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

Risposta accettata

Torsten
Torsten il 9 Ott 2018
Modificato: Torsten il 9 Ott 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 Commenti
Torsten
Torsten il 10 Ott 2018
For an explanation, see
https://stats.stackexchange.com/questions/26858/how-to-generate-numbers-based-on-an-arbitrary-discrete-distribution

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Più risposte (3)

Bruno Luong
Bruno Luong il 9 Ott 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
  1 Commento
Clarisha Nijman
Clarisha Nijman il 19 Ott 2018
This answer works for me the best. I need this to do random column sampling (sampling some columns of a very big matrix A)

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Jeff Miller
Jeff Miller il 20 Ott 2018

With Cupid you could write:

v = [3 7 10];       % the values
p = [0.2 0.3 0.5];  % their probabilities
rv = List(v,p);     % a random variable with those values & probabilities
n = 10000;
randoms = rv.Random(n,1);  % generate n random values of the random variable
  3 Commenti
Jeff Miller
Jeff Miller il 20 Ott 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.

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PARTHEEBAN R
PARTHEEBAN R il 22 Mag 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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