以下是引用juedistar在2008-7-22 10:47:00的发言:请教一下关于期权定价的问题:在B-S公式中,看涨期权的S(T)大于K(T)和看跌时S(T)小于K(T)的概率怎么算?或者说S(T)是什么分布?
另一个,有个可以在T也可以在T/2时执行的期权,S(T/2)大于a的概率和S(T)大于b的概率一样么。我原来是学别的专业的,但愿我说清楚了。。。
拜谢答复!
continues的情况下,ST=St*exp(mu-.5sigma^2+sigma*t^.5*z), where z follows Normal(0,1), this is from the GBM sde dS=S(mu*dt+sigma*dWt)
therefore Pr(ST>K) =Pr(St*exp(mu-.5sigma^2+sigma*t^.5*z>K) (1),
make z in one side in the inequality and use standard nor cdf to find the value of probability.
Actually log diff of St follows normal, in other word St-St-1 follows lognormal distribution
As to the second one, you can calculate them by (1). and because the increment of Wt is independent, i.e. WT/2 and WT is indep since W0=0, therefore you can also find the P( A,B)= P(A)P(B)
If you mean not specify the T/2 and T be the only two stopping times, I can show you a formula for stopping time t = inf(t: Wt=a or Wt=-b), denotes Ta if Wt Wiener process touch a, and T-b for hitting -b. where a and b are positive real numbers.
Besides we know that E(Wt)=0, Then Pr (Ta<T-b), standing for the wiener process first touch a not b. by def of prob,
we have P(Ta<T-b) + P (Ta<T-b) =1,
and by means of Expectation, we have a*P+b*(1-P)=0, we can get P.
PS: in your situation St will hit a and b, rather Wt. So you need to do little algebra to find proper a and b for Wt.
I hope this is clear for you.