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
the submartingales
||M (t)||
^{2} for M
(t)
∈ℳ_{T}^{2}
(H)
where M 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.
Theorem 62.7.1Let
{Xn}
be a submartingale. Then there exists a uniquestochastic process,
{An }
and martingale,
{Mn }
such that
A_{n}
(ω)
≤ A_{n+1}
(ω)
,A_{1}
(ω)
= 0,
A_{n}is ℱ_{n−1}adapted for all n ≥ 1 where ℱ_{0}≡ℱ_{1}.
= E (Xn+1 |ℱn )− E (An+1 − An |ℱn )− An
= E (Xn+1 |ℱn )− E (E (Xn+1 − Xn|ℱn)|ℱn) − An
= E (Xn+1 |ℱn )− E (Xn+1 − Xn |ℱn )− An
= E (Xn |ℱn )− An
= Xn − An ≡ Mn
This proves the existence part.
It remains to verify uniqueness. Suppose then that
X = M + A = M ′+ A ′
n n n n n
where
{An}
and
′
{An}
both satisfy the conditions of the theorem and
{Mn }
and
′
{M n}
are both martingales. Then
′ ′
Mn − M n = An − An
and so, since A_{n}^{′}− A_{n} is ℱ_{n−1} measurable and
{Mn − Mn′}
is a martingale,
M − M ′ = E(M − M ′|ℱ )
n−1 n−1 n′ n n−1
= E(A n − An|ℱn−1)
= A′n − An = Mn − Mn′.
Continuing this way shows M_{n}−M_{n}^{′} is a constant. However, since A_{1}^{′}−A_{1} = 0 = M_{1}−M_{1}^{′},
it follows M_{n} = M_{n}^{′} and this proves uniqueness. This proves the theorem.
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.
Then the first term equals zero because since A_{j} is ℱ_{j−1} measurable,
∫ ∫ ∫ ∫
A M dP − A M = A E (M |ℱ )dP − A M dP
Ω j j−1 Ω j j ∫Ω j j j−1 ∫Ω j j
= Ω E (AjMj|ℱj−1)dP − Ω AjMjdP
∫ ∫
= AjMjdP − AjMjdP = 0.
Ω Ω
The last term equals
∫ ∫
Mn −1AndP = E (Mn |ℱn− 1)AndP
Ω ∫Ω
= E (M A |ℱ )dP = E (M A ).
Ω n n n−1 n n
This proves the lemma.
Definition 62.7.4Let A be an increasing function defined on ℝ. By Theorem 3.3.4on Page 89there exists a positive linear functional, L defined on C_{c}
(ℝ)
givenby
∫
b
Lf ≡ a fdA where spt(f) ⊆ [a,b]
where the integral is just the Riemann Stieltjes integral. Then by the Riesz representationtheorem, Theorem 10.3.2on Page 779, there exists a unique Radon measure, μ whichextends this functional, as described in the Riesz representation theorem. Then for B ameasurable set, I will write either
∫ ∫
B fdμ or B fdA
to denote the Lebesgue integral,
∫
XBf dμ.
Lemma 62.7.5Let f be right continuous. Then f is Borel measurable. Also,if the limit from the left exists, then f_{−}
(x)
≡ f
(x)
_{−}≡ lim_{y→x−}f
(y)
is alsoBorel measurable. If A is an increasing right continuous function and f is rightcontinuous and f_{−}, the left limit function exists, then if f is bounded, on
, denote by I_{x} the union of all intervals containing x such
that f
(y)
is larger than a for all y in the interval. Since f is right continuous,
each I_{x} has positive length. Now if I_{x} and I_{y} are two of these intervals, then
either they must have empty intersection or they are the same interval. Thus
f^{−1}
((a,∞ ))
is of the form ∪_{x∈f−1((a,∞ ))
}I_{x} and there can only be countably many
distinct intervals because each has positive length and ℝ is separable. Hence
f^{−1}
((a,∞ ))
equals the countable union of intervals and is therefore, Borel measurable.
Now
f− (x) = lim f (x− rn) ≡ lim frn (x )
n→ ∞ n→ ∞
where r_{n} is a decreasing sequence converging to 0. Now each f_{rn} is Borel measurable by
the first part of the proof because it is right continuous and so it follows the same is true
of f_{−}.
