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.

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 to skellam.pmf(k - loc, mu1, mu2).

Examples

>>> 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()
../../_images/scipy-stats-skellam-1_00_00.png

Check accuracy of cdf and ppf:

>>> 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)

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.