如何在一幅图中绘制拟合的伽玛分布图和实际图？

加载所需的包。

library(ggplot2)
library(MASS)

生成10,000个适合伽玛分布的数字。

x <- round(rgamma(100000,shape = 2,rate = 0.2),1)
x <- x[which(x>0)]

假设我们不知道x符合哪个分布，则绘制概率密度函数。

t1 <- as.data.frame(table(x))
names(t1) <- c("x","y")
t1 <- transform(t1,x=as.numeric(as.character(x)))
t1$y <- t1$y/sum(t1[,2])
ggplot() + 
  geom_point(data = t1,aes(x = x,y = y)) + 
  theme_classic()

从图中可以看出，x的分布与伽马分布非常相似，因此fitdistr()在包中使用它MASS可以获取形状和伽马分布速率的参数。

fitdistr(x,"gamma") 
##       output 
##       shape           rate    
##   2.0108224880   0.2011198260 
##  (0.0083543575) (0.0009483429)

在同一图中绘制实际点（黑点）和拟合图（红线），这是问题所在，请先查看该图。

ggplot() + 
  geom_point(data = t1,aes(x = x,y = y)) +     
  geom_line(aes(x=t1[,1],y=dgamma(t1[,1],2,0.2)),color="red") + 
  theme_classic()

我有两个问题：

1. 真正的参数shape=2，rate=0.2和我用函数的参数fitdistr()来获得的shape=2.01，rate=0.20。这两个几乎是相同的，但是为什么拟合图不能很好地拟合实际点，所以拟合图中一定存在错误，或者我绘制拟合图和实际点的方式完全错误，我该怎么办？

2. 得到模型的参数后，可以用哪种方式评估模型，例如线性模型的RSS（残差平方和），还是的p值shapiro.test()，ks.test()以及其他检验？

我的统计知识很差，请您帮我一下吗？

ps：我在Google，stackoverflow和CV中搜索了很多次，但没有发现与此问题相关的信息

问题1

手工计算密度的方法似乎是错误的。无需对来自伽玛分布的随机数进行四舍五入。正如@Pascal指出的那样，您可以使用直方图来绘制点的密度。在下面的示例中，我使用该函数density估算密度并将其绘制为点。我用点和直方图表示拟合：

library(ggplot2)
library(MASS)

# Generate gamma rvs

x <- rgamma(100000, shape = 2, rate = 0.2)

den <- density(x)

dat <- data.frame(x = den$x, y = den$y)

# Plot density as points

ggplot(data = dat, aes(x = x, y = y)) + 
  geom_point(size = 3) +
  theme_classic()

# Fit parameters (to avoid errors, set lower bounds to zero)

fit.params <- fitdistr(x, "gamma", lower = c(0, 0))

# Plot using density points

ggplot(data = dat, aes(x = x,y = y)) + 
  geom_point(size = 3) +     
  geom_line(aes(x=dat$x, y=dgamma(dat$x,fit.params$estimate["shape"], fit.params$estimate["rate"])), color="red", size = 1) + 
  theme_classic()

# Plot using histograms

ggplot(data = dat) +
  geom_histogram(data = as.data.frame(x), aes(x=x, y=..density..)) +
  geom_line(aes(x=dat$x, y=dgamma(dat$x,fit.params$estimate["shape"], fit.params$estimate["rate"])), color="red", size = 1) + 
  theme_classic()

这是@Pascal提供的解决方案：

h <- hist(x, 1000, plot = FALSE)
t1 <- data.frame(x = h$mids, y = h$density)

ggplot(data = t1, aes(x = x, y = y)) + 
  geom_point(size = 3) +     
  geom_line(aes(x=t1$x, y=dgamma(t1$x,fit.params$estimate["shape"], fit.params$estimate["rate"])), color="red", size = 1) + 
  theme_classic()

问题2

为了评估合身性，我推荐包装fitdistrplus。这是可用于拟合两个分布并以图形和数字方式比较其拟合的方法。该命令gofstat打印出一些度量，例如AIC，BIC和一些gof统计信息，例如KS-Test等。这些度量主要用于比较不同分布的拟合（在这种情况下为gamma和Weibull）。更多信息可以在我的答案中找到：

library(fitdistrplus)

x <- c(37.50,46.79,48.30,46.04,43.40,39.25,38.49,49.51,40.38,36.98,40.00,
       38.49,37.74,47.92,44.53,44.91,44.91,40.00,41.51,47.92,36.98,43.40,
       42.26,41.89,38.87,43.02,39.25,40.38,42.64,36.98,44.15,44.91,43.40,
       49.81,38.87,40.00,52.45,53.13,47.92,52.45,44.91,29.54,27.13,35.60,
       45.34,43.37,54.15,42.77,42.88,44.26,27.14,39.31,24.80,16.62,30.30,
       36.39,28.60,28.53,35.84,31.10,34.55,52.65,48.81,43.42,52.49,38.00,
       38.65,34.54,37.70,38.11,43.05,29.95,32.48,24.63,35.33,41.34)

fit.weibull <- fitdist(x, "weibull")
fit.gamma <- fitdist(x, "gamma", lower = c(0, 0))

# Compare fits 

graphically

par(mfrow = c(2, 2))
plot.legend <- c("Weibull", "Gamma")
denscomp(list(fit.weibull, fit.gamma), fitcol = c("red", "blue"), legendtext = plot.legend)
qqcomp(list(fit.weibull, fit.gamma), fitcol = c("red", "blue"), legendtext = plot.legend)
cdfcomp(list(fit.weibull, fit.gamma), fitcol = c("red", "blue"), legendtext = plot.legend)
ppcomp(list(fit.weibull, fit.gamma), fitcol = c("red", "blue"), legendtext = plot.legend)

@NickCox正确地建议QQ图（右上图）是判断和比较拟合的最佳单图。拟合密度很难比较。为了完整起见，我还包括其他图形。

# Compare goodness of fit

gofstat(list(fit.weibull, fit.gamma))

Goodness-of-fit statistics
                             1-mle-weibull 2-mle-gamma
Kolmogorov-Smirnov statistic    0.06863193   0.1204876
Cramer-von Mises statistic      0.05673634   0.2060789
Anderson-Darling statistic      0.38619340   1.2031051

Goodness-of-fit criteria
                               1-mle-weibull 2-mle-gamma
Aikake's Information Criterion      519.8537    531.5180
Bayesian Information Criterion      524.5151    536.1795