Many thanks! It worked. ... View more

execution time is 3.65 seconds; CPU time is 0.15 seconds ... View more

Thanks very much much.   I need the negative branch of Lambert function. Here is quantile function expressed in terms of the negative branch of the Lambert function.   Quantile function (Q): Q(p|alpha)=-1-(1/alpha)-(1/alpha)W_ 1 [(1+alpha)*(p-1)*exp(-(1+alpha))]; 0<p<1; Where W_ 1 where denotes the negative branch of the Lambert W function P is essentially a unif(0,1)   Thanks again!   J ... View more

Thank you very much, Rick!   I must say that you are not from this earth! You are a genius to have figured that the distribution is a mixture by merely looking at the information I provided. It is indeed a mixture of exponential an gamma(2). In fact, you got the missing probability exactly right.     But this is simplest form of the distribution that we work with. It is the flexibility of this distribution that makes it appealing. We work with several generalization of this distribution. The generalization allows for flexibility in modeling; the generalizations cannot easily be expressed  as a mixture. For instance, there is a generalization in which the CDF is raised to a power, so that when the power=1 then we have the parent distribution. In such cases, one has to result to CDF or Lambert function.  That is the reason we want a SAS program that implements Lambert function and CDF that we can easily modify when we develop generalizations of this distribution.      Again, you are great thinker. And thanks for sharing your knowledge.   J ... View more

Here are the CDF and the Qauntile function expressed in terms of Lambert function.   The goal is to generate the random variables t using1)  inverse CDF and 2) Lambert function method.   Distribution function:  alphat=(alpha*t/(1+alpha)); expalpha=exp(-alpha*t); F(t|alpha)=1-(1+alphat)*expalpha;   Quantile function (Q): Q(P|alpha)=-1-(1/alpha)-(1/alpha)W(-(1+alpha)*(1-p)*exp(-(1+alpha)); 0<p<1; Where W is the Lambert function. P is essentially a unif(0,1)   I will also like to track the computer time using both method. I want SAS to win   Thanks,   J     ... View more

Thank you very Rick. I use SAS for my random number generation. Actually, I am trying to convince some colleagues who is are fan of another software. We are developing a new density that has not been listed. I am recommending we use SAS. They have been using Lambert function implemented in another software for genetation since the density is not supported by software. Since Lambert function is the method they have used for many years for generating data from unlisted, I want to demonstrate that it can be done in SAS. And show that the variate generatedd using Lambert function is no differrent that generated using inverse CDF solve using numerical methods. Thanks J ... View more

Thanks Rick.    Yes. I want to it use for generating random variates from a new function with a specified parameter.   I am generating data from a function that I created, E.g: f(x)=theta*exp(-theta*x). I want to generate random variable from f(x). When theta is given say 0.1.    I like to specifically use Lambert function for the generation   Thanks,   J ... View more

You are a savior!!! Thank you sooo much!!! Unbelievable. Speed with which write these SAS program to execute these complex task is remarkable and facinating. Thanks again for your assistance and patience. J ... View more

data have1;input id $ Sampleid reps x y;datalines;01 1 1 2 202 1 1 2 503 1 1 6 504 1 1 7 305 1 1 4 706 1 1 6 407 1 1 5 308 1 1 4 609 1 1 2 510 1 1 1 3 21 1 2 2 222 1 2 2 523 1 2 6 524 1 2 7 325 1 2 4 726 1 2 6 427 1 2 5 328 1 2 4 629 1 2 2 530 1 2 1 3 61 1 3 2 262 1 3 2 563 1 3 6 564 1 3 7 365 1 3 4 766 1 3 6 467 1 3 5 368 1 3 4 669 1 3 2 570 1 3 1 3 91 2 4 2 292 2 4 2 593 2 4 6 594 2 4 7 395 2 4 4 796 2 4 6 497 2 4 5 398 2 4 4 699 2 4 2 5100 2 4 1 3 31 2 1 2 232 2 1 2 533 2 1 6 534 2 1 7 335 2 1 4 736 2 1 6 437 2 1 5 338 2 1 4 639 2 1 2 540 2 1 1 3 41 2 2 2 242 2 2 2 543 2 2 6 544 2 2 7 345 2 2 4 746 2 2 6 447 2 2 5 348 2 2 4 649 2 2 2 550 2 2 1 3 71 2 3 2 272 2 3 2 573 2 3 6 574 2 3 7 375 2 3 4 776 2 3 6 477 2 3 5 378 2 3 4 679 2 3 2 580 2 3 1 3 41 2 4 2 242 2 4 2 543 2 4 6 544 2 4 7 345 2 4 4 746 2 4 6 447 2 4 5 348 2 4 4 649 2 4 2 550 2 4 1 3;run; data have2;input id $ x y;datalines;11 6 512 7 413 8 714 5 615 5 4;run; /*Result will look like this:*/ sampleid reps distance1          1   1.4041          2   1.4041          3   1.4041          4   1.4042          1   1.4042          2   1.4042          3   1.4042          4   1.404 Each sample will have observation 1 to N; within each sample, there is reps=1 to k ; each reps  will have its distance from HAVE2. each reps within reps with each sample will have their respective distance from HAVE2. its essentially the same as the program you have written so far. The only difference is that program you written so far is for sampleid=1. With real data the value the distance will be expected to be different. It's the same in this example because the samples are same.   Thanks. ... View more

That is we have three variable: sampleID, replicate distance. I am realy very grateful for your patients and assistance. I have been trying to figure this for over 3 years now. Thanks again J ... View more