Generate random doubles that follow a specific distribution - c

Question

So I have the distribution f(x) = (4-2x)/3, for x in [0,1], and f(x) = 0 elsewhere.
I now want to generate, say n = 100 random numbers from this distribution. I tried to follow this example, and this is my own code:
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <math.h>
#include <stdlib.h>
//pdf(x) = 1 if x>1
// = 0 if x<0
// = 1/3*(4-2x) otherwise
int N = 10;
int var1(int argc, char **argv) {
int p = 0, i;
for (i = 0; i < N; i++) {
p = (double)(rand() % 100)/100; // Generates 100 numbers in [0,1]
if (p > 1)
printf("%d ", 0);
else if (p < 0)
printf("%d ", 0);
else
printf("%f ", p * (4-2*0.1)/3);
}
printf("...");
return 0;
}
However my output is just zeroes:
0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 …
I have two questions:
What am I missing?
In the link I provided, why has he replaced x by 0.1? I just did the same thing but I don't understand why.

Answer 1

First, you're not returning the range of random numbers you think you are:
p = (double)(rand() % 100)/100;
The right hand size will give you a number in the range ( 0.00, 0.01, 0.02 ... 0.99 ). But then you assign that value to p, which is of type int. The fractional part is truncated, so the assigned value will always be 0.
Change the type of p to double to store the proper values. Also, if you want a wider range of random values, use this instead:
p = (double)rand() / RAND_MAX;
As for this:
printf("%f ", p * (4-2*0.1)/3);
Compared to the linked code:
printf("%f ", p * 0.1 / 360);
Their code isn't putting 0.1 in place of x. p is the same as x, but for some reason they're adding the extra 0.1 factor.
For you to be doing f(x) = (4-2x)/3, you need this:
printf("%f ", (4-2*p)/3);

Answer 2

Preamble
The function in the question, f(x) = (4-2x)/3 for x in [0, 1], is a probability density function, not a cumulative distribution function. If the distribution were drawn as a curve with unit area between the x axis and the curve, the probability density function is the slope of that curve, while the cumulative distribution function is the amount of area under the curve up to a certain x.
We can see (4-2x)/3 is not a cumulative distribution function because the cumulative distribution must be 1 at the end of the distribution (∞ in general, but 1 here since the function is zero beyond 1, but (4-2x)/3 is 2/3 at 1. We can see it is a probability density function because its integral (shown below) is 1 at the end of the distribution.
The link given in the question uses a cumulative distribution function. In fact, it is a cumulative distribution function for a uniform distribution over [0, 360]. Because of that, solving the necessary equation (see below) is merely a matter of scaling. This is trivial and does not serve as a general example for generating samples for an arbitrary distribution.
Solution
Given a probability density function, f(x), the corresponding cumulative distribution function, F(x), is the integral of f from −∞ to x. We can use the cumulative distribution function to convert a uniform distribution to the desired distribution.
The integral of (4-2x)/3 is (4x−x2)/3 (plus a constant). Since this is zero at x=0 and f(x) is zero for x < 0, F(x) is also (4x−x2)/3 in [0, 1].
If we have a sample p from a uniform distribution in [0, 1], the point x where the amount of the desired distribution at or below x equals the amount of the uniform distribution at or below p satisfies F(x) = p.
Thus (4x−x2)/3 = p, so x = 2 + sqrt(4−3p).
Thus, given p generated with p = (double) rand() / RAND_MAX, we can find the sample for the desired distribution as 2 + sqrt(4-3*p).