Custom Uniform Random Distribution

Say I have a uniform random distribution in matlab using the rand() function.

How can I change the distribution for a given function such as 1/sqrt(x), so the distribution follows this curve?

I've tried to reject values but had no luck.

[John BG]

Hi Matlabkid602

this is John BG <mailto:[e-mail] [e-mail]>

1.

the main constraint is that whatever probability density function you choose, it must have area 1, integrating from -Inf to +Inf.

2.

Since you choose 1/sqrt(x), you have to truncate, let's say with a continuous interval [a b], a<b.

The new pdf should have the shape

pdf2(x<a)=0

pdf2(x>b)=0

pfg2(0<x<b)=1/sqrt(x)

3. make sure that for instance 0<b<a, otherwise and undesired discontinuity will be included in the function intended to behave as pdf and it will not work.

4.

The values of the interval [a b] are related in the following way:

sqrt(b)-sqrt(a)=0.5

Answer 1

[Image Analyst]

See Walter's recent answer: https://www.mathworks.com/matlabcentral/answers/384302-how-to-generate-a-random-sequence-with-some-specific-probability#answer_306604

Answer 2

[Roger Stafford]

If you want to obtain a density distribution proportional to 1/sqrt(x) for x in some finite interval [a,b], you can proceed as follows. The cumulative probability function must be:

cdf(x) = (sqrt(x)-sqrt(a))/(sqrt(b)-sqrt(a))

which you get by integrating 1/sqrt(x) from a to x and adjusting the proportionality constant to get a cdf(b)=1 for the entire interval from a to b.

To generate this using rand, set the above cdf(x) to r = rand and solve for x:

r = rand;
%Solve for x in (sqrt(x)-sqrt(a))/(sqrt(b)-sqrt(a)) = r:
x = (sqrt(a)+(sqrt(b)-sqrt(a))*r).^2;

This illustrates how one would proceed to use rand for generating any given probability density function. It depends on being able to solve for x in an equation cdf(x) = r.