c**********e
发帖数: 2007 | 1 【Probability】some MS written test questions
by xiaoyuesixi (bigredbear),
5. find the coefficients before df/dx in the B-S equation for the price of
an option which pays off max(S_T^2-K,0) where S_T is the usual stock
price.
( my answer is (r+sigma^2/2)S^2).
I do not think the answer is correct. Anybody has any comment? |
l******i
发帖数: 1404 | 2 你看的不仔细啊，我在回帖里给了自己的答案：
http://www.mitbbs.com/article/Quant/31309627_0.html
第五题是红皮书2.6的变形。
S_t^2 is a geometric Brownian motion with diffusion 2*sigma and drift
2r+sigma^2.
the original coefficient before df/dx in the B-S equation is rx,
so the new coefficient should be (2r+sigma^2)*(x^2).
price of
【在 c**********e 的大作中提到】
: 【Probability】some MS written test questions
: by xiaoyuesixi (bigredbear),
: 5. find the coefficients before df/dx in the B-S equation for the price of
: an option which pays off max(S_T^2-K,0) where S_T is the usual stock
: price.
: ( my answer is (r+sigma^2/2)S^2).
: I do not think the answer is correct. Anybody has any comment?
|
c**********e
发帖数: 2007 | 3 The B-S equation should not change, only the boundary condition changes. Don
't you think so?
【在 l******i 的大作中提到】
: 你看的不仔细啊，我在回帖里给了自己的答案：
: http://www.mitbbs.com/article/Quant/31309627_0.html
: 第五题是红皮书2.6的变形。
: S_t^2 is a geometric Brownian motion with diffusion 2*sigma and drift
: 2r+sigma^2.
: the original coefficient before df/dx in the B-S equation is rx,
: so the new coefficient should be (2r+sigma^2)*(x^2).
:
: price of
|
l******i
发帖数: 1404 | 4 你找纯码工职位的吧，怎么也要弄这个？
Don
【在 c**********e 的大作中提到】
: The B-S equation should not change, only the boundary condition changes. Don
: 't you think so?
|
h**********y
发帖数: 41 | 5 dS=mu* S*dt+sigma*S*dw, mu为S的drift那么
d(S^2)=(2mu+ sigma ^2)* (S^2)*dt+2*sigma* (S^2)*dw
根据risk-neutral pricing 的理论D[exp(-rt)*S^2]必须为martingale
在此情况下
d(S^2)=r* (S^2)*dt+2*sigma* (S^2)*dw'
令y= S^2,则
D[exp(-rt)*C(y,t)]为martingale,所以
(-rC+Ct)*dt+Cy*dy+0.5*Cyy*dy*dy=0
所以Cy前的系数为r* (S^2)
不知这样理解对不对，请大家指正 |
c**********e
发帖数: 2007 | 6
不对.
E_t[exp(-rT)*g(S_T)] is a martingale for any nice g().
But exp(-rt))*S_t^2 is not a martingale. Nor is E_0[exp(-rt))*S_t^2].
【在 h**********y 的大作中提到】
: dS=mu* S*dt+sigma*S*dw, mu为S的drift那么
: d(S^2)=(2mu+ sigma ^2)* (S^2)*dt+2*sigma* (S^2)*dw
: 根据risk-neutral pricing 的理论D[exp(-rt)*S^2]必须为martingale
: 在此情况下
: d(S^2)=r* (S^2)*dt+2*sigma* (S^2)*dw'
: 令y= S^2,则
: D[exp(-rt)*C(y,t)]为martingale,所以
: (-rC+Ct)*dt+Cy*dy+0.5*Cyy*dy*dy=0
: 所以Cy前的系数为r* (S^2)
: 不知这样理解对不对，请大家指正
|
k*****y
发帖数: 744 | 7 想问一个naive的问题：如果S_T只有一个factor的GBM，max(S_T-K, 0)和max(S_T^2-L,
0)同时是tradable的话，有没有arbitrage? |
h**********y
发帖数: 41 | 8 根据Steven E. Shreve的书
令f(t,S_T)=E_t[exp(-r(T-t))*g(S_T)]
exp(-rt)f(t,S_T)才是martingale吧。
我本人对risk-neutral pricing的理解现在还不透彻，希望有人能给这题一个详细的解
答。
谢谢
【在 c**********e 的大作中提到】
:
: 不对.
