About Peaw

Peaw · ‎07-13-2018

I am using BASE sas fitting distribution for the data. and the result showed as I posted the picture above. as I know 'Alpha' parameter of pareto distributon is usually not less than 0 but, the result show like that. I would like to know Is it work for this data (regardless of p-value)?

Peaw · ‎07-25-2017

Thank you very much for your kind suggestion above. However, after I did as per your suggestion, I found out many issues as following: First of all, please let me clarify you that the data range of this fitted Pareto distribution is 300,000-800,000 (with theta =300,000). And then I simulated Pareto distribution without identifying theta. After I used the coding to add theta as per your suggestion, I think that the output statistic seems worse. Kindly find the compared statistic as below details: Fitting Pareto distribution -> output1 Simulation without theta -> output2 (after cut off data, so the data range of is 300,000-800,000) Simulation with theta -> output3 (after cut off data, so the data range of is 300,000-800,000) As you can see that, the statistic of simulation with theta are different from the fitted Pareto quite much. I am not sure if these output are acceptable or not? If not, please let have your further suggestion.

Peaw · ‎07-18-2017

I am working on modeling data. By using "Proc univariate" to fit data, distribution that I get is Gamma with theta=60500 and Pareto with theta=300000. then, I simulate data from these distributions by using code as follow: %let N=3500; %let NumSamples = 120; data simu.pareto1(keep=ID X); do ID = 1 to &NumSamples; a = 0.383045; k = 222548.1; call streaminit(1234); do i = 1 to &N; U = rand("Uniform"); X = k / U**(1/a); output; end; end; run; but the output seems deviate from the fitted data quite much. I'm not sure if it's relevant with the unspecify theta on my code or not? If yes, please let me know the code for simulation with specify theta (for both Gamma and Pareto).

Peaw · ‎06-28-2017

I am sorry to bother you. But after following your suggestion I got the attached output. How can I know this model fits the data or not.

Peaw · ‎06-26-2017

could you please tell me that how can I know p of this data from fitting the distribution. because when I fitted this distribution I fitted from raw data of frequency of claim of each auto policy. That's why I do not know P of this data.

Peaw · ‎06-25-2017

I am using base sas 9.4. Now I am working on Simulating data. After I fitted data to negative binomial distribution using the following code ; proc countreg data=claim; model y = / dist=negbin; output out=exp_nb prob=prob_nb; freq freq; run; proc format; value yfmt 7-high = ">=7"; run; proc means sum nway data=exp_nb; class y; var prob_nb; format y yfmt.; output out=exp_nb sum=_testp_; run; data exp_nb; set exp_nb; sumexp + _testp_; sumtolast = lag(sumexp); if y=7 then _testp_ = 1 - sumtolast; run; proc freq data=claims; table y / chisq(testp=exp_nb df=-2 lrchisq); format y yfmt.; weight freq; run; Then I got only p-value from fitting this data. Now I want to simulate data from negative binomial distribution. How can I do that ? please help (version of sas program do not have Proc iml procedure)

Peaw · ‎05-03-2017

I use this following code to fit poisson and negative binomial distribution but, I am not sure Is it correct?? please let me know. data ec.no; input y freq; datalines; 0 3475 1 229 2 28 3 6 5 1 ; proc countreg data=ec.no; model y = / dist=poisson; output out=exp_poi prob=prob_poi; freq freq; run; proc print; id y; var prob_poi; run; proc format; value yfmt 3-high = "3+"; run; 　 proc means sum nway data=exp_poi; class y; var prob_poi; format y yfmt.; output out=exp_poi sum=_testp_; run; proc freq data=ec.no; table y / chisq(testp=exp_poi df=-1 lrchisq); format y yfmt.; weight freq; run; 　 /*fit nagative binomial*/ proc countreg data=ec.no; model y = / dist=negbin; output out=exp_nb prob=prob_nb; freq freq; run; proc format; value yfmt 3-high = ">=3"; run; proc means sum nway data=exp_nb; class y; var prob_nb; format y yfmt.; output out=exp_nb sum=_testp_; run; data exp_nb; set exp_nb; sumexp + _testp_; sumtolast = lag(sumexp); if y=3 then _testp_ = 1 - sumtolast; run; proc freq data=ec.no; table y / chisq(testp=exp_nb df=-2 lrchisq); format y yfmt.; weight freq; run;

Peaw · ‎05-01-2017

Reference is made to my questions on the SAS communities regarding 2 or 3 parameters distribution. As I’m still not sure about the output, could you please let me have more clarification as following? The data that I used for fitting distribution is loss data (claim) with range [50 to 524] (after transforming data by square root). Because, I don’t know how to set the threshold, I ran SAS as “threshold = est”. So I got the threshold value which is 46.9. After that, I specified the mentioned threshold value of 46.9 in the histogram statement again. I think I got the 2 parameter log normal distribution (with p-value 0.017) as per suggestion. However, could you please let me have your confirmation if this model is valid? Additionally, I am not sure for the next step of “simulation”. Please advise us how to simulate the 2 parameter distribution with specify threshold like this case?

Peaw · ‎04-20-2017

Thank you so much for your kindness. it helps a lot.

Peaw · ‎04-18-2017

thank you so much ,it helps a lot. but may I ask you for more information. After i run the statement as per your suggestion, please find my distribution output as attached. title 'Lognormal dist.; ods select Histogram Lognormal.ParameterEstimates Lognormal.GoodnessOfFit FitQuantiles; proc univariate data=ec.uy2013; var root_avg_jt; where root_avg_jt ge 55 and root_avg_jt le 496; histogram / lognormal(w=3 threshold=46.7) odstitle = title; inset n mean (5.3) std='Std Dev' (5.3) skewness (5.3) / pos = ne header = 'Summary Statistics'; run; I am just curious: 1. I want to make sure that the output is 2-parameters lognormal (not 3 parameters). 2. how do we know the lower bound/theta? should I start from considering the data range (minimum value)?

Peaw · ‎04-18-2017

http://support.sas.com/documentation/cdl/en/procstat/70116/HTML/default/viewer.htm#procstat_univariate_examples27.htm why this example specifies theta=est but the result of this is two-parameter lognormal distribution not three.

Peaw · ‎04-17-2017

I am working on fitting distributon to the data and now I am so confuse about the code. I have found the example of creating a histogram to display lognormal fit and use code as follow; title 'Lognormal dist. '; ods select Histogram Lognormal.ParameterEstimates Lognormal.GoodnessOfFit FitQuantiles; proc univariate data=uy2013; var avg_claim; histogram / lognormal(w=3 theta=est) odstitle = title; inset n mean (5.3) std='Std Dev' (5.3) skewness (5.3) / pos = ne header = 'Summary Statistics'; run; I would like to know 'Is this code for fitting two-parameter lognormal distribution?' if it is, what theta=est is used for ??? the data that I used start from -0.100 but the threshold in the result is -772.2 why the threshold is that? (I am using Base SAS 9.4)

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Re: simulation from negative binomail distribution

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Re: proc univariate

Re: proc univariate

Re: proc univariate

Re: proc univariate

proc univariate