62.7 Doob Meyer Decomposition
This section is on the Doob Meyer decomposition which is a way of starting with a
submartingale and writing it as the sum of a martingale and an increasing adapted
stochastic process of a certain form. This is more general than what was done above with
is a continuous martingale.
There are two forms for this theorem, one for discrete martingales and one for
martingales defined on an interval of the real line which is much harder. According to [?]
this material is found in [?]
however, I am following [?]
for the continuous version of this
Theorem 62.7.1 Let
be a submartingale. Then there exists a unique
,A1 = 0
- An is ℱn−1 adapted for all n ≥ 1 where ℱ0 ≡ℱ1.
and also Xn = Mn + An.
Proof: Let A1 ≡ 0 and define
It follows An is ℱn−1 measurable. Since
is a submartingale,
It is a submartingale because
Now let Mn be defined by
Then from 62.7.27,
This proves the existence part.
It remains to verify uniqueness. Suppose then that
both satisfy the conditions of the theorem and
are both martingales. Then
and so, since An′− An is ℱn−1 measurable and
is a martingale,
Continuing this way shows Mn−Mn′
is a constant. However, since A1′−A1
= 0 = M1 −M1′,
it follows Mn
and this proves uniqueness. This proves the theorem.
Definition 62.7.2 A stochastic process,
which satisfies the conditions of
where ℱ0 ≡ℱ1 is said to be natural.
The Doob Meyer theorem needs to be extended to continuous submartingales and this
will require another description of what it means for a stochastic process to be natural.
To get an idea of what this condition should be, here is a lemma.
Lemma 62.7.3 Let a stochastic process,
be natural. Then for every martingale,
Proof: Start with the right side.
Then the first term equals zero because since Aj
The last term equals
This proves the lemma.
Definition 62.7.4 Let A be an increasing function defined on ℝ. By Theorem 3.3.4
on Page 89 there exists a positive linear functional, L defined on Cc
where the integral is just the Riemann Stieltjes integral. Then by the Riesz representation
theorem, Theorem 10.3.2 on Page 779, there exists a unique Radon measure, μ which
extends this functional, as described in the Riesz representation theorem. Then for B a
measurable set, I will write either
to denote the Lebesgue integral,
Lemma 62.7.5 Let f be right continuous. Then f is Borel measurable. Also,
if the limit from the left exists, then f−
Borel measurable. If A is an increasing right continuous function and f is right
continuous and f−, the left limit function exists, then if f is bounded, on
is a sequence of partitions of
More generally, let
Then 62.7.29 holds.
Proof: For x ∈ f−1
denote by Ix
the union of all intervals containing x
is larger than
for all y
in the interval. Since f
is right continuous,
has positive length. Now if Ix
are two of these intervals, then
either they must have empty intersection or they are the same interval. Thus
is of the form
and there can only be countably many
distinct intervals because each has positive length and ℝ
is separable. Hence
equals the countable union of intervals and is therefore, Borel measurable.
where rn is a decreasing sequence converging to 0. Now each frn is Borel measurable by
the first part of the proof because it is right continuous and so it follows the same is true
Finally consider the claim about the integral. Since A is right continuous, a simple
argument involving the dominated convergence theorem and approximating (c,d] with a
piecewise linear continuous function nonzero only on
will show that for μ
the measure of Definition 62.7.4
Therefore, the sum in 62.7.29 is of the form
and by 62.7.28
for each x ∈ (a,b]. Therefore, since f is bounded, 62.7.29 follows from the dominated
convergence theorem. The last claim follows the same way. This proves the
Definition 62.7.6 An increasing stochastic process,
which is right continuous is
said to be natural if A
is a bounded right continuous
a.e. where D is a countable dense subset of
. By Corollary 61.8.2 the right side of
62.7.30 is not dependent on the choice of D since if ξ− is computed using two different
dense subsets, the two random variables are equal a.e.
Some discussion is in order for this definition. Pick ω ∈ Ω. Then since A is right
continuous, the function t → A
is increasing and right continuous. Therefore, one
can do the Lebesgue Stieltjes integral defined in Definition
for each ω
is Borel measurable and bounded. Now it is assumed
is bounded and right
continuous. By Lemma
is measurable and by this
is a sequence of partitions of
and D ≡∪p=1∞∪k=1np
Also, if t → A
is right continuous, hence Borel measurable, then for
the above bounded right continuous martingale, it follows it makes sense to
Consider the right sum,
and by right continuity, it follows
and so the dominated convergence theorem applies and it follows
where this is a random variable. Thus
Now as mentioned above,
and since A is increasing, this is bounded above by an expression of the form CA
function in L1.
Therefore, by the dominated convergence theorem, 62.7.32
is a martingale,
and so in 62.7.33
the term with the sum equals 0 and it reduces to
This is sufficiently interesting to state as a lemma.
Lemma 62.7.7 Let A be an increasing adapted stochastic process which is right
continuous. Also let ξ
be a bounded right continuous martingale. Then