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Custom Uniform Random Distribution

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matlabkid602 on 23 Feb 2018
Answered: Roger Stafford on 24 Feb 2018
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.
  1 Comment
John BG
John BG on 23 Feb 2018
Hi Matlabkid602
the main constraint is that whatever probability density function you choose, it must have area 1, integrating from -Inf to +Inf.
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
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.
The values of the interval [a b] are related in the following way:

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Answers (2)

Image Analyst
Image Analyst on 23 Feb 2018

Roger Stafford
Roger Stafford on 24 Feb 2018
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.

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