Simple scheme to sample from the Bernoulli distribution

Question

I was implementing a simple scheme of Bernoulli distribution sampler. $ X \sim B(p) $. I have a function that generates a uniform random number $r \in (0,1)$. Then, I set $ X = 1 $ if $p > r $, and $X =0$ otherwise. Is this correct?

Answer 1

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): $$\begin{align} 1.&\quad x \leftarrow 0, k \leftarrow 0 \\ 2.&\quad \text{Repeat} \\ &\quad\quad \text{Generate } u\sim\mathcal{U}(0,1), k\leftarrow k + 1 \\ &\quad\quad \text{if }u\le p\text{ then } x \leftarrow x + 1\\ &\quad\text{Until }k = n \\ 3.&\text{Return}\end{align}$$ 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