1
$\begingroup$

I have a set of insurance data. Development factors Development factors fall with period, so follow Weibull distribution. I want to estimate Weibull parameters and smooth Development Factors. If I estimate parameters of Weibull using fitdistr() code in R I get shape parameter = 0.9567142 and scale parameter = 1.8321954. With this parameters I want to smooth the Development Factors given x-values equal to Period, but this gives me very low numbers. So for Period = 1 and these parameters, it returns 0.3061, which is way lower than original 20.27199. What am I doing wrong? Thank you in advance

$\endgroup$
1
  • 3
    $\begingroup$ 1. While development factors "fall" that doesn't make them Weibull; in fact they cannot be a multiple of a Weibull density because they converge to 1 (from above) while a Weibull density converges to 0. 2. fitdistr works on random samples from a distribution, not from density or cdf values. 3 You also cannot reasonably use nonlinear least squares to fit some curve to the development factors since development factor values are highly dependent and strongly heteroskedastic. $\endgroup$
    – Glen_b
    Jan 14, 2019 at 13:07

1 Answer 1

0
$\begingroup$

If it might be of some use, I got an OK fit to factor as a function of period using an exponential equation with offset, "factor = a * exp(b * period) + Offset", with parameters a = 1.2148647556770337E+02, b = -1.8447383394431971E+00, and Offset = 1.0678593236739169E+00 yielding R-squared = 0.9997 and RMSE = 0.0876.

plot

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.