How to generate numbers from probability mass function?
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Hallo,
Given a probability mass function defined as P(X=3)=0.2, P(X=7)=0.3 and P(X=10)=0.5, I want to generate randomly 30 numbers (values for X) with this probability mass function as base. But I really have no idea how and where to start.
Can somebody help me?
Thank you in advance
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採用された回答
Torsten
2018 年 10 月 9 日
編集済み: Torsten
2018 年 10 月 9 日
n = 30;
X = zeros(n,1);
x = rand(n,1);
X(x <= 0.5) = 10;
X(x > 0.5 & x <= 0.8) = 7;
X(x > 0.8) = 3;
3 件のコメント
Torsten
2018 年 10 月 10 日
For an explanation, see
https://stats.stackexchange.com/questions/26858/how-to-generate-numbers-based-on-an-arbitrary-discrete-distribution
その他の回答 (3 件)
Bruno Luong
2018 年 10 月 9 日
A more generic method:
p = [0.2 0.3 0.5];
v = [3 7 10];
n = 10000;
c = cumsum([0,p(:).']);
c = c/c(end); % make sur the cumulative is 1
[~,i] = histc(rand(1,n),c);
r = v(i); % map to v values
Jeff Miller
2018 年 10 月 20 日
3 件のコメント
Jeff Miller
2018 年 10 月 20 日
Did you download the Cupid files (see the link in my answer)? These define the List class (which handles the cumulative distribution behind the scene). Do the other Cupid demos run correctly?
Well, Cupid may be overkill for your problem, but it does have a lot of flexibility.
PARTHEEBAN R
2021 年 5 月 22 日
A random variable X has cdf F(x) = { 0 , if x < − 1 a(1 + x ) , if − 1 < < 1 1 , if x ≥ 1 . Find (1) the value of a, (2) P(X > 1/4 ) and P ( − 0 . 5 ≤ X ≤ 0 ) .
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