How to generate numbers from probability mass function?
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Clarisha Nijman
el 9 de Oct. de 2018
Respondida: PARTHEEBAN R
el 22 de Mayo de 2021
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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Respuesta aceptada
Torsten
el 9 de Oct. de 2018
Editada: Torsten
el 9 de Oct. de 2018
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 comentarios
Torsten
el 10 de Oct. de 2018
For an explanation, see
https://stats.stackexchange.com/questions/26858/how-to-generate-numbers-based-on-an-arbitrary-discrete-distribution
Más respuestas (3)
Bruno Luong
el 9 de Oct. de 2018
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
el 20 de Oct. de 2018
3 comentarios
Jeff Miller
el 20 de Oct. de 2018
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
el 22 de Mayo de 2021
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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