伽马分布拟合误差 [英] Error in fitdist with gamma distribution

问题描述

以下是我的代码:

library(fitdistrplus)

s <- c(11, 4, 2, 9, 3, 1, 2, 2, 3, 2, 2, 5, 8,3, 15, 3, 9, 22, 0, 4, 10, 1, 9, 10, 11,
 2, 8, 2, 6, 0, 15, 0 , 2, 11, 0, 6, 3, 5, 0, 7, 6, 0, 7, 1, 0, 6, 4, 1, 3, 5,
 2, 6, 0, 10, 6, 4, 1, 17, 0, 1, 0, 6, 6, 1, 5, 4, 8, 0, 1, 1, 5, 15, 14, 8, 1,
 3, 2, 9, 4, 4, 1, 2, 18, 0, 0, 10, 5, 0, 5, 0, 1, 2, 0, 5, 1, 1, 2, 3, 7)

o <- fitdist(s, "gamma", method = "mle")
summary(o)
plot(o)

<小时>

并且错误说:

---

and the error says:

fitdist(s, "gamma", method = "mle") 中的错误:函数 mle未能估计参数，错误代码为 100

Error in fitdist(s, "gamma", method = "mle") : the function mle failed to estimate the parameters, with the error code 100

推荐答案

Gamma 分布不允许零值(对于 0 的响应，似然将评估为零，对数似然将是无限的)，除非形状参数正好是 1.0(即指数分布 - 见下文)......这是一个统计/数学问题，而不是一个编程问题.您将不得不为零值找到一些明智的做法.根据对您的应用程序有意义的内容，您可以(例如)

The Gamma distribution doesn't allow zero values (the likelihood will evaluate to zero, and the log-likelihood will be infinite, for a response of 0) unless the shape parameter is exactly 1.0 (i.e., an exponential distribution - see below) ... that's a statistical/mathematical problem, not a programming problem. You're going to have to find something sensible to do about the zero values. Depending on what makes sense for your application, you could (for example)

• 选择不同的发行版进行测试
• 排除零值 (fitdist(s[s>0], ...))
• 将零值设置为一些合理的非零值 (fitdist(replace(s,which(s==0),0.1),...)

其中哪个(如果有)最好取决于您的应用.

@Sandipan Dey 的第一个答案(在数据集中留下零)看似是有道理的，但实际上它卡在形状参数等于 1 处.

@Sandipan Dey's first answer (leaving the zeros in the data set) appears to make sense, but in fact it gets stuck at the shape parameter equal to 1.

o <- fitdist(s, "exp", method = "mle")

给出与@Sandipan 代码相同的答案(除了它估计速率=0.2161572，为 Gamma 分布估计的尺度参数的倒数 = 4.626262 - 这只是参数化的变化).如果您选择拟合指数而不是 Gamma，那很好 - 但您应该有目的地这样做，而不是偶然...

gives the same answer as @Sandipan's code (except that it estimates rate=0.2161572, the inverse of the scale parameter=4.626262 that's estimated for the Gamma distribution - this is just a change in parameterization). If you choose to fit an exponential instead of a Gamma, that's fine - but you should do it on purpose, not by accident ...

为了说明包含零的拟合可能无法按预期工作，我将构建自己的负对数似然函数并显示每种情况的似然面.

To illustrate that the zeros-included fit may not be working as expected, I'll construct my own negative log-likelihood function and display the likelihood surface for each case.

mfun <- function(sh,sc,dd=s) {
  -sum(dgamma(dd,shape=sh,scale=sc,log=TRUE))
}

library(emdbook)  ## for curve3d() helper function

包含零的曲面:

cc1 <- curve3d(mfun(x,y),
      ## set up "shape" limits" so we evaluate
      ## exactly shape=1.000 ...
        xlim=c(0.55,3.55),
        n=c(41,41),
        ylim=c(2,5),
        sys3d="none")


png("gammazero1.png")
with(cc1,image(x,y,z))
dev.off()

在这种情况下，表面仅定义为 shape=1(即指数分布)；白色区域代表无限对数似然.不是 shape=1 是最佳合身，而是唯一合身...

In this case the surface is only defined at shape=1 (i.e. an exponential distribution); the white regions represent infinite log-likelihoods. It's not that shape=1 is the best fit, it's that it's the only fit ...

排除零的曲面:

cc2 <- curve3d(mfun(x,y,dd=s[s>0]),
               ## set up "shape" limits" so we evaluate
               ## exactly shape=1.000 ...
               xlim=c(0.55,3.55),
               n=c(41,41),
               ylim=c(2,5),
               sys3d="none")


png("gammazero2.png")
with(cc2,image(x,y,z))
with(cc2,contour(x,y,z,add=TRUE))
abline(v=1.0,lwd=2,lty=2)
dev.off()

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