w*****e
发帖数: 197 | 1 Perpetual American Call Option under Zero Interest Rate
This is actually a homework problem from Mark Joshi's famous book.
If we strcitly follow BS formula
C = S*N(d1) - K*exp(-rT)*N(d2)
When T goes to +infinity, d1 goes to +infinity and d2 goes to
-infinity, so we get C = S.
But here comes a replication based method, which shows that C should
be worth no more than S - K.
We can always borrow K in cash and long a share and short a perpetual
american call. Note, we have completely hedged our risk | w*****e
发帖数: 197 | 2 Why can't I read the post from koox? | b********u
发帖数: 63 | 3 the payoff of your replication is -K+S-max(S-K,0), which will never be
positive. so you cost -K+S-c<=0 => c>=S-K. | w*****e
发帖数: 197 | 4 The reverse direction works as well.
Suppose C = S - K + e, e > 0
Short C, long S and borrow K - e, you portfolio
value will be -C + S - ( K - e ) = 0
At the end of exercise, you will have exactly e > 0
left, which is an arbitrage.
【在 b********u 的大作中提到】
: the payoff of your replication is -K+S-max(S-K,0), which will never be
: positive. so you cost -K+S-c<=0 => c>=S-K.
| b********u
发帖数: 63 | 5
what if it's not exercised?
【在 w*****e 的大作中提到】
: The reverse direction works as well.
: Suppose C = S - K + e, e > 0
: Short C, long S and borrow K - e, you portfolio
: value will be -C + S - ( K - e ) = 0
: At the end of exercise, you will have exactly e > 0
: left, which is an arbitrage.
| z****i
发帖数: 406 | 6 The probability that the option will be exercised is 1, right?
【在 b********u 的大作中提到】
:
: what if it's not exercised?
| m*********g
发帖数: 646 | 7 恩，这个似问题的关键，俺刚才也迷惑了
【在 z****i 的大作中提到】
: The probability that the option will be exercised is 1, right?
| m*********g
发帖数: 646 | 8 REPLICATE的意义在于完全复制原先的PAYOFF,在任何时刻都保持原先的ＰＡＹＯＦＦ一
致才是ＲＥＰＬＩＣＡＴＩＴＯＮ ＰＯＲＴＦＯＬＩＯ。
你的ＰＯＲＴＦＯＬＩＯ只有在s>k的时刻里是成功的，在所有s
格卖低了。如果看到这里你明白了，那不用往下看了。不要被这个ＯＰＴＩＯＮ肯定执
行迷惑了，肯定执行不等于在每一个时刻都执行，而ＲＥＰＬＩＣＡＴＥＤ ＰＯＲＴ
ＦＯＬＩＯ的意义在于任何时刻任何情况。举例，ＰＵＴ－ＣＡＬＬ ＰＡＲＩＴＹ就
是一个任何时刻任何情况都一致的ＲＥＰＬＩＣＡＴＩＯＮ. （你可能会认为你这个CALL在s(t)
你这里试图REPL | Q***5
发帖数: 994 | 9 "不要被这个ＯＰＴＩＯＮ肯定执
行迷惑了，肯定执行不等于在每一个时刻都执行"
I think the main problem is that with probability >0 the option might never
be exercised. ( under the risk neutral prob, the drifting of log(S) is
negative)
If it had been the case that with probability =1 the option will be
exercised, then LZ's argument is right: the short call can always be covered
, it does not matter whether the total portfolio value will become negative
before the option is exercised. | b********u
发帖数: 63 | 10 hmmmm, can you explain why, since the drift is negative (r=0)?
what's more, to replicate an american call, you need to match at each time
point t, right?
【在 z****i 的大作中提到】
: The probability that the option will be exercised is 1, right?
| | | m*********g
发帖数: 646 | 11 我补充了原文，如果对于这一点还是不清楚的话可以参考一下。
大概就是 这个OPTION是流通的，所以你不能考虑它在s(t)
个option在 s(t)
payoff这个谁都同意吧？假设楼主的定价是arbi. free的，那么我们采用这个方法定价
，价格是 s(t)-k<0 意味着，你白送这个OPTION+FREE MONEY?这肯定是ARBI. OPPO.,
我拿了这个OPTION，不管将来怎么样，我都白赚了CASH.这也就说明了，在所有的s(t)<
k的情况下，楼主的方法都低估了该OPTION的价值。
OPTION是流通的，这个是关键。如果加入了这个OPTION不能流通，在购买后和所有的非
执行时刻该OPTION都不能做市场被交易，那我们可以再讨论。
never
covered
negative
【在 Q***5 的大作中提到】
: "不要被这个ＯＰＴＩＯＮ肯定执
: 行迷惑了，肯定执行不等于在每一个时刻都执行"
: I think the main problem is that with probability >0 the option might never
: be exercised. ( under the risk neutral prob, the drifting of log(S) is
: negative)
: If it had been the case that with probability =1 the option will be
: exercised, then LZ's argument is right: the short call can always be covered
: , it does not matter whether the total portfolio value will become negative
: before the option is exercised.
| J****g
发帖数: 103 | 12 "borrow K in cash and long a share and short a perpetual
american call"， 这个portfolio初始值就不是0啊。 | z****i
发帖数: 406 | 13 我说错了。。 想当然地想成 arithmetic BM了，drift是 -1/2 sigma^2 ， 不好意思
我仔细想想
【在 b********u 的大作中提到】
: hmmmm, can you explain why, since the drift is negative (r=0)?
