L****o
发帖数: 166 | 1 dx=dt+f(x)dW(t)
假设f(0)=0，以及初值x(0)>0，怎么证明x(t)>0 for all t>0？ | o*q
发帖数: 630 | 2 好像是以前学过的stochastic inequality
这个不缺条件么？最好f(x)increasing 什么的
如果不是的话，你自己玩吧 | M****i
发帖数: 58 | 3 To illustrate the main idea, assume that f is Lipschitz.
Let dy(t)=(f(x_t)/x(t))*1_{x(t)=\neq 0}dw(t) with y(0)=0,
then the original SDE becomes (note that f(0)=0)
dx(t)=dt+x(t)dy(t).
Now choose the integrating factor
z(t)=exp([y](t)/2-y(t)), ([y] is the quadratic variation of y) and use Ito's
formula to get
d(z(t)x(t))=z(t)dt. So that, because x(0)>0,
x(t)=z(t)^{-1}(x(0)+\int_0^t z(s) ds)>0.
In fact, this is an easy case of comparison theorems for SDEs. You may find
the latter in standard SDE books. | L****o
发帖数: 166 | 4 多谢！！！给你发信了，呵呵
's
find
【在 M****i 的大作中提到】
: To illustrate the main idea, assume that f is Lipschitz.
: Let dy(t)=(f(x_t)/x(t))*1_{x(t)=\neq 0}dw(t) with y(0)=0,
: then the original SDE becomes (note that f(0)=0)
: dx(t)=dt+x(t)dy(t).
: Now choose the integrating factor
: z(t)=exp([y](t)/2-y(t)), ([y] is the quadratic variation of y) and use Ito's
: formula to get
: d(z(t)x(t))=z(t)dt. So that, because x(0)>0,
: x(t)=z(t)^{-1}(x(0)+\int_0^t z(s) ds)>0.
: In fact, this is an easy case of comparison theorems for SDEs. You may find
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