c**a
发帖数: 316 | 1 Problem: B(t) a standard Brownian motion, how is the expecting time it first
hits either -1 or 1.
Solution:
X(t)=exp(B(t)-0.5t) is a martingale. By option sampling theorem, X(t)
stopped at B(t)=-1 or 1 is also a martingale.
Since X(0)=1, we have 0.5*X(1)+0.5*X(-1)=1.
Or exp(1-0.5t)+exp(-1-0.5t)=2,
Or exp(-0.5t)=2/(e+1/e)
Or t = sqrt(-2 ln(2/(e+1/e)).
What is wrong? | N*******A
发帖数: 65 | 2 stopping time Tau=inf{t: B(t)=-1 or B(t)=1}
E[X(Tau)]=X(0)=1. By optional sampling theorem, X(t)
stopped at Tau is also a martingale.
E[X(Tau)]=E[exp{B(Tau)-0.5*Tau}]
=P(B(Tau)= 1)* E[exp( 1-0.5*Tau)|B(Tau)= 1]
+P(B(Tau)=-1)* E[exp(-1-0.5*Tau)|B(Tau)=-1]
=0.5*E[exp( 1-0.5*Tau)|B(Tau)= 1]
+0.5*E[exp(-1-0.5*Tau)|B(Tau)=-1]
Then you jumped to
=0.5* exp( 1-0.5*E(Tau))
+0.5* exp(-1-0.5*E(Tau))
without justification?
first
【在 c**a 的大作中提到】
: Problem: B(t) a standard Brownian motion, how is the expecting time it first
: hits either -1 or 1.
: Solution:
: X(t)=exp(B(t)-0.5t) is a martingale. By option sampling theorem, X(t)
: stopped at B(t)=-1 or 1 is also a martingale.
: Since X(0)=1, we have 0.5*X(1)+0.5*X(-1)=1.
: Or exp(1-0.5t)+exp(-1-0.5t)=2,
: Or exp(-0.5t)=2/(e+1/e)
: Or t = sqrt(-2 ln(2/(e+1/e)).
: What is wrong?
| i*****k
发帖数: 5 | | t********t
发帖数: 1264 | 4 "Since X(0)=1, we have 0.5*X(1)+0.5*X(-1)=1."
this sentence is wrong | B*******8
发帖数: 8 | 5 I think the right way is:
B(t)^2 - t is martingale.
E(B(tau)^2 - tau) = 0
E(tau) = 1 | s***d
发帖数: 6 | 6 The sqrt on the last row shouldn't be there... Otherwise looks good? |
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