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99 lines
2.7 KiB
Python
99 lines
2.7 KiB
Python
from scipy.stats import betabinom, hypergeom, nhypergeom, bernoulli, boltzmann
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import numpy as np
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from numpy.testing import assert_almost_equal, assert_equal, assert_allclose
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def test_hypergeom_logpmf():
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# symmetries test
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# f(k,N,K,n) = f(n-k,N,N-K,n) = f(K-k,N,K,N-n) = f(k,N,n,K)
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k = 5
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N = 50
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K = 10
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n = 5
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logpmf1 = hypergeom.logpmf(k, N, K, n)
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logpmf2 = hypergeom.logpmf(n - k, N, N - K, n)
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logpmf3 = hypergeom.logpmf(K - k, N, K, N - n)
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logpmf4 = hypergeom.logpmf(k, N, n, K)
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assert_almost_equal(logpmf1, logpmf2, decimal=12)
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assert_almost_equal(logpmf1, logpmf3, decimal=12)
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assert_almost_equal(logpmf1, logpmf4, decimal=12)
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# test related distribution
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# Bernoulli distribution if n = 1
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k = 1
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N = 10
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K = 7
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n = 1
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hypergeom_logpmf = hypergeom.logpmf(k, N, K, n)
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bernoulli_logpmf = bernoulli.logpmf(k, K/N)
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assert_almost_equal(hypergeom_logpmf, bernoulli_logpmf, decimal=12)
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def test_nhypergeom_pmf():
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# test with hypergeom
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M, n, r = 45, 13, 8
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k = 6
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NHG = nhypergeom.pmf(k, M, n, r)
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HG = hypergeom.pmf(k, M, n, k+r-1) * (M - n - (r-1)) / (M - (k+r-1))
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assert_allclose(HG, NHG, rtol=1e-10)
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def test_nhypergeom_pmfcdf():
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# test pmf and cdf with arbitrary values.
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M = 8
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n = 3
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r = 4
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support = np.arange(n+1)
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pmf = nhypergeom.pmf(support, M, n, r)
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cdf = nhypergeom.cdf(support, M, n, r)
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assert_allclose(pmf, [1/14, 3/14, 5/14, 5/14], rtol=1e-13)
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assert_allclose(cdf, [1/14, 4/14, 9/14, 1.0], rtol=1e-13)
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def test_nhypergeom_r0():
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# test with `r = 0`.
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M = 10
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n = 3
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r = 0
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pmf = nhypergeom.pmf([[0, 1, 2, 0], [1, 2, 0, 3]], M, n, r)
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assert_allclose(pmf, [[1, 0, 0, 1], [0, 0, 1, 0]], rtol=1e-13)
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def test_boltzmann_upper_bound():
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k = np.arange(-3, 5)
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N = 1
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p = boltzmann.pmf(k, 0.123, N)
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expected = k == 0
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assert_equal(p, expected)
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lam = np.log(2)
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N = 3
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p = boltzmann.pmf(k, lam, N)
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expected = [0, 0, 0, 4/7, 2/7, 1/7, 0, 0]
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assert_allclose(p, expected, rtol=1e-13)
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c = boltzmann.cdf(k, lam, N)
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expected = [0, 0, 0, 4/7, 6/7, 1, 1, 1]
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assert_allclose(c, expected, rtol=1e-13)
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def test_betabinom_a_and_b_unity():
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# test limiting case that betabinom(n, 1, 1) is a discrete uniform
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# distribution from 0 to n
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n = 20
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k = np.arange(n + 1)
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p = betabinom(n, 1, 1).pmf(k)
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expected = np.repeat(1 / (n + 1), n + 1)
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assert_almost_equal(p, expected)
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def test_betabinom_bernoulli():
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# test limiting case that betabinom(1, a, b) = bernoulli(a / (a + b))
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a = 2.3
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b = 0.63
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k = np.arange(2)
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p = betabinom(1, a, b).pmf(k)
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expected = bernoulli(a / (a + b)).pmf(k)
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assert_almost_equal(p, expected)
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