x******a
发帖数: 6336 | 1 int_0^\infty I_{B_t>0}I_{t<1} dB_t
I_{X} is the charactersitc function of set X.
B_t is a Brownian motion |
J**********g
发帖数: 213 | 2 I doubt it has an explicit expression for the ito integral.
It's clear that it equals to
\int_0^1 I_{B_t>0}dB_t (*)
but I don't know how to keep going. I might be wrong, but there might not be
an explicit expression for (*). |
p*****k
发帖数: 318 | 3 to get an explicit expression, seems one should consider the local time and use tanaka's formula:
http://en.wikipedia.org/wiki/Tanaka%27s_formula
which gives:
int_0^1 [sgn(B_t)+1]/2 dB_t = (|B_1|+B_1-L_1)/2 |
x******a
发帖数: 6336 | 4
and use
tanaka's formula:
这样考虑怎么样,错在哪里?
{B_t>0} is a open set of t hence a countable union of open intervals since B
_t is
continuous. on each interval the integral is 0 except for the one have 1 as
an
interior point. it follows the integral should be max(B_1,0). where is the
mistake?
thanks.
【在 p*****k 的大作中提到】
: to get an explicit expression, seems one should consider the local time and use tanaka's formula:
: http://en.wikipedia.org/wiki/Tanaka%27s_formula
: which gives:
: int_0^1 [sgn(B_t)+1]/2 dB_t = (|B_1|+B_1-L_1)/2
|
p*****k
发帖数: 318 | 5 xiaojiya, note max(B_1,0)=(|B_1|+B_1)/2. the local time part is subtle
and i will leave it to expert/textbook |
c****o
发帖数: 1280 | 6
B
on each such interval, the integral is B{t+1}-B{t} instead of 0.
except for the one have 1 as
【在 x******a 的大作中提到】
:
: and use
: tanaka's formula:
: 这样考虑怎么样,错在哪里?
: {B_t>0} is a open set of t hence a countable union of open intervals since B
: _t is
: continuous. on each interval the integral is 0 except for the one have 1 as
: an
: interior point. it follows the integral should be max(B_1,0). where is the
: mistake?
|
x******a
发帖数: 6336 | 7 may i ask why B_{t+1}-B_t?
【在 c****o 的大作中提到】
:
: B
: on each such interval, the integral is B{t+1}-B{t} instead of 0.
: except for the one have 1 as
|
c****o
发帖数: 1280 | 8 There is some subtle point in your argument, any open set can be expressed
as union of open interval, but the size of such interval may goes to 0,
think about the complement of cantor set.
【在 x******a 的大作中提到】
: may i ask why B_{t+1}-B_t?
|
x******a
发帖数: 6336 | 9 but we know the number of the intervals is countable.
【在 c****o 的大作中提到】
: There is some subtle point in your argument, any open set can be expressed
: as union of open interval, but the size of such interval may goes to 0,
: think about the complement of cantor set.
|
M****i
发帖数: 58 | 10 I don't think that this integral can be computed explicitely, but another
form can be given by using Meyer-Ito formula (P214,Th 6.22 in GTM113 by
Karatzas & Shreve). For this, assume that B_0 = 0, choose f(x) = max{x,0} in
Meyer-Ito formula and observe that the left derivative of f is 1_{x>0}.
Hence
int_0^t 1_{B_s>0}dB_s = max{B_t,0} - L_t(0)/2, where L_t(0) is the Brownian
local time at 0. For your integral, it suffices to choose t = 1. |
c****o
发帖数: 1280 | 11 The complement of cantor set is also open and countable.
countable union of countable set is still countable.
【在 x******a 的大作中提到】
: but we know the number of the intervals is countable.
|
