A**u
发帖数: 2458 | 1 Up and out put, with barrier H = 100, striked at 100. Current stock
price 50, no interest rate. what is the price? (Hint, using hedge) Asked
twice, different hedging strategy.
求讲解 | x******a
发帖数: 6336 | | A**u
发帖数: 2458 | 3 大牛，你确定绿皮书有
我翻了半天，也没找到
【在 x******a 的大作中提到】
: green book.
| s*****u
发帖数: 164 | 4 147页，D
【在 A**u 的大作中提到】
: 大牛，你确定绿皮书有
: 我翻了半天，也没找到
| A**u
发帖数: 2458 | 5 大牛 谢谢查找
但是这两题目不大相同
难道真要找stopping time，硬算
有办法replicate没
【在 s*****u 的大作中提到】
: 147页，D
| C***m
发帖数: 120 | 6 Put(S=50,K=100) - Call(S=50,K=100)??
上面是 vanilla put 和 call。S = H =100时，put（100，100）= call(100,100)（没
有interest rate）。 | f********k
发帖数: 136 | 7 What if S>H？ Put will be 0 but call has value. The portfolio will have a
negative value, not 0. So, it's not a replication.
【在 C***m 的大作中提到】
: Put(S=50,K=100) - Call(S=50,K=100)??
: 上面是 vanilla put 和 call。S = H =100时，put（100，100）= call(100,100)（没
: 有interest rate）。
| f********k
发帖数: 136 | 8 The option can be replicated by the following portfolio
Put(S=50,K=100)-Put(S=50,K=100)*I(K=100), where I(K=100) is an American
style digital call option strike at 100, which pays 1 when S hit 100 for the
first time.
Thus, if S is always smaller than 100, I(K=100) pays nothing, and the final
payoff of the portfolio is same as a vanilla put option. If S hit 100, the
porforlio becomes 0 immediately, which replicates the knock out condition.
According to Green Book p.147 D, I(K=100)=50/100=1/2.
So, finally, the up and out option becomes 0.5*Put(S=50,K=100).
Correct me if I'm wrong. | A**u
发帖数: 2458 | 9 还可以这么replicate啊
一直认为，应该看看T=0的pay, T=t的时候pay off.
the
final
【在 f********k 的大作中提到】
: The option can be replicated by the following portfolio
: Put(S=50,K=100)-Put(S=50,K=100)*I(K=100), where I(K=100) is an American
: style digital call option strike at 100, which pays 1 when S hit 100 for the
: first time.
: Thus, if S is always smaller than 100, I(K=100) pays nothing, and the final
: payoff of the portfolio is same as a vanilla put option. If S hit 100, the
: porforlio becomes 0 immediately, which replicates the knock out condition.
: According to Green Book p.147 D, I(K=100)=50/100=1/2.
: So, finally, the up and out option becomes 0.5*Put(S=50,K=100).
: Correct me if I'm wrong.
| h****y
发帖数: 49 | 10 You can only replicate by using linear combination of other options, because
generally E[f(x)] is not equal to f(E[x]).
Replicating works because the conditional expectation is a linear operator.
the
final
【在 f********k 的大作中提到】
: The option can be replicated by the following portfolio
: Put(S=50,K=100)-Put(S=50,K=100)*I(K=100), where I(K=100) is an American
: style digital call option strike at 100, which pays 1 when S hit 100 for the
: first time.
: Thus, if S is always smaller than 100, I(K=100) pays nothing, and the final
: payoff of the portfolio is same as a vanilla put option. If S hit 100, the
: porforlio becomes 0 immediately, which replicates the knock out condition.
: According to Green Book p.147 D, I(K=100)=50/100=1/2.
: So, finally, the up and out option becomes 0.5*Put(S=50,K=100).
: Correct me if I'm wrong.
| | | f*********5
发帖数: 367 | 11 嗯，put*I的payoff的相对的当前价值可能不是(value of put)*(value of I)那么简单
。之前有人问过http://www.mitbbs.com/article_t/Quant/31340413.html
不过好像也没结论。有大牛指点一下吗 | s*****u
发帖数: 164 | 12 Mark Joshi, The Concepts and Practice of Mathematical Finance, Sect. 10.6.2.
