Poisson and Binomial QuestionsDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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