numpy.random.Generator.negative_binomial

method

random.Generator.negative_binomial(n, p, size=None)

Draw samples from a negative binomial distribution.

Samples are drawn from a negative binomial distribution with specified parameters, n successes and p probability of success where n is > 0 and p is in the interval (0, 1].

Parameters

nfloat or array_like of floats

Parameter of the distribution, > 0.

pfloat or array_like of floats

Parameter of the distribution. Must satisfy 0 < p <= 1.

sizeint or tuple of ints, optional

Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn. If size is None (default), a single value is returned if n and p are both scalars. Otherwise, np.broadcast(n, p).size samples are drawn.

Returns

outndarray or scalar

Drawn samples from the parameterized negative binomial distribution, where each sample is equal to N, the number of failures that occurred before a total of n successes was reached.

Notes

The probability mass function of the negative binomial distribution is

\[P(N;n,p) = \frac{\Gamma(N+n)}{N!\Gamma(n)}p^{n}(1-p)^{N},\]

where \(n\) is the number of successes, \(p\) is the probability of success, \(N+n\) is the number of trials, and \(\Gamma\) is the gamma function. When \(n\) is an integer, \(\frac{\Gamma(N+n)}{N!\Gamma(n)} = \binom{N+n-1}{N}\), which is the more common form of this term in the the pmf. The negative binomial distribution gives the probability of N failures given n successes, with a success on the last trial.

If one throws a die repeatedly until the third time a “1” appears, then the probability distribution of the number of non-“1”s that appear before the third “1” is a negative binomial distribution.

References

1

Weisstein, Eric W. “Negative Binomial Distribution.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/NegativeBinomialDistribution.html

2

Wikipedia, “Negative binomial distribution”, https://en.wikipedia.org/wiki/Negative_binomial_distribution

Examples

Draw samples from the distribution:

A real world example. A company drills wild-cat oil exploration wells, each with an estimated probability of success of 0.1. What is the probability of having one success for each successive well, that is what is the probability of a single success after drilling 5 wells, after 6 wells, etc.?

>>> s = np.random.default_rng().negative_binomial(1, 0.1, 100000)
>>> for i in range(1, 11): 
...    probability = sum(s<i) / 100000.
...    print(i, "wells drilled, probability of one success =", probability)

numpy.random.Generator.multivariate_normal numpy.random.Generator.noncentral_chisquare