Hypergeometric distribution

In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the probability of successes in draws, without replacement, from a finite population of size that contains exactly objects with that feature, where in each draw is either a success or a failure. In contrast, the binomial distribution describes the probability of successes in draws with replacement.

Definitions

Probability mass function

The following conditions characterize the hypergeometric distribution:

The result of each draw can be classified into one of two mutually exclusive categories.
The probability of a success changes on each draw, as each draw decreases the population.

A random variable follows the hypergeometric distribution if its probability mass function is given by
where

is the population size,
is the number of success states in the population,
is the number of draws,
is the number of observed successes,
is a binomial coefficient.

The is positive when.
A random variable distributed hypergeometrically with parameters, and is written and has probability mass function above.

Combinatorial identities

As required, we have
which essentially follows from Vandermonde's identity from combinatorics.
Also note that
This identity can be shown by expressing the binomial coefficients in terms of factorials and rearranging the latter. Additionally, it
follows from the symmetry of the problem, described in two different but interchangeable ways.
For example, consider two rounds of drawing without replacement. In the first round, out of neutral marbles are drawn from an urn without replacement and coloured green. Then the colored marbles are put back. In the second round, marbles are drawn without replacement and colored red. Then, the number of marbles with both colors on them has the hypergeometric distribution. The symmetry in and stems from the fact that the two rounds are independent, and one could have started by drawing balls and colouring them red first.
Note that we are interested in the probability of successes in draws without replacement, since the probability of success on each trial is not the same, as the size of the remaining population changes as we remove each marble. Keep in mind not to confuse with the binomial distribution, which describes the probability of successes in draws 'with replacement.'''''

Properties

Working example

The classical application of the hypergeometric distribution is sampling without replacement. Think of an urn with two colors of marbles, red and green. Define drawing a green marble as a success and drawing a red marble as a failure. Let N describe the number of all marbles in the urn and K describe the number of green marbles, then N − K corresponds to the number of red marbles. Now, standing next to the urn, you close your eyes and draw n marbles without replacement. Define X as a random variable whose outcome is k, the number of green marbles drawn in the experiment. This situation is illustrated by the following contingency table:

	drawn	not drawn	total
green marbles	k	K − k	K
red marbles	n − k	N + k − n − K	N − K
total	n	N − n	N

Indeed, we are interested in calculating the probability of drawing k green marbles in n draws, given that there are K green marbles out of a total of N marbles. For this example, assume that there are 5 green and 45 red marbles in the urn. Standing next to the urn, you close your eyes and draw 10 marbles without replacement. What is the probability that exactly 4 of the 10 are green?
This problem is summarized by the following contingency table:

	drawn	not drawn	total
green marbles	k = 4	K − k = 1	K = 5
red marbles	n − k = 6	N + k − n − K = 39	N − K = 45
total	n = 10	N − n = 40	N = 50

To find the probability of drawing k green marbles in exactly n draws out of N total draws, we identify X as a hyper-geometric random variable to use the formula
To intuitively explain the given formula, consider the two symmetric problems represented by the identity

left-hand side - drawing a total of only n marbles out of the urn. We want to find the probability of the outcome of drawing k green marbles out of K total green marbles, and drawing n-k red marbles out of N-K red marbles, in these n rounds.
right hand side - alternatively, drawing all N marbles out of the urn. We want to find the probability of the outcome of drawing k green marbles in n draws out of the total N draws, and K-k green marbles in the rest N-n draws.

Back to the calculations, we use the formula above to calculate the probability of drawing exactly k green marbles
Intuitively we would expect it to be even more unlikely that all 5 green marbles will be among the 10 drawn.
As expected, the probability of drawing 5 green marbles is roughly 35 times less likely than that of drawing 4.

Symmetries

Swapping the roles of green and red marbles:
Swapping the roles of drawn and not drawn marbles:
Swapping the roles of green and drawn marbles:
These symmetries generate the dihedral group.

Order of draws

The probability of drawing any set of green and red marbles depends only on the numbers of green and red marbles, not on the order in which they appear; i.e., it is an exchangeable distribution. As a result, the probability of drawing a green marble in the draw is
This is an ex ante probability—that is, it is based on not knowing the results of the previous draws.

Tail bounds

Let and. Then for we can derive the following bounds:
where
is the Kullback–Leibler divergence and it is used that.
Note: In order to derive the previous bounds, one has to start by observing that where are dependent random variables with a specific distribution. Because most of the theorems about bounds in sum of random variables are concerned with independent sequences of them, one has to first create a sequence of independent random variables with the same distribution and apply the theorems on. Then, it is proved from Hoeffding that the results and bounds obtained via this process hold for as well.
If n is larger than N/2, it can be useful to apply symmetry to "invert" the bounds, which give you the following:

Statistical Inference

Hypergeometric test

The hypergeometric test uses the hypergeometric distribution to measure the statistical significance of having drawn a sample consisting of a specific number of successes from a population of size containing successes. In a test for over-representation of successes in the sample, the hypergeometric p-value is calculated as the probability of randomly drawing or more successes from the population in total draws. In a test for under-representation, the p-value is the probability of randomly drawing or fewer successes.
The test based on the hypergeometric distribution is identical to the corresponding one-tailed version of Fisher's exact test. Reciprocally, the p-value of a two-sided Fisher's exact test can be calculated as the sum of two appropriate hypergeometric tests.
The test is often used to identify which sub-populations are over- or under-represented in a sample. This test has a wide range of applications. For example, a marketing group could use the test to understand their customer base by testing a set of known customers for over-representation of various demographic subgroups.

Related distributions

Let and.

If then has a Bernoulli distribution with parameter.
Let have a binomial distribution with parameters and ; this models the number of successes in the analogous sampling problem with replacement. If and are large compared to, and is not close to 0 or 1, then and have similar distributions, i.e.,.
If is large, and are large compared to, and is not close to 0 or 1, then

where is the standard normal distribution function

If the probabilities of drawing a green or red marble are not equal then has a noncentral hypergeometric distribution
The beta-binomial distribution is a conjugate prior for the hypergeometric distribution.

The following table describes four distributions related to the number of successes in a sequence of draws:

	With replacements	No replacements
Given number of draws	binomial distribution	hypergeometric distribution
Given number of failures	negative binomial distribution	negative hypergeometric distribution

Multivariate hypergeometric distribution

The model of an urn with green and red marbles can be extended to the case where there are more than two colors of marbles. If there are K_i marbles of color i in the urn and you take n marbles at random without replacement, then the number of marbles of each color in the sample has the multivariate hypergeometric distribution:
This has the same relationship to the multinomial distribution that the hypergeometric distribution has to the binomial distribution—the multinomial distribution is the "with-replacement" distribution and the multivariate hypergeometric is the "without-replacement" distribution.
The properties of this distribution are given in the adjacent table, where c is the number of different colors and is the total number of marbles in the urn.