Q. 10.4 - A First Course in Probability

Question

Present a method for simulating a random variable having distribution function 

$F\left(x\right)=\left\{\begin{array}{ll}0& x\le -3\\ \frac{1}{2}+\frac{x}{6}& -3<x<0\\ \frac{1}{2}+\frac{x^2}{32}& 0<x\le 4\\ 1& x>4\end{array}\right.$

Step-by-Step Solution

Answer

The universality of uniform is used to obtain the required method.

Step 1: Given Information

We have given the function

$F\left(x\right)=\left\{\begin{array}{ll}0& x\le -3\\ \frac{1}{2}+\frac{x}{6}& -3<x<0\\ \frac{1}{2}+\frac{x^2}{32}& 0<x\le 4\\ 1& x>4\end{array}\right.$

Step 2: Simplify

Finding the value $F^{-1}$. for$y\in (0,1/2)$ we have

 $y=\frac{1}{2}+\frac{x}{6}$

$x=6y-3$

and for$y\in$(1/2,1) we have

$y=\frac{1}{2}+\frac{x^2}{32}$

$x=4\sqrt{2y-1}$

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

$$\begin{gathered} y-\frac{1}{2}+\frac{x}{6} \\ x = 6y-3 \end{gathered}$$$$\begin{gathered} y=\frac{1}{2}+\frac{x}{6} \\ x=6y-3 \end{gathered}$$