In R, I want to generate a random sample of a discrete random variable: X, where: P(X=a)=P(X=-a)=1/2. I have been searching for a function online, but there seems no direct function doing this.
I think you are looking to generate samples of a Bernoulli random variable. A Bernoulli random variable is a special case of a binomial random variable. Therefore, you can try rbinom(N,1,p). This will generate N samples, with value 1 with probability p, value 0 with probability (1-p). To get values of a and -a you can use a*(2*rbinom(N,1,p)-1).
1) If you use sample, this is sufficient:
sample(c(-a,a),1)
e.g.:
a <- 10
sample(c(-a,a),1)
[1] -10
Try another couple:
> sample(c(-a,a),1)
[1] -10
> sample(c(-a,a),1)
[1] 10
Works.
If you need to sample more than one element, then set replace=TRUE ... here we sample 12 times:
sample(c(-a,a),12,replace=TRUE)
[1] 10 10 -10 10 10 10 -10 -10 10 -10 10 -10
2) you can use runif; here's a sample of size 9:
a <- 1
ifelse(runif(9)<.5,-a,a)
[1] -1 1 -1 1 -1 1 -1 1 1
3) you can use rbinom; here's a sample of size 4:
a <- 6
ifelse(rbinom(4,1,.5),-a,a)
[1] -6 6 -6 6
Or this:
> n=10
> X=rep(0,n)
> Y=rbinom(n,1,1/2)
> #Since they the probability is 1/2 for both cases, I assigned "a" when Y=1 and "-a" otherwise.
> X[Y==1]="a"
> X[Y==0]="-a"
> X
[1] "a" "-a" "a" "a" "a" "-a" "a" "-a" "-a" "-a"
> Y
[1] 1 0 1 1 1 0 1 0 0 0
>
index <- sample(1,c(1,2),replace=T)
if (index == 1) {xx = a} else {xx = -a}
Each distribution generating procedure begins with using $\text{uniform}(0,1)$. Since discrete distributions are much easier to generate with $\text{uniform}(0,1)$, people don't wrap up a function for them. However, you can write your own function and just pick them up next time you're going to use them.
来源：https://stackoverflow.com/questions/22893089/r-how-to-generate-random-sample-of-a-discrete-random-variables