Generate random number from a two-parameter exponential distribution
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Hello everybody,
I am looking for a way to easily generate random numbers from a two-parameter exponential distribution. The probability density function is: 
. I am aware of expand() but it does not seem to include the two-parameter distribution.
. I am aware of expand() but it does not seem to include the two-parameter distribution.Do you have any idea how I could achieve that?
Thank you very much.
Alexis
4 个评论
Torsten
2019-3-21
Your function f does not depend on t ?
Alexis Stevenson
2019-3-21
Torsten
2019-3-21
What is the support of f ?
I ask because if support(f) = IR+, f is only a probability density function if gamma = 0.
Alexis Stevenson
2019-3-21
采纳的回答
Torsten
2019-3-21
0 个投票
gam = ...;
lambda = ...;
n = 100;
x = gam - log(1-rand(n,1))/lambda
give you 100 random numbers distributed according to your two-parameter distribution.
1 个评论
Alexis Stevenson
2019-3-21
编辑：Alexis Stevenson
2019-3-21
Oh yes thank you so much !
I just realized the cdf was 
and I had to invert it.
and I had to invert it. 更多回答（0 个）
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