The problem to consider first is to define an integral
∫ t
fdM
0
where f has values in H^{′} and M is a continuous martingale having values in H. For the
sake of simplicity assume M
(0)
= 0. The process of definition is the same as before. First
consider an elementary function
m∑−1
f (t) ≡ fkX(tk,tk+1](t) (62.6.22)
k=0
(62.6.22)
where f_{k} is measurable into H^{′} with respect to ℱ_{tk}. Then define
∫ t m∑−1
fdM ≡ fk(M (t∧ tk+1) − M (t∧tk)) ∈ ℝ (62.6.23)
0 k=0
(62.6.23)
Lemma 62.6.1The k^{th}term in the above sum is a martingale and the integral isalso a martingale.
Proof: Let σ be a stopping time with two values. Then
E (fk(M (σ ∧tk+1)− M (σ ∧tk)))
= E (E (fk (M (σ∧ tk+1) − M (σ∧ tk))|ℱt ))
k
= E (fkE ((M (σ ∧ tk+1)− M (σ ∧ tk))|ℱtk)) = 0
and it works the same with σ replaced with t. Hence by the lemma about recognizing
martingales, Lemma 62.1.1, each term is a martingale and so it follows that the integral
∫_{0}^{t}fdM is also a martingale. ■
Note also that, since M is continuous, this is a continuous martingale.
As before, it is important to estimate this.
(|∫ t |2)
E || fdM || ≤?
| 0 |
Consider a mixed term. For j < k, it follows from measurability considerations
that
E ((fk(M (t∧ tk+1 )− M (t∧tk)))(fj(M (t∧ tj+1)− M (t∧ tj))))
= E (E [(fk(M (t∧ tk+1)− M (t ∧tk)))(fj(M (t∧ tj+1)− M (t ∧tj)))|ℱtk])
= E ((fj(M (t∧tj+1)− M (t∧ tj)))fkE [(M (t∧ tk+1) − M (t∧tk))|ℱtk]) = 0
Therefore,
( ) ( )
||∫ t ||2 m∑−1 2
E || fdM || = E |fk(M (t∧ tk+1)− M (t∧ tk))|
0 k=0
(m∑−1 )
≤ E ∥fk∥2|M (t∧tk+1)− M (t∧ tk)|2
k=0
( )
m∑−1 2 ([ tk+1 tk] )
= E ∥fk∥ M − M (t)+ Nk (t)
k=0
(m −1 )
= E ∑ ∥f ∥2([M tk+1](t)− [M tk](t)+ N (t))
k=0 k k
( )
m∑ −1 2
= E ∥fk∥ ([M ](t∧tk+1)− [M ](t∧ tk)+ Nk (t))
k=0
where N_{k} is a martingale which equals 0 for t ≤ t_{k}. The above equals
(∫ t 2 ) ( ∫ t 2 )
E ∥f∥ d[M ] ≡ E ∥f∥ dν
0 0
the integral inside being the ordinary Lebesgue Stieltjes integral for the step function
where ν is the measure determined by the positive linear functional
∫
T
Λg = 0 gd [M ]
where the integral on the right is the ordinary Stieltjes integral. Thus, the following
inequality is obtained.
( )
||∫ t ||2 ( ∫ t )
E || fdM || ≤ E ∥f∥2d[M ], (62.6.24)
0 0
(62.6.24)
Now what would it take for
(||∫ t ||2)
E || fdM || (62.6.25)
0
(62.6.25)
to be well defined? A convenient condition would be to insist that each
∥fk∥
M^{∗} is in
L^{2}
(Ω )
where
M ∗(ω) ≡ sup |M (t)(ω)|
t∈[0,T] H
Is this condition also sufficient for the above integral 62.6.25 to be finite? From the
above, that integral equals
( )
m∑−1 2 2
E ∥fk∥ |M (t∧tk+1)− M (t∧ tk)|
k=0
( m∑−1 )
≤ E 4 ∥fk∥2(M ∗)2
k=0
Thus the condition that for each k,
∥fk∥
M^{∗}∈ L^{2}
(Ω)
is sufficient for all of the above to
consist of real numbers and be well defined.
