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上面的那道题怎么做呢？或者说思路是什么？最近感觉学习SDE都要被虐哭了

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关键词：随机微分方程 微分方程 随机微分 牛人解 SDE

代入ITo公式，可以吧

guowy67 发表于 2014-5-29 13:47
代入ITo公式，可以吧

额，能说详细一点吗？

万帅 发表于 2014-5-29 13:52
额，能说详细一点吗？

is it E(Y|X) or E(Y/X)?

if it is E(Y/X) just applying Ito lemma to Z=Y/X and put dY and dX into it. You will get the process that dZ should follow and the expectation of Z is easily obtained.

intuitively, rho stands for the projection of X on Y. The higher the rho, the larger Y's mean is offset, thus smaller E(Y/X).

Chemist_MZ 发表于 2014-5-29 14:39
is it E(Y|X) or E(Y/X)?

if it is E(Y/X) just applying Ito lemma to Z=Y/X and put dY and dX into ...

谢谢你啊，但是Z是一个二元函数，是用偏导数求dz吗？

万帅 发表于 2014-5-29 19:07
谢谢你啊，但是Z是一个二元函数，是用偏导数求dz吗？

Yes, the same as you learned in the calculus class.

Chemist_MZ 发表于 2014-5-29 22:38
Yes, the same as you learned in the calculus class.

恩，能不能帮我看一下上面我做的对不对呢？如果对的话，接下来怎么能通过dz求Ez？

万帅 发表于 2014-5-30 10:46
恩，能不能帮我看一下上面我做的对不对呢？如果对的话，接下来怎么能通过dz求Ez？

You have to include the second order terms.

d(y/x)=1/x*dy-y/x^2*dx+y/x^3 (dx)^2-1/x^2(dxdy) (I may write it wrong, you need to check yourself)

Z will be a lognormal, you can express Z as exp(a) where "a" is a normal variable N(mu,sigma^2). so E(Z) will be exp(mu+0.5*sigma^2)

Chemist_MZ 发表于 2014-5-30 10:55
You have to include the second order terms.

d(y/x)=1/x*dy-y/x^2*dx+y/x^3 (dx)^2-1/x^2(dxdy) (I  ...

恩，我已经知道怎么做了，实在是太感谢你了！！！！！
但是最后一个问题，为啥要求二次倒数呢？

万帅 发表于 2014-5-30 12:12
恩，我已经知道怎么做了，实在是太感谢你了！！！！！
但是最后一个问题，为啥要求二次倒数呢？

because the quadratic variation of a brownian motion W(t) is t, in a more understandable form (dw)^2=dt.

and you know we write the formula in the differential form but actually it is in the integral form (can be viewed as an abbreviation) . Thus, all the integral with repeat to (dwdt) or (dt)^2 are zero (this can be proved). So they can be discarded. But only (dw)^2=dt is retained,because it is not zero so that this term can not be ignored.

Thus we need to include the second order terms to include the (dw)^2 terms.