# Finding a way to simulate random numbers for this distribution

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

• Not enough time for a complete answer, but you can check algorithms of Importance Sampling, as an alternative. Feb 6, 2019 at 12:28
• it is not a textbook exercise, I only stipulated the constraint because it is a reasonable assumption for my data Feb 6, 2019 at 16:00
• 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). Feb 6, 2019 at 20:48

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: • really cool solution! 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
}

mean(c(left,right))
}

set.seed(1)
results <- Vectorize(simulate,"prob")(runif(10000))
hist(results)

xx <- seq(0,max(results),by=.01)
plot(ecdf(results))
lines(xx,cdf(xx),col="red") 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){
reepeat=TRUE
while (reepeat){
if (runif(1)<.5) x=rexp(1,a) else
x=rexp(1,b/(p+1))^(1/(p+1))
reepeat=(runif(1)>(a+b*x^p)*exp(-a*x-b*x^(p+1)/(p+1))/
(a*exp(-a*x)+b*x^p*exp(-b*x^(p+1)/(p+1))))}
return(x)}


and for an n-sample:

simuF <- function(n,a,b,p){
sampl=NULL
while (length(sampl)<n){
x=u=sample(0:1,n,rep=TRUE)
x[u==0]=rexp(sum(u==0),b/(p+1))^(1/(p+1))
x[u==1]=rexp(sum(u==1),a)
sampl=c(sampl,x[runif(n)<(a+b*x^p)*exp(-a*x-b*x^(p+1)/(p+1))/
(a*exp(-a*x)+b*x^p*exp(-b*x^(p+1)/(p+1)))])
}
return(sampl[1:n])}


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