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Why are the geometric distribution and hypergeometric distribution called "geometric" and "hypergoemetric" respectively?

Is it because their pmfs take some special form? Thanks!

whuber
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Tim
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2 Answers2

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Yes, the terms refer to the probability mass functions (pmfs).

2,500 years ago, Euclid (in Books VIII and IV of his Elements) studied sequences of lengths having common proportions.. At some point such sequences came to be known as "geometric progressions" (although the term "geometric" could for a similar reason just as easily have been applied to many other regular series, including those now called "arithmetic").

The probability mass function of a geometric distribution with parameter $p$ forms a geometric progression

$$p, p(1-p), p(1-p)^2, \ldots, p(1-p)^n, \ldots.$$

Here the common proportion is $1-p$.

Several hundred years ago a vast generalization of such progressions became important in the studies of elliptic curves, differential equations, and many other deeply interconnected areas of mathematics. The generalization supposes that the relative proportions among successive terms at positions $k$ and $k+1$ could vary, but it limits the nature of that variation: the proportions must be a given rational function of $k$. Because these go "over" or "beyond" the geometric progression (for which the rational function is constant), they were termed hypergeometric from the ancient Greek prefix $\grave\upsilon^\prime\pi\varepsilon\rho$ ("hyper").

The probability mass function of a hypergeometric function with parameters $N, K,$ and $n$ has the form

$$p(k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$

for suitable $k$. The ratio of successive probabilities therefore equals

$$\frac{p(k+1)}{p(k)} = \frac{(K-k)(n-k)}{(k+1)(N-K-n+k+1)},$$

a rational function of $k$ of degree $(2,2)$. This places the probabilities into a (particular kind of) hypergeometric progression.

whuber
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  • Thanks! Are there other distributions whose pmfs also form geometric or hypergeometric progressions? – Tim Mar 20 '14 at 00:45
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    If a pmf forms a geometric progression then it must be a shifted, rescaled, and/or truncated geometric distribution. If it forms a hypergeometric progression of degree (2,2) then a similar conclusion holds. There are distributions associated with *any* series that sums to a finite value, and so the hypergeometric distribution generalizes to many other distributions (by using different rational functions). Most of them do not have names. One exception is the [negative binomial distribution](http://en.wikipedia.org/wiki/Negative_binomial_distribution) whose pmf is hypergeometric of degree (1,1). – whuber Mar 20 '14 at 14:37
  • Thanks! Does the pmf of Poisson distribution form some special series/progression? Given a Poission distribution with rate parameter $\lambda$, then $p(k+1)/p(k) = \lambda/(k+1)$. Does the pmf form some special series or progression? – Tim Mar 20 '14 at 22:58
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    Yep, that's a rational function of degree (0,1), so it fits the general definition of a hypergeometric progression. – whuber Mar 21 '14 at 00:52
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According to one source, it is because for the geometric distribution pmf(k) is the geometric mean of pmf(k-1) and pmf(k+1). The geometric mean of two numbers A and B is $\sqrt{A B}$. Classically this problem was interpreted as finding the length of the the sides of a square with area equal to a rectangle with sides of length A and B, a geometric problem.

veryshuai
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    Your source resorts to the kind of speculation I was referring to (somewhat elliptically) at the beginning of my answer. The internet is full of people who make the same assertion, but because it is equally easy geometrically to find an *arithmetic mean* as a geometric mean, in the end this property (of having a "geometric" construction) does not appear to explain anything. It would be very interesting to find an authority who can track down the actual historical uses of "geometric" and "arithmetic" to help us understand how these terms really arose. – whuber Mar 19 '14 at 18:59