scipy.special.bdtrik#

scipy.special.bdtrik(y, n, p, out=None) = <ufunc 'bdtrik'>#

Binomial distribution quantile.

Finds the number of successes k such that the sum of the terms 0 through k of the Binomial probability density for n events with probability p is equal to the given cumulative probability y.

Parameters:
yarray_like

Cumulative probability (probability of k or fewer successes in n events).

narray_like

Number of events (float).

parray_like

Success probability (float).

outndarray, optional

Optional output array for the function values

Returns:
kscalar or ndarray

The number of successes k such that bdtr(k, n, p) = y.

See also

bdtr

Binomial distribution cumulative distribution function

Notes

Formula 26.5.24 of [1] is used to reduce the binomial distribution to the cumulative incomplete beta distribution.

This function uses routines from the Boost.Math C++ library [3] which rely on numerical inversion of the binomial distribution CDF [4].

References

[1]

Milton Abramowitz and Irene A. Stegun, eds. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. New York: Dover, 1972.

[2]

NIST Digital Library of Mathematical Functions https://dlmf.nist.gov/8.17.5#E5

[3]

The Boost Developers. “Boost C++ Libraries”. https://www.boost.org/.

Examples

We have a coin for which the probability of showing heads when flipped is 0.525. The coin is flipped 8 times. Find the largest value of k such that the probability that X <= k is not greater than 0.2, where X is the number of heads.

In fact, there is no integer value of k that will give a probability of exactly 0.2, as this plot of the cumulative distribution function shows.

>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> from scipy.special import bdtr, bdtrik

>>> n = 8
>>> p = 0.525
>>> k = np.arange(0, n + 1)
>>> plt.plot(k, bdtr(k, n, p), 'o')
>>> plt.grid(True, alpha=0.5)
>>> plt.xlabel('k')
>>> plt.axhline(0.2, linestyle='--', color='k', alpha=0.5)
>>> plt.title(f"bdtr(k, {n}, {p})")
>>> plt.show()
../../_images/scipy-special-bdtrik-1_00_00.png

From the graph we can see that we would choose k = 2. The function bdtrik lets us find this value directly.

bdtrik returns a floating point value that is like a continuous extension of k. This computes k as a noninteger floating point value:

>>> bdtrik(0.2, n, p)
np.float64(2.508332751475262)

For our final answer we need an integer k, and since we want to ensure that the probability at k does not exceed 0.2, we truncate the fractional part of this value with np.floor:

>>> np.floor(bdtrik(0.2, n, p))
np.float64(2.0)
On this page