Random.nextgaussian()可以从均值和标准差不同的分布中采样值吗? [英] Can Random.nextgaussian() sample values from a distribution with different mean and standard deviation?

问题描述

这是Java和基本数学问题的组合. Random.nextGaussian()的文档指出，它是从均值0和标准偏差为1的正态分布中抽样的.如果我想从均值和方差不同的正态分布中抽样，该怎么办?

This is a combined Java and basic math question. The documentation from Random.nextGaussian() states that it samples from a normal distribution with mean 0 and standard deviation 1. What if I wanted to sample from a normal distribution with a different mean and variance?

推荐答案

简短的答案是

Random r = new Random();
double mySample = r.nextGaussian()*desiredStandardDeviation+desiredMean;

例如，此答案在此处给出: http://www.javamex.com/tutorials/random_numbers/gaussian_distribution_2 .shtml

For example this answer is given here: http://www.javamex.com/tutorials/random_numbers/gaussian_distribution_2.shtml

我真的不明白为什么这样做，但是仔细研究一下后，我想我已经明白了.采样点的平均值为0，标准偏差为1；这意味着原始样本也是其自己的z得分( https://en.wikipedia.org/wiki/Standard_score).引自维基百科:"z的绝对值表示原始分数和总体平均值之间的距离，以标准差为单位".公式为z =(x-mean)/stdev，因此默认值为z = x.如果我们想保留样本的z分数，但更改均值和标准差，我们将怎么办?

I didn't really understand why this worked, but after looking into it a bit I think I figured it out. The mean of the sample point is 0, and the standard deviation is 1; that means that the original sample is also its own z-score ( https://en.wikipedia.org/wiki/Standard_score ). To quote from wikipedia "The absolute value of z represents the distance between the raw score and the population mean in units of the standard deviation". The formula is z=(x-mean)/stdev, so with the default values z=x. If we wanted to retain the z score for the sample but change the mean and stdev what would we do?

z * stdev +均值= x'，其中z = x，x'表示具有期望均值和标准差的分布样本.

z*stdev + mean = x' where z=x, and x' represents the sample from the distribution with the desired mean and standard deviation.

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