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This semester, I teach an introduction to probability course tailored for students with no science background and so with very very little prerequisites. We started with the basics of analytic combinatorics then moved on to random variables and the study of common laws (binomial, hypergeometric, geometric, Poisson). The audience being what it is, I try to avoid as much as possible calculus derivations of probability facts.

For some aspects of the course, it worked out well (for instance the derivation of the expectation of the binomial law) but because I am barely more knowledgeable than my students when it comes to probability, I have been unable to answer this question:

Is there a set of natural probability properties which characterize the discrete Poisson law?

If yes, then I could use this as a definition of the Poisson law, which would suit my students better than saying "it's the law such that $P(X=k)=e^{-\lambda}\frac{\lambda^{k}}{k!}$". By natural above, I want to convey the meaning that I hope they can be formulated using natural language (like, say, memorylessness) rather than using analytic objects.

More precisely, what I have in mind is the following:

Is there such a set of properties which would make it at least a little intuitively plausible that the sum of two variables following Poisson law also follows Poisson law?

Of course, the proof of the above fact is completely elementary, but it would still be above the level of everyone in the audience except perhaps the 3 top students.

Note that I would be happy even if proving that this set of properties characterize Poisson law turned out to be much harder than anything I will do in this course (or even much harder than anything I know myself about probability), because what I am looking for is not logical rigour but rather psychological efficiency: in 10 years, my students will have completely forgotten what a derivative is, but I would like them to be able to recollect something if confronted with an epidemiological survey using random variables (at least my most successful students use this course to strengthen their math knowledge before studying medicine).

I realize this question is very elementary, and would understand if it is deemed inappropriate, but the standard references I might consult on the subject will invariably (and with good reasons) develop much more calculus that my students will ever know before dealing with such questions (typically, they will characterize the Poisson law as the limit of the binomial law via Stirling's formula).

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4 Answers 4

up vote 4 down vote accepted

What about the characterization of Poisson point processes ?

Let us consider a counting process $(N(t))_{t \ge 0}$. That is, $N(0)=0$, $N(t)$ only increases by jump of height $1$, and is right continuous. You can see $N(t)$ as the number of points of a random set in $]0,t]$.

Then $N(t)_t$ is a homogeneous poisson point process if and only if :

1) the increments are independent

2) the increments are stationary : $N(t+s)-N(t)$ has the same law as $N(s)$.

(Maybe there is a further regularity assumption).

This implies that the increments are Poisson distributed : there exists $\lambda$ such that $N(s)$ is distributed according to Poisson with parameter $s\lambda$ for all $s$. This shows that under seemingly general conditions the Poisson distribution appears.

You can also see this from a more geometrical point of view, by considering more general point process than on the line.

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Thanks a lot, Camomille. –  Olivier Apr 24 '11 at 21:36
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We can't expect a completely finite way for the Poisson distribution to arise, since the number $e$ must come from somewhere. On the other hand, it should definitely not be necessary to introduce Stirling's formula.

I think the most natural approach is to define Poisson($\lambda$) as the limit distribution of the number of heads in a sequence of $N$ independent flips of a biased coin with probability $\lambda/N$ of heads.

This must be accompanied by some derivation showing that there is such a limit, and leading to the formula you state. Such a derivation will use binomial coefficients and the definition of the number $e$, but not much more.

But even without the derivation, if we just assume that the limit exists, it shows why the sum of two independent Poisson variables is again Poisson.

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Related to camomille's answer, I also like to think about Poisson processes, but in another way.

The exponential distribution is perhaps not too hard to motivate, due to its memoryless property. If you have "arrivals" that are going to happen "unexpectedly", then it's reasonable to guess that the inter-arrival times should be iid exponential with some rate $\lambda$. If you ask for $N(t)$, the number of arrivals by time $t$, you find that it's Poisson with parameter $\lambda t$.

This also plays nicely into the sum of independent Poissons being Poisson: if you have exponential arrivals of two types occurring independently with rates $\lambda, \mu$, it's easy to see that the time until the next arrival (of either type) is also exponential, with rate $\lambda + \mu$. Thus the number of arrivals of either type by time $1$ (i.e. the superposition of the two Poisson processes) is on the one hand the sum of independent Poissons with parameters $\lambda, \mu$, and on the other hand Poisson with parameter $\lambda + \mu$.

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Thanks a lot; I'm going to try this. –  Olivier Apr 24 '11 at 21:35
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The following heuristic argument is finitary apart from the sum of the exponential series.

Consider a birth clinic observing on average $\lambda$ male births and $\lambda$ female births per day. Let $p_k$ be the probability that on a given day we have exactly $k$ boys; then this same $p_k$ is also the probability that we have exactly $k$ girls. Consider the days where together $n$ kids are born. On such days the boys and girls are binomially distributed among them, i.e., we have $${\bf P}[B=k\ \wedge\ G=n-k]={1\over 2^n}{n\choose k}\ .$$ As boys and girls are born independently of each other when their time has come the left side of this equation has the value $p_k p_{n-k}/P_n$ where $P_n={\bf P}[B+G=n]$. It follows that $$p_k p_{n-k}={P_n\over 2^n}{n\choose k}$$ and in particular $${p_0 p_n\over p_1 p_{n-1}}={1\over n}\ .$$ Since this holds for any $n\geq1$ we conclude that $$p_n={\mu^n\over n!}p_0\quad (n\geq1)\ ,\qquad \mu:={p_1\over p_0}\ .$$ The condition $\sum_{n\geq 0} p_n=1$ gives $p_0=e^{-\mu}$, so $$p_n={\mu^n\over n!}e^{-\mu}\qquad(n\geq0)\ .$$ Now we observe $$\lambda={\bf E}(B)=\sum_{k=0}^\infty k p_k=e^{-\mu}\sum_{k=1}^\infty {k\over k!}\mu^k =\mu\ ,$$ so that definitively $p_k={\lambda^k\over k!}e^{-\lambda}$.

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