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### 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
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Probability

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