Brownian Motion and Stochastic Calculus
Xiongzhi Chen
University of Hawaii at Manoa
Department of Mathematics
August 3, 2008
Contents
1 Dsicrete Time Martingales
1
2 Continuous-Time Martingales
5
Brownian Motion and Stochastic Calculus
Basic Properties of Continuous-Time Martingales
1
Dsicrete Time Martingales
Lemma 1 Let z = z (!) be the integrable random variable (E (jz (!)j) < 1) and be a Markov
time relative to the system fFt g ; t 2 T: Then on the set f! : = tg the conditional mathematical
expectation E (zjF ) coincides with E (zjFt ) ;i.e.,
E (zjF ) = E (zjFt ) ; P -a:s: on f! :
= tg
Proof. It must be shown that
P (f = tg
or, equivalently,
f =tg E
Since
f =tg
:=
T
fE (zjF ) 6= E (zjFt )g) = 0
(zjF ) =
f =tg E
(zjFt ) ; P -a:s:
is both Ft - and F -measurable, then
f =tg E
(zjF ) = E ( zjF ) ;
f =tg E
(zjFt ) = E ( zjFt )
and it su¢ ces to show
E ( zjFt ) = E ( zjF ) ; P -a:s:
First of all, note that E ( zjFt ) is F -measurable. Actually, let s 2 T and a 2 R: Then if t
s;evidently
T
fE ( zjFt ) ag f
sg 2 Fs
since fE ( zjFt )
ag 2 Ft : But if t > s;then the set
B = fE ( zjFt )
ag
T
T
f
sg
= f E (zjFt ) ag f
sg
?
if a < 0
=
f
sg if a 0
1
since f =tg := = 0 on f
sg ; i.e., f
sg f E (zjFt ) = 0g for all t > s:
Further, according to the de…nition of the c.m.e for all A 2 F
Z
Z
Z
zdP
zdP =
E ( zjF ) dP =
Since A
f = tg 2 Ft ;then
Z
Z
zdP =
A\f =tg
A\f =tg
A
A
T
Z
E (zjFt ) dP =
E (zjFt ) dP =
E ( zjFt ) dP
A
A
A\f =tg
Z
The arbitrariness of A 2 F and the F -measurability of E (zjFt ) implies that
E ( zjFt ) = E ( zjF ) ; P -a:s:
as desired.
Remark 2 For a right-continuous distribution function F ( ) according to the law P ((a; b]) =
F (b) F (a) and a random variable (x) the Lebesgue-Stieltijes integral
Z
b
(x) P (dx) :=
a
Z
b
(x) dF (x)
a
can be reduced to an integral with respect to Lebesgue measure P (dt) = dt by letting
c (t) = inf fx : F (x)
Then
Z
b
(x) dF (x) =
a
Theorem 3 For the family f
Z
(x)
tg
F (b)
(c (t)) dt
F (a)
:
2 Ug ;the following holds
8
R
>
sup j j d < 1
>
>
R
>
2U
>
<
limc!1 sup fj j>cg j j d = 0
(integrally uniformly
R bnded)
2U
()
lim
sup
sup
j jd = 0
>
>
Uniform Integrability
!0+ fA2F : (A) g 2U A
>
>
>
: (integrally equi-absolute continuity)
where d is the Lebesgue measure on Rn .
Now that the more general settings will be
R
8
sup E jf j d < 1
>
>
>
2U
>
>
8" > R0; 9g 0; g 2 L1 (E) :
>
(integrally uniformly bnded)
<
1
sup Efjf j>gg jf j d < " ()
R 8" > 0; 9h 0; h 2 LR (E) ; 9 > 0 :
2U
>
>
>
>
Uniform Integrability
e h (x) dx < =) sup e jf j d < "; 8e
>
>
2U
:
(integrally equi-absolute continuity)
2
0 and
Proof. Necessity. Pick " = 1 and the corresponding g 0; g 2 L1 (E) ;then
Z
Z
Z
jf j d
jf j d + sup
jf j d
sup
sup
E
2U
Efj
2U
j>gg
Efj
2U
j gg
1 + kg1E k1 < 1
For any " > 0; choose the corresponding g
0; g 2 L1 (E) ;
such that (e) < and
Z
gd < "
> 0 and any measurble set e
E
e
Thus
sup
2U
Z
e
jf j d
sup
Z
eEfjf j>gg
2U
sup
2U
Z
Z
"+
fjf j>gg
j
j
j d + sup
2U
Z
j d + sup
Z
eEfjf j gg
j
jd
gd
eEfjf j gg
2U
gd = 2"
e
Su¢ ciency: Let M = sup
2U
such that
R
E
jf j d < 1: For any "; > 0;pick h
Z
e
and pick bounded set E1 such that
hd < ; 8 (e) <
