The Poisson Distribution

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Our analysis of The Drunkard痴 Walk gave us the normal Gaussian distribution of probabilities. But we have other probability distributions that come into play under certain circumstances.

(Eq地 1)

which we derived from our analysis of the Drunkard痴 Walk. If an event has a probability p of occurring in a single trial, then in N trials the event has a probability PN(K) of occurring K times. If we allow N to take on very large values, Equation 1 segues into the equivalent description of the Gaussian probability distribution, the famous bell-shaped curve when plotted on a graph.

We have another way of applying Equation 1. We have filled a large vat with x black pebbles and y white pebbles, which we draw from the vat one at a time without looking at them. The probability p of drawing a black pebble thus equals x/(x+y). After drawing a pebble we toss it back into the vat, thereby making all of our drawings of pebbles independent of one another. PN(K) then represents the probability that we will draw a black pebble K times out of N drawings.

Now consider the situation in which p<<1. PN(K) will differ significantly from zero only when K<<N, so we consider only those cases. We then let p approach zero as N approaches infinity in such a way that pN=A, a constant. We can accomplish that by adding only white pebbles to the vat. The low probability condition enables us to use two approximations to rewrite Equation 1:

1. We can write

(Eq地 2)

But we know that

(Eq地 3)

When p 0, ln(1-p) -p, so for cases in which K<<N we have

(Eq地 4)

2. When K<<N

(Eq地 5)

because only the K highest factors of N! survive the division and, for very large values of N, they are all approximately equal to N.

Substituting from Equations 4 and 5 into Equation 1 then gives us

(Eq地 6)

which describes the probabilities of the Poisson distribution. Note that because

(Eq地 7)

the probability function is automatically normalized.

As an example of an application of the Poisson distribution to describe some feature of Reality consider radioactive decay. Take a sample that consists of a very large number M of a single species of unstable atom. We consider a time interval T divided into brief intervals of duration t such that the probability of two decays occurring in a given interval t is negligible. The decays occur at random, so the N=T/t trials are all independent of each other. The probability of a decay occurring in a given interval t must stand in direct proportion to the size of the interval and of the number of atoms available to decay. So we have the probability p=ctM, in which c represents a constant that encodes those properties of the atom that make it decay. We want to know how rapidly the material decays.

Start by calculating the average number of decays that occurred in the period T. For that we have simply

(Eq地 8)

In that description I exploited the fact that in the last step the term for K=0 does not exist, so the summation of AK-1/(K-1)! still yields the exponential of A. The average rate at which the decays occur, then, conforms to

(Eq地 9)

Rearranging the variables in that equation gives us

(Eq地 10)

which integrates to

(Eq地 11)

and gives us

(Eq地 12)

the law of exponential decay, which experiments with rapidly decaying species of atoms have confirmed. In Equation 11 lnM0 represents the constant of integration, a to-be-determined number.

Thus we see one use of the Poisson distribution and are well prepared to find others.

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