Tim Tim - 2 months ago 6
R Question

Random generation code translated from R fails in C++

I am working on code implementing random generation algorithm for sampling from tails of normal distribution proposed by Christian Robert. The problem is that while code in R worked properly, then after translating it to C++ if fails. I can't see any reason for that and I'd be grateful for explaining me what went wrong and why.

Notice that the code below is far from elegant and efficient, it is simplified to make reproducible example.

Here is the function in R:

rtnormR <- function(mean = 0, sd = 1, lower = -Inf, upper = Inf) {
lower <- (lower - mean) / sd
upper <- (upper - mean) / sd

if (lower < upper && lower >= 0) {
while (TRUE) {
astar <- (lower + sqrt(lower^2 + 4)) / 2
z <- rexp(1, astar) + lower
u <- runif(1)
if ((u <= exp(-(z - astar)^2 / 2)) && (z <= upper)) break
}
} else {
z <- NaN
}
z*sd + mean
}


and here C++ version:

#include <Rcpp.h>
using namespace Rcpp;

// [[Rcpp::export]]

double rtnormCpp(double mean, double sd, double lower, double upper) {
double z_lower = (lower - mean) / sd;
double z_upper = (upper - mean) / sd;
bool stop = false;
double astar, z, u;

if (z_lower < z_upper && z_lower >= 0) {
while (!stop) {
astar = (z_lower + std::sqrt(std::pow(z_lower, 2) + 4)) / 2;
z = R::exp_rand() * astar + z_lower;
u = R::unif_rand();
if ((u <= std::exp(-std::pow(z-astar, 2) / 2)) && (z <= z_upper))
stop = true;
}
} else {
z = NAN;
}
return z*sd + mean;
}


Now compare the samples obtained using both functions (they are compared to
dtnorm
function from msm library):

xx = seq(-6, 6, by = 0.001)
hist(replicate(5000, rtnormR(mean = 0, sd = 1, lower = 3, upper = 5)), freq= FALSE, ylab = "", xlab = "", main = "rtnormR")
lines(xx, msm::dtnorm(xx, mean = 0, sd = 1, lower = 3, upper = 5), col = "red")
hist(replicate(5000, rtnormCpp(mean = 0, sd = 1, lower = 3, upper = 5)), freq= FALSE, ylab = "", xlab = "", main = "rtnormCpp")
lines(xx, msm::dtnorm(xx, mean = 0, sd = 1, lower = 3, upper = 5), col = "red")


enter image description here

As you can see,
rtnormCpp
returns biased samples. Do you have any ideas why?

Answer

While one can use either scale or rate in rexp(), the default parameterization is rate - so rexp(1,astar) has a mean of 1/astar, not astar.

If you change the relevant line of C++ code to

z = R::exp_rand() / astar + z_lower;

everything seems to work fine.