On Sun, May 9, 2010 at 1:01 AM, T J <
tjhnson@gmail.com> wrote:
> 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.
>
> ----
>
> 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.