Minimum of Two Exponential

Search…
Suppose we want to simulate pseudo random numbers from the following distribution, which I came across in r - Finding a way to simulate random numbers for this distribution - Cross Validated,
F(x)=1−exp⁡(−ax−bp+1xp+1) x≥0F(x) = 1-\exp\left(-ax - \frac{b}{p+1}x^{p+1}\right)\,\quad x\ge 0F(x)=1−exp(−ax−p+1b​xp+1)x≥0
where a,b>0a, b>0a,b>0
 and p∈(0,1)p\in (0,1)p∈(0,1)
.
​Xi'an provided a very elegant solution.
Note that
(1−F(x))=exp⁡{−ax−bp+1xp+1}=exp⁡{−ax}⏟1−F1(x)exp⁡{−bp+1xp+1}⏟1−F2(x)(1-F(x))=\exp\left\{-ax-\frac{b}{p+1}x^{p+1}\right\}=\underbrace{\exp\left\{-ax\right\}}_{1-F_1(x)}\underbrace{\exp\left\{-\frac{b}{p+1}x^{p+1}\right\}}_{1-F_2(x)}(1−F(x))=exp{−ax−p+1b​xp+1}=1−F1​(x)exp{−ax}​​1−F2​(x)exp{−p+1b​xp+1}​​
the distribution FFF
 is the distribution of
X=min⁡{X1,X2}X1∼F1 ,X2∼F2X=\min\{X_1,X_2\}\qquad X_1\sim F_1\,,X_2\sim F_2X=min{X1​,X2​}X1​∼F1​,X2​∼F2​
since
F(x)=P(X≤x)=1−P(X>x)=1−P(X1>x)P(X2>x)=1−(1−F1(X))(1−F2(X)) .F(x) = P(X\le x) = 1- P(X > x)  = 1-P(X_1 > x)P(X_2 > x) = 1-(1-F_1(X))(1-F_2(X))\,.F(x)=P(X≤x)=1−P(X>x)=1−P(X1​>x)P(X2​>x)=1−(1−F1​(X))(1−F2​(X)).
Thus, the R code to simulate is simple to be
1
x = pmin(rexp(n, a), rexp(n, b/(p+1))^(1/(p+1)))
Copied!
References
​Finding a way to simulate random numbers for this distribution​
​simulation fodder for future exams​
Copulas
Gibbs
Last modified 2yr ago
Copy link
Contents
References