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)
, thenm * n * k
samples are drawn. If size isNone
(default), a single value is returned ifn
andp
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 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.
Because this method internally calls
Generator.poisson
with an intermediate random value, a ValueError is raised when the choice of \(n\) and \(p\) would result in the mean + 10 sigma of the sampled intermediate distribution exceeding the max acceptable value of theGenerator.poisson
method. This happens when \(p\) is too low (a lot of failures happen for every success) and \(n\) is too big ( a lot of successes are allowed). Therefore, the \(n\) and \(p\) values must satisfy the constraint:\[n\frac{1-p}{p}+10n\sqrt{n}\frac{1-p}{p}<2^{63}-1-10\sqrt{2^{63}-1},\]Where the left side of the equation is the derived mean + 10 sigma of a sample from the gamma distribution internally used as the \(lam\) parameter of a poisson sample, and the right side of the equation is the constraint for maximum value of \(lam\) in
Generator.poisson
.References
[1]Weisstein, Eric W. “Negative Binomial Distribution.” From MathWorld–A Wolfram Web Resource. https://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.?
>>> rng = np.random.default_rng() >>> s = 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)