Poisson and Binomial Questions

Date: 06/13/98 at 11:56:23
From: Kenneth W. Hodgkinson
Subject: Probability

1) Why does the waiting time for an event in a Poisson process exceed 
   t with the probability e^(-Lambda*t)?

2) Can a binomial variable can be thought of as a sum? Please explain 
   this, and use this idea to find the mean of a binomial variable.

Date: 06/13/98 at 17:04:04
From: Doctor Anthony
Subject: Re: Probability

Problem 1:

I will illustrate with an example. Suppose the number of accidents per 
week on a certain stretch of road follows a Poisson distribution with 
parameter 2.

If the average is 2 per week, then in time t weeks, the average would 
be 2t.

The probability of no accident P(0) in time t is therefore e^(-2t)

The probability:

   P(T>t) = probability that there are no events in time 0 to t  
          = e^(-2t)

   P(T<t) =  1 - e^(-2t)

This is the cumulative distribution function (cdf) of the time to 
first event.

To get the probability density function (pdf) we differentiate this 
and get:

   f(t) = 2e^(-2t)    

This is pdf of time interval between events. It is called the 
exponential distribution.

Problem 2:

Suppose we have probability p of success at each trial and probability 
q of failure. The expectation in one trial is given by:

   E(X) = p.1 + q.0  = p

Suppose now we have n trials, let X1, X2, X3,  ......., Xn be the 
results of these trials with any Xr taking values 0 or 1 with 
probability q and p respectively.

If:

   X = X1 + X2 + X3 + X4 + ...... + Xn  
    (= total successes in n trials)

then:

   E(X) = E(X1) + E(X2) + E(X3) + .... + E(Xn)

        =  p + p + p + p + .........+ p

        = np

So the mean of the binomial distribution with parametrs n and p is np.  

-Doctor Anthony,  The Math Forum
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