通过新分布的最大似然法进行数据拟合 [英] Data fitting by the method of maximum likelihood for a new distribution

问题描述

我想知道使用MLE的方法如何使任何分布适合给定的一组数据.作为一个特定示例，当

I would like to know how one is able to fit any distribution to a given set of data using the method of MLEs. As a particular example, could anyone suggest a working code that would give the correct results for the MLEs for $\theta$ and $\beta$ when the generalised Lindley distribution described in https://rivista-statistica.unibo.it/article/viewFile/6836/7039 is applied to the data: 5.1, 6.3, 10.8, 12.1, 18.5, 19.7, 22.2, 23, 30.6, 37.3, 46.3, 53.9, 59.8, 66.2 on pg. 156? How can this then be used to fit the distribution over a histogram?

推荐答案

基于

I'm going to answer a now-deleted question of yours that is very similar, based on a problem in Parari et al. (data on air conditioning from Proschan (Table 7.2, results in Table 7.5)

通常，您需要知道合理的起始值才能进行通用的最大似然估计(这无疑是鸡和蛋的问题，但是有很多解决方案[使用矩量法，根据问题选择合理的值，根据图表选择值等.]在这种情况下，由于作者给出了答案，因此您可以使用这些参数来选择正确数量级的起始值. ，您需要了解一些有关合理数量级的知识才能开始使用.

In general you need to know reasonable starting values in order to be able to do general-purpose maximum-likelihood estimation (this is admittedly a chicken-and-egg problem, but there are lots of solutions [use the method of moments, pick reasonable values based on the problem, eyeball a graph to pick values, etc.]. In this case, since the authors gave you the answers, you can use those parameters to pick starting values of the right order of magnitude. More generally, you need to know something about reasonable orders of magnitude in order to get started.

请注意，有很多挑剔的数字问题具有一般的最大似然估计:请参阅第7章此处 ...

Note that there are lots of fussy numerical issues with general maximum likelihood estimation: e.g. see chapter 7 here ...

数据(x)和似然函数(dBEPL)在下面定义.我正在定义密度函数，并使用公式接口来mle2() ...

Data (x) and likelihood function (dBEPL) are defined below. I am defining the density function and using the formula interface to mle2() ...

dd  <- data.frame(x)
library(bbmle)
## parameters from table
ovec <- list(alpha=.7945,beta=0.1509,omega=6.7278,
              a=0.2035,b=0.2303)
## starting vals -- right order of magnitude
svec <- list(alpha=0.8,beta=0.2,omega=10,a=0.2,b=0.2)
m1 <- mle2( x ~ dBEPL(alpha,beta,omega,a,b),
      data=dd,
      start=svec,
      control=list(parscale=unlist(svec)))
coef(m1)  ## NOT the same as reported, but close
par(las=1,bty="l")
hist(x,col="gray",freq=FALSE,breaks=40,ylim=c(0.,0.014))
with(as.list(coef(m1)),curve(dBEPL(x,alpha,beta,omega,a,b,log=FALSE),add=TRUE,
                             col=2,lwd=2,n=201))

x <- scan(textConnection("
194 413 90 74 55 23 97 50 359 50 130 487 57 102 
15 14 10 57 320 261 51 44 9 254 493 33 18 209 41 
58 60 48 56 87 11 102 12 5 14 14 29 37 186 29 104 
7 4 72 270 283 7 61 100 61 502 220 120 141 22 603 35 
98 54 100 11 181 65 49 12 239 14 18 39 3 12 5 32 9 438 
43 134 184 20 386 182 71 80 188 230 152 5 36 79 59 33 
246 1 79 3 27 201 84 27 156 21 16 88 130 14 118 44 15 42 
106 46 230 26 59 153 104 20 206 566 34 29 26 35 5 82 31 
118 326 12 54 36 34 18 25 120 31 22 18 216 139 67 310 
346 210 57 76 14 111 97 62 39 30 7 44 11 63 23 22 23 14 
18 13 34 16 18 130 90 163 208 124 70 16 101 52 
208 95 62 11 191 14 71"))


dBEPL <- function(x,alpha,beta,omega,a,b,log=TRUE) {
    r <- log(alpha*beta^2*omega/(beta(a,b)*(beta+1)))+
           log(1+x^alpha)+(alpha-1)*log( x)-beta* x^alpha+(omega*a-1) *
           log(1-(1+beta* x^alpha/(beta+1))*exp(-beta* x^alpha))+
           (b-1)*log(1-(1-(1+beta* x^alpha/(beta+1))*exp(-beta* x^alpha))^omega)
    if (log) return(r) else return(exp(r))
}

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