Definition 62.6.2A function f is called an elementary function if it is a step functionof the form given in 62.6.22where each f_{k}is ℱ_{tk}measurable and for eachk,
∥fk∥
M^{∗}∈ L^{2}
(Ω)
. Define G_{M}to be the collection of functions f having values in H^{′}which have the property that there exists a sequence of elementary functions
{fn}
withf_{n}→ f in the space
L2(Ω;L2 ([0,T],ν))
Then picking such an approximating sequence,
∫ t ∫ t
fdM ≡ lim fndM
0 n→∞ 0
the convergence happening in L^{2}
(Ω )
.
The inequality 62.6.24 shows that this definition is well defined. So what are the
properties of the integral just defined? Each ∫_{0}^{t}f_{n}dM is a continuous martingale
because it is the sum of continuous martingales. Since convergence happens in L^{2}
(Ω)
, it
follows that ∫_{0}^{t}fdM is also a martingale. Is it continuous? By the maximal inequality
Theorem 61.9.4, it follows that
( )
( [ ||∫ t ∫ t || ]) 1 ||∫ T ||2
P sup || fmdM − fndM || > λ ≤ -2E ( || (fm − fn)dM || )
t∈[0,T] 0 0 λ |0 |
( )
1- ∫ T 2
≤ λ2E 0 ∥fm − fn∥ d[M ]
and it follows that there exists a subsequence, still called n such that for all p
positive,
([ ])
||∫ t ∫ t || 1 −n
P ts∈u[0p,T]|| 0 fn+pdM − 0 fndM || > n- < 2
By the Borel Cantelli lemma, there exists a set of measure zero N such that for ω
∕∈
N,
{∫t }
0 fndM
is a Cauchy sequence. Thus, what it converges to is continuous in t for each
ω
∕∈
N and for each t, it equals ∫_{0}^{t}fdM a.e. Hence we can regard ∫_{0}^{t}fdM as this
continuous version.
What is an example of such a function in G_{M}?
Lemma 62.6.3Let R : H → H^{′}be the Riesz map.
〈Rf,g〉 ≡ (f,g)H .
Also suppose M is a uniformly bounded continuous martingale with values in H. ThenRM ∈G_{M}.
Proof: I need to exhibit an approximating sequence of elementary functions as
described above. Consider
mn∑− 1
Mn (t) ≡ M (ti)X (tni,tni+1](t)
i=0
Then clearly RM_{n}
(ti)
M^{∗}∈ L^{∞}
(Ω)
and so in particular it is in L^{2}
(Ω)
. Here
{| | }
lnim→∞ max |tni − tni+1|,i = 0,⋅⋅⋅,mn = 0.
Say M^{∗}
(ω)
≤ C. Furthermore, I claim that
( )
∫ T 2
nli→m∞E 0 ∥RMn − RM ∥ d [M ] = 0. (62.6.26)
(62.6.26)
This requires a little proof. Recall the description of
[M ]
(t)
. It was as follows. You
considered
∑ (( ( n ) n ) n )
Pn (t) ≡ 2 M t∧τk+1 − M (t ∧τk) ,M (t∧ τk)
k≥0
where the stopping times were defined such that τ_{k+1}^{n} is the first time t > τ_{k}^{n} such that
|M (t)− M (τnk)|
^{2} = 2^{−n}and τ_{0}^{n} = 0. Recall that lim_{k→∞}τ_{k}^{n} = ∞ or T in the way it
was formulated earlier. Then it was shown that P_{n}
(t)
converged to a martingale P
(t)
in
L^{1}
(Ω )
. Then by the usual procedure using the Borel Cantelli lemma, a subsequence
converges to P
(t)
uniformly off a set of measure zero. It is easy to estimate
P_{n}
. In case M is just a continuous local martingale, theabove limit happens in probability.
Proof: First suppose M is uniformly bounded.
mn∑−1| ( n ) n |2
|M t∧ tk+1 − M (t∧ tk)|H
k=0
m −1 m − 1
= n∑ ||M (t∧ tn )||2 − |M (t∧ tn)|2 − 2 ∑n (M (t∧ tn),M (t∧ tn )− M (t∧ tn))
k=0 k+1 k k=0 k k+1 k
2 mn∑− 1( n ( n ) n)
= |M (t)|H − 2 M (t∧ tk),M t ∧tk+1 − M (t∧ tk)
k=0
2 mn∑− 1 n ( ( n ) n )
= |M (t)|H − 2 RM (t∧ tk) M t∧ tk+1 − M (t∧tk)
k=0
2 mn∑− 1 n ( ( n ) n )
= |M (t)|H − 2 RM (tk) M t∧ tk+1 − M (t∧ tk)
k=0