The docstring for np.pareto says:
This is a simplified version of the Generalized Pareto distribution
(available in SciPy), with the scale set to one and the location set to
zero. Most authors default the location to one.
and also:
The probability density for the Pareto distribution is
.. math:: p(x) = \frac{am^a}{x^{a+1}}
where :math:`a` is the shape and :math:`m` the location
These two statements seem to be in contradiction. I think what was
meant is that m is the scale, rather than the location. For if m were
equal to zero, as the first portion of the docstring states, then the
entire pdf would be zero for all shapes a>0.
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Also, I'm not quite understanding how the stated pdf is actually the
same as the pdf for the generalized pareto with the scale=1 and
location=0. By the wikipedia definition of the generalized Pareto
distribution, if we take \sigma=1 (scale equal to one) and \mu=0
(location equal to zero), then we get:
(1 + a x)^(-1/a - 1)
which is normalized over $x \in (0, \infty)$. If we compare this to
the distribution stated in the docstring (with m=1)
a x^{-a-1}
we see that it is normalized over $x \in (1, \infty)$. And indeed,
the distribution requires x > scale = 1.
If we integrate the generalized Pareto (with scale=1, location=0) over
$x \in (1, \infty)$ then we have to re-normalize. So should the
docstring say:
This is a simplified version of the Generalized Pareto distribution
(available in Scipy), with the scale set to one, the location set to zero,
and the distribution re-normalized over the range (1, \infty). Most
authors default the location to one.