Waiting Time Distributions

Date: 06/08/97 at 21:13:23
From: Nick
Subject: probability

I've tried talking to my teacher, but I need another person to explain 
waiting time distribution tables and how to solve them. 

Date: 06/09/97 at 12:48:50
From: Doctor Anthony
Subject: Re: Probability

Waiting time distributions are based on the exponential distribution, 
which in turn is derived from the Poisson distribution. Whilst the 
Poisson distribution governs the occurrence of random events in space 
or time, the intervals between such events is governed by the 
exponential distribution.

Consider a problem such as the emission of radioactive particles. The 
number of emissions during any particular unit of time is governed by 
the Poisson distribution with mean L per unit time. We will consider 
the distribution of time intervals between successive emissions and 
find the probability that there is a time-interval of length t between 
successive emissions.

We divide t into increments dt in length such that there is a small 
probability p of the occurrence of an emission during dt.  q is the 
probability of no occurrence of an emission in that interval, where 
p+q = 1.  dt is assumed small enough to make the probability of more 
than one emission in dt negligible.

The probability, denoted by dP, that there are n incremental intervals 
between successive emissions is given by:

   dP = q^n*p      That is n intervals with no emissions, and in the        
                   (n+1)th interval we have an emission.

So dP = (1-p)^n*p.

Now if there are L emissions per unit time, the mean number of 
emissions in time dt is L*dt. So using the formula, mean = number of 
trials times the probability of success at any trial, we get:
  
            L*dt = 1 x p

So             p = L*dt.

This gives us dt = p/L.

Writing     n*dt = t,     we also have dt = t/n.

Equating expressions for dt we have:  p/L = t/n
                                        p = Lt/n

Then     dP = (1 - Lt/n)^n*L*dt

      dP/dt = [1 - Lt/n]^n*L

As n -> infinity, the expression in brackets tends to e^(-Lt)

      dP/dt = L*e^(-Lt)

The righthand side is the exponential distribution, and is the 
probability density function for the interval between successive 
events, i.e. it is the probability that the time interval lies between 
t and t+dt.

The mean is given by INT(0 to infinity)[L*e^(-Lt)*t*dt].

Integrating by parts gives Mean = 1/L.

This result is what we should expect.  If L is the expected number of 
events in unit time (from the original Poisson distribution), then we 
should also expect that the mean interval between events is 1/L.

-Doctor Anthony,  The Math Forum
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