使随机数趋向于特定值的平均值 [英] Make Random Numbers Tend / Average to a Specific Value

问题描述

我如何生成一个0和1之间的随机数列表，但它们的平均值为0.8？

How can I for example generate a list of random numbers between 0 and 1, but have them avarage at 0.8?

我在C ++中写了这个小脚本，它会告诉你输出的是什么数字。这个问题不是真正的C ++相关的。

I have written this little script in C++ that'll tell you what numbers got output. This question is not really C++ related though.

#include <iostream>
#include <random>
#include <time.h>
int main(int argCount, char** argVector) {
    std::cout << "Generating Randoms" << std::endl;

    float avarage = 0.F;
    srand(rand() + (int) time(NULL));
    float ceiling = 0;
    float bottom = 1;
    for(unsigned int i = 0; i < 1000000; i++) {
        float random = (float) (rand() % 101) / 100;
        if(random > ceiling)
            ceiling = random;
        else if(random < bottom)
            bottom = random;
        avarage += random;
    }
    std::cout << "Avarage: " << avarage/1000000 << std::endl;
    std::cout << "Ceiling: " << ceiling << std::endl;
    std::cout << "Bottom: " << bottom << std::endl;
    return 0;
}

此输出：

This outputs:

Generating Randoms
Avarage: 0.499287
Ceiling: 1
Bottom: 0

我希望天花板和底部仍然是0和1，但能够改变平均值。该算法应当优选地也是有效的。

I would like the ceiling and bottom to be still 0 and 1, but be able to change the average. The algorithm should preferably be efficient too.

再次，我现在发布C ++代码，但任何语言都可以。

Once again, I'm now posting C++ code, but any language will do.

推荐答案

NolanPower有一个伟大的想法使用权力，但他建议选择电源的机制是关闭。如果随机数 U 是均匀的（0,1）无意识统计学家的法则说，我们可以得到任何函数的期望值 g（U） as Integral [g从：0到：1] 。如果我们的函数 g（U）是一个多项式，即 U ** c c> c ，计算积分产生一般解 1 /（c + 1）作为期望值。设置这等于所需的平均值 m 并求解，我们得到 c =（1 / m） - 1 。

NolanPower had a great idea using powers, but the mechanism he recommended for choosing the power is off. If the random numbers U are uniform(0,1) the law of the unconscious statistician says we can derive the expected value of any function g(U) as Integral[g(U) from: 0 to: 1]. If our function g(U) is a polynomial, i.e., U**c for some constant c, evaluating the integral yields the general solution 1 / (c + 1) as the expected value. Setting this equal to the desired mean m and solving, we get that c = (1 / m) - 1.

为了得到预期值0.8， c =（1 / 0.8） - 1 = 0.25 结束 U ** 0.25 。为了得到预期值0.2， c =（1 / 0.2） - 1 = 4 ，即使用 U ** 4 。

To get an expected value of 0.8, c = (1 / 0.8) - 1 = 0.25, i.e., crank out U**0.25. To get an expected value of 0.2, c = (1 / 0.2) - 1 = 4, i.e., generate values using U**4.

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