SAS如何生成服从Weibull分布的随机数？有区间限制呢？

各位高手，有两个问题请教：

（1）SAS如何生成服从Weibull分布的随机数(已解决，见下面回帖，第二个问题继续寻求答案)；

（2）如何生成某个区间内的随机数，如服从正态分布，但随机数的范围在a和b之间；

谢谢各位！

bwnsm 发表于 2012-3-23 10:29
Since the distribution function for the Weibull is closed, you can solve
it for the variable and pl ...

SAS provides such a random function.

Weibull Distribution
x = RAND('WEIBULL',a,b)
Arguments
x
is an observation from the distribution with the following probability density function:

Range:x ≥ 0
a
is a numeric shape parameter.
Range:a > 0
b
is a numeric scale parameter.
Range:b > 0

攒经验 帮顶

Since the distribution function for the Weibull is closed, you can solve
it for the variable and plug a uniformally distributed random number in
as an argument.  The Weibull distribution is --
     F(y) = 1 - exp(-(y-a)/b)**c)
Solving for y gives --
     y = [-ln(1-F(y))]**(1/c)+a
So you can generate random Weibull numbers with --
     y=(-log(1-ranuni(0)))**(1/c);
(Using 1-ranuni(0) or ranuni(0) is immaterial.)

以上内容来自于http://listserv.uga.edu/cgi-bin/ ... amp;D=0&P=45818

代码：

data a;
a=2; b=20; c=6;/*a=Location;b=Scale;c=Shape ,in CrystalBall. */
do i=1 to 10000;   /* Do 10000 times. ie. generate 10000 random numbers */
    x=a+b*(-log(1-uniform(0)))**(1/c);     /* Calculate Weibull Random Number. Using 1-ranuni(0) or ranuni(0) is immaterial. */
    /*An argument of 0 -- uses your computor clock as a seed. */
    /*Functions RANUNI and UNIFORM are identical. Function UNIFORM cannot be utilized as a CALL routine.*/
output;
end;
run;

生成某个区间内的随机数代码：

data work.trunc_norm;
mean=0;
sd=1;
max=2;
min=-2;
do i = 1 to 10000;
random=max+1;
do while (random<min or random>max);
random=rannor(-1)*sd+mean;
end;
output;
end;
run;

参考：http://confounding.net/2010/12/0 ... ormal-distribution/

learning ing ....

（2）如何生成某个区间内的随机数，如服从正态分布，但随机数的范围在a和b之间.

There is now way you can do that. The support of normal distribution is from -00 to +00. But you can let any number from below a be a and any number from u b be b. It is called a truncated normal distribution.

bwnsm 发表于 2012-3-23 10:29
Since the distribution function for the Weibull is closed, you can solve
it for the variable and pl ...

学习了

bobguy 发表于 2012-3-25 11:48
SAS provides such a random function.

Weibull Distribution

谢谢，这个我也试过了，但这个函数只能包含shape和Scale参数，不能输Location参数

bwnsm 发表于 2012-3-25 19:16
谢谢，这个我也试过了，但这个函数只能包含shape和Scale参数，不能输Location参数

Because it does not have a location parameter. For details check the link below,

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