: E_t[exp(-rT)*g(S_T)] is a martingale for any nice g().
: But exp(-rt))*S_t^2 is not a martingale. Nor is E_0[exp(-rt))*S_t^2].
|
c**********e
发帖数: 2007 | 9 You are right. I made a mistake. I have corrected mine.
But I think your notation f(t,S_T) should be f(t,S_t). It is the price of
the derivative at t.
【在 h**********y 的大作中提到】
: 根据Steven E. Shreve的书
: 令f(t,S_T)=E_t[exp(-r(T-t))*g(S_T)]
: exp(-rt)f(t,S_T)才是martingale吧。
: 我本人对risk-neutral pricing的理解现在还不透彻，希望有人能给这题一个详细的解
: 答。
: 谢谢
|
c**********e
发帖数: 2007 | 10 What do you mean max(S_T-K, 0) tradable?
(1) If you mean the derivatives which pay max(S_T-K, 0) and max(S_T^2-L, 0),
then there is no arbitrage opportunity, regardless how many derivatives are
tradeable on the market.
(2) If you mean max(S_t-K, 0) tradable, the arbitrage opportunity is obvious
. When max(S_t-K, 0) is 0, you long it. Then when it is in the money, you
have a riskfree profit.
L,
【在 k*****y 的大作中提到】
: 想问一个naive的问题：如果S_T只有一个factor的GBM，max(S_T-K, 0)和max(S_T^2-L,
: 0)同时是tradable的话，有没有arbitrage?
|
k*****y
发帖数: 744 | 11 Thanks. Let me rephrase my question.
假设有两个stock：S和T，S是GBM过程，T=S^2.
那么尽管不存在同一个risk neutral的measure，但是如果按各自的risk neutral
measure来price各自的call/put option和hedging，这样似乎还是no arbitrage的，是
不是这样？
这样的话，是不是说risk neutral measure只是对每个asset来说的，market上所有
tradable的asset不一定能找到一个simultaneous的risk neutral measure？ |
c**********e
发帖数: 2007 | 12 If both St and St^2 are tradable, then there is an arbitrage opportunity.
Consider the portfolio (St-S0)^2 = St^2 - 2*S0*St + S0^2. In this portfolio,
you long 1 share of St^2, -2*S0 share of St, and S0^2 dollar risk-free
asset.
It worths 0 at time 0. But at time t, its value is
St^2 - 2*S0*St + S0^2*exp(rt) = (St-S0)^2 + S0^2(exp(rt)-1).
It is always positive! So St and St^2 can not be both tradable.
【在 k*****y 的大作中提到】
: Thanks. Let me rephrase my question.
: 假设有两个stock：S和T，S是GBM过程，T=S^2.
: 那么尽管不存在同一个risk neutral的measure，但是如果按各自的risk neutral
: measure来price各自的call/put option和hedging，这样似乎还是no arbitrage的，是
: 不是这样？
: 这样的话，是不是说risk neutral measure只是对每个asset来说的，market上所有
: tradable的asset不一定能找到一个simultaneous的risk neutral measure？
|
k*****y
发帖数: 744 | 13 Many thanks.
【在 c**********e 的大作中提到】
: If both St and St^2 are tradable, then there is an arbitrage opportunity.
: Consider the portfolio (St-S0)^2 = St^2 - 2*S0*St + S0^2. In this portfolio,
: you long 1 share of St^2, -2*S0 share of St, and S0^2 dollar risk-free
: asset.
: It worths 0 at time 0. But at time t, its value is
: St^2 - 2*S0*St + S0^2*exp(rt) = (St-S0)^2 + S0^2(exp(rt)-1).
: It is always positive! So St and St^2 can not be both tradable.
|
