同时估计R中两个数据集的威布尔分布的参数

Question

我试图估计三参数威布尔分布的参数(转换参数β=-0.5(。问题是我必须同时拟合两组数据。使用nlc(请参阅下面的代码(，我能够单独估计每组数据的分布参数，但不能同时估计。GAMMA有点像一个共享参数(估计的GAMMA在两个nlc估计中必须相同(。

我的数据如下：

x = seq(from =0, to =10, by =1)
y = c(0.1315, 0.2368, 0.2631, 0.1578, 0.1578, 0.0000, 0.0526, 0.0000, 0.0000, 0.0000, 0.0000)
z = c(0.3684, 0.3157, 0.2105, 0.0789, 0.0263, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000)

这是我的代码

# WEIBULL FUNCTION WITH ALPHA_GS and beta=-0.5
weibull_GS = function(x, GAMMA, ALPHA_GS){
(GAMMA/ALPHA_GS)*(((x-(-0.5))/ALPHA_GS)^(GAMMA-1))*exp(-((x-(-0.5))/ALPHA_GS)^GAMMA)
}
#ESTIMATE ALPHA_GS
nlc <- nls.control(maxiter=100)
n <- nls(y ~ weibull_GS(x, GAMMA, ALPHA_GS), control="nlc",
start = list(GAMMA=2, ALPHA_GS=3), trace=T, model=F)
summary(n)
# WEIBULL FUNCTION WITH ALPHA_GA beta=-0.5
weibull_GA = function(x, GAMMA, ALPHA_GA){
(GAMMA/ALPHA_GA)*(((x-(-0.5))/ALPHA_GA)^(GAMMA-1))*exp(-((x-(-0.5))/ALPHA_GA)^GAMMA)
}
# ESTIMATE ALPHA_GA
nlc <- nls.control(maxiter=100)
m <- nls(z ~ weibull_GA(x, GAMMA, ALPHA_GA), control="nlc",
start = list(GAMMA=2, ALPHA_GA=3), trace=T, model=F)
summary(m)

Answer 1

您正在做的是用威布尔分布的密度函数f拟合非线性回归模型y = f(x) + error。这与样本的威布尔分布拟合无关。

如果这真的是你想做的，下面是如何回答你的问题：

f <- function(x1, x2, GAMMA, ALPHA_GS, ALPHA_GA){
c(weibull_GS(x1, GAMMA, ALPHA_GS), weibull_GA(x2, GAMMA, ALPHA_GA))
}
Sample <- c(y, z)
nls(Sample ~ f(x, x, GAMMA, ALPHA_GS, ALPHA_GA), ......)