How to generate random numbers with exponential di

I am trying to generate exponentially distributed random number with mean equal to 1. I know how to get random number for normal distribution with mean and standard deviation. We can get it by normal(mean, standard_deviation), but I don't know how to get random number for exponential distribution.

Can anyone help me with this?

回答1:

With C++11 the standard actually guarantees that there is a RNG following the requirements of exponential-distribution available in the STL, and fittingly the object-type has a very descriptive name.

The mean in an exponentially distributed random generator is calculated by the formula E[X] = 1 / lambda¹.

std::exponential_distribution has a constructor taking lambda as an argument, so we can easily create an object following your rules by calculating the value of lambda and passing this to our generator.

std::exponential_distribution rng (1/1); // lambda = 1 / E[X]

^Footnotes
^{^1. according to en.wikipedia.org - Exponential distribution > Mean, variance, moments and median}

Distribution as readable ascii chart

#include <iomanip>
#include <random>
#include <map>
#include <iostream>

int
main (int argc, char *argv[])
{
  double const exp_dist_mean   = 1;
  double const exp_dist_lambda = 1 / exp_dist_mean;

  std::random_device rd; 

  std::exponential_distribution<> rng (exp_dist_lambda);
  std::mt19937 rnd_gen (rd ());

  /* ... */

  std::map<int, int> result_set;

  for (int i =0; i < 100000; ++i)
    ++result_set[rng (rnd_gen) * 4]; 

  for (auto& v : result_set) {
    std::cout << std::setprecision (2) << std::fixed;

    std::cout << v.first/4.f << " - " << (v.first+1)/4.f << " -> ";
    std::cout << std::string (v.second/400, '.') << std::endl;

    if (v.second/400 == 0)
      break;
  }
}

0.00 - 0.25 -> ........................................................
0.25 - 0.50 -> ...........................................
0.50 - 0.75 -> .................................
0.75 - 1.00 -> .........................
1.00 - 1.25 -> ....................
1.25 - 1.50 -> ...............
1.50 - 1.75 -> ............
1.75 - 2.00 -> .........
2.00 - 2.25 -> .......
2.25 - 2.50 -> .....
2.50 - 2.75 -> ....
2.75 - 3.00 -> ...
3.00 - 3.25 -> ..
3.25 - 3.50 -> ..
3.50 - 3.75 -> .
3.75 - 4.00 -> .
4.00 - 4.25 -> .
4.25 - 4.50 ->

回答2:

Generating an exponential distribution random variable can be done by:

-ln(U)/lambda (where U~Uniform(0,1)).

More information can be found in this wikipedia article

In exponential distribution: lamda = 1/mean, so it gets you:

myVar = -ln(U) * mean (where U~Uniform(0,1)).