# Problem with Pareto distribution and R

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.

library(PtProcess)
dd<-rpareto(10000,1.5,0.01)
cdf<-ecdf(dd)
df<-density(dd)
ff<-(1-cdf(df$x))/df$y

If i plot ff

plot(df$x,ff) I do not obtain the correct straight line. I guess that this is due at the way density() and ecdf() works. I need this form of the test (an a posteriori evaluation of fd and cdf) in order to perform the same test on a sample of data of unknown orgin. I guess that i need a way to binning the ecdf() function in the same way as hist() is the binning version of density. So my question is: • Does there exist an equivalent binned function of ecdf() as hist() is the binned function of density()? • or can I simulate ecdf() with hist()? • It looks like ff<-(1-cdf(df$x))/df$x is calculating$P(X>x)/x$not$P(X>x)/f(x)$Jun 22, 2012 at 16:51 • yes! you right! thanx :) Jun 22, 2012 at 17:49 • but ff<-(1-cdf(df$x))/df$y seems does not works too. Jun 22, 2012 at 17:56 • @emauele, there are probably many points in your estimated density that are close to 0 which may cause numerically unstable results (I noticed this when pasting your code). Beyond that, I don't have much insight into the problem. Jun 22, 2012 at 17:57 ## 1 Answer By using ecdf and density, you're not actually doing the Pareto calculations, but instead using estimates based on a sample that are, by their non-parametric nature, not guaranteed (read: not going to) have the desired property. Try the following: x <- seq(0.1,10,by=0.1) fx <- dpareto(x, 1.5, 0.05) Fx <- ppareto(x, 1.5, 0.05) plot((1-Fx)/fx ~ x) You'll get the nice straight line out: • Good, but i need that form in order to perform the same test on a sample of data of unknown origin. Jun 22, 2012 at 23:00 • Since the property characterizes the Pareto distribution, i.e., no other distribution has that property, you could just use a goodness of fit test on the data. That's fully equivalent to testing for that property, since$(1-F(x))/f(x) = x/a \leftrightarrow x \sim \text{Pareto}$. Not sure how you'd test for the property directly, w/o going through the Pareto, though. Jun 22, 2012 at 23:13 • Actually the$\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$-stable distribution. Jun 22, 2012 at 23:27 • The$\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.) Jun 22, 2012 at 23:42
• Yes i know. Where is the problem? You have to see the tails of the distribution. If it is possible, of course. Jun 23, 2012 at 7:23