I am trying to write a program in R that simulates pseudo random numbers from a distribution with the cumulative distribution function:

$$F(x)= 1-\exp \left(-ax-\frac{b}{p+1}x^{p+1}\right), \quad x \geq 0$$

where $a,b>0, p \in (0,1)$

I tried inverse transform sampling but the inverse does not seem to be analytically solvable. I would be glad if you could suggest a solution to this problem

  • 1
    $\begingroup$ Not enough time for a complete answer, but you can check algorithms of Importance Sampling, as an alternative. $\endgroup$
    – chuse
    Commented Feb 6, 2019 at 12:28
  • 1
    $\begingroup$ it is not a textbook exercise, I only stipulated the constraint because it is a reasonable assumption for my data $\endgroup$
    – Sebastian
    Commented Feb 6, 2019 at 16:00
  • 6
    $\begingroup$ I am then surprised at the "miraculous" normalisation by $(p+1)^{-1}$ that turns the distribution into a perfect power of an Exponential, but miracles do happen (with small probability). $\endgroup$
    – Xi'an
    Commented Feb 6, 2019 at 20:48

3 Answers 3


There is a straightforward (and if I may add, elegant) solution to this exercise: since $1-F(x)$ appears like a product of two survival distributions: $$(1-F(x))=\exp\left\{-ax-\frac{b}{p+1}x^{p+1}\right\}=\underbrace{\exp\left\{-ax\right\}}_{1-F_1(x)}\underbrace{\exp\left\{-\frac{b}{p+1}x^{p+1}\right\}}_{1-F_2(x)}$$ the distribution $F$ is the distribution of $$X=\min\{X_1,X_2\}\qquad X_1\sim F_1\,,X_2\sim F_2$$ In this case $F_1$ is the Exponential $\mathcal{E}(a)$ distribution and $F_2$ is the $1/(p+1)$-th power of an Exponential $\mathcal{E}(b/(p+1))$ distribution.

The associated R code is as simple as it gets

x=pmin(rexp(n,a),rexp(n,b/(p+1))^(1/(p+1))) #simulating an n-sample

and it is definitely much faster than the inverse pdf and accept-reject resolutions:

> n=1e6
> system.time(results <- Vectorize(simulate,"prob")(runif(n)))
utilisateur     système      écoulé 
    89.060       0.072      89.124 
> system.time(x <- simuF(n,1,2,3))
utilisateur     système      écoulé 
     1.080       0.020       1.103 
> system.time(x <- pmin(rexp(n,a),rexp(n,b/(p+1))^(1/(p+1))))
utilisateur     système      écoulé 
     0.160       0.000       0.163 

with an unsurprisingly perfect fit:

enter image description here

  • 5
    $\begingroup$ really cool solution! $\endgroup$
    – Sebastian
    Commented Feb 6, 2019 at 13:34

You can always numerically solve the inverse transformation.

Below, I do a very simple bisection search. For a given input probability $q$ (I use $q$ since you already have a $p$ in your formula), I start with $x_L=0$ and $x_R=1$. Then I double $x_R$ until $F(x_R)>q$. Finally, I iteratively bisect the interval $[x_L,x_R]$ until its length is shorter than $\epsilon$ and its middle point $x_M$ satisfies $F(x_M)\approx q$.

The ECDF fits your $F$ well enough for my choices of $a$ and $b$, and it's reasonably fast. You could probably speed this up by using some Newton-type optimization instead of the simple bisection search.

aa <- 2
bb <- 1
pp <- 0.1

cdf <- function(x) 1-exp(-aa*x-bb*x^(pp+1)/(pp+1))

simulate <- function(prob,epsilon=1e-5) {
    left <- 0
    right <- 1
    while ( cdf(right) < prob ) right <- 2*right

    while ( right-left>epsilon ) {
        middle <- mean(c(left,right))
        value_middle <- cdf(middle)
        if ( value_middle < prob ) left <- middle else right <- middle


results <- Vectorize(simulate,"prob")(runif(10000))

xx <- seq(0,max(results),by=.01)



There is a somewhat convoluted if direct resolution by accept-reject. First, a simple differentiation shows that the pdf of the distribution is $$f(x)=(a+bx^p)\exp\left\{-ax-\frac{b}{p+1}x^{p+1}\right\}$$ Second, since $$f(x)=ae^{-ax}\underbrace{e^{-bx^{p+1}/(p+1)}}_{\le 1}+bx^pe^{-bx^{p+1}/(p+1)}\underbrace{e^{-ax}}_{\le 1}$$ we have the upper bound $$f(x)\le g(x)=ae^{-ax}+bx^pe^{-bx^{p+1}/(p+1)}$$ Third, considering the second term in $g$, take the change of variable $\xi=x^{p+1}$, i.e., $x=\xi^{1/(p+1)}$. Then$$\dfrac{\text{d}x}{\text{d}\xi}=\dfrac{1}{p+1}\xi^{\frac{1}{p+1}-1}=\dfrac{1}{p+1}\xi^{\frac{-p}{p+1}}$$ is the Jacobian of the change of variable. If $X$ has a density of the form $\kappa bx^pe^{-bx^{p+1}/(p+1)}$ where $\kappa$ is the normalising constant, then $\Xi=X^{1/(p+1)}$ has the density $$\kappa b\xi^{\frac{p}{p+1}}e^{-b\xi/(p+1)}\,\dfrac{1}{p+1}\xi^{\frac{-p}{p+1}}=\kappa \dfrac{b}{p+1}e^{-b\xi/(p+1)}$$ which means that (i) $\Xi$ is distributed as an Exponential $\mathcal{E}(b/(p+1))$ variate and (ii) the constant $\kappa$ is equal to one. Therefore, $g(x)$ ends up being equal to the equally weighted mixture of an Exponential $\mathcal{E}(a)$ distribution and the $1/(p+1)$-th power of an Exponential $\mathcal{E}(b/(p+1))$ distribution, modulo a missing multiplicative constant of $2$ to account for the weights: $$f(x)\le g(x)=2\left(\frac{1}{2} ae^{-ax}+\frac{1}{2} bx^pe^{-bx^{p+1}/(p+1)}\right)$$ And $g$ is straightforward to simulate as a mixture.

An R rendering of the accept-reject algorithm is thus

simuF <- function(a,b,p){
  while (reepeat){
   if (runif(1)<.5) x=rexp(1,a) else

and for an n-sample:

simuF <- function(n,a,b,p){
  while (length(sampl)<n){

Here is an illustration for a=1, b=2, p=3:

enter image description here


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