61.1 Fundamental Definitions And Properties
Here E will be a separable Banach space and ℬ
will be the Borel sets of
be a probability space and
will be an interval of ℝ
. A set of E
variables, one for each t ∈ I,
is called a stochastic process
. Thus for each t,
is a measurable function of
Ω. Set X
Functions t → X
are called trajectories.
Thus there is a trajectory for each ω ∈
Ω. A stochastic
is called a version
or a modification
of a stochastic process, X
if for all
t ∈ I,
There are several descriptions of stochastic processes.
- X is measurable if X :
I × Ω → E is B
×ℱ measurable. Note that a
stochastic process, X is not necessarily measurable.
- X is stochastically continuous at t0 ∈ I means: for all ε > 0 and δ > 0 there exists
ρ > 0 such that
Note the above condition says that for each ε > 0,
- X is stochastically continuous if it is stochastically continuous at every
t ∈ I.
- X is stochastically uniformly continuous if for every ε,δ > 0 there exists ρ > 0 such
that whenever s,t ∈ I with
< ρ, it follows
- X is mean square continuous at t0 ∈ I if
- X is mean square continuous in I if it is mean square continuous at every point of
- X is continuous with probability 1 or continuous if t → X is continuous for all
ω outside some set of measure 0.
- X is Hölder continuous if t → X is Hölder continuous for a.e.
Lemma 61.1.1 A stochastically continuous process on
≡ I is uniformly
stochastically continuous on
Proof: If this is not so, there exists ε,δ > 0 and points of I,sn,tn such that even
Taking a subsequence, still denoted by sn and tn there exists t ∈ I such that the above
But the sum of the last two terms converges to 0 as n →∞
by stochastic continuity of X
, violating 61.1.1
for all n
large enough. This proves the lemma.
For a stochastically continuous process defined on a closed and bounded interval,
there always exists a measurable version. This is significant because then you can do
things with product measure and iterated integrals.
Proposition 61.1.2 Let X be a stochastically continuous process defined on a
closed interval, I ≡
. Then there exists a measurable version of X.
Proof: By Lemma 61.1.1 X is uniformly stochastically continuous and so
there exists a sequence of positive numbers,
such that if
be a partition of
is obviously B
measurable because it is the sum of functions which are.
Consider the set, A
is a Cauchy sequence. This set is of the
and so it is a B
measurable set. Now define
I claim Y
To see this, consider 61.1.2
. From the construction of
it follows that for each t,
Also, for a fixed t, if Xn
fails to converge to
must be in infinitely
many of the sets,
which is a set of measure zero by the Borel Cantelli lemma. Recall why this is
Therefore, for each t,
for a.e. ω.
a.e. and so
measurable version of X
Lemma 61.1.3 Let D be a dense subset of an interval, I =
: D → E satisfies
for all d′,d ∈ D. Then X extends uniquely to a continuous Y defined on
Proof: Let t ∈ I and let dk → t where dk ∈ D. Then
is a Cauchy sequence
converges. The thing it
converges to will be called
Note this is well defined, giving X
t ∈ D
. Also, if
dk → t
and dk′→ t,
and so X
converge to the same thing. Therefore, it makes sense to define
only remains to verify the estimate. But letting
be small enough,
is arbitrary, this proves the existence part of the lemma. Uniqueness follows from
observing that Y
must equal lim
. This proves the lemma.