Remanding integer from exponential distribution with mean
1 view (last 30 days)
Show older comments
Hi, I found out exprnd() can generate random number from exponential distribution function with mean. Could anyone suggest how to generate integers instead of numbers from exponential distribution with mean? Thanks!
2 Comments
Walter Roberson
on 1 Jan 2023
Is the mean integer? Is the random selection required to have exactly that mean or is it statistical? Exactly that mean is equivalent to fixed sum which is the direction DGM took it.
Accepted Answer
DGM
on 1 Jan 2023
Edited: DGM
on 1 Jan 2023
If I recall, there are other ways to do this.
MIMT has a tool for generating random integers with sum and range constraints. The above example is more memory-intensive than the method used in randisum().
N = 1000; % number of values to generate
av = 20; % mean of all values
R = randisum([0 NaN],av*N,[N 1],'exponential');
histogram(R,'binmethod','integers')
mean(R) % the mean is as specified
nnz(mod(R,1)) % all values are integers
Note that the constraint here is actually the sum, not the mean. While av can be a non-integer, the value av*N must be an integer. This also means that (unlike exprnd()), av is the sample mean, not the mean of the distribution. I don't know if that's sufficient for your needs.
0 Comments
More Answers (0)
See Also
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!