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51 lines
1.4 KiB
Python
51 lines
1.4 KiB
Python
5 years ago
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from __future__ import division, print_function, absolute_import
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from scipy.stats import hypergeom, 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_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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