Simulating a divide area random variable

Question

Suppose that we have a random variable with pdf like $f = \left\{ \begin{array}{ c l } x+1, x\in [-1,0] \\ -x+1, x\in (0,1] \end{array} \right.$ and we have to simulate it.

It's pdf is $F(x) = \left\{ \begin{array}{ c l } x^2/2+x+0.5, x\in [-1,0] \\ -x^2/2+x+0.5, x\in (0,1] \end{array} \right.$

two methods to simulate this can be found here http://web.ics.purdue.edu/~hwan/IE680/Lectures/Chap08Slides.pdf page 12.

My question is if we can simulate problems like that by fist simulate a random U(0,1) and

if it is smaller than the probability $P(x\in [-1,0])=1/2$ simulate from the first part with inverse transformation $\sqrt{2U}-1$ but for a $U\sim Unif(0,1/2=P(x\in [-1,0]))$

and if it is bigger simulate with inverse transformation $1-\sqrt{(2(1-U))}$ using a $U\sim Unif(1/2,1)$.

When I make the histogram it seems to working, but is it correct?

[code]rand=function(){
if(runif(1,0,1)<=1/2){x=sqrt(2*runif(1,0,1/2))-1}else{
x=1-sqrt(2-2*runif(1,1/2,1))
}
return(x)
}


pri=NULL


for(i in 1:10000)
{
pri=c(pri,rand())
}


hist(pri,prob=T)[/code]

Answer 1

Yes, your code generates the correct results, but as you can see from the calculations below, it's hard to understand. I suggest sticking with the linked lecture notes and use runif(1,0,1), which represents $\mathrm{Unif}(0,1)$.

The composition algorithm in the linked notes:

1. Generate two independent $U_1, U_2 \sim \mathrm{Unif}(0,1)$
2. • If $U_1 < 1/2$, return $X = \sqrt{U_2}-1$
   • Otherwise, return $X = 1 - \sqrt{1-U_2}$

The composition algorithm in the question body contains a rescaled version of $U_2$.

1. Generate $U_1 \sim \mathrm{Unif}(0,1)$
2. • $U_1 < 1/2$, return $X = \sqrt{2U}-1$ for $U\sim \mathrm{Unif}(0,1/2)$
     In fact, $2U \stackrel{(d)}{=} U_2$.
   • Otherwise, return $X = 1 - \sqrt{2-2U}$ for $U\sim \mathrm{Unif}(1/2,1)$
     In fact, $2U \sim \mathrm{Unif}(1,2)$, so $2-2U \sim \mathrm{Unif}(0,1)$ and $2-2U \stackrel{(d)}{=} 1-U_2$.

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These figures illustrates the how the code works.

If we increase the sample size from 10k to 50k, the histogram looks more similar to the triangular distribution on $[-1,1]$.

We increasing the class number from 20 to 50 in order to take a closer look of the data.

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My code

rand=function(){
if(runif(1,0,1)<=1/2){x=sqrt(2*runif(1,0,1/2))-1}else{
x=1-sqrt(2-2*runif(1,1/2,1))
}
return(x)
}

pri=NULL
for(i in 1:50000)
{
pri=c(pri,rand())
}

hist(pri,breaks=50,prob=T)