: what's more, to replicate an american call, you need to match at each time
: point t, right?
| p*****k
发帖数: 318 | 14 i don't understand everything moonsspring said, but he definitely
had a point that if S < K now, a contract with a non-negative payoff
cannot have a negative price, so the replication cannot be exact.
there are couple of ways to get the price for a perpetual american
call option - the easiest is to solve the steady-state black-scholes,
though the boundary condition is a little tricky.
(see e.g., wilmott's book, and it's indeed S)
also the optimal exercise boundary tends to infinity, so technical | Q***5
发帖数: 994 | 15 I think the price should be S, because a perpectual american should be at
least the value of an american call with same strike but with arbitrarily
large maturity, and by LZ's argument, it should be S.
【在 p*****k 的大作中提到】
: i don't understand everything moonsspring said, but he definitely
: had a point that if S < K now, a contract with a non-negative payoff
: cannot have a negative price, so the replication cannot be exact.
: there are couple of ways to get the price for a perpetual american
: call option - the easiest is to solve the steady-state black-scholes,
: though the boundary condition is a little tricky.
: (see e.g., wilmott's book, and it's indeed S)
: also the optimal exercise boundary tends to infinity, so technical
| z****i
发帖数: 406 | 16 steady-state black-scholes 解下来是C(S) = A*S + B*S^(-r/2sigma^2), A和B是两
个代定常数。 C(0) = 0, 所以B=0. 另外一个boundary condition 莫非是说the limit
of C(S)/S as S approaches infinity is 1, 所以A=1?
【在 p*****k 的大作中提到】
: i don't understand everything moonsspring said, but he definitely
: had a point that if S < K now, a contract with a non-negative payoff
: cannot have a negative price, so the replication cannot be exact.
: there are couple of ways to get the price for a perpetual american
: call option - the easiest is to solve the steady-state black-scholes,
: though the boundary condition is a little tricky.
: (see e.g., wilmott's book, and it's indeed S)
: also the optimal exercise boundary tends to infinity, so technical
| Q***5
发帖数: 994 | 17 I guess one thing we need to be careful when talk about the argument of
replicating strategy of perpetual option: we may easily introduce arbitrage
if we are allowed to borrow indefinitely.
Consider the following strategy: At time one, I borrow $1 ( and I
immediately spend it on an ice cream, and enjoy it). At time 2, I borrow $1
and pay back the $1 I borrowed at time one. At time 3, I borrow $1 and pay
back the $1 I borrowed at time two, ....-- so I get a free ice cream.
The problem with LZ's a | m*********g
发帖数: 646 | 18 Sorry,没有看仔细+刚在别处吵架有点火。已经删了
【在 z****i 的大作中提到】
: steady-state black-scholes 解下来是C(S) = A*S + B*S^(-r/2sigma^2), A和B是两
: 个代定常数。 C(0) = 0, 所以B=0. 另外一个boundary condition 莫非是说the limit
: of C(S)/S as S approaches infinity is 1, 所以A=1?
| Q***5
发帖数: 994 | 19 I guess one thing we need to be careful when talk about the argument of
replicating strategy of perpetual option: we may easily introduce arbitrage
if we are allowed to borrow indefinitely.
Consider the following strategy: At time one, I borrow $1 ( and I
immediately spend it on an ice cream, and enjoy it). At time 2, I borrow $1
and pay back the $1 I borrowed at time one. At time 3, I borrow $1 and pay
back the $1 I borrowed at time two, ....-- so I get a free ice cream.
The problem with LZ's a | w*****e
发帖数: 197 | 20 Thanks, this makes sense.
But zero interest rate really makes many things weird,
because with zero interest rate, a $ tomorrow is exactly
the same as a $ today. But I guess the point here is:
a $ today is definitely better than a $ that may never
come. I.e., the option might never be exercised in this case.
arbitrage
1
"
【在 Q***5 的大作中提到】
: I guess one thing we need to be careful when talk about the argument of
: replicating strategy of perpetual option: we may easily introduce arbitrage
: if we are allowed to borrow indefinitely.
: Consider the following strategy: At time one, I borrow $1 ( and I
: immediately spend it on an ice cream, and enjoy it). At time 2, I borrow $1
: and pay back the $1 I borrowed at time one. At time 3, I borrow $1 and pay
: back the $1 I borrowed at time two, ....-- so I get a free ice cream.
: The problem with LZ's a
| | | Q***5
发帖数: 994 | 21 On second thought, it can not be S either: because if some sucker pays me S
dollars for the American call, I will just use the S dollars to buy the
stock -- and the short position is covered, and if by luck, the sucker
exercise the option, I will get K dollars for free -- a perfect arbitrage.
So, I guess there is no arbitrage free price for such an option -- perhaps
due to the `free ice cream' problem I mentioned earlier.
【在 Q***5 的大作中提到】
: I think the price should be S, because a perpectual american should be at
: least the value of an american call with same strike but with arbitrarily
: large maturity, and by LZ's argument, it should be S.
| w*****e
发帖数: 197 | 22 What you are talking about is very similar to what I am saying in the
original post. With zero interest rate, borrowing is essentially free. So we
can roll over the debt as long as we want, so any price bigger than S - K
can be exploited in this way.
S
【在 Q***5 的大作中提到】
: On second thought, it can not be S either: because if some sucker pays me S
: dollars for the American call, I will just use the S dollars to buy the
: stock -- and the short position is covered, and if by luck, the sucker
: exercise the option, I will get K dollars for free -- a perfect arbitrage.
: So, I guess there is no arbitrage free price for such an option -- perhaps
: due to the `free ice cream' problem I mentioned earlier.
| p*****k
发帖数: 318 | 23 zhucai, try wilmott's book (1st volume) for the boundary condition
- he calls it "smooth pasting condition", which is essentially the
continuity of the option price and delta. |
|
