Remanding integer from exponential distribution with mean
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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!
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Walter Roberson
2023-1-1
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.
HYZ
2023-1-1
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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.
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