L****o
发帖数: 166 | | M****i
发帖数: 58 | 2 This kind of equation can be solved by the method of integrating factor.
For your problem, choose the integrating factor z(t)=exp(-w(t)+t/2) and then
use Ito's formula to get d(x(t)z(t))=z(t)dt (this is only an ODE). So that
the solution of your equation is given by
x(t)=z(t)^{-1}(x(0)z(0)+\int_0^t z(s) ds).
In fact, the same idea can be used to sovle the SDEs of the form
dx(t)=f(t,x(t))dt+c(t)x(t)dw(t),
where f and c are continuous functions.
In this general case, the integrating factor is
z(t)=exp(-\int_0^t c(s) dw(s)+(1/2)*\int_0^t c^2(s) ds).
Use Ito's formula to get
d(x(t)z(t))=z(t)f(t,x(t))dt.
Now let y(t)=x(t)z(t), then y(t) verifies the ODE
dy(t)=z(t)f(t,y(t)/z(t))dt.
Finally, it suffices to solve this ODE to get x(t).
See P79, Exercise 5.16 of the book by B. Oksendal: Stochastic Differential
Equations, 6th edition (Universitext). Springer 2005. | L****o
发帖数: 166 | 3 very helpful, many thanks
i also found it's been addressed by pcasnik in an old post
http://mitbbs.com/article1/Quant/31267285_3_0.html
i guess the integrating factor is found by solving an auxiliary GBM
but your post mentioned a very general case
dx(t)=f(t,x(t))dt+c(t)x(t)dw(t)
which is quite good.
i would appreciate if you could give some hints on the following questions
as well:
<1> how would we solove the ODE?
dy(t)=z(t)f(t,y(t)/z(t))dt
where $z(t)$ is a process depending on w(t)
<2> with $z(t)=exp(-w(t)+t/2)$
would it be possible to further evaluate
$\int_0^t z(s) ds$ in the final solution?
<3> what if the diffusion term is only c(t)dW(t) instead of c(t)x(t)dW(t)
can it still be solved in a smilar way?
thanks! any hint will be much appreciated!!
then
that
【在 M****i 的大作中提到】
: This kind of equation can be solved by the method of integrating factor.
: For your problem, choose the integrating factor z(t)=exp(-w(t)+t/2) and then
: use Ito's formula to get d(x(t)z(t))=z(t)dt (this is only an ODE). So that
: the solution of your equation is given by
: x(t)=z(t)^{-1}(x(0)z(0)+\int_0^t z(s) ds).
: In fact, the same idea can be used to sovle the SDEs of the form
: dx(t)=f(t,x(t))dt+c(t)x(t)dw(t),
: where f and c are continuous functions.
: In this general case, the integrating factor is
: z(t)=exp(-\int_0^t c(s) dw(s)+(1/2)*\int_0^t c^2(s) ds).
| M****i
发帖数: 58 | 4 You are welcome.
An advantage of the above generic method is that it can be used to solve
some nonlinear SDEs. For example:
dx(t)=x(t)^rdt+ax(t)dw(t), x_0=x>0,
Where r and a are constants.
(This is a good example to understand
how does the method work.)
Some ideas to your questions:
<1> The ODE dy(t)=z(t)f(t,y(t)/z(t))dt should be considered pathwisely, i.e.
for every fixed sample point $\omega$,
we regard it as a deterministic ODE and try to solve it.
So it doesn't matter that the BM involes. For some particular function f,
this equation can be solved explicitely but for general f, I don't think so.
Whatever, this is a question of ODE, so you may consult some ODE books to
get more information.
<2> In my opinion, the expression $\int_0^t z(s) ds$ is already a simple one
. It may be transformed equivalently but the results seem longer.
<3> For general function f, I have no useful idea. Maybe a possible way is to guess a form of x(t) then use Ito's formula to the candidate form and compare with the original equation.
For the linear equation dx(t)=a(t)x(t)dt+c(t)dw(t), with a, c deterministic functions, the integrating factor is exp(-\int_0^t a(s) ds).
Further discussions are welcome. |
|