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
(c,d + h)
which approximates X_{(c,d]}
will show that for μ the measure of Definition 62.7.4
n
∑p ( p )
pl→im∞ f xk−1 X(xk−1,xk](x) = f− (x)
k=1
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
lemma.
which is right continuous issaid to be naturalif A
(0)
= 0 and whenever
{ξ(t)}
is a bounded right continuousmartingale,
( )
∫
E (A (t)ξ(t)) = E (0,t]ξ− (s)dA(s) . (62.7.30)
(62.7.30)
Here
ξ− (s,ω ) ≡ lim ξ(r,ω)
r→s−,r∈D
a.e. where D is a countable dense subset of
[0,t]
. By Corollary 61.8.2the right side of62.7.30is not dependent on the choice of D since if ξ_{−}is computed using two differentdense 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
(t,ω)
is increasing and right continuous. Therefore, one
can do the Lebesgue Stieltjes integral defined in Definition 62.7.4 for each ω whenever f
is Borel measurable and bounded. Now it is assumed
{ξ (t)}
is bounded and right
continuous. By Lemma 62.7.5ξ_{−}
(t)
≡ lim_{r→t−,r∈D}ξ
(r)
is measurable and by this
lemma,
∫ np
ξ− (s)dA (s) = lim ∑ ξ(tp )(A (tp)− A (tp ))
(0,t] p→∞ k=1 k−1 k k−1
where
{tp}
k
_{k=1}^{np} is a sequence of partitions of
[0,t]
such that
lim max {||tp− tp || : k = 1,2,⋅⋅⋅,np} = 0. (62.7.31)
p→ ∞ k k−1
(62.7.31)
and D ≡∪_{p=1}^{∞}∪_{k=1}^{np}
p
{tk}
_{k=1}^{np}.
Also, if t → A
(t,ω )
is right continuous, hence Borel measurable, then for ξ
(t)
the above bounded right continuous martingale, it follows it makes sense to
write
∫
ξ (s)dA (s).
(0,t]
Consider the right sum,
∑np ( ( ))
ξ(tpk) A (tpk)− A tpk−1
k=1
This equals
∫ n∑p p
(0,t] ξ(tk)X (tpk−1,tpk](s)dA(s)
k=1
and by right continuity, it follows
∑np p
pli→m∞ ξ(tk)X(tpk−1,tpk](s) = ξ(s)
k=1
and so the dominated convergence theorem applies and it follows
np ∫
∑ p ( p ( p ))
lp→im∞ ξ(tk) A (tk) − A tk−1 = (0,t]ξ(s)dA (s)
k=1
where this is a random variable. Thus
(∫ ) ∫ ( ∫ np )
E ξ (s)dA (s) = lim ∑ ξ(tp)X p p(s)dA (s) dP
(0,t] Ω p→ ∞ (0,t]k=1 k (tk−1,tk]
(62.7.32)
(62.7.32)
Now as mentioned above,
∫ n∑p n∑p ( ( ))
ξ(tpk)X (tp ,tp](s)dA (s) = ξ(tpk) A (tpk)− A tpk−1
(0,t]k=1 k−1 k k=1
and since A is increasing, this is bounded above by an expression of the form CA
(t)
, a
function in L^{1}. Therefore, by the dominated convergence theorem, 62.7.32 reduces to
∫ ∫ ∑np p
pli→m∞ Ω (0,t] ξ (tk)X(tpk−1,tpk](s)dA (s)dP
∫ n k=1
∑p p ( p ( p ))
= pli→m∞ Ω ξ(tk) A(tk)− A tk−1 dP
∫ k(=1n n −1 )
∑ p p p p∑ (p ) p
= pli→m∞ Ω ξ (tk)A (tk)− ξ tk+1 A (tk) dP
k=1 k=0
np∑−1∫ ( p (p )) p ∫
= pli→m∞ Ω ξ(tk) − ξ tk+1 A (tk)dP + Ω ξ(t)A (t)dP.(62.7.33)
k=1
Since ξ is a martingale,
∫ ∫
( p ) p ( (p ) p p)
Ωξ tk+1 A (tk)dP = ΩE ξ tk+1 A (tk)|ℱtk dP
∫ p ( (p ) )
= A (tk)E ξ tk+1 |ℱtpk dP
∫Ω
= A (tpk)ξ(tpk)dP
Ω
and so in 62.7.33 the term with the sum equals 0 and it reduces to
E (ξ(t)A(t)).
This is sufficiently interesting to state as a lemma.
Lemma 62.7.7Let A be an increasing adapted stochastic process which is rightcontinuous. Also let ξ