t*****n
发帖数: 58 | 1 Given two stocks at time T, one has to pay the difference of their values
unless they had the same value (which is considered to be continuous in time
) at some time before T, in which case there is no obligation. How much
does this derivative worth now? What if we drop the constraint of not
having the same value?
我不明白这里的constraint "unless they had the same value"， 有区别吗？另外，
第一问，是不是考虑S_1(T)>S_2(T), or S_1(T)
谢谢！ | t**********a
发帖数: 166 | 2 This is essentially a special case of KO barrier options, in which there is
very simple solution if r=0 and B=K. If r=0, max(S1,S2)-min(S1,S2).
time
【在 t*****n 的大作中提到】
: Given two stocks at time T, one has to pay the difference of their values
: unless they had the same value (which is considered to be continuous in time
: ) at some time before T, in which case there is no obligation. How much
: does this derivative worth now? What if we drop the constraint of not
: having the same value?
: 我不明白这里的constraint "unless they had the same value"， 有区别吗？另外，
: 第一问，是不是考虑S_1(T)>S_2(T), or S_1(T): 谢谢！
| t*****n
发帖数: 58 | 3 谢谢解答，不过我觉得应该N(d_1)max(S1,S2)-N(d_2)min(S1,S2),另外有什么好方法算
吗？另外那个条件有什么用啊？谢谢。 | r***n
发帖数: 6 | 4 how do you define d_1 and d_2?
【在 t*****n 的大作中提到】
: 谢谢解答，不过我觉得应该N(d_1)max(S1,S2)-N(d_2)min(S1,S2),另外有什么好方法算
: 吗？另外那个条件有什么用啊？谢谢。
| l*******1
发帖数: 113 | 5 max(S1-S2,0) + Max(S2-S1,0)
Use exchange option formula | t**********a
发帖数: 166 | 6 You didn't get it...
the answer is if S1,0>S2,0, the option is S1,0-S2,0
There is not much dynamic hedging, hold static position and liquidate the
position if S1t=S2t
【在 t*****n 的大作中提到】
: 谢谢解答，不过我觉得应该N(d_1)max(S1,S2)-N(d_2)min(S1,S2),另外有什么好方法算
: 吗？另外那个条件有什么用啊？谢谢。
| a********e
发帖数: 508 | 7 nice! it looks like your static hedging works when r>0 too
So the answer is the same no matter what r is, right?
【在 t**********a 的大作中提到】
: You didn't get it...
: the answer is if S1,0>S2,0, the option is S1,0-S2,0
: There is not much dynamic hedging, hold static position and liquidate the
: position if S1t=S2t
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