scipy.special.bdtrin#

scipy.special.bdtrin(k, y, p, out=None) = <wrapped_ufunc 'bdtrin'>#

Inverse function to bdtr with respect to n.

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

Parameters:

karray_like: Number of successes (float).
yarray_like: Cumulative probability (probability of k or fewer successes in n events).
parray_like: Success probability (float).
outndarray, optional: Optional output array for the function values

Returns:

nscalar or ndarray: The number of events n such that bdtr(k, n, p) = y.

See also

bdtr

Notes

This function uses the find_minimum_number_of_trials method of the binomial_distribution class of the Boost.Math C++ library [1].

Array API Standard Support

bdtrin has support for Python Array API Standard compatible backends in addition to NumPy. The following combinations of backend and device (or other capability) are supported.

Library	CPU	GPU
NumPy	✅	n/a
CuPy	n/a	⛔
PyTorch	✅	⛔
JAX	✅	⛔
Dask	✅	n/a

For the NumPy backend, this function supports all NumPy ufunc keyword arguments. Other backends may support out, but typically none of the other ufunc kwargs. out is typically supported for CuPy and PyTorch, but not currently in cases where SciPy relies on a generic Array API implementation or, for PyTorch on CPU, falls back to the NumPy backend. out is never supported for JAX because JAX arrays are immutable. bdtrin does not currently support out for the PyTorch backend.

See Support for the array API standard for more information.

References

[1]

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

Examples

What is the largest number of times we can flip a fair coin and still have at least a 90% chance of getting 10 or fewer heads? bdtrin answers this question:

>>> import scipy.special as sc
>>> k = 10  # maximum number of heads
>>> p = 0.5  # probability of flipping heads
>>> y = 0.9  # cumulative probability
>>> result = sc.bdtrin(k, y, p)
>>> result
np.float64(15.90442928275109)

To verify, compute the cumulative probability of getting 10 or fewer successes in 15 trials with probability 0.5 using the binomial distribution from scipy.stats. Since the cumulative probability decreases as the number of trials increases, we round down to the nearest integer:

>>> from scipy.stats import Binomial
>>> Binomial(n=15, p=p).cdf(k)
np.float64(0.940765380859375)