【在 A**u 的大作中提到】
: Up and out put, with barrier H = 100, striked at 100. Current stock
: price 50, no interest rate. what is the price? (Hint, using hedge) Asked
: twice, different hedging strategy.
: 求讲解
| C***m
发帖数: 120 | 13 Formula is in Bjork 3rd Ed. P275 (18.20).
Check it when r = 0, it is as simple as K-S0. | s*******0
发帖数: 3461 | | A**u
发帖数: 2458 | 15 Arbitrage Theory in Continuous Time?
这书也太复杂了吧
【在 C***m 的大作中提到】
: Formula is in Bjork 3rd Ed. P275 (18.20).
: Check it when r = 0, it is as simple as K-S0.
| s*****u
发帖数: 164 | 16 还好吧，Shreve 第二卷看完，这本书还是看的下去的。
我看你问面试的问题，你这是书都没看在押宝还是最后
做有针对性的准备？要是前者的话，很容易被面试官撂
倒的。
【在 A**u 的大作中提到】
: Arbitrage Theory in Continuous Time?
: 这书也太复杂了吧
| A**u
发帖数: 2458 | 17 shreve我看了前8章
大牛你好厉害...
后面 change Numéraire 就没再看了
【在 s*****u 的大作中提到】
: 还好吧，Shreve 第二卷看完，这本书还是看的下去的。
: 我看你问面试的问题，你这是书都没看在押宝还是最后
: 做有针对性的准备？要是前者的话，很容易被面试官撂
: 倒的。
| n******m
发帖数: 169 | 18 能不能这样
A = up and out put
B = put - call
senario 1: barrier not hit, end up below 100. payoff(A) = payoff(B)
senario 2: barrier hit, end up belwo 100. payoff(A)=0, payoff(B)=K-S
senario 3: barrier hit, end up above 100. payoff(A)=0, payoff(B)=S-K
注意到senario2和3的对称性，B 的价值应该跟A一样。
然后parity, 价值是100-50=50 | w**********9
发帖数: 5 | 19 agree it is 50.
【在 n******m 的大作中提到】
: 能不能这样
: A = up and out put
: B = put - call
: senario 1: barrier not hit, end up below 100. payoff(A) = payoff(B)
: senario 2: barrier hit, end up belwo 100. payoff(A)=0, payoff(B)=K-S
: senario 3: barrier hit, end up above 100. payoff(A)=0, payoff(B)=S-K
: 注意到senario2和3的对称性，B 的价值应该跟A一样。
: 然后parity, 价值是100-50=50
| s*******0
发帖数: 3461 | 20 change numeriare 据说是重点 需要很仔细看
我一个银行的人说的
【在 A**u 的大作中提到】
: shreve我看了前8章
: 大牛你好厉害...
: 后面 change Numéraire 就没再看了
| | | r**a
发帖数: 536 | 21 price is 50. Two hedge strats: one is the famours static hedge, you can
Google it. The other is buy a forword with exercise price at ur H. but need
the underlying stock pays dividends same as ur r, which is 0 in this case.
【在 A**u 的大作中提到】
: Up and out put, with barrier H = 100, striked at 100. Current stock
: price 50, no interest rate. what is the price? (Hint, using hedge) Asked
: twice, different hedging strategy.
: 求讲解
| x********9
发帖数: 31 | 22 You can simply hedge with a future contract with settlement price = 100.
The strategy would be:
1. Hold the future to maturity if the barrier is not hit. Then the put will
end up in the money, and the future delivers the same payoff.
2. If the barrier is hit, clear the future position. if r=0, then the
clearing does not incur any extra cost.
The value of the future contract is 100 - 50 = 50. | g****n
发帖数: 6 | 23 why? what is the solution to the question?
hard to understand why it is the case p = K-S_t.
it is a good way to explain. But it is not true.
