s****n
发帖数: 1237 | 1 要求price a call of FX exchange rate in local currency。一开始不懂题目意思，
后来搞懂了其实是类似于要求price一个call，payoff是VT=max(ST^2-ST*K,0),ST是GBM
。因为是电面，我说把VT写出来，像call一样积分，然后discount e(-rT)就是C0的价
格。他说这个可能行，但是如果用change of numeraire怎么做，我没有很好的算出来。
事后我自己按两种方法做了一下，发现结果不太一样，但是找不出原因，请指教。
assume dS=rSdt+\sigmaSdW
1.直接积分，然后discount e(-rT)我得到
C0=S0^2*e(r+\sigma^2)N(d1+\sigma*sqrt(T))-S0*K*N(d2+\sigma*sqrt(T)).
2.用change of numeraire (exactly follow Zhou's book P161-P162)
ST_1=ST^2 dS1=(2r+\sigma^2)*S1dt+2*\sigma*S1dW1
ST_2=S |
p*****k
发帖数: 318 | 2 sunson, i was not able to follow your 2nd approach.
the standard way via change of numeraire:
V0/S0 = E[VT/ST], with dS = (r+sig^2)*S*dt + sig*S*dW.
thus V0 = S0 * E[(ST^2-ST*K)^+/ST] = S0 * E[(ST-K)^+]
the last term is exactly the payoff a vanilla call, which
gives the standard B-S formula, except all "r" being
replaced by "r+sigma^2" |
s****n
发帖数: 1237 | 3 the 2nd approach is folloiwng Zhou's book P161-P162. Maybe there's something
wrong in his method?
However, if just use change of measure, the price as you mentioned is
S0 * E[(ST-K)^+], with "r" => "r+sigma^2", we will get
V0 = S0^2*N(d1+sigma*sqrt(T)) - S0*K*e(-r-sigma^2)N(d2+sigma*sqrt(T))
which is different with the direct intergration (offset by e(-r-sigma^2)).
【在 p*****k 的大作中提到】
: sunson, i was not able to follow your 2nd approach.
: the standard way via change of numeraire:
: V0/S0 = E[VT/ST], with dS = (r+sig^2)*S*dt + sig*S*dW.
: thus V0 = S0 * E[(ST^2-ST*K)^+/ST] = S0 * E[(ST-K)^+]
: the last term is exactly the payoff a vanilla call, which
: gives the standard B-S formula, except all "r" being
: replaced by "r+sigma^2"
|
w******t
发帖数: 13 | 4 Shouldn't the drift of S_t under numeraire measure be r+0.5*sigma^2 to make
(B_t/S_t) a martingale where B_t is the bank account?
【在 p*****k 的大作中提到】
: sunson, i was not able to follow your 2nd approach.
: the standard way via change of numeraire:
: V0/S0 = E[VT/ST], with dS = (r+sig^2)*S*dt + sig*S*dW.
: thus V0 = S0 * E[(ST^2-ST*K)^+/ST] = S0 * E[(ST-K)^+]
: the last term is exactly the payoff a vanilla call, which
: gives the standard B-S formula, except all "r" being
: replaced by "r+sigma^2"
|
S*********g
发帖数: 5298 | 5 你的计算我没看。我有几个comment：
1. FX option 的payoff 应该是 (1-K/S)吧
2. zhou的书关于numeraire讲的不是很严谨。你可以看看John Hull的书，还有Baxter
的书。这两本书我觉得讲得要清楚的多
3.这里有个关于foreign exchange的lecture note，可能会有帮助
http://www.stat.uchicago.edu/~lalley/Courses/390/Lecture9.pdf
something
【在 s****n 的大作中提到】
: the 2nd approach is folloiwng Zhou's book P161-P162. Maybe there's something
: wrong in his method?
: However, if just use change of measure, the price as you mentioned is
: S0 * E[(ST-K)^+], with "r" => "r+sigma^2", we will get
: V0 = S0^2*N(d1+sigma*sqrt(T)) - S0*K*e(-r-sigma^2)N(d2+sigma*sqrt(T))
: which is different with the direct intergration (offset by e(-r-sigma^2)).
|
p*****k
发帖数: 318 | 6 wtgscott, yes, log(S) indeed has drift of r+sigma^2/2,
but not S. |
s****n
发帖数: 1237 | 7 多谢，我有空看看。
Baxter
【在 S*********g 的大作中提到】
: 你的计算我没看。我有几个comment：
: 1. FX option 的payoff 应该是 (1-K/S)吧
: 2. zhou的书关于numeraire讲的不是很严谨。你可以看看John Hull的书，还有Baxter
: 的书。这两本书我觉得讲得要清楚的多
: 3.这里有个关于foreign exchange的lecture note，可能会有帮助
: http://www.stat.uchicago.edu/~lalley/Courses/390/Lecture9.pdf
:
: something
|
z****i
发帖数: 406 | 8 恩，刚刚粗读了下那个FX的notes，看起来讲得不错。多谢啦
Baxter
【在 S*********g 的大作中提到】
: 你的计算我没看。我有几个comment：
: 1. FX option 的payoff 应该是 (1-K/S)吧
: 2. zhou的书关于numeraire讲的不是很严谨。你可以看看John Hull的书，还有Baxter
: 的书。这两本书我觉得讲得要清楚的多
: 3.这里有个关于foreign exchange的lecture note，可能会有帮助
: http://www.stat.uchicago.edu/~lalley/Courses/390/Lecture9.pdf
:
: something
|
s****n
发帖数: 1237 | 9
Why for the normal call, we need discount e^(-rT) and get C0=S0*N(d1)-K*N(d2)
but no discount for this one?
【在 p*****k 的大作中提到】
: wtgscott, yes, log(S) indeed has drift of r+sigma^2/2,
: but not S.
|
p*****k
发帖数: 318 | 10 the "discount"ing (prob abusing words here) is by S0
in the stock measure.
note in the original r.n. measure,
V0 = B0 * E[VT/BT]
with B0=e^(-r*T) and BT=1 |
