在 R 中评估非常大的阶乘

Question

我需要集成一个在表达式中有阶乘的函数。但是，如果您尝试评估阶乘，当 n > 170 时，R returns Inf.

我发现有很多包可以让您计算非常大的数字，但是，它们总是 returns 来自 class 的对象，我无法集成。积分的最终结果总是一个小数。

这是我的代码：

integrand <- function(n, i, x) {
    (factorial(n) / (factorial(i - 1) * factorial(n - i))) *
        x^(i - 1) * (1 - x)^(n - i)
}

forder <- function(Fx, x, n, i, ...) {
    lower <- sapply(x - 1, Fx, ...)
    upper <- sapply(x, Fx, ...)
     integrate(integrand,
               lower = lower, upper = upper, n = n, i = i,
               stop.on.error = FALSE)$value
}
forder <- Vectorize(forder, "x")

##------------------------------------------------------------------------------
## Some example
y <- sort(rpois(100, 1))

## Works fine
forder(ppois, y, 170, 10, lambda = 1)

## Does not work
forder(ppois, y, 171, 10, lambda = 1)
##------------------------------------------------------------------------------

Answer 1

将 integrand 更改为使用对数，这两个调用都有效。我还 post 函数 @StéphaneLaurent 关于使用 choose/lchoose 和 pbeta/beta 函数的想法。

integrand <- function(n, i, x) {
  y <- lfactorial(n) - lfactorial(i - 1) - lfactorial(n - i) +
    (i - 1)*log(x) + log(1 - x)*(n - i)
  exp(y)
}

forder <- function(Fx, x, n, i, ...) {
  lower <- sapply(x - 1, Fx, ...)
  upper <- sapply(x, Fx, ...)
  integrate(integrand,
            lower = lower, upper = upper, n = n, i = i,
            stop.on.error = FALSE)$value
}
forder <- Vectorize(forder, "x")

integrandSL <- function(n, i, x) {
  y <- log(i) + lchoose(n, i) + (i - 1)*log(x) + log(1 - x)*(n - i)
  exp(y)
}

forderSL <- function(Fx, x, n, i, ...) {
  lower <- sapply(x - 1, Fx, ...)
  upper <- sapply(x, Fx, ...)
  integrate(integrandSL,
            lower = lower, upper = upper, n = n, i = i,
            stop.on.error = FALSE)$value
}
forderSL <- Vectorize(forderSL, "x")

forderSL2 <- function(Fx, x, n, i, ...) {
  lower <- sapply(x - 1, Fx, ...)
  upper <- sapply(x, Fx, ...)
  lg <- log(i) + lchoose(n, i) + 
    log(pbeta(upper, i, n-i+1) - pbeta(lower, i, n-i+1)) + lbeta(i, n-i+1)
  exp(lg)
}

现在是测试。所有结果都是all.equal.

##-------------------------------------------------
set.seed(1234)
y <- sort(rpois(100, 1))

res_170 <- forder(ppois, y, 170, 10, lambda = 1)
res_171 <- forder(ppois, y, 171, 10, lambda = 1)

resSL_170 <- forderSL(ppois, y, 170, 10, lambda = 1)
resSL_171 <- forderSL(ppois, y, 171, 10, lambda = 1)

resSL2_170 <- forderSL2(ppois, y, 170, 10, lambda = 1)
resSL2_171 <- forderSL2(ppois, y, 171, 10, lambda = 1)

all.equal(res_170, resSL_170)    # TRUE
all.equal(res_170, resSL2_170)   # TRUE
all.equal(res_171, resSL_171)    # TRUE
all.equal(res_171, resSL2_171)   # TRUE

Answer 2

正如我在评论中所说，您可以将 (factorial(n) / (factorial(i - 1) * factorial(n - i))) 替换为 i*choose(n, i)。这两个数量相等，但 choose(n,i) 允许更高的值 n。

或者您可以使用 pbeta 函数而不是进行数值积分：

forder <- function(Fx, x, n, i, ...) {
  lower <- sapply(x - 1, Fx, ...)
  upper <- sapply(x, Fx, ...)
  i*choose(n, i) * (pbeta(upper, i, n-i+1) - pbeta(lower, i, n-i+1)) * beta(i, n-i+1)
}

更好的是，使用对数：

forder <- function(Fx, x, n, i, ...) {
  lower <- sapply(x - 1, Fx, ...)
  upper <- sapply(x, Fx, ...)
  lg <- log(i) + lchoose(n, i) + 
    log(pbeta(upper, i, n-i+1) - pbeta(lower, i, n-i+1)) + lbeta(i, n-i+1)
  exp(lg)
}

编辑

我没有注意到这种简化：i*choose(n, i) * beta(i, n-i+1) = 1。所以你可以简单地做：

forder <- function(Fx, x, n, i, ...) {
  lower <- sapply(x - 1, Fx, ...)
  upper <- sapply(x, Fx, ...)
  pbeta(upper, i, n-i+1) - pbeta(lower, i, n-i+1)
}

在 R 中评估非常大的阶乘

Eval very large factorial in R

r

factorial

numerical-integration

largenumber

编辑