Take the 2-minute tour ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

I have a dataset containing information about a bunch of wireless devices. Specifically, the number of other wireless devices each device encounters. Based on other researcher's prior work in the same area, I have reason to believe my distribution may potentially be well fitted by the not so well known biPareto distribution, whose CCDF is defined as:

$1 - F(x) = (x/k)^{-\alpha}(\frac{x+kb}{k + kb})^{\alpha-\beta}$, for $ x > k$

and simply

$1 - F(x) = 1$, for $x \le k$.

Assume I already know $k = 1$. This leaves a function to fit with the three parameters $\alpha$,$b$,$\beta$, with the restriction that each parameter is $> 0$.

What I am unsure about is how one would go about determining the most suitable fitting parameter values, ideally in R.

I am quite the novice when it comes to fitting data, so aside from the syntactical considerations in R, I am also interested in any advice regarding what specific methods of fitting would be most appropriate for this particular function. For example:

  1. Does this function qualify for linear regression?
  2. Is least squares appropriate/applicable here? What about minimum mean squared error (MMSE)?
  3. How sensitive is any given fitting method to any "guiding" initial parameter values I give it (iterative methods?), or is the result determined analytically (always produces the same answer in the end?)

If it's at all relevant, I'm ultimately looking to use the Kolmogorov-Smirnov test on the data to see how well it matches the parameterized biPareto distribution.

share|improve this question
    
I don't quite see how linear regression could help you estimate $\alpha, b, \beta$. I suppose least squares could help you to set of a moment-based estimating equation if you know the moments of this distribution as a function of $\alpha, b, \beta$, but I'd tend to go with maximum likelihood as the default choice. You have the CDF there, so you can get the log-likelihood easily and optimize it. Is there some reason you don't want to do that? –  Macro Jan 17 '13 at 14:26
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.