b***k
发帖数: 2673 | 1 ☆─────────────────────────────────────☆
trenchant (N/A) 于 (Tue Jan 8 00:13:29 2008) 提到:
Nt (t>=0) is a Poisson process with event
arrival rate lambda . P(lambda)
Consider a process
Xt = exp(Nt - at), (t>=0)
What is the condition expectation
E[Xt|Xs] for 0≤s≤t .
and: When is this process a martingale?
谢谢。。。
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finalguy (答案) 于 (Tue Jan 8 00:19:20 2008) 提到:
a=lambda*(e-1)
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trenchant (N/A) 于 (Tue Jan 8 0 | w********r
发帖数: 290 | 2 E[Xt|Xs]=E[Xs*(Xt/Xs)|Xs]=Xs*E[(Xt/Xs)|Xs]
=Xs*E[exp(Nt-Ns)/exp(a(t-s))|Xs]=Xs*E[exp(Nt-Ns)|Xs]/exp(a(t-s))
=Xs*E[exp(Nt-s)]/exp(a(t-s))=Xs*M_(Nt-s)(1)/exp(a(t-s))
=Xs*exp(lambda*(t-s)*(e-1))/exp(a(t-s))
=Xs*exp[(lambda*(e-1)-a)*(t-s)]
When a=lambda*(e-1), i.e. exp[(lambda*(e-1)-a)*(t-s)]=1, E[Xt|Xs]=Xs Xt is a
martingale.
where M_(Nu)(t) is the mgf of Poisson distribution Nu (u>0)
Reference: http://en.wikipedia.org/wiki/Poisson_distribution | i******d
发帖数: 54 | |
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