有没有办法在 R 中使用 beta 分布参数找到中位数？

Question

我正在使用名为 productQuality 的 CSV 数据集，其中每一行代表一种焊接类型和该特定焊接的 beta 分布参数（α 和 β）。我想知道是否有办法计算和列出每种焊接类型的中值？这是我的数据集的输入：

structure(list(weld.type.ID = 1:33, weld.type = structure(c(29L, 
11L, 16L, 4L, 28L, 17L, 19L, 5L, 24L, 27L, 21L, 32L, 12L, 20L, 
26L, 25L, 3L, 7L, 13L, 22L, 33L, 1L, 9L, 10L, 18L, 15L, 31L, 
8L, 23L, 2L, 14L, 6L, 30L), .Label = c("1,40,Material A", "1,40S,Material C", 
"1,80,Material A", "1,STD,Material A", "1,XS,Material A", "10,10S,Material C", 
"10,160,Material A", "10,40,Material A", "10,40S,Material C", 
"10,80,Material A", "10,STD,Material A", "10,XS,Material A", 
"13,40,Material A", "13,40S,Material C", "13,80,Material A", 
"13,STD,Material A", "13,XS,Material A", "14,40,Material A", 
"14,STD,Material A", "14,XS,Material A", "15,STD,Material A", 
"15,XS,Material A", "2,10S,Material C", "2,160,Material A", "2,40,Material A", 
"2,40S,Material C", "2,80,Material A", "2,STD,Material A", "2,XS,Material A", 
"4,80,Material A", "4,STD,Material A", "6,STD,Material A", "6,XS,Material A"
), class = "factor"), alpha = c(281L, 196L, 59L, 96L, 442L, 98L, 
66L, 30L, 68L, 43L, 35L, 44L, 23L, 14L, 24L, 38L, 8L, 8L, 5L, 
19L, 37L, 38L, 6L, 11L, 29L, 6L, 16L, 6L, 16L, 3L, 4L, 9L, 12L
), beta = c(7194L, 4298L, 3457L, 2982L, 4280L, 3605L, 2229L, 
1744L, 2234L, 1012L, 1096L, 1023L, 1461L, 1303L, 531L, 233L, 
630L, 502L, 328L, 509L, 629L, 554L, 358L, 501L, 422L, 566L, 403L, 
211L, 159L, 268L, 167L, 140L, 621L)), class = "data.frame", row.names = c(NA, 
-33L))

Answer 1

根据Wikipedia，alpha、beta >1 的中位数有一个近似解，但没有通用的封闭式解。下面我实现暴力精确解和近似解：

## I_{1/2}^{-1}(alpha,beta)
med_exact0 <- function(alpha,beta,eps=1e-12) {
    uniroot(function(x) pbeta(x,alpha,beta)-1/2,
            interval=c(eps,1-eps))$root
}
med_exact <- Vectorize(med_exact0, vectorize.args=c("alpha","beta"))
med_approx <- function(alpha,beta) (alpha-1/3)/(alpha+beta-2/3)

edit 评论指出逆 ('brute force') 解决方案已经在基础 R 中实现为 qbeta(p=0.5,...)！几乎可以肯定比我的解决方案更健壮且计算效率更高...

我调用了你的数据dd:

evals <- with(dd,med_exact(alpha,beta))
avals <- with(dd,med_approx(alpha,beta))
evals2 <- with(dd,qbeta(0.5,alpha,beta))
max(abs((evals-avals)/evals))  ## 0.0057

在您的数据中最坏的情况下，精确解和近似解相差约 0.6% ...

有没有办法在 R 中使用 beta 分布参数找到中位数？

Is there a way to find the median using beta distribution parameters in R?

r

median

beta-distribution