Horst,

Thanks a lot.

Fan


"Horst Kraemer" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> 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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