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)

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.

You can also solve for X and Y first, and then compute E(Y/X) directly.

楼上的解释都非常到位！

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