一道从Hull书上来的题，请教各位高手

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关键词：hull hul

So, what is your question.

Chemist_MZ 发表于 2013-7-13 06:22
So, what is your question.

Show u, d = ...

Ok，sorry I didn't realize this is a question but some straightforward derivation.

You can expand u and d by taylor series showing below. Then put u and d into (3) rearrange the terms and ignore the terms that has a higher order than dt. Then you are done.

best,

Chemist_MZ 发表于 2013-7-13 13:11
Ok，sorry I didn't realize this is a question but some straightforward derivation.

You can expand ...

还请多指点一下。

pertain 发表于 2013-7-14 13:25
还请多指点一下。

ok

first apply taylor expansion to the exponential function in (3) and you just need to keep the term up to order dt. The left side is therefore:

(1+mu*dt)(1+sigma*sqrt(dt)+sigma^2*dt/2+1-sigma*sqrt(dt)+sigma^2*dt/2)-ud-(1+2*mu*dt)

ud=1

(1+mu*dt)(2+sigma^2*dt)-1-(1+2*mu*dt)

2+sigma^2*dt+2*mu*dt+mu*sigma^2*dt^2-1-1-2*mu*dt

dt^2 is ignored

so the left hand side is sigma^2*dt which is the same as the right hand side.

This is a straightforward way to do it. Another way is to solve two equations:

(3)

ud=1

one of my classmate has solved it and he shows that you can get the solution (perhaps you also need some approximation)

best,

我的解答：
把（1）代带入到（2），经过组合化简得到（3）；
（3）里令时间delta-t趋于0，考虑ud=1，可知道u，d趋于1；
令u=exp（x），带入（3）；
u+d展开到2阶，其余的展开到1阶；
化简为x^2 * (1 + mu * delta-t) = sigma^2 * delta-t;
x^2和sigma^2 * delta-t是同阶无穷小，所以可以考虑令
x = sigma * sqrt(delta-t)
u和d也就得到了