Question

Use the rejection method with g(x) = 1, 0 < x < 1, to determine an algorithm for simulating a random variable having density function

$f\left(x\right)\left\{\begin{array}{ll}60x^3{(1-x)}^2& 0<x<1\\ 0& otherwise\end{array}\right.$

Short answer

The algorithm is generate$Y~g$(which is unform) and take random number$U\in (0,1)$.

Step-by-step solution

Given Information

We have given the density function

$f\left(x\right)\left\{\begin{array}{ll}60x^3{(1-x)}^2& 0<x<1\\ 0& otherwise\end{array}\right.$

Simplify

Finding the upper bound of $f$on the interval $(0,1)$. Using the differentiation, we have

$f^1\left(x\right)=180x^2{(1-x)}^2-120x^3(1-x)=0$

which implies the equality

$3(1-x)=2x\Rightarrow x=\frac{3}