如何根据拒绝标准从分布中生成目标样本数 [英] How to generate target number of samples from a distribution under a rejection criterion
问题描述
我正在尝试 rnbinom
x<- rnbinom(500, mu = 4, size = .1)
xtrunc <- x[x>0]
然后我只得到125个观测值。
then I just get 125 observations.
但是，我要进行500个观测值，但排除0（零）且条件相同（ mu = 4，size = .1 ）。
However, I want to make 500 observations excluding 0 (zero) with same condition (mu = 4, size =.1).
推荐答案
这可以完成工作：
N <- 500 ## target number of samples
## set seed for reproducibility
set.seed(0)
## first try
x <- rnbinom(N, mu = 4, size = .1)
p_accept <- mean(success <- x > 0) ## estimated probability of accepting samples
xtrunc <- x[success]
## estimated further sampling times
n_further <- (N - length(xtrunc)) / p_accept
## further sampling
alpha <- 1.5 ## inflation factor
x_further <- rnbinom(round(alpha * n_further), mu = 4, size = .1)
## filter and combine
xtrunc <- c(xtrunc, (x_further[x_further > 0])[seq_len(N - length(xtrunc))])
## checking
length(xtrunc)
# [1] 500
summary(xtrunc)
# Min. 1st Qu. Median Mean 3rd Qu. Max.
# 1.00 2.00 5.00 12.99 16.00 131.00
在上面，采样分为两个阶段。初始阶段的结果用于估计接受率的概率，以指导第二阶段的采样。
In above, sampling takes two stages. The result of the initial stage is used to estimate probability of acceptance rate to guide the second stage sampling.
但是，由于基本分布是已知的，因此理论上的概率接受率是已知的。因此，在这种情况下，无需执行两阶段方法。尝试：
However, since the underlying distribution is explicitly known, the theoretical probability of acceptance rate is know. Therefore, there is no need to perform a two-stage approach in this case. Try:
p <- 1 - pnbinom(0, mu = 4, size = .1) ## theoretical probability
alpha <- 1.5
n_try <- round(alpha * N / p)
set.seed(0)
x <- rnbinom(n_try, mu = 4, size = .1)
xtrunc <- (x[x > 0])[1:N]
## checking
length(xtrunc)
# [1] 500
summary(xtrunc)
# Min. 1st Qu. Median Mean 3rd Qu. Max.
# 1.00 2.00 5.00 12.99 16.00 131.00
背后的想法是几何分布理论。 我的回答与此密切相关。阅读更有效的矢量化方法部分以获取详细说明。
The idea behind is the theory of geometric distribution. My answer here is closely related. Read the "More efficient vectorized method" section for detailed explanation.
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