scipy.stats.kstat#

scipy.stats.kstat(data, n=2)[source]#

Return the nth k-statistic (1<=n<=4 so far).

The nth k-statistic k_n is the unique symmetric unbiased estimator of the nth cumulant kappa_n.

Parameters

dataarray_like: Input array. Note that n-D input gets flattened.
nint, {1, 2, 3, 4}, optional: Default is equal to 2.

Returns

kstatfloat: The nth k-statistic.

See also

kstatvar: Returns an unbiased estimator of the variance of the k-statistic.
moment: Returns the n-th central moment about the mean for a sample.

Notes

For a sample size n, the first few k-statistics are given by:

\[k_{1} = \mu k_{2} = \frac{n}{n-1} m_{2} k_{3} = \frac{ n^{2} } {(n-1) (n-2)} m_{3} k_{4} = \frac{ n^{2} [(n + 1)m_{4} - 3(n - 1) m^2_{2}]} {(n-1) (n-2) (n-3)}\]

where \(\mu\) is the sample mean, \(m_2\) is the sample variance, and \(m_i\) is the i-th sample central moment.

References

http://mathworld.wolfram.com/k-Statistic.html

http://mathworld.wolfram.com/Cumulant.html

Examples

>>> from scipy import stats
>>> from numpy.random import default_rng
>>> rng = default_rng()

As sample size increases, n-th moment and n-th k-statistic converge to the same number (although they aren’t identical). In the case of the normal distribution, they converge to zero.

>>> for n in [2, 3, 4, 5, 6, 7]:
...     x = rng.normal(size=10**n)
...     m, k = stats.moment(x, 3), stats.kstat(x, 3)
...     print("%.3g %.3g %.3g" % (m, k, m-k))
-0.631 -0.651 0.0194  # random
0.0282 0.0283 -8.49e-05
-0.0454 -0.0454 1.36e-05
7.53e-05 7.53e-05 -2.26e-09
0.00166 0.00166 -4.99e-09
-2.88e-06 -2.88e-06 8.63e-13

scipy.stats.skew

scipy.stats.kstatvar