Z
0; h 2 L1 (E) and
1
hd <
EnE1
Now de…ne
2M
g=
Then g
0; g 2 L1 (E) and
Z
jfk j d
=
if
if
1
0
Z
Efjfk j>ggE1
fjfk j>gg
Z
Efjfk j>ggE1
Z
Efjfk j>ggE1
x 2 E1
x 2 EnE1
jfk j d +
jfk j d +
Z
Efjfk j>ggE1C
Z
EE1C
jfk j d
jfk j d
jfk j d + "
But
2M
=
Z
1
(E fjfk j > gg E1 )
Z
2M
d
Efjfk j>ggE1
1
Efjfk j>ggE1
jfk j d
implies
(E fjfk j > gg E1 )
3
1
2
Z
E
jfk j d
M
1
>0
Thus
R
Efjfk j>ggE1
hd < ;which implies
Z
R
jfk j d < " and
Efjfk j>ggE1
fjfk j>gg
jfk j d
2"
Theorem 4 For the family fn 2 L1 (E) : n 2 N such that fn ! f -a:s: The following holds
Z
Z
Z
jfn j d = 0
fn d ! f d () lim sup
c!1
n2N
fj
n j>cg
Proof. Su¢ ciency. From the above proved theorem. It’s clear that
Z
Z
Z
jf j d =
limn jfn j d
jfn j d
M = sup fkfn k1 g
limn
E
E
For any "; > 0; choose h
E
0; h 2 L1 (E) ;
Z
n2N
> 0 and bounded set E1 such that
1
hd <
EnE1
and
Z
e
jf j d < ";
Z
e
hd < ; 8 (e) <
By Egorov Theorem, there exist a subset F
uniformly to f on F: Thus
Z
lim sup
jfn
F
n2N
Z
sup jfn
lim
F
Further, since
R
E1 with
E1 nF
sup
n2N
1
(E1 nF ) <
1
such that fn converges
fj d
fj d = 0
n2N
hd < ; then
Z
E1 nF
jfn
fj d
sup
n2N
Z
E1 nF
jfn j d +
Z
E1 nF
jf j d
" + " = 2"
Consequently
kfn k ! kf k
Necissity. kfn k ! kf k implies M = sup fkfn kg < 1: For any " > 0; there is some k0 2 N such
n2N
that
sup kfn
fk < "
n k0
P 0
Now let g = ki=1
jfi j + 2 jf j : Then g 0 and g 2 L1 (E) and
Z
sup
jfn j d
n
fjfn j>gg
Z
Z
Z
sup
jfn f j d + sup
jf j d + sup
n<k0
= 0
"
fjfn j>gg
since fjfn j > gg = ? all n
n<k0
n k0
fjfn j>gg
fjfn j>gg
jfn
k0 and n < k0 by the fact that lim fn = f
4
f j d + sup
n k0
Z
fjfn j>gg
jf j d
2
Continuous-Time Martingales
Problem 5 Let T1 ; T2 ;
with parameter > 0 :
be a sequence of independent, exponentially distributed random variables
P [Ti 2 dt] = e
Pn
t
dt; t
0
Let S0 = 0; Sn = i=1 Ti ; n
1 (We may think of Sn as the time at which the n-th customer
arrives in a queue, and of the random variables Ti ; i = 1; 2;
as the interarrival times.) De…ne a
continuous-time, integer-valued RCLL process
Nt = max fn
0; Sn
tg ; 0
t<1
(we may regard Nt as the number of customers who arrive up to time t). (i) Show that for 0
we have
P SNs +1 > tjFsN = e (t s) ; P a:s:
(ii) Show that for 0
of FsN :
s < t; Nt Ns is a Poisson random variable with parameter
Proof. (i) Firstly, we’ll show that
n
T
T
A = A~ 2 FsN : A~ fNs = ng = A fNs = ng for some A 2
(Ti ; 1
(t
s) ;independent
i
o
n)
is actually FsN ;which is supported by (1) ?; 2 A (2) A~ 2 A implies
T
A~C fNs = ng
C
T ~T
T T
A fNs = ng
= fNs = ng
= fNs = ng (A fNs = ng)C
T
= AC fNs = ng
So, A 2
(Ti ; 1
i
n) implies A~C 2 A. (3) Suppose A~m 2 A; 1
n
n
S
T
S
A~m
fNs = ng =
m=1
=
n
S
m=1
which implies
(Am
T
fNs = ng) =
m=1
n
S
m
s<t
n; then
T
A~m fNs = ng
Am
m=1
n
S
~ Then
A~m 2 A:(4) Suppose A~m 2 A; A~m " B:
T
fNs = ng
m=1
~ T fNs = ng
B
T
T
fNs = ng = lim A~m fNs = ng
=
lim A~m
m
m
T
T
fNs = ng
= lim (Am fNs = ng) = lim Am
m
m
~ 2 A: Consequentely, A is a -algebra. To
Let limm Am = B:Then B 2 (Ti ; 1 i n) :Thus B
show A = FsN ; it su¢ ces to show A FsN :
Let Aa;b = (a; b) ; a; b 2 Q (here a; b may be negative/positive in…nity and a = b is also allow2 d.
De…ne
D = Nt 1 (Aa;b ) : t 2 [0; s] ; Aa;b R
5
If it’s shown that D A; then FsN = (D) A and the assertion is proved.
Pick any Nt 1 (Aa;b ) 2 D:Obviously Nt 1 (Aa;b ) 2 A if Aa;b = ? or ( 1; +1) : So, let’s suppose
Aa;b 6= ?; R and denote W = Nt 1 (Aa;b ) : Then
W
= fa < Nt (!) < bg
(1)
= fc + 1
Nt (!) d 1g
T
= fSc+1 (!) tg fSd (!) > tg
where c = [a] and d = ]b[ : Consequently
T
W fNs = ng
(2)
T
T
T
= fSd (!) > tg fSc+1 (!) tg fSn (!) sg fSn+1 (!) > sg
T
T
where in (2) A = ? is set if W fNs = ng = ?: Else if W fNs = ng =
6 ? in (2) then by the nondecreasing property of Nt with respect to t 2 [0; 1); it’s clear that c + 1 n since t s;which further
gives (1) if d n; then W 2 Sd := (T1 ;
; Td ) since fSc+1 (!) tg 2 Sc+1 := (T1 ;
; Tc+1 )
and fSd (!) > tg 2 Sd := (T1 ;
; Td ) and with Sc+1
Sd
Sn := (Ti ; 1 i n) ; that is,
W itself is the needed set A; or (2) if d > n; then fSn+1 (!) > sg fSd (!) > tg (since additionally
t s) and
T
W fNs = ng
T
T
= fSc+1 (!) tg fSn (!) sg fSn+1 (!) > sg
T
= fSc+1 (!) tg fNs = ng
which sets A = fSc+1 (!) tg 2 Sn as the desired set.
Whence D A: Finally from
(D) =
=
=
Nt
1
Nt
1
Nt
1
(Aa;b ) : t 2 [0; s] ; a; b 2 Q
( (Aa;b ; ; a; b 2 Q))
(B (R)) ; t 2 [0; s]
it’s implied that
FsN =
(D)
A
FsN
as the desired assertion. (With your hints, I didn’t useTtrans…nite induction,
did I miss anything?)
T
Now, for any A~ 2 FsN choose A 2 Sn such that A~ fNs = ng = A fNs = ng ; then
Z
P Sn+1 > tjFsN dP
~
A\fN
s =ng
Z
Z
=
1fSn+1 >tg (!) dP =
1A 1fSn sg 1fSn+1 >tg (!) dP
A\fNs =ng
Z
=
1A 1[r2Q \[0;s] (fSn =rg\fTn+1 >t rg) dP
X Z
=
dP
r2Q \[0;s] AfSn =rg\fTn+1 >t rg
Note that the above decomposition is usually what supposes, which is unfruitful.
6
Remark 6 De…ne
C=
S
(Nt ; t 2 I)
I [0;s];jIj<1
which is a sub-algebra of FsN such that FsN =
(C) : Thus C
A yields A = FsN :
Proof. For any C 2 C;it’s clear that C 2
assertion will be proved if we can show
(Nt ; t 2 I) for some I
[0; s] with jIj < 1: the
Theorem 7 Let 0 = A0 A1
; where random variables An are Fn -measurable and E (A1 ) <
1: In order that An are Fn 1 -measurable, n 1; it’s necessary and su¢ cient that for each bounded
martingale Y = (yn ; Fn ) ; n = 0; 1;
: (the bracket means each yn is Fn -measurable.)
!
1
X
E
yn 1 (An An 1 ) = E (y1 A1 )
(3)
n=1
where y1 = lim yn
Proof. Necessity: let An be Fn
1 -measurable,
E (yn An
1)
E (A1 ) < 1: Since
= E (yn An ) ; n
1
(4)
therefore
E
1
X
yn
1 (An
!
An
1)
n=1
=
=
lim E
N !1
N
X
(yn
= lim E
N !1
y n ) An + y N AN
1
n=1
!
N
X
yn
1 (An
An
n=1
!
1)
lim E (yN AN ) = E (y1 A1 )
N !1
Su¢ ciency: Suppose (3) is satis…ed. Then
E (y1 A1 ) = E
=
=
lim E
N !1
lim E
N !1
N
X
n=1
N
X
yn
1
X
yn
1 (An
n=1
N
X1
1 An
y n An
n=1
(yn
An
1
yn ) An
n=1
!
!
!
1)
= lim E
= lim E
N !1
E
N
X
n=1
yn
1 (An
An
1)
n=1
(yn
1
!
y n ) An + y N AN
!
+ lim E (yN AN )
N !1
implies
1
X
N !1
N
X
(yn
y n ) An
1
n=1
!
=0
(5)
for any bounded martingale Y = (yn ; Fn ) ; n 0: Now make use of the fact that if Y = (yn ; Fn ) ; n
0;is a martingale, then Y = (yn ; Fn ) ; n 0 is a martingale, then the "stopped" sequence (yn^ ; Fn ) ; n
0 will be also a martingale for any stopping time : Taking time
1 and appyling (5) to the
martingale (yn^1 ; Fn ) ;we infer that
E [A1 (y0
y1 )] = 0
7
(6)
Similarly consederations on
2;
3;etc, lead to the fac that if (5) is correct, then there
exist the equalities given by (4) for any bounded martingale Y = (yn ; Fn ) ; n 0:
From (4) it follows that
E f(yn
yn
1 ) [An
E (An jFn
1 )]g
=0
(7)
(This above equalitity is what I don’t understand since it requires necessarily
E f(yn
yn
1) E
(An jFn
1 )g
=0
(8)
or
E [yn E (An jFn
I don’t see how to get this from E (yn An
Ok, this is the way to do it:
1)
E f(yn
= E (yn
1 An )
= E (yn An ) ; n
yn
= E fE [(yn
1 )]
1) E
yn
= E fE (An jFn
= E fE (An jFn
(An jFn
1) E
1 )g
(An jFn
1) E
[(yn
1)
0g
1)
yn
1 )] jFn 1 g
1 ) jFn 1 ]g
= E (0) = 0
Let yn+m = yn ; m
…nd
0; yn = sign [An
0 = E f[sign [An
= E fsign [An
= E (jAn
from which An = E (An jFn
1)
E (An jFn
E (An jFn
E (An jFn
E (An jFn
1 )]
1 )] ; yk
yn
1 )] [An
1 )j)
(P -a.s), i.e., An is Fn
= E (yn jFk ) ; k < n:Then from (7), we
1 ] [An
E (An jFn
E (An jFn
1 -measurable
1 )]g
1 )]g
for each n:
Problem 8 Show that the compensated Poisson process fMt = Nt
t; Ft ; 0 t < 1g is a martingale.
o
n
(1)
(d)
; Ft ; 0 t < 1 be a vector of martingales, and ' : Rd !
Problem 9 Let Xt = Xt ;
; Xt
R a convex function with E j' (X)j < 1 valid for every t 0: Then f' (Xt ) ; Ft ; 0 t < 1g is a
submartingale; in particular fkXt k ; Ft ; 0 t < 1g is a submartingale
Proof. For ' : R ! R is convex then
'0+ (x) = inf
y>x
' (y)
y
' (x) 0
' (x)
; ' (x) = sup
x
x
x<y
' (y)
y
and
'0 (x)
'0+ (x)
and '0 ; '0+ are increasing. And ' is continuous. For y
x; it’s clear that ' (y)
' (x) +
'0+ (x) (y x) and for y < x;it’s clear that ' (y) ' (x) + '0 (x) (y x). Thus for '0 (x) ax
'0+ (x) we have
' (y) ' (x) + ax (y x) =: Lx (y) ; 8y 2 I
8
and
' (x) = sup fLs (x)g
s2I
Now for any x 2 I let rn 2 Q and rn ! x: Then ' (x) + ar (x
Lr (y) g (y) and
'0 (r) ar '0+ (r)
r) =: Lr (x) ! ' (x) with
For ' : Rd ! R convex. De…ne
g (t) = ' (ty + (1
t) x)
Then g is convex and
g 0 (t) = hr' (ty + (1
t) x) ; y
xi
and
g (1)
0
g (0) + g+
(0)
which implies
' (y)
' (x) + grad+ '; y
x ;y
x
' (y)
' (x) + grad '; y
x ;y
x
and
Same procedure again
' (x) = sup fLs (x)g
s2I
Problem 10 Let fFn g1
1g
n=1 be a decreasing sequence of sub- -algebras of F and let fXn ; Fn ; n
be a backward submartingale; i.e., E jXn j < 1; Xn is Fn -measurable, and E (Xn jFn+1 )
Xn+1
a.s. P for every n 1: Then l := lim E (Xn ) > 1 implies that the sequence fXn g1
n=1 is uniformly
integrable.
Hints of these above 2 problems are given in the textbook.
9