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Proof:

P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n}

(Bayes theorem)

=P{T>n+m}/P{T>n}

=((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m

(1-p)^m = {T>m}

So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.

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โˆ™ 2011-03-13 15:37:21
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