sungmoo 发表于 2011-10-14 12:35 
并帖
https://bbs.pinggu.org/thread-1107862-1-1.html
哇~~~~~我都没搜索到~~!
~>.<~
好吧,谢谢版主~~
我把结果粘过来好了……
其中有例子(左截点-2,右截点2,方差1/4,期望0):
set obs 10000
gen c = -2 /*left truncation point*/
gen d = 2 /*right truncation point*/
gen stdv = 0.5 /*standard deviation*/
gen v = uniform()
gen c_bar = c/stdv
gen d_bar = d/stdv
gen v_bar = (normal(c_bar) + (normal(d_bar) - normal(c_bar))*v)
gen u = stdv*invnorm(v_bar)
下载
http://fmwww.bc.edu/repec/bocode/g/gentrun.ado
http://fmwww.bc.edu/repec/bocode/g/gentrun.hlp
可生成左右截断的标准正态分布随机数。
*上述过程可以简化为一步:
g u=0.5*invnormal(normal(-2/0.5)+(normal(2/0.5)-normal(-2/0.5))*uniform())
以上解答来自于sungmoo