Pseudorandom Number Generator - Exponential Distribution

Since you have access to a uniform random number generator, generating a random number distributed with other distribution whose CDF you know is easy using the inversion method.

So, generate a uniform random number, u, in [0,1), then calculate x by:

x = log(1-u)/(-λ),

where λ is the rate parameter of the exponential distribution. Now, x is a random number with an exponential distribution. Note that log above is ln, the natural logarithm.

The Fundamental Theorem of Sampling holds that if you can normalize, integrate and invert the desired distribution you are home free.

If you have a desired distribution F(x) normalized on [a,b]. You compute

C(y) = \int_a^y F(x) dx

invert that to get C^{-1}, throw z uniformly on [0,1) and find

x_i = C^{-1}(z_i)

which will have the desired distribution.

In your case: F(x) = ke^{-kx} and I will assume that you want [0,infinity]. We get :

C(y) = 1 - e^{-ky}

which is invertable to give

x = -1/k  ln(1 - z)

for z thrown uniformly on [0,1).

But, frankly, using a well debugged library is smarter unless you're doing this for your own edification.

This is the formula I found on Wikipedia :

T = -Ln(u) / λ

We create a random number with a uniform distribution (u) in [0,1]and we get x :

Random R = new Random();

double u = R. NextDouble();

double x = -Math.Log(u)/(λ);