Sas help里的一个例子-----用NLP作Maximum Likelihood Weibull Estimatio

shiyiming

sas help里搜索Example 6.6: Maximum Likelihood Weibull Estimation，这个例子里，第三个程序，proc nlp中，为什么会把s 求最大值呢，不是应该整个log函数求吗？

proc nlp data=pike tech=tr inest=par1 outest=opar1   outmodel=model cov=2 vardef=n pcov phes;      
[color=#BF00BF:18rbghgd]max logf;   [/color:18rbghgd]
parms sig c;      
profile sig c / alpha = .9 to .1 by -.1 .09 to .01 by -.01;
x_th = days - theta;
s    = - (x_th / sig)**c;      
if cens=0 then s + log(c) - c*log(sig) + (c-1)*log(x_th);      
[color=#BF00FF:18rbghgd]logf = s;    [/color:18rbghgd]         /*只把s的值赋给logf,而对logf求最大值？？*/
run;

shiyiming

对NLP还是不太理解

shiyiming

In fact, here s = logf = log(Likelihood) = l, for an observation;
MAX logf; --->maximize the equation of summarization of [logf]i where i from 1 to m if you have m observations
另外，非常羡慕你学习sas两个月就可以做mathematical modeling.

shiyiming

对 我知道max logf是对所有的变量求和之后再求最大值，
但是似然函数不应该是 s + log(c) - c*log(sig) + (c-1)*log(x_th)么 ，而不单是s:

proc nlp data=pike tech=tr inest=par1 outest=opar1 outmodel=model cov=2 vardef=n pcov phes;
max logf;
parms sig c;
profile sig c / alpha = .9 to .1 by -.1 .09 to .01 by -.01;
x_th = days - theta;
s = - (x_th / sig)**c;
if cens=0 then[color=#FF40FF:15q998no] f= s + log(c) - c*log(sig) + (c-1)*log(x_th)[/color:15q998no];
[color=#FF40FF:15q998no]logf = f[/color:15q998no];
run;

另外，非常高兴你的回复，但你不是讽我呢吧，呵呵。接触SAS很久了，不过没专门研究过，学统计很久了哈。

shiyiming

总共19个记录。所以maximize（l1+l2+...+l19);
前17个都是事件(cens = 0)，所以 l =  - (x_th / sig)**c + log(c) - c*log(sig) + (c-1)*log(x_th) ， for i = 1, 2, ..., 17;
后两个是censors(cens = 1)，所以 l = - (x_th / sig)**c, for i = 18, 19；
也就是说19个记录生成19个loglikelihood，按照以上的方法。
你的改动有些问题，就是两个censored的值没有考虑其中。
如下的改动应该是等价的：

[code:4m505uxm]proc nlp data=pike tech=tr inest=par1 outest=opar1
      outmodel=model cov=2 vardef=n pcov phes;
/*   max logf;*/
                max s;
   parms sig c;
   profile sig c / alpha = .9 to .1 by -.1 .09 to .01 by -.01;

   x_th = days - theta;
   s    = - (x_th / sig)**c;
   if cens=0 then s + log(c) - c*log(sig) + (c-1)*log(x_th);
/*   logf = s;*/
run;[/code:4m505uxm]

shiyiming

还是不太明白，为什么cens = 1，所以 l = - (x_th / sig)**c,  而且也没写 else logf = s,
那前17个 l = - (x_th / sig)**c + log(c) - c*log(sig) + (c-1)*log(x_th) 也没有求最大值 : max l  ??