如何从 R 中向左移动的卡方分布生成随机数? [英] how to generate random numbers from a shifted to the left chi square distribution in R?

问题描述

我想从具有 3 个自由度但向左移动的卡方分布生成随机数.我的意思是移位分布函数 f(x-a) a 是移位量.

I want to generate random numbers from chi-square distribution with 3 degrees of freedom but shifted to the left . I mean the shifted distribution function f(x-a) a is the amount of shifting.

在 r 中表示非中心性参数必须为非负.

in r it is said the non centrality parameter must be non negative.

推荐答案

让我们看一下具有 3 个自由度的卡方分布:

Let's look at the Chi-square distribution with 3 degrees of freedom:

x_vals <- seq(0, 10, 0.1)

plot(x_vals, dchisq(x_vals, 3), type = "l", 
     main = "Chi Squared distribution of x with 3 DOF")

现在让我们将它向左移动一个常量 a.我们将在 x = 0 处绘制一条垂直线以强调偏移:

Now let's shift it to the left by a constant a. We'll plot a vertical line at x = 0 to emphasize the shift:

a <- 2
plot(x_vals - a, dchisq(x_vals, 3), type = "l",
     main = "Chi Squared distribution of x - 2 with 3 DOF")
abline(v = 0, lty = 2)

这是您希望从中抽样的分布.既然如此，我们只需要从卡方分布中采样，然后从每个绘制的元素中减去 a.在 R 中，这就像执行 rchisq(n, 3) - a 一样简单，其中 n 是所需的样本大小.

This is the distribution from which you wish to sample. That being the case, we need only sample from the Chi-square distribution and subtract a from each element drawn. In R this is as easy as doing rchisq(n, 3) - a where n is the desired sample size.

为了演示，以下是从该分布中抽取的 10,000 个样本的直方图:

To demonstrate, here is a histogram of 10,000 samples drawn from this distribution:

hist(rchisq(10000, 3) - a, breaks = 100, xlim = c(-2, 8), 
     main = "10,000 samples from Chi Square distribution of (x - 2) with 3 DOF")

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