Remanding integer from exponential distribution with mean

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HYZ 2023-1-1

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https://ww2.mathworks.cn/matlabcentral/answers/1886897-remanding-integer-from-exponential-distribution-with-mean

评论： HYZ 2023-1-1

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

The mean is integer. The random selection isn't statistical.

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Answer 1

DGM 2023-1-1

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https://ww2.mathworks.cn/matlabcentral/answers/1886897-remanding-integer-from-exponential-distribution-with-mean#answer_1139212

编辑：DGM 2023-1-1

在 MATLAB Online 中打开

If I recall, there are other ways to do this.

https://www.mathworks.com/matlabcentral/answers/327656-conditional-random-number-generation#answer_257296

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

ans = 20

nnz(mod(R,1)) % all values are integers

ans = 0

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.