From - Mon Aug 17 18:41:02 1998
From: horst.kraemer@snafu.de (Horst Kraemer)
Newsgroups: sci.stat.math
Subject: Re: classical poisson process problem
Date: Mon, 17 Aug 1998 14:28:21 GMT
Message-ID: <35d7fd94.341134614@news.snafu.de>
References: <6r2vr1$f0q$1@lwnws01.ne.highway1.com>
Lines: 62

On Fri, 14 Aug 1998 23:38:09 -0400, "Arthur M. Schneiderman"
<art@schneiderman.com> wrote:

>Trains arrive randomly at a station at an average rate of 1 per x minutes
> or y per hour).
>
>(a)  You arrive at the station and there is no train in sight.  What is the
>probability that the next train arrives in T minutes?
>
>(b)  You arrive at the station just as a train pulls away.  Now what is the
>answer to (a).
>
>I remember this question on an exam I took 20 years ago.  I also remember
>that I got it wrong and didn't understand why.  What I do remember is that
>(b) considers the conditional probability given that the event just
>occurred.  Can anyone help me?  Can anyone give me a reference?
>
>PS. No, it hasn't kept me up for 20 years.  I need to know the answer for a
>similar problem that I'm working on.


By definition a process is a POISSON point process with intensity m
(in your case for example m = 10, 10 trains per hour, the unit if T is
"hours") if the probability that there are exactly k events (arrivals)
in any interval [t0,t0+T] has a POISSON distribution with mean T*m.

                 (mT)^k 
	Pr(k) =  -------  * exp (-mT)
                    k!

This distribution is the same for any t0 and any T>=0 and the
distributions for disjoint time intervals [t0,t1], [t2,t3], t1<t2, are
stochastically independent.

Therefore the probability that there is at least one arrival in
[t0,t0+T] is

	P(T)  =  1 - Pr(0) = 1- exp(-mT)

if you your observation starts at t0.

This distribution which expresses the "waiting time" for the next
arrival

	f(T) = m * exp (-mT)       ( density,pdf )

	F(T) = 1 - exp (-mT)       ( distribution, cdf)

is an exponential distribution. It has "no memory", i.e. the fact how
many events occured _before_ t0 has no impact on the probability of
events occurring in the interval [t0,t0+T].


Thus - as far as I understand the question - the answer to a) and b)
is the same, i.e.  p=1-exp(-Tm). If m=10 trains per hour, the
probability of at least one arrival in the next 1/2 hour is
1-exp(-10*(1/2)) = 1-exp(-5).


Regards
Horst