Solved – Simple scheme to sample from the Bernoulli distribution

If $X$ is a Bernoulli random variable then $E[X]=p$ and $V[X]=p(1-p)$. For example:

> x <- rbinom(1000,1,0.3) > mean(x) [1] 0.302 > var(x) [1] 0.211007 

The most basic way to generate a Bernoulli sample is (Kachitvichyanukul and Schmeise): $$begin{align} 1.&quad x leftarrow 0, k leftarrow 0 \ 2.&quad text{Repeat} \ &quadquad text{Generate } usimmathcal{U}(0,1), kleftarrow k + 1 \ &quadquad text{if }ule ptext{ then } x leftarrow x + 1\ &quadtext{Until }k = n \ 3.&text{Return}end{align}$$ This algorithm generates $x$ successes out of $n$ trials, but can be slightly modified to generate a sample form the Bernoulli distribution. In R:

> rbernoulli <- function(n, p) { +     x <- c() +     for (i in 1:n) { +         u <- runif(1,0,1) +         if (u <= p) +             x <- c(x, 1) +         else +             x <- c(x, 0) +     } +     return (x) + } > x <- rbernoulli(1000, 0.3) > mean(x) [1] 0.314 > var(x) [1] 0.2156196