Because prob( senario 2 ) is not equal to prob(senario 3).
prob(senario 3) = N( d_1)
d_1 = [ log( S_t/ S_t) + (r+ 1/2*vol^2) T ] / [ vol * sqrt( T) ]
= [ 0 + 0 + 1/2*vol^2 T ] / [ vol * sqrt( T) ]
= 1/2 vol * sqrt( T) >0
prob(senario 3) = N( d_1) > 1/2
but prob(senario 2) = 1 - N( d_1) < 1/2
发信人: nevergum (em), 信区: Quant
标 题: Re: 问道题目
发信站: BBS 未名空间站 (Tue Mar 26 00:13:31 2013, 美东)
能不能这样
A = up and out put
B = put - call
senario 1: barrier not hit, end up below 100. payoff(A) = payoff(B)
senario 2: barrier hit, end up belwo 100. payoff(A)=0, payoff(B)=K-S
senario 3: barrier hit, end up above 100. payoff(A)=0, payoff(B)=S-K
注意到senario2和3的对称性，B 的价值应该跟A一样。 | g****n
发帖数: 6 | 24 why? what is the solution to the question?
hard to understand why it is the case p = K-S_t.
it is a good way to explain. But it is not true.
Because prob( senario 2 ) is not equal to prob(senario 3).
prob(senario 3) = N( d_1)
d_1 = [ log( S_t/ S_t) + (r+ 1/2*vol^2) T ] / [ vol * sqrt( T) ]
= [ 0 + 0 + 1/2*vol^2 T ] / [ vol * sqrt( T) ]
= 1/2 vol * sqrt( T) >0
prob(senario 3) = N( d_1) > 1/2
but prob(senario 2) = 1 - N( d_1) < 1/2
发信人: nevergum (em), 信区: Quant
标 题: Re: 问道题目
发信站: BBS 未名空间站 (Tue Mar 26 00:13:31 2013, 美东)
能不能这样
A = up and out put
B = put - call
senario 1: barrier not hit, end up below 100. payoff(A) = payoff(B)
senario 2: barrier hit, end up belwo 100. payoff(A)=0, payoff(B)=K-S
senario 3: barrier hit, end up above 100. payoff(A)=0, payoff(B)=S-K
注意到senario2和3的对称性，B 的价值应该跟A一样。 | n******m
发帖数: 169 | 25 先回答 ghlian：
概率是不一样，但是payoff的分布也不一样。当然，说“对称”是不妥的。
不过这两种情况确实是抵消的。
假设在某个时间 t1, barrier hit了，此时 S(t1)=H, 考虑后续发展，可能最后 S(T)
高于 H,（senario2) 也可能低于 H （senario3)
现在你对所有后续路径积分，integrand = payoff = (S(T)-H) (in both case), 积分
的结果就是
forward(S(t1))-H = S(t1)-H = 0
正解见 xuzhikai19 和 rrua，的 replication.
不过我现在被这种有决策参与的replication搞晕了，
你能用一个exercise price=100的forward replicate出这个barier option的payoff,
好像只能说明这个forward比这个option值钱阿？？就像你可以选择永远不early
exercise, 那么american option的payoff也完全跟european的一样，但这不说明两者
同价。
是我哪个地方没弄明白呢？
【在 g****n 的大作中提到】
: why? what is the solution to the question?
: hard to understand why it is the case p = K-S_t.
: it is a good way to explain. But it is not true.
: Because prob( senario 2 ) is not equal to prob(senario 3).
: prob(senario 3) = N( d_1)
: d_1 = [ log( S_t/ S_t) + (r+ 1/2*vol^2) T ] / [ vol * sqrt( T) ]
: = [ 0 + 0 + 1/2*vol^2 T ] / [ vol * sqrt( T) ]
: = 1/2 vol * sqrt( T) >0
: prob(senario 3) = N( d_1) > 1/2
: but prob(senario 2) = 1 - N( d_1) < 1/2
| m******e
发帖数: 45 | 26 Besides the static replicating, you can use standard BS, considering r=0 and
time invariance/perpetual, we get dV/dt = 0, d^2V/dS^2=0, which means V(S,t
)=a+bS with a/b constant.
V(0,t)=100 (stock will stuck at 0) ==> a = 100
V(100,t)=0 ==> a + b*100 = 0 ==> b=-1
V(S_0,0) = 100 - S_0 = 50 |
|
