关于微分方程问题的求解

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假设X(t)满足下面随机微分方程，dx(t)=x的三次方（t）dt—x的平方(t)*db(t),x(0)=0.2
其中b(t)为标准布朗运动。
当布朗运动B(t)首次达到-3时，X(t)的取值为多少？
菜鸟希望得到大家的帮助，多谢啦。

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关键词：微分方程 随机微分方程 布朗运动 随机微分

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I think the answer should be an expectation. X(t) is a random number since first passage time is random.

Chemist_MZ 发表于 2013-4-10 01:05 I think the answer should be an expectation. X(t) is a random number since first passage time is ran ...

How can I do it ?

I just the give the general step which may not work for your case. But anyway, it is hint.

Well the first step is of course solve this SDE. I assume you can do that.

You then get X(t) expressed in time t and B(t). So what you need to find is just the time when B(t) first come to -3. Let's say that time=tao.

The first passage time' expectation should be included in the text book. E(tao|B(tao)=-3), so put tao and B(tao)=-3 into X(t), you get the result.

I can't give you the detailed solution since that is not good for you.

Thanks a lot for answering my so many questions .You are a very good boy.