枚举一系列不同概率的伯努利试验的所有可能组合概率

Question

假设我有一系列 n 个独立伯努利试验成功的概率，p1 到 pn 使得 p1 != p2 != ... != pn。为每个试验指定一个唯一的名称。

    p <- c(0.5, 0.12, 0.7, 0.8, .02)
    a <- c("A","B","C","D","E")

我通过搜索堆栈交换（例如，here and here）知道我可以使用泊松二项分布函数找到 cdf、pmf 等。

我感兴趣的是每种可能的成功与失败组合的确切概率。（例如，如果我画了一棵概率树，我想知道每个分支结束时的概率。）

    all <- prod(p)
    all
    [1] 0.000672
    o1 <- (0.5 * (1-0.12) * 0.7 * 0.8 * .02)
    o1
    [1] 0.004928
    o2 <- (0.5 * 0.12 * (1-0.7) * 0.8 * .02)
    o2
    [1] 0.000288

...对于 success/failure.

的所有 2^5 种可能组合

在 R 中执行此操作的有效方法是什么？

就我的实际数据集而言，试验次数为 19，因此我们在概率树上讨论的路径总数为 2^19。

Answer 1

快速计算的关键是以对数概率 space 进行计算，以便树的每个分支的乘积是一个总和，可以计算为矩阵乘法的内部总和.以这种方式，所有分支都可以以矢量化的方式一起计算。

首先，我们构造一个所有分支的枚举。为此，我们使用 R.utils 包中的 intToBin 函数：

library(R.utils)
enum.branches <- unlist(strsplit(intToBin(seq_len(2^n)-1),split=""))

其中 n 是伯努利变量的数量。例如，n=5:

matrix(enum.branches, nrow=n)
##     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17]
##[1,] "0"  "0"  "0"  "0"  "0"  "0"  "0"  "0"  "0"  "0"   "0"   "0"   "0"   "0"   "0"   "0"   "1"  
##[2,] "0"  "0"  "0"  "0"  "0"  "0"  "0"  "0"  "1"  "1"   "1"   "1"   "1"   "1"   "1"   "1"   "0"  
##[3,] "0"  "0"  "0"  "0"  "1"  "1"  "1"  "1"  "0"  "0"   "0"   "0"   "1"   "1"   "1"   "1"   "0"  
##[4,] "0"  "0"  "1"  "1"  "0"  "0"  "1"  "1"  "0"  "0"   "1"   "1"   "0"   "0"   "1"   "1"   "0"  
##[5,] "0"  "1"  "0"  "1"  "0"  "1"  "0"  "1"  "0"  "1"   "0"   "1"   "0"   "1"   "0"   "1"   "0"  
##     [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26] [,27] [,28] [,29] [,30] [,31] [,32]
##[1,] "1"   "1"   "1"   "1"   "1"   "1"   "1"   "1"   "1"   "1"   "1"   "1"   "1"   "1"   "1"  
##[2,] "0"   "0"   "0"   "0"   "0"   "0"   "0"   "1"   "1"   "1"   "1"   "1"   "1"   "1"   "1"  
##[3,] "0"   "0"   "0"   "1"   "1"   "1"   "1"   "0"   "0"   "0"   "0"   "1"   "1"   "1"   "1"  
##[4,] "0"   "1"   "1"   "0"   "0"   "1"   "1"   "0"   "0"   "1"   "1"   "0"   "0"   "1"   "1"  
##[5,] "1"   "0"   "1"   "0"   "1"   "0"   "1"   "0"   "1"   "0"   "1"   "0"   "1"   "0"   "1"

生成一个矩阵，其中每一列都是概率树分支的结果。

现在，使用它来构造一个与 enum.branches 大小相同的对数概率矩阵，其中如果 enum.branches=="1" 则值为 log(p)，否则为 log(1-p)。对于您的数据，p <- c(0.5, 0.12, 0.7, 0.8, .02)，这是：

logp <- matrix(ifelse(enum.branches == "1", rep(log(p), 2^n), rep(log(1-p), 2^n)), nrow=n)

然后，对对数概率求和并取指数得到概率的乘积：

result <- exp(rep(1,n) %*% logp)
##         [,1]     [,2]     [,3]     [,4]     [,5]     [,6]     [,7]     [,8]     [,9]   [,10]
##[1,] 0.025872 0.000528 0.103488 0.002112 0.060368 0.001232 0.241472 0.004928 0.003528 7.2e-05
        [,11]    [,12]    [,13]    [,14]    [,15]    [,16]    [,17]    [,18]    [,19]    [,20]
##[1,] 0.014112 0.000288 0.008232 0.000168 0.032928 0.000672 0.025872 0.000528 0.103488 0.002112
        [,21]    [,22]    [,23]    [,24]    [,25]   [,26]    [,27]    [,28]    [,29]    [,30]
##[1,] 0.060368 0.001232 0.241472 0.004928 0.003528 7.2e-05 0.014112 0.000288 0.008232 0.000168
        [,31]    [,32]
##[1,] 0.032928 0.000672

result 将与 enum.branches 中的分支编号顺序相同。

我们可以把计算封装成一个函数：

enum.prob.product <- function(n, p) {
  enum.branches <- unlist(strsplit(intToBin(seq_len(2^n)-1),split=""))
  exp(rep(1,n) %*% matrix(ifelse(enum.branches == "1", rep(log(p), 2^n), rep(log(1-p), 2^n)), nrow=n))
}

用 19 个独立的伯努利变量来计时：

n <- 19
p <- runif(n)
system.time(enum.prob.product(n,p))
##   user  system elapsed 
## 24.064   1.470  26.082

这是在我的 2 GHz MacBook（大约 2009 年）上。应该注意的是计算本身是相当快的；它是概率树分支的枚举（我猜是其中的 unlist）占用了大部分时间。我们将不胜感激社区对其他方法的任何建议。

Answer 2

只需在 base R 中尝试这个：

p <- c(0.5, 0.12, 0.7, 0.8, .02)
a <- c("A","B","C","D","E")
n <- length(p)
apply(expand.grid(replicate(n,list(0:1)))[n:1], 1, 
                  function(x) prod(p[which(x==1)])*prod(1-p[which(x==0)]))

#[1] 0.025872 0.000528 0.103488 0.002112 0.060368 0.001232 0.241472 0.004928 0.003528 0.000072 0.014112 0.000288 0.008232 0.000168 0.032928 0.000672 0.025872
#[18] 0.000528 0.103488 0.002112 0.060368 0.001232 0.241472 0.004928 0.003528 0.000072 0.014112 0.000288 0.008232 0.000168 0.032928 0.000672

枚举一系列不同概率的伯努利试验的所有可能组合概率

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