C:随机游走(布朗运动)程序没有返回预期的理论值。为什么？

Question

这只是基于《费曼物理讲义》第6-3节的实验:

In its simplest version, we imagine a “game” in which a “player” starts at the point x=0 and at each “move” is required to take a step either forward (toward +x) or backward (toward −x). The choice is to be made randomly, determined, for example, by the toss of a coin.

来源:http://www.feynmanlectures.caltech.edu/I_06.html#Ch6-S3

我的目标是计算距起始点的预期距离。所以，我假设每一步等于一个距离单位。我写了一个简单的 C 程序来模拟 30 个随机步骤，然后计算离起点的最终距离。重复一百万次，程序对距离进行平均以获得预期距离。

理论上，期望距离应该是步数的平方根。那应该是大约 sqrt(30) = 5.48。

但是，该程序运行了几次并不断返回接近 4.33 的值(更准确地说，是 4.33461、4.33453 和 4.34045)。为什么它甚至不接近 5.48 左右的理论值？

这是我的代码:

#include    <time.h>
#include    <stdlib.h>
#include    <stdio.h>

int main ( int argc, char *argv[] )
{

  int number_of_steps = 30;
  int repetition = 1000000;
  int distance = 0;
  int total_distance = 0;
  double expected_distance;
  int i, j;

  srand(time(NULL));

  for ( i = 0; i < repetition; i++ ) {

    for ( j = 0; j < number_of_steps; j++) {
      distance += rand() & 1 ? -1 : 1;
    }

    total_distance += abs(distance);
    distance = 0;

  }

  expected_distance = (float) total_distance / i;

  printf ( "%g\n", expected_distance );
  return EXIT_SUCCESS;
}       /* ----------  end of function main  ---------- */

Answer 1

根据您链接的讲座，您的理论期望基于 root mean square ，这不同于 arithmetic mean ，这是您编码的内容。通过将算法从一种更改为另一种，the code现在给你预期的结果。

for ( i = 0; i < repetition; i++ ) {

    for ( j = 0; j < number_of_steps; j++) {
      distance += rand() & 1 ? -1 : 1;
    }

    total_distance += distance * distance;
    distance = 0;

  }

  expected_distance = sqrt((float) total_distance / repetition);

  printf ( "%g\n", expected_distance );
  return EXIT_SUCCESS;
}