On Sat, 8 Nov 2003 16:43:30 -0600, "Fan Yang" <[EMAIL PROTECTED]>
wrote:
> Suppose arrivals come as a rate lambda Poisson process. Let Au be
> the time since the last arrival up to time u. Precisely, if
> 0 = T0 < T1 < T2 < T3 < ... where {Ti : i >= 1} are the arrival times
> and Ti <= u < Ti+1, then Au = u-Ti which is a number in [0, u]. In
> particular,
> if there are no arrivals in [0, u] then we set At = t.
You mean "...then we set Au = u"
This somewhat artificial definition can be interpreted more naturally
as: the first arrival takes place at t=0 with probabilty 1, i.e.
N(0)=1.
> The notation Au
> refers to
> age, namely the age of the most recent arrival.
>
> Find the probability distribution of Au, by giving the cumulative
> distribution function Gu(t) = P[Au <= t] for all real t.
>
> **********************************************************
> The following is my solution.
>
> P[A(u)>t] = P[u-A(u)<u-t] = P[T(i) < u-t]
> = P[no arriavals in (u-t, u)]
> = P[N(u)-N(u-t) = 0]
> = exp(-lamba*t)
>
> Then P[A(u)<=t] = 1-exp(-lamba*t)
>
> Is it right?
Partially ;-)
F_past[u](t) = 1-exp(-lamba*t) for 0<=t<u
1 for t>=u
It is crucial to note that the cdf is *discontinuous* for t=u due to
the fact that we force an arrival t=0
It is different from the waiting time for the next event in the future
which has the distribution
F_future[u](t) = 1-exp(-lambda*t) 0<=t<oo
A more natural distribution would be the age of the last arrival up to
t=u under the condition that there was at last one arrival in the past
using the standard definition N(0)=0.
1
F[u](t) = ---------------- * (1-exp(-lambda*t)
1-exp(-lambda*u)
which increases continuously from 0 to 1 in [0,1].
--
Horst
.
.
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