Typically, the estimates of parameters are those values (of the parameters) for which the observed data is most likely. This is called maximum likelihood estimation (MLE).

You can find a lot in the internet if you search for "maximum likelihood estimation".

In short, the likelihood is a function of the parameter (p), given the data (x). This is the same like the (joint) probability of the data, given the parameter, but with the emphasis on the fact that the parameter value is varied (while referring always to the same data).

The probability of the data, given the parameter, is P(x|p). These probabilities are from the probability model (binomial, exponential, normal, whatever). The letters are bold to indicate that both can be vectors. The data typically is a vector, i.e. it consists of several observations x1, x2, ... xn (a typical sample of n values). When these values are independent then the resulting probability is simply the product: P(x1 and x2 and x3...|p) = P(x1|p)*P(x2|p)*.... what is the joint probability of getting all data (x1, x2,...) together (at the same time, so to say).

The likelihood is now simply

L(p) = P(x1|p)*P(x2|p)*.... 

given the known data.

The MLE is then the value of p for which L(p) is maximal.

For some distributions there are analytical solutions available. For instance, the MLE for the parameter p of the binomial distribution is k/n (with k being the number of "successes" and n the number of trials). For other distributions the parameter value(s) for which the maximum value of the likelihood is obtained has to be searched. A description of an algorithm is given here, for instance:

http://www.statistik.lmu.de/~chris/bowlpaper/bowlpaper/node14.html

http://www.utstat.toronto.edu/~brunner/oldclass/312f12/lectures/312f12NumericalMLE.pdf

http://math.berkeley.edu/~jrodrig/sisthe/thesis.pdf

Not sure if this anwers your question, but never mind :)

If your are using R, fitdistr" function in MASS package helps you fit distribution to your data.

Here's a simple worked out example for estimating the parameters of your data based on exponential distribution.

## Worked out example for estimating parameters based on Exponential Distribution

require(MASS) 

## Suppose your data looks like this. 

set.seed(10) # For reproducibility

y

Dear Fausto, 

thanks for your response.. this, I attach the file that I am stil analyze such that I share a question.

Hi Fausto, 

yes you are right. Actually, the primary data consist of date of occurance (Cj), date of payment (Tj), and the amount of klaim for each occurance. However, data from occurance to payment Lj=Tj-Cj. This is I attach the file for you more completely.

Hi Fausto

Firstly, thanks for your response,

• You are right I am looking for computing the "occurrance rate" of time between date of occurrance (Cj). That is time of the j-1th loss to jth loss (Cj-Cj-1).

Moreover, I am looking for "the rate of occurence to payment".That is (Tj-Cj ) or ( Lj).

• Yes, I just need their parameters, those are  time of the j-1th loss to jth loss (Cj-Cj-1), the rate of occurence to payment ( Lj), and certainly, distribution of claim (Xj). Furtheremore, I generate data based on each their distribution (Exponential, Weibull, Pareto) with parameters themselves. And then, by the particular formula, I compute the value premium.
• During the time flow. The insurer needs to check the policy and the righteousness of the occurence. Such that, the first occurence not always get the money first.

Yes, I think so that the model is much more important than the data, although the parameters of data are obtained from data.

If we drop away the 12th observation of Cj-Cj-1, since it involves a zero value and the R function does not work, then I found:

Cj-Cj-1 Weibull distributed:

shape               scale

0.790509772    0.021135088

(0.096655497) (0.004923457)

Without dropping the 12th observation for Lj:

Lj Weibull distributed:

shape           scale

2.8366599    0.2209851

(0.3290431) (0.0142385)

I was lazy and did not do the Pareto, you can trust Fausto.

Hi Demetris..

thanks for your response. However for observation of Cj-Cj-1, I need exponential not weibull distribution. And for Lj, I do it also with R, the result is same. But I don't know how to find parameters in my pareto distribution for Xj data. This Pareto distribution is type II or Lomax distribution. Actually I used library(Renext) in R, since CV (coefficience ov variance) less than 1 so I can't find the parameters. So that, what suggestion about this? thanks.

Did you try package 'actuar'? I think it has pareto distribution, try also 'cm' function.

As for weibull and exponential, Fausto gave the answer: exponential is a sub-case of weibull, look both:

http://en.wikipedia.org/wiki/Exponential_distribution

http://en.wikipedia.org/wiki/Weibull_distribution

There exist not a problem here. Pareto is the difficult case...

@Fausto

thanks for your response

what do you mean with beta and eta for exponential distributed

in Cj-Cj-1? Instead, exponential distribution has just a parameter (theta),doesn't it?

I tried to find parameters for weibull distributed in Lj by R programming. The syntax is

library(MASS)

fitdistr(dataselang,"weibull")

and the output is

shape scale

2.8366599 0.2209851

(0.3290431) (0.0142385)

Warning messages:

1: In densfun(x, parm[1], parm[2], ...) : NaNs produced

2: In densfun(x, parm[1], parm[2], ...) : NaNs produced

I think it is almost as well as yours.

for the last, what program do you use for finding these values?

@ Demetris,

So, how to use actuar package to find parameters in Pareto, since ther is no menu there..

Try the help html page, I had not time to find it, but look at 'cm' function.

@Fausto

@Demetris

Thanks for your responses and your helps,,,...

I will try your suggestions