Generation of random numbers
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How can I generate uniformly distributed random numbers in the range 10^(-4) to 10^(+4)? I tried using rand function but it does not give me the required range. This is what I tried:
dd = (10^(4) - 10^(-4)).*rand(1000,1) + 10^(-4); plot(dd);
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Accepted Answer
Mahdi
on 27 May 2014
Edited: Mahdi
on 27 May 2014
I think what you're trying to do is done using randn, so that
dd=(10^(4) - 10^(-4)).*randn(1000,1) + 10^(-4)
hist(dd,1000) # To see it in a histogram with 1000 bins
Note the range of x-axis
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John D'Errico
on 27 May 2014
NO. randn is NOT the solution here, since it generates numbers that can fall ANYWHERE in theory.
More Answers (3)
Shashank Prasanna
on 27 May 2014
That sounds about right. Take a look at the first example in the documentation:
Roger Wohlwend
on 27 May 2014
You use the correct method. When your range is from 10^(-4) to 10^4 it is quite normal that values below 10^(-1) are very, rare.
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John D'Errico
on 27 May 2014
I think the problem is that while you SAY you want numbers that are uniformly distributed over that interval, you don't really WANT them uniformly distributed.
The probability is actually quite small that for an interval like that, that you will see numbers less than say 1. In fact, suppose we did generate a uniform randm number in the interval [1e-4,1e4]?
What is the probability that any given such deviate will be in the sub-interval [1e-4,1]? This is easy to compute.
(1 - 1e-4)/(1e4 - 1e-4)
ans =
9.999e-05
Essentially, you need to have many such samples before you ever see a SINGLE such deviate in the sub-interval.
So while you SAY you want uniform, I think you really want numbers that are uniformly distributed in a log domain. That is, the logs of your deviates will be uniformly distributed.
% these numbers will be uniform over the interval [-4,4]
E = 8*rand(1,10000) - 4;
% The numbers in R will have the distribution you desire
R = 10.^E;
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