Question

Give a technique for simulating a random variable having the probability density function

$f\left(x\right)\left\{\begin{array}{ll}\frac{1}{2}(x-2)& 2\le x\le 3\\ \frac{1}{2}(2-\frac{x}{3})& 3<x\le 6\\ 0& otherwise\end{array}\right.$

Short answer

The technique for simulating a random variable is the universality of unform.

Step-by-step solution

Given Information

We have given the function

$f\left(x\right)\left\{\begin{array}{ll}\frac{1}{2}(x-2)& 2\le x\le 3\\ \frac{1}{2}(2-\frac{x}{3})& 3<x\le 6\\ 0& otherwise\end{array}\right.$

Simplify

We have

$F\left(x\right)={\int }_2^x\frac{1}{2}(s-2)ds=\frac{{(x-2)}^2}{4}$

and for $3<x\le 6$we have

$F\left(x\right)={\int }_2^3\frac{1}{2}(s-2)ds+{\int }_3^x\frac{1}{2}(2-\frac{s}{3}ds=\frac{1}{4}-\frac{(x-9)(x-3)}{12}$

Let's find $F^{-1}$. For $y\in (0,1/4)$we have

$$\begin{gathered} y=\frac{{(x-2)}^2}{4} \\ x=2\sqrt{y}+2 \end{gathered}$$

and for$y\in (1/4,1)$we have

$$\begin{gathered} y=\frac{1}{4}-\frac{(x-9)(x-3)}{12} \\ x=6-2\sqrt{3-3y} \end{gathered}$$

Now, choose a random number (call it $y$) from the interval $(0,1)$. If it is less than $\frac{1}{4}$, declare $x=2\sqrt{y}+2.$if it is greater than $\frac{1}{4}$, declare $x=6-2\sqrt{3-3y}$. From the universality of the uniform, we have the $x$follows the distribution of $X$.