Question

Develop a technique for simulating a random variable having density function

$f\left(x\right)\left\{\begin{array}{ll}{e}^{2x}& -\infty <x<0\\ e^{-2x}& 0<x<\infty \end{array}\right.$

Short answer

The information for the function will get by using the universality of uniform distribution.

Step-by-step solution

Given Information

We have given a function

$f\left(x\right)\left\{\begin{array}{ll}{e}^{2x}& -\infty <x<0\\ e^{-2x}& 0<x<\infty \end{array}\right.$.

Simplify

We have

$F\left(x\right)={\int }_{-\infty }^x{e}^{2_s}ds=\frac{e^{2x}}{2}$

and for $x>0$we have

$F\left(x\right)={\int }_{-\infty }^0{e}^{2_s}ds+{\int }_0^x{e}^{-2_s}ds=1-\frac{e^{-2x}}{2}$

Calculating the value$F^{-1}$. For $\mathcal{Y}\in \left(0,1/2\right)$we have

$\mathcal{Y}=\frac{e^{2x}}{2}\Rightarrow x=\frac{1}{2}\log \left(2\mathcal{Y}\right)$

and for $\mathcal{Y}\in \left(1/2,1\right)$we have that

$\mathcal{Y}=1-\frac{e^{-2x}}{2}\Rightarrow x=-\frac{1}{2}\log (2-2\mathcal{Y})$

Choose random number (call it $\mathcal{Y}$) from the interval $\left(0,1\right)$. if it is less than$\frac{1}{2}$, declare$x=\frac{1}{2}\log \left(2\mathcal{Y}\right)$.If it is greater than $\frac{1}{2}$, declare $x=-\frac{1}{2}\log (2-2\mathcal{Y})$. From the universality of the Uniform, we have $x$follows the distribution of $X$.