Negative Hypergeometric Distribution

An experiment of sampling of objects without replacement till a fixed number of objects having a certain characteristic is obtained is known as negative hypergeometric experiment. Thus, a random variable X denoting the number of trials or drawings required to get a fixed number of objects with a specified characteristics is called negative hypergeometric random variable and its distribution is called negative hypergeometric distribution.

Defination

A random variable X is said to follow a negative hypergeometric ddistribution with three parameters P, Q, and r, if its probability mass function is given by

$$p(x; P, Q, r) \ = \ \frac{ \binom{x+r-1}{x} \ \binom{N-r-x}{Q-x}}{ \binom{N}{Q}} \ \ \ \ ; \ x \ = \ 0, \ 1, \ 2, ..., \ Q$$

where, N = P+Q and r∈ { 1, 2, 3, ..., P}.

Derivation of Negative Hypergeometric Distribution

Suppose that a population contains N objects of which P objects have a certain characteristic called P successes and Q = N - P objects do not have the specified characteristics called Q failures. Suppose a random sample of Q objects is drawn without replacement from N objects. The objects are drawn at random one by one without replacement from the population until r successes are obtained.

Let X denotes the number of failures in the sample drawn. Then, in this case the last object drawn is the success. So, the x failures out of x + r - 1 objects can be obtained in $$\binom{x+r-1}{x} \ ways$$ and Q - x failures can be obtained out of N - r - x objects in $$\binom{N-r-x}{Q-x} \ ways.$$

$$Again, \ the \ Q \ failures \ out \ of \ N \ objects \ can \ be obtained \ in \ \binom{N}{Q} \ ways.$$

Then the probability of x failures in obtaining r successes is given by

$$p(x) \ = \ \frac{ \binom{r+x-1}{x} \ \binom{N-r-x}{Q-r}}{ \binom{N}{Q}} \ \ \ \ ; \ x \ = \ 0, \ 1, \ 2, \ ..., \ Q$$

Here, N = P + Q and r∈ { 1, 2, 3, ..., P}.

The random variable X follows negative hypergeometric distribution with three parameters P, Q and r is denoated by X ~ NH { P, Q, r }.

It can be shown that

$$\sum_{x=0}^Q \ p(x; P, Q, r) \ = \sum_{x=0}^Q \ \frac{ \binom{r+x-1}{x} \ \binom{N-r-x}{Q-x}}{ \binom{N}{Q}}$$

$$= \ \frac{ \binom{N}{Q}}{ \binom{N}{Q}}$$

$$\sum_{x=0}^Q p(x; P, Q, r) \ = \ 1$$