How to find the inverse for the inverse sampling method in R - r

Question

Generally for the inverse sampling method, we have a density and we would like to sample from it. A first step is to find the the cumulative density function for the density. Then to find it's inverse, and finally to find the inverse function for a randomly sampled value from the uniform distribution.
For example, I have this function y= ((3/2)/(1+x)^2) so the cdf equals (3x)/2(x+1) and the inverse of the cdf is ((3/2)*u)/(1-(3/2)*u)
To do this in R, I wrote
f<-function(x){
y= ((3/2)/(1+x)^2)
return(y)
}
cdf <- function(x){
integrate(f, -Inf, x)$value
}
invcdf <- function(q){
uniroot(function(x){cdf(x) - q}, range(x))$root
}
U <- runif(1e6)
X <- invcdf(U)
I have two problem! First: the code returns the function and not the samples.
The second: is there another simple way to do this work? for example to find the cdf and inverse in more simple ways?
I would like to add that I am not looking for efficiency of the code. I am just interested of a code that could be written by a beginner.

Answer 1

You could try a numerical approach to inverse sampling. As per your request, this is more about transparency of method than efficiency.
This function will numerically integrate a given function over the given range (though it will trim infinite values)
cdf <- function(f, lower_bound, upper_bound)
{
if(lower_bound < -10000) lower_bound <- -10000 # Trim large negatives
if(upper_bound > 10000) upper_bound <- 10000 # Trim large positive
x <- seq(lower_bound, upper_bound, length.out = 100001) # Finely divide x axis
delta <- mean(diff(x)) # Get delta x (i.e. dx)
mid_x <- (x[-1] + x[-length(x)])/2 # Get the mid point of each slice
result <- cumsum(delta * f(mid_x)) # sum f(x) dx
result <- result / max(result) # normalize
list(x = mid_x, cdf = result) # return both x and f(x) in list
}
And to get the inverse, we find the closest value in the cdf of a random number drawn from the uniform distribution between 0 and 1. We then see which value of x corresponds to that value of the cdf. We want to be able to do this for n samples at a time so we use sapply:
inverse_sample <- function(f, n = 1, lower_bound = -1000, upper_bound = 1000)
{
CDF <- cdf(f, lower_bound, upper_bound)
samples <- runif(n)
sapply(samples, function(s) CDF$x[which.min(abs(s - CDF$cdf))])
}
We can test it by drawing histograms of the results. We'll start with the normal distribution's density function (dnorm in R), drawing 1000 samples and plotting their distribution:
hist(inv_sample(dnorm, 1000))
And we can do the same for the exponential distribution, this time setting the limits of integration between 0 and 100:
hist(inv_sample(dexp, 1000, 0, 100))
And finally we can do the same with your own example:
f <- function(x) 3/2/(1 + x)^2
hist(inv_sample(f, 1000, 0, 10))