GaussianCopula.Random returns samples uniformly distributed about copulae HOT 3 CLOSED

Yup, it's meant to be the case. Assuming H is the function you are studying

Then by Sklar's Theorem, it can also be expressed as a copula, C, with univariate margins, F.

Since the copula does not store the function that generated each univariate margin, you have to apply the inverse transformation to recover the original marginals. In most cases, it's a simple call to the quantile function (or distribution.ppf(x)).

from copulae.

Thanks -- I was a bit confused by the fact this is not implicitly done by GaussianCopula.Random. So if I understand this correctly in copulae you always assume this to be a post-processing step that is carried out by the user?

from copulae.

Yup. A short example to clarify

import numpy as np
from scipy.stats import norm, expon

u = np.random.uniform(size=(100, 2))   # assume this to be generated from the copula

# assume these are you marginal distributions
d1 =norm(0.5, 2)
d2 = expon(1)

# then your marginals
m1 = d1.ppf(u[:, 0])
m2 = d2.ppf(u[:, 1])

from copulae.