r - R - Levenberg Marquardt 中的非线性最小二乘法拟合 Heligman Pollard 模型参数

Question

我正在尝试重现 Kostakis 的论文解决方案。在本文中，使用 de Heligman-Pollard 模型将简化的死亡率表扩展为完整的生命表。该模型有 8 个必须拟合的参数。作者使用了修改后的 Gauss-Newton 算法；该算法 (E04FDF) 是 NAG 计算机程序库的一部分。Levenberg Marquardt 不应该产生相同的参数集吗？我的代码或 LM 算法的应用有什么问题？

library(minpack.lm)


## Heligman-Pollard is used to expand an abridged table.
## nonlinear least squares algorithm is used to fit the parameters on nqx observed over 5 year   intervals (5qx)
AGE <- c(0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70)
MORTALITY <- c(0.010384069, 0.001469140, 0.001309318, 0.003814265, 0.005378395, 0.005985625,     0.006741766, 0.009325056, 0.014149626, 0.021601755, 0.034271934, 0.053836246, 0.085287751, 0.136549522, 0.215953304)

## The start parameters for de Heligman-Pollard Formula (Converged set a=0.0005893,b=0.0043836,c=0.0828424,d=0.000706,e=9.927863,f=22.197312,g=0.00004948,h=1.10003)
## I modified a random parameter "a" in order to have a start values. The converged set is listed above. 
parStart <- list(a=0.0008893,b=0.0043836,c=0.0828424,d=0.000706,e=9.927863,f=22.197312,g=0.00004948,h=1.10003)

## The Heligman-Pollard Formula (HP8) = qx/px = ...8 parameter equation
HP8 <-function(parS,x)
ifelse(x==0, parS$a^((x+parS$b)^parS$c) + parS$g*parS$h^x, 
             parS$a^((x+parS$b)^parS$c) + parS$d*exp(-parS$e*(log(x/parS$f))^2) +
                 parS$g*parS$h^x)

## Define qx = HP8/(1+HP8)
qxPred <- function(parS,x) HP8(parS,x)/(1+HP8(parS,x))

## Calculate nqx predicted by HP8 model (nqxPred(parStart,x))
nqxPred <- function(parS,x)
(1 -(1-qxPred(parS,x)) * (1-qxPred(parS,x+1)) *
    (1-qxPred(parS,x+2)) * (1-qxPred(parS,x+3)) *
    (1-qxPred(parS,x+4))) 

##Define Residual Function, the relative squared distance is minimized  
ResidFun <- function(parS, Observed,x) (nqxPred(parS,x)/Observed-1)^2

## Applying the nls.lm algo. 
nls.out <- nls.lm(par=parStart, fn = ResidFun, Observed = MORTALITY, x = AGE,
                  control = nls.lm.control(nprint=1,
                                           ftol = .Machine$double.eps,
                                           ptol = .Machine$double.eps,
                                           maxfev=10000, maxiter = 500))

summary(nls.out)


## The author used a modified Gauss-Newton algorithm, this alogorithm (E04FDF) is part of the NAG library of computer programs
## Should not Levenberg Marquardt yield the same set of parameters

score 13 · Accepted Answer

这里的底线是@Roland 绝对正确，这是一个非常不适定的问题，您不必期望得到可靠的答案。下面我有

用一些小方法清理代码（这只是美学）
更改ResidFun为返回残差，而不是平方残差。（前者是正确的，但这并没有太大区别。）
探索了几个不同优化器的结果。实际上，您得到的答案看起来比您上面列出的“收敛参数”要好，我假设这是原始研究中的参数（您能提供参考吗？）。

加载包：

library(minpack.lm)

数据，作为数据框：

d <- data.frame(
   AGE = seq(0,70,by=5),
   MORTALITY=c(0.010384069, 0.001469140, 0.001309318, 0.003814265,
               0.005378395, 0.005985625, 0.006741766, 0.009325056,
               0.014149626, 0.021601755, 0.034271934, 0.053836246,
               0.085287751, 0.136549522, 0.215953304))

数据的第一个视图：

library(ggplot2)
(g1 <- ggplot(d,aes(AGE,MORTALITY))+geom_point())
g1+geom_smooth()  ## with loess fit

参数选择：

大概这些是原始论文中的参数...

parConv <- c(a=0.0005893,b=0.0043836,c=0.0828424,
             d=0.000706,e=9.927863,f=22.197312,g=0.00004948,h=1.10003)

扰动参数：

parStart <- parConv
parStart["a"] <- parStart["a"]+3e-4

公式：

HP8 <-function(parS,x)
    with(as.list(parS),
         ifelse(x==0, a^((x+b)^c) + g*h^x, 
                a^((x+b)^c) + d*exp(-e*(log(x/f))^2) + g*h^x))
## Define qx = HP8/(1+HP8)
qxPred <- function(parS,x) {
    h <- HP8(parS,x)
    h/(1+h)
}
## Calculate nqx predicted by HP8 model (nqxPred(parStart,x))
nqxPred <- function(parS,x)
    (1 -(1-qxPred(parS,x)) * (1-qxPred(parS,x+1)) *
     (1-qxPred(parS,x+2)) * (1-qxPred(parS,x+3)) *
     (1-qxPred(parS,x+4))) 
##Define Residual Function, the relative squared distance is minimized  
ResidFun <- function(parS, Observed,x) (nqxPred(parS,x)/Observed-1)

注意，这与 OP 的版本略有不同；nls.lm想要残差，而不是平方残差。

与其他优化器一起使用的平方和函数：

ssqfun <- function(parS, Observed, x) {
   sum(ResidFun(parS, Observed, x)^2)
}

申请nls.lm。（不知道为什么ftol 并且ptol从降低sqrt(.Machine$double.eps)到.Machine$double.eps- 前者通常是对精度的实际限制......

nls.out <- nls.lm(par=parStart, fn = ResidFun,
                  Observed = d$MORTALITY, x = d$AGE,
                  control = nls.lm.control(nprint=0,
                                           ftol = .Machine$double.eps,
                                           ptol = .Machine$double.eps,
                                           maxfev=10000, maxiter = 1000))

parNLS <- coef(nls.out)

pred0 <- nqxPred(as.list(parConv),d$AGE)
pred1 <- nqxPred(as.list(parNLS),d$AGE)

dPred <- with(d,rbind(data.frame(AGE,MORTALITY=pred0,w="conv"),
               data.frame(AGE,MORTALITY=pred1,w="nls")))

g1 + geom_line(data=dPred,aes(colour=w))

这些线无法区分，但参数有一些很大的不同：

round(cbind(parNLS,parConv),5)
##     parNLS  parConv
## a  1.00000  0.00059
## b 50.46708  0.00438
## c  3.56799  0.08284
## d  0.00072  0.00071
## e  6.05200  9.92786
## f 21.82347 22.19731
## g  0.00005  0.00005
## h  1.10026  1.10003

d,f,g,h 很接近，但 a,b,c 相差几个数量级，e 相差 50%。

查看原始方程，这里发生的事情a^((x+b)^c)是被设置为一个常数，因为a接近 1：一次a大约是 1，b并且c本质上是不相关的。

让我们检查一下相关性（我们需要一个广义逆，因为矩阵的相关性很强）：

obj <- nls.out
vcov  <- with(obj,deviance/(length(fvec) - length(par)) * 
              MASS::ginv(hessian))

cmat <- round(cov2cor(vcov),1)
dimnames(cmat) <- list(letters[1:8],letters[1:8])

##      a    b    c    d    e    f    g    h
## a  1.0  0.0  0.0  0.0  0.0  0.0 -0.1  0.0
## b  0.0  1.0 -1.0  1.0 -1.0 -1.0 -0.4 -1.0
## c  0.0 -1.0  1.0 -1.0  1.0  1.0  0.4  1.0
## d  0.0  1.0 -1.0  1.0 -1.0 -1.0 -0.4 -1.0
## e  0.0 -1.0  1.0 -1.0  1.0  1.0  0.4  1.0
## f  0.0 -1.0  1.0 -1.0  1.0  1.0  0.4  1.0
## g -0.1 -0.4  0.4 -0.4  0.4  0.4  1.0  0.4
## h  0.0 -1.0  1.0 -1.0  1.0  1.0  0.4  1.0

这实际上并没有那么有用——它实际上只是证实了许多变量是强相关的......

library(optimx)
mvec <- c('Nelder-Mead','BFGS','CG','L-BFGS-B',
          'nlm','nlminb','spg','ucminf')
opt1 <- optimx(par=parStart, fn = ssqfun,
         Observed = d$MORTALITY, x = d$AGE,
               itnmax=5000,
               method=mvec,control=list(kkt=TRUE))
               ## control=list(all.methods=TRUE,kkt=TRUE)) ## Boom!

##         fvalues      method fns  grs itns conv KKT1 KKT2 xtimes
## 2 8.988466e+307        BFGS  NA NULL NULL 9999   NA   NA      0
## 3 8.988466e+307          CG  NA NULL NULL 9999   NA   NA      0
## 4 8.988466e+307    L-BFGS-B  NA NULL NULL 9999   NA   NA      0
## 5 8.988466e+307         nlm  NA   NA   NA 9999   NA   NA      0
## 7     0.3400858         spg   1   NA    1    3   NA   NA  0.064
## 8     0.3400858      ucminf   1    1 NULL    0   NA   NA  0.032
## 1    0.06099295 Nelder-Mead 501   NA NULL    1   NA   NA  0.252
## 6   0.009275733      nlminb 200 1204  145    1   NA   NA  0.708

这警告了不良缩放，并且还找到了各种不同的答案：只ucminf声称已经收敛，但nlminb得到了更好的答案——而且itnmax参数似乎被忽略了......

opt2 <- nlminb(start=parStart, objective = ssqfun,
         Observed = d$MORTALITY, x = d$AGE,                   
               control= list(eval.max=5000,iter.max=5000))

parNLM <- opt2$par

完成，但出现错误的收敛警告......

round(cbind(parNLS,parConv,parNLM),5)

##     parNLS  parConv   parNLM
## a  1.00000  0.00059  1.00000
## b 50.46708  0.00438 55.37270
## c  3.56799  0.08284  3.89162
## d  0.00072  0.00071  0.00072
## e  6.05200  9.92786  6.04416
## f 21.82347 22.19731 21.82292
## g  0.00005  0.00005  0.00005
## h  1.10026  1.10003  1.10026

sapply(list(parNLS,parConv,parNLM),
       ssqfun,Observed=d$MORTALITY,x=d$AGE)
## [1] 0.006346250 0.049972367 0.006315034

它看起来nlminb并且minpack.lm正在得到类似的答案，并且实际上比最初声明的参数做得更好（相当多）：

pred2 <- nqxPred(as.list(parNLM),d$AGE)

dPred <- with(d,rbind(dPred,
               data.frame(AGE,MORTALITY=pred2,w="nlminb")))

g1 + geom_line(data=dPred,aes(colour=w))
ggsave("cmpplot.png")

在此处输入图像描述

ggplot(data=dPred,aes(x=AGE,y=MORTALITY-d$MORTALITY,colour=w))+
   geom_line()+geom_point(aes(shape=w),alpha=0.3)
ggsave("residplot.png")

在此处输入图像描述

可以尝试的其他事情是：

适当的缩放——尽管对此进行快速测试似乎没有太大帮助
提供分析梯度
使用 AD 模型生成器
使用slice函数 frombbmle来探索新旧参数是否似乎代表不同的最小值，或者旧参数是否只是一个错误的收敛...
optimx从用于类似检查的相关软件包或相关软件包中获取 KKT (Karsh-Kuhn-Tucker) 标准计算器

PS：最大的偏差（到目前为止）是针对最老的年龄组的，它们可能也有小样本。从统计的角度来看，可能值得做一个由各个点的精度加权的拟合......

score 0 · Accepted Answer

@BenBolker，用整个数据集（基础 qx）值拟合参数。仍然无法重现参数

library(minpack.lm)

library(ggplot2)

library(optimx)

getwd()

d <- data.frame(AGE = seq(0,74), MORTALITY=c(869,58,40,37,36,35,32,28,29,23,24,22,24,28,
                                           33,52,57,77,93,103,103,109,105,114,108,112,119,
                                           125,117,127,125,134,134,131,152,179,173,182,199,
                                           203,232,245,296,315,335,356,405,438,445,535,594,
                                           623,693,749,816,915,994,1128,1172,1294,1473,
                                           1544,1721,1967,2129,2331,2559,2901,3203,3470,
                                           3782,4348,4714,5245,5646))


d$MORTALITY <- d$MORTALITY/100000

ggplot(d,aes(AGE,MORTALITY))+geom_point()  

##Not allowed to post Images

g1 <- ggplot(d,aes(AGE,MORTALITY))+geom_point()

g1+geom_smooth()## with loess fit

报告参数：

parConv <- c(a=0.0005893,b=0.0043836,c=0.0828424,d=0.000706,e=9.927863,f=22.197312,
             g=0.00004948,h=1.10003)

parStart <- parConv

parStart["a"] <- parStart["a"]+3e-4


## Define qx = HP8/(1+HP8)

HP8 <-function(parS,x)
with(as.list(parS),
ifelse(x==0, a^((x+b)^c) + g*h^x, a^((x+b)^c) + d*exp(-e*(log(x/f))^2) + g*h^x))



qxPred <- function(parS,x) {
  h <- HP8(parS,x)
  h/(1+h)
}



##Define Residual Function, the relative squared distance is minimized,
ResidFun <- function(parS, Observed,x) (qxPred(parS,x)/Observed-1)

ssqfun <- function(parS, Observed, x) {
  sum(ResidFun(parS, Observed, x)^2)
}

nls.out <- nls.lm(par=parStart, fn = ResidFun, Observed = d$MORTALITY, x = d$AGE, 
                  control = nls.lm.control(nprint=1, ftol = sqrt(.Machine$double.eps), 
                  ptol = sqrt(.Machine$double.eps), maxfev=1000, maxiter=1000))


parNLS <- coef(nls.out)

pred0 <- qxPred(as.list(parConv),d$AGE)
pred1 <- qxPred(as.list(parNLS),d$AGE)


#Binds Row wise the dataframes from pred0 and pred1
dPred <- with(d,rbind(data.frame(AGE,MORTALITY=pred0,w="conv"),
      data.frame(AGE,MORTALITY=pred1,w="nls")))


g1 + geom_line(data=dPred,aes(colour=w))

round(cbind(parNLS,parConv),7)

mvec <- c('Nelder-Mead','BFGS','CG','L-BFGS-B','nlm','nlminb','spg','ucminf')
opt1 <- optimx(par=parStart, fn = ssqfun,
    Observed = d$MORTALITY, x = d$AGE,
    itnmax=5000,
    method=mvec, control=list(all.methods=TRUE,kkt=TRUE,)
## control=list(all.methods=TRUE,kkt=TRUE)) ## Boom

get.result(opt1, attribute= c("fvalues","method", "grs", "itns",
           "conv", "KKT1", "KKT2", "xtimes"))

##       method       fvalues  grs itns conv KKT1 KKT2 xtimes
##5         nlm 8.988466e+307   NA   NA 9999   NA   NA      0
##4    L-BFGS-B 8.988466e+307 NULL NULL 9999   NA   NA      0
##2          CG 8.988466e+307 NULL NULL 9999   NA   NA   0.02
##1        BFGS 8.988466e+307 NULL NULL 9999   NA   NA      0
##3 Nelder-Mead     0.5673864   NA NULL    0   NA   NA   0.42
##6      nlminb     0.4127198  546   62    0   NA   NA   0.17


opt2 <- nlminb(start=parStart, objective = ssqfun,
    Observed = d$MORTALITY, x = d$AGE,
    control= list(eval.max=5000,iter.max=5000))

parNLM <- opt2$par

检查参数：

round(cbind(parNLS,parConv,parNLM),5)

##    parNLS  parConv   parNLM
##a  0.00058  0.00059  0.00058
##b  0.00369  0.00438  0.00369
##c  0.08065  0.08284  0.08065
##d  0.00070  0.00071  0.00070
##e  9.30948  9.92786  9.30970
##f 22.30769 22.19731 22.30769
##g  0.00005  0.00005  0.00005
##h  1.10084  1.10003  1.10084

上交所审查：

sapply(list(parNLS,parConv,parNLM),
  ssqfun,Observed=d$MORTALITY,x=d$AGE)  

 ##[1] 0.4127198 0.4169513 0.4127198

无法上传图表，但代码在这里。当使用完整的死亡率数据（未删节或子集）时，文章中发现的参数似乎仍然不是最合适的

##pred2 <- qxPred(as.list(parNLM),d$AGE)

##dPred <- with(d,rbind(dPred,
    data.frame(AGE,MORTALITY=pred2,w="nlminb")))

##g1 + geom_line(data=dPred,aes(colour=w))
ggplot(data=dPred,aes(x=AGE,y=MORTALITY-d$MORTALITY,colour=w))
        + geom_line()+geom_point(aes(shape=w),alpha=0.3)

r - R - Levenberg Marquardt 中的非线性最小二乘法拟合 Heligman Pollard 模型参数

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