m******g
发帖数: 12 | 1 1. Let B_t be a Brownian motion. Let
A = \int_0^T B_t d t
B = \int_0^T t d B_t
What's the correlation between A and B?
2. Consider a Brownian motion starting at x where x is randomly drawn from
[-1, 1]. What's the expected time it takes for the Brownian motion to
leave the region [-1, 1]? |
z****i
发帖数: 406 | |
m******g
发帖数: 12 | |
z****i
发帖数: 406 | 4
好。两道都不对？
【在 m******g 的大作中提到】
: 再想想。。
|
m******g
发帖数: 12 | |
z****i
发帖数: 406 | 6 1.
Var(A) = Var(B) = T^3/3.
Cov(A,B) = T^3/6.
rho = 1/2.
2.
For fixed staring point x, the expected time is 1-x^2.
If you mean that x is uniformed distributed in [-1,1], then the expected
exit time is 2/3 ?
【在 m******g 的大作中提到】
: 再想想。。
|
m******g
发帖数: 12 | |
j*****4
发帖数: 292 | 8 how to get cov(A,B)?
【在 z****i 的大作中提到】
: 1.
: Var(A) = Var(B) = T^3/3.
: Cov(A,B) = T^3/6.
: rho = 1/2.
: 2.
: For fixed staring point x, the expected time is 1-x^2.
: If you mean that x is uniformed distributed in [-1,1], then the expected
: exit time is 2/3 ?
|
|
z****i
发帖数: 406 | 9 T*B_T = A+B
【在 j*****4 的大作中提到】
: how to get cov(A,B)?
|
p******5
发帖数: 138 | 10 How do you solve the second problem?
THX |
|
|
m******g
发帖数: 12 | 11 consider the martingales B_t and B_t^2 - t. and use optional stopping
theorem. |
t********s
发帖数: 54 | 12 Would you mind explain a bit more here?
Or do you mean since T is known value, so is B_T, therefore var(T*B_T) = 0?
Thanks.
【在 z****i 的大作中提到】
: T*B_T = A+B
|
t*******y
发帖数: 637 | 13 Var(T*B_T)=var(A)+VAR(B)+2COV(A,B)
VAR(T*B_T)=T^3
VAR(A)=T^3/3
VAR(B)=T^3/3
SO COV(A,B)=T^3/6
【在 t********s 的大作中提到】
: Would you mind explain a bit more here?
: Or do you mean since T is known value, so is B_T, therefore var(T*B_T) = 0?
: Thanks.
|
t********s
发帖数: 54 | 14 thanks for your explanation, tennisboy! |
p*******t
发帖数: 213 | |
o********n
发帖数: 100 | 16 could you please explain a little bit about how to get Var(A) = Var(B) = T^3
/3. ?
sorry for the ignorance... |
p*****k
发帖数: 318 | |
