( )
E U M ≤ --1--(E (Y (M )− a) + a− a)+ 1
[a,b] b − a +
= --1--E |Y (M )|+ -1--|a|+ 1
b − a b− a
With this it is easy to prove a continuous submartingale convergence theorem.
Theorem 61.10.3Let
{X (t)}
be a continuous submartingale adapted to a normalfiltration such that
sup{E (|X (t)|)} = C < ∞.
t
Then there exists X_{∞}∈ L^{1}
(Ω)
such that
lim X (t)(ω) = X∞ (ω) a.e. ω.
t→ ∞
Proof: Let U_{}
[a,b]
be defined by
U[a,b] = lim UM[a,b].
M→ ∞
Thus the random variable U_{}
[a,b]
is an upper bound for the number of upcrossings. From
Lemma 61.10.2 and the assumption of this theorem, there exists a constant C
independent of M such that
( )
E UM ≤ -C--+ 1.
[a,b] b− a
Letting M →∞, it follows from monotone convergence theorem that
( ) C
E U[a,b] ≤ ----+ 1
b− a
also. Therefore, there exists a set of measure 0 N_{ab} such that if ω
∕∈
N_{ab}, then
U_{[a,b]
}
(ω)
< ∞. That is, there are only finitely many upcrossings. Now let
N = ∪{N : a,b ∈ ℚ} .
ab
It follows that for ω
∈∕
N, it cannot happen that
lim sup X (t) (ω )− lim inf X (t)(ω) > 0
t→∞ t→ ∞
because if this expression is positive, there would be arbitrarily large values of t where
X
(t)
(ω)
> b and arbitrarily large values of t where X
(t)
(ω )
< a where a,b are rational
numbers chosen such that
lim sup X (t)(ω) > b > a > lim it→nf∞ X (t)(ω)
t→∞
Thus there would be infinitely many upcrossings which is not allowed for ω
∕∈
N.
Therefore, the limit lim_{t→∞}X
(t)
(ω )
exists for a.e. ω. Let X_{∞}
(ω)
equal this limit for
ω
∕∈
N and let X_{∞}
(ω)
= 0 for ω ∈ N. Then X_{∞} is measurable and by Fatou’s
lemma,