I am trying to test this property of pareto distribution: Let f(x) be a pareto distribution

$$ f(x)=\alpha \frac{x_m^\alpha}{x^{\alpha+1}} $$

so we have the cdf that is

$$ CDF(x)=\int_{x_m}^{x}\alpha \frac{t_m^\alpha}{t^{\alpha+1}}dt=1-\frac{x_m^\alpha}{x^\alpha} $$

then the probability that $x>x_0$ is

$$ P(x>x_0)=1-CDF(x)=\frac{x_m^\alpha}{x^\alpha} $$

and so we have

$$ \frac{P(x>x_0)}{f(x)}=\frac{x}{\alpha} $$

Now i am trying to test it with R.


If i plot ff


I do not obtain the correct straight line. I guess that this is due at the way density() and ecdf() works. I need this form of the test (an a posteriori evaluation of fd and cdf) in order to perform the same test on a sample of data of unknown orgin. I guess that i need a way to binning the ecdf() function in the same way as hist() is the binning version of density.

So my question is:

  • Does there exist an equivalent binned function of ecdf() as hist() is the binned function of density()?
  • or can I simulate ecdf() with hist()?
  • 1
    $\begingroup$ It looks like ff<-(1-cdf(df$x))/df$x is calculating $P(X>x)/x$ not $P(X>x)/f(x)$ $\endgroup$ – Macro Jun 22 '12 at 16:51
  • $\begingroup$ yes! you right! thanx :) $\endgroup$ – emanuele Jun 22 '12 at 17:49
  • $\begingroup$ but ff<-(1-cdf(df$x))/df$y seems does not works too. $\endgroup$ – emanuele Jun 22 '12 at 17:56
  • $\begingroup$ @emauele, there are probably many points in your estimated density that are close to 0 which may cause numerically unstable results (I noticed this when pasting your code). Beyond that, I don't have much insight into the problem. $\endgroup$ – Macro Jun 22 '12 at 17:57

By using ecdf and density, you're not actually doing the Pareto calculations, but instead using estimates based on a sample that are, by their non-parametric nature, not guaranteed (read: not going to) have the desired property.

Try the following:

x <- seq(0.1,10,by=0.1)
fx <- dpareto(x, 1.5, 0.05)
Fx <- ppareto(x, 1.5, 0.05)
plot((1-Fx)/fx ~ x)

You'll get the nice straight line out: enter image description here

  • $\begingroup$ Good, but i need that form in order to perform the same test on a sample of data of unknown origin. $\endgroup$ – emanuele Jun 22 '12 at 23:00
  • $\begingroup$ Since the property characterizes the Pareto distribution, i.e., no other distribution has that property, you could just use a goodness of fit test on the data. That's fully equivalent to testing for that property, since $(1-F(x))/f(x) = x/a \leftrightarrow x \sim \text{Pareto}$. Not sure how you'd test for the property directly, w/o going through the Pareto, though. $\endgroup$ – jbowman Jun 22 '12 at 23:13
  • $\begingroup$ Actually the $\alpha$-stable distributions share the same tail behaviour. I would like to use this way because i think that this is the better way for a straightforward measure of the $\alpha$ parameter in a generic $\alpha$-stable distribution. $\endgroup$ – emanuele Jun 22 '12 at 23:27
  • $\begingroup$ The $\alpha$-stable distributions only share that behavior asymptotically as $x \to \infty$, I'm afraid, see eldorado.tu-dortmund.de/bitstream/2003/5219/1/47_02.pdf for example, also Johnson, Balakrishnan & Kotz, Continuous Univariate Dist'ns Vol. 1, pp. 603-604 (sorry, Amazon's "look inside" doesn't let you look inside those pages.) $\endgroup$ – jbowman Jun 22 '12 at 23:42
  • $\begingroup$ Yes i know. Where is the problem? You have to see the tails of the distribution. If it is possible, of course. $\endgroup$ – emanuele Jun 23 '12 at 7:23

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