Remanding integer from exponential distribution with mean
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
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 Kommentare
Walter Roberson
am 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.
Akzeptierte Antwort
DGM
am 1 Jan. 2023
Bearbeitet: DGM
am 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 Kommentare
Weitere Antworten (0)
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!