def test_mean_var(self):
poisson_mean, poisson_var = poisson.mean(12), poisson.var(12)
zipoisson_mean = sm.distributions.zipoisson.mean(12, 0)
zipoisson_var = sm.distributions.zipoisson.var(12, 0)
assert_allclose(poisson_mean, zipoisson_mean, rtol=1e-10)
assert_allclose(poisson_var, zipoisson_var, rtol=1e-10)
m = np.array([1, 5, 10])
poisson_mean, poisson_var = poisson.mean(m), poisson.var(m)
zipoisson_mean = sm.distributions.zipoisson._mean(m, 0)
zipoisson_var = sm.distributions.zipoisson._var(m, 0.0)
assert_allclose(poisson_mean, zipoisson_mean, rtol=1e-10)
assert_allclose(poisson_var, zipoisson_var, rtol=1e-10)
def test_mean_var(self):
# compare with Poisson special case
m = np.array([1, 5, 10])
poisson_mean, poisson_var = poisson.mean(m), poisson.var(m)
zigenpoisson_mean = sm.distributions.zigenpoisson._mean(m, 0, 1, 0)
zigenpoisson_var = sm.distributions.zigenpoisson._var(m, 0.0, 1, 0)
assert_allclose(poisson_mean, zigenpoisson_mean, rtol=1e-10)
assert_allclose(poisson_var, zigenpoisson_var, rtol=1e-10)
def test_substitutions_changes(self):
"""Test that substitions occur at the expected rate."""
cfg = AppSettings()
reps = 1000
deltas = []
for _ in range(0, reps):
seq = "a" * 100
dna = Chromosome(seq)
dna.substitutions()
deltas.append(sum(1 for a, b in zip(seq, dna.sequence) if a != b))
# Expand the conf_99 to compensate for repeated mutations in the same place
expected_delta = cfg.genetics.mutation_rate * 100 * \
(1 - 1/len(Chromosome.nucleotides()))
# Because there is a little slop around synonymous substitions I multiply
# the confidence by 10 just to limit the number of failing tests.
conf_99 = ((poisson.var(cfg.genetics.mutation_rate * 100) / 1000)
**(1 / 2)) * 10
observed_delta = sum(deltas) / reps
assert (expected_delta - conf_99) < observed_delta < (expected_delta +
conf_99)
def var(self, dist):
return poisson.var(*self._get_params(dist))
def var(self, n, p):
var = poisson.var(self, n, p)
return var
from scipy.stats import poisson
import numpy as np
# Assume the number of typo errors on a single page of a book follows
# Poisson distribution with parameter 1/3. Calculate the probability
# that on one page there are
# no typos (pmf(0,1/3) = 0.7165313105737893)
# exactly two typos = (pmf(0,1/3) = 0.03980729503187717)
k = 1
mu = 2.5
expect = poisson.expect(args=(mu,), loc=0)
mean = poisson.mean(mu)
var = poisson.var(mu)
sigma = poisson.std(mu)
pmf = poisson.pmf(k, mu, loc=0)
cdf = poisson.cdf(k, mu, loc=0)
e_x_squared = mu**2 + mu
print('expected = ', expect)
print('mean = ', mean)
print('variance = ', var)
print('std. dev. = ',sigma)
print('pmf = ', pmf)
print('cdf = ', cdf)
print('E[X]^2 = ', e_x_squared)
def test_mean_var(self):
poisson_mean, poisson_var = poisson.mean(12), poisson.var(12)
zipoisson_mean = sm.distributions.zipoisson.mean(12, 0)
zipoisson_var = sm.distributions.zipoisson.mean(12, 0)
assert_allclose(poisson_mean, zipoisson_mean, rtol=1e-10)
assert_allclose(poisson_var, zipoisson_var, rtol=1e-10)