4.7 brownian bridge(part 2)
DESCRIPTION
4.7 Brownian Bridge(part 2). 報告人:李振綱. Outline. 4.7.4 Multidimensional Distribution of the Brownian Bridge 4.7.5 Brownian Bridge as a Conditioned Brownian Motion. 4.7.4 Multidimensional Distribution of the Brownian Bridge. - PowerPoint PPT PresentationTRANSCRIPT
4.7 Brownian Bridge(part 2)
報告人:李振綱
OutlineOutline
4.7.4 Multidimensional Distribution of the Brownian Bridge4.7.4 Multidimensional Distribution of the Brownian Bridge
4.7.5 Brownian Bridge as a Conditioned Brownian Motion4.7.5 Brownian Bridge as a Conditioned Brownian Motion
4.7.4 Multidimensional Distribution of the Brownian 4.7.4 Multidimensional Distribution of the Brownian BridgeBridge
We fix and and let denote the Brownian bridge from a to b on . We also fix . We compute the joint density of .
We recall that the Brownian bridge from a to b has the mean function
(P.176)and covariance function
(P.176) When , we may write this as
To simplify notation, we set so that .
a b ( )a bX t
0,T0 1 20 nt t t t T
1( ), , ( )a b a bnX t X t
( ) ( )( )a b b a t T t a bt
m t aT T T
s t
( )( , ) , 0
st s T tc s t s s t T
T T
( , )st
c s t s tT
j jT t 0 T
We define random variable
Because are jointly normal, so that are jointly normal. We compute , and .
1
1
( ) ( )a b a bj j
jj j
X t X tZ
1( ), , ( )a b a bnX t X t
1( ), , ( )nZ t Z tjEZ ( )jVar Z ( , )i jCov Z Z
Brownian Bridge as Gaussian Process (4.7.2)
-1 -1 -1 -11 -1 -1 1
-1 -1 -1 -1
-1 -1 -1 -1
1 1 1 1( , ) ( , ) - ( , ) - ( , ) ( , )
( ) ( ) ( ) ( ) - - 0.
i j i j i j i j i ji j i j i j i j
i j i j i j i j
i j i j i j i j
Cov Z Z c t t c t t c t t c t t
t T t t T t t T t t T t
T T T T
-1 1 -11
1 -1 -1
1
-1
( ) ( )1 1( ) ( ) ( )
( ) .
j j j j j ja b a bj j j
j j j j j j
j j
j j
bt bt bt T t bt T ta aE Z EX t EX t
T T T T T
b t t
1 12 21 1
1 1 12 21 1
-1 -1 1 -1
-1 -1
1 2 1( ) ( ( )) ( ( ), ( )) ( ( ))
1 2 1 ( , ) ( , ) ( , )
2 ( ) 2 (
a b a b a b a bj j j j j
j jj j
j j j j j jj jj j
j j j j j j
j j j
Var Z Var X t Cov X t X t Var X t
c t t c t t c t t
t t t t T t t
T T T
-1 1
-1 -1
) ( ).j j j j j
j j j j
T t t T t t t
T
X,Y are independent Cov(X,Y)=0
( )j jT t a bt
T T
1 1( )j jT t a bt
T T
1 1 11 1
( )( , ) j j j j j j
j j j
t t t T t tc t t t
T T T
If X, Y are normal distribution
課本符號有錯
i j
So we conclude that the normal random variable are independent, and we can write down their joint density, which is
we make the change of variables
1
2
1
-1
( ),..., ( ) 111 1
-1-1
2
1
-1
11
-1
( )
1 1( ,..., ) exp
22
( )
1 exp -
2
n
j jj
nj j
Z t Z t nj jj j j
j jj j
j jj
nj j
j jj
j j
b t tz
f z zt tt t
b t tz
t t
1 1
-1
1.
2
n
j j j
j j
t t
1
1
, 1,..., ,j jj
j j
x xz j n
1 , , nZ Z
mean
VarianceVariance
接下來把 z的部份做變數變換
Where , to find joint density for . We work first on the sum in the exponent to see the effect of this change of variables. We have
Now
2
1
-1
11
-1
2
-1 1 1
1 1 1 -1
2 2 2 2-1 1 1 1 1 1
2 2 2 2 21 11 1 1
( )
( )
( ) 2 2 ( ) 2
j jj
nj j
j jj
j j
nj j j j j j
j j j j j j j
j j j j j j j j j j j j
j j j jj j j j j j
b t tz
t t
x x b t t
t t
x x b t t x x x b t t x
t t
1
21 1
2 2 2-1 1 1 1 1
1 1 1 1 1 1 1
2 21 1 1
1 1 1
( )
( ) 2 2 2=
( ) ( )
- - 21 1-
nj j
j j j
nj j j j j j j j j j
j j j j j j j j j j j j j
j j j j j j
j j j j j j
b t t
x x b t t x x x b x b
t t t t t t
x x
t t t t
1 12
1 1 11 1 1
1 12 .
n n nj j j j
j j jj j j j j j
x x x xb b
t t
1 1 1( ) ( )j j j j j jT t T t t t
0x a 1( ), , ( )a b a bnX t X t
So this last expression is equal to
To change s density, we also need to account for the Jacobian of the change of variables. In this case, we have
and all other partial derivatives are zero. This leads to the Jacobian matrix
2 2 2 21 1 1 12
1 1 1 11 1 1 1
2 2 2n1 2
j=1 1
21
2 1 1 2
( ) 1 1= ( ) 2 ( )
( )
n n n nj j j j j j j j
j j j jj j j j j j j j
j j n n
j j n n n
j j
j j
x x x x x x x xb b
t t
x x x xa ab b
t t T t T T t T T t T
x x
t t
2 2n
j=1 1
( ) ( ).n
n
b x b a
T t T
1 1
1 , 1,..., ,
1 , 2,..., ,
j
j j
j
j j
zj n
x
zj n
x
1
1 2
10 0
1 10
10 0
n
J
1
1n
j j
J
Whose determinant is . Multiplying by this
determinant and using the change of variables worked out above, we
obtain the density for ,
11( ),..., ( )
2 2 21
1 1 11 1
-1
2 21 1
11 11
( ,..., )
( ) ( )1 ( ) 1 1exp
2 2( ) 22
( ) ( )1 1exp
2 2( )2 ( )
a b a bn
nX t X t
n nnj j n
j j jj j n j j j
j j
n nj j j n
jj j j j nj j
f x x
x x b x b a
t t T t T t t
x x b x
t t T tt t
2
2 2 21
11 11
1 11
( )
2
( ) ( )1 1 ( )exp
2 2( ) 22 ( )
( - , , )( , , )
( , , )
n nj j n
jjn j j nj j
nn n
j j j jj
b a
T
x x b xT b a
T t t t T t Tt t
p T t x bp t t x x
p T a b
2
(4.7.6)
1 ( ) ( , , ) exp .
22
y xwhere p x y is the transition density for Brownian motion
1
1n
jj
1( ),..., ( ) 1( ,..., )nZ t Z t nf z z
1( ), , ( )a b a bnX t X t
J
Ch3-5 P.108
4.7.5 Brownian Bridge as a Conditioned Brownian 4.7.5 Brownian Bridge as a Conditioned Brownian MotionMotion The joint density (4.7.6) for permits us to give one
more interpretation for Brownian bridge from a to b on . It is a Brownian motion on this time interval, starting at and conditioned to arrive at b at time T (i.e., conditioned on ). Let be given. The joint density of is
This is because is the density for the Brownian motion going from to in the time between and . Similarly, is the density for going from to between time and . The joint density for and is then the product
1( ),..., ( ), ( ) 1 1 11
0
( ,..., , ) ( , , ) ( , , )
, (0)
n
n
W t W t W T n n n j j j jj
f x x b p T t x b p t t x x
where W x a
1 1 2 1 1 2( , , ) ( , , ).p t a x p t t x x
1( ), , ( )a b a bnX t X t
0,T( )W t (0)W a
( )W T b0 1 20 nt t t t T
1( ), , ( ), ( )nW t W t W T
1 0 0 1 1 1( , , ) ( , , )p t t x x p t a x (0)W a 1 1( )W t x 0t
1t t2 1 1 2( , , )p t t x x 1 1( )W t x 2 2( )W t x
1t t
2t t 1( )W t 2( )W t
??
1 2 1 2 1 1 2 1( ), ( ) 1 2 ( )| (0) 1 ( )| ( ) 2 1 ( ) 1 ( )| ( ) 2 1( , ) ( | ) ( | ) ( ) ( | )W t W t W t W W t W t W t W t W tf x x f x a f x x f x f x x
So, the density of conditioned on is thus the quotient
and this is of (4.7.6).
Finally, let us define
to be the maximum value obtained by the Brownian bridge from a to b on . This random variable has the following distribution.
1 11
( - , , )( , , )
( , , )
nn n
j j j jj
p T t x bp t t x x
p T a b
0( ) max ( )a b a b
t TM T X t
1( ), , ( )nW t W t ( )W T b
11( ),..., ( )
( ,..., )a b a bn
nX t X tf x x
0,T
Corollary 4.7.7.
The density of is
Proof : Because the Brownian bridge from 0 to on is a Brownian motion conditioned on , the maximum of on is the maximum of on conditioned on . Therefore, the density of was computed in Corollary 3.7.4 and is
The density of can be obtained by translating from the initial condition to and using (4.7.9). In particular, in (4.7.9) we replace by and replace by . This result in (4.7.8).
2( )( )
( )
2(2 )( ) , max{ , }. (4.7.8)a b
y a y bT
M T
y b af y e y a b
T
0
2 ( )
( )
2(2 )( ) , , 0. (4.7.9)w
m m w
TM T
m wf m e w m m
T
( )a bM T
0,T( )W T w
w0 wX
0,T 0,T
0 ( )wM TW ( )W T w
( )( )a bM T
f y
(0)W a (0) 0W m y a w b a
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