[quote]
引用第29楼hexm26于2007-08-31 16:06发表的“”:
这其实并没有什么不好理解的。如果说是不同的数据来估计不同的模型,likelihood当然是不适合。现在楼主的前提是“两者结构完全相同,只是相关参数的估计不同”,更进一步说,两个样本只是代表相关的指标值有所不同,而你举的例子是“the fixed effect structure is different”,所以和楼主想表达的特定环境下的定义并不同。
换个角度想likelihood,它其实是条件概率:p(x|parameter)=p(parameter|x),将数据和被估计的参数换个方向考虑就可以了,当然前提是模型结构完全相同。不过必须承认,即使用likelihood来比较数据还是太过牵强,我个人的意见是这种情况就没有一个统一的“量化”比较,还是从数据本身代表的指标值考虑吧。[/quote]
My example is only to demonstrate this principle is clearly wrong. I'm not saying the mixed model example
is exactly the situation here.
To see why likelihood cannot be compared, even if the model is the same, think of these two examples further:
1) Y1,Y2,...,Y10~i.i.d. N(0,1); X1,X2,...,X1000~i.i.d. N(0,1). That is, the only difference is the sample size. Then the likelihood for X's is much smaller than that of Y's, as here, the more data we have, the smaller the likelihood is. But clearly, the data set X's is no worse than the data set Y's.
> n1=rnorm(10)<br />
> n2=rnorm(1000)<br />
> sum(dnorm(n1,log=T))<br />
[1] -13.90256<br />
> sum(dnorm(n2,log=T))<br />
[1] -1392.248
2) Y1,Y2,...,Y100~i.i.d. N(0,1); X1,X2,...,X100~i.i.d N(0,0.01). Assume the same correct model N(mu,sigma), then the likelihood for data set X's should be a log larger then that on Y's, but clearly the two data set is of the same quality, neither being better than the other. One can plot the density to see why.
> y1=rnorm(100)<br />
> y2=rnorm(100,,.1)<br />
> sum(dnorm(y1,mean(y1),sqrt(var(y1)*99/100),log=T))<br />
[1] -130.2282<br />
> sum(dnorm(y2,mean(y2),sqrt(var(y2)*99/100),log=T))<br />
[1] 86.12534