scipy.stats.skellam#
- scipy.stats.skellam = <scipy.stats._discrete_distns.skellam_gen object>[source]#
A Skellam discrete random variable.
As an instance of the
rv_discrete
class,skellam
object inherits from it a collection of generic methods (see below for the full list), and completes them with details specific for this particular distribution.Methods
rvs(mu1, mu2, loc=0, size=1, random_state=None)
Random variates.
pmf(k, mu1, mu2, loc=0)
Probability mass function.
logpmf(k, mu1, mu2, loc=0)
Log of the probability mass function.
cdf(k, mu1, mu2, loc=0)
Cumulative distribution function.
logcdf(k, mu1, mu2, loc=0)
Log of the cumulative distribution function.
sf(k, mu1, mu2, loc=0)
Survival function (also defined as
1 - cdf
, but sf is sometimes more accurate).logsf(k, mu1, mu2, loc=0)
Log of the survival function.
ppf(q, mu1, mu2, loc=0)
Percent point function (inverse of
cdf
— percentiles).isf(q, mu1, mu2, loc=0)
Inverse survival function (inverse of
sf
).stats(mu1, mu2, loc=0, moments=’mv’)
Mean(‘m’), variance(‘v’), skew(‘s’), and/or kurtosis(‘k’).
entropy(mu1, mu2, loc=0)
(Differential) entropy of the RV.
expect(func, args=(mu1, mu2), loc=0, lb=None, ub=None, conditional=False)
Expected value of a function (of one argument) with respect to the distribution.
median(mu1, mu2, loc=0)
Median of the distribution.
mean(mu1, mu2, loc=0)
Mean of the distribution.
var(mu1, mu2, loc=0)
Variance of the distribution.
std(mu1, mu2, loc=0)
Standard deviation of the distribution.
interval(confidence, mu1, mu2, loc=0)
Confidence interval with equal areas around the median.
Notes
Probability distribution of the difference of two correlated or uncorrelated Poisson random variables.
Let \(k_1\) and \(k_2\) be two Poisson-distributed r.v. with expected values \(\lambda_1\) and \(\lambda_2\). Then, \(k_1 - k_2\) follows a Skellam distribution with parameters \(\mu_1 = \lambda_1 - \rho \sqrt{\lambda_1 \lambda_2}\) and \(\mu_2 = \lambda_2 - \rho \sqrt{\lambda_1 \lambda_2}\), where \(\rho\) is the correlation coefficient between \(k_1\) and \(k_2\). If the two Poisson-distributed r.v. are independent then \(\rho = 0\).
Parameters \(\mu_1\) and \(\mu_2\) must be strictly positive.
For details see: https://en.wikipedia.org/wiki/Skellam_distribution
skellam
takes \(\mu_1\) and \(\mu_2\) as shape parameters.The probability mass function above is defined in the “standardized” form. To shift distribution use the
loc
parameter. Specifically,skellam.pmf(k, mu1, mu2, loc)
is identically equivalent toskellam.pmf(k - loc, mu1, mu2)
.Examples
>>> import numpy as np >>> from scipy.stats import skellam >>> import matplotlib.pyplot as plt >>> fig, ax = plt.subplots(1, 1)
Calculate the first four moments:
>>> mu1, mu2 = 15, 8 >>> mean, var, skew, kurt = skellam.stats(mu1, mu2, moments='mvsk')
Display the probability mass function (
pmf
):>>> x = np.arange(skellam.ppf(0.01, mu1, mu2), ... skellam.ppf(0.99, mu1, mu2)) >>> ax.plot(x, skellam.pmf(x, mu1, mu2), 'bo', ms=8, label='skellam pmf') >>> ax.vlines(x, 0, skellam.pmf(x, mu1, mu2), colors='b', lw=5, alpha=0.5)
Alternatively, the distribution object can be called (as a function) to fix the shape and location. This returns a “frozen” RV object holding the given parameters fixed.
Freeze the distribution and display the frozen
pmf
:>>> rv = skellam(mu1, mu2) >>> ax.vlines(x, 0, rv.pmf(x), colors='k', linestyles='-', lw=1, ... label='frozen pmf') >>> ax.legend(loc='best', frameon=False) >>> plt.show()
Check accuracy of
cdf
andppf
:>>> prob = skellam.cdf(x, mu1, mu2) >>> np.allclose(x, skellam.ppf(prob, mu1, mu2)) True
Generate random numbers:
>>> r = skellam.rvs(mu1, mu2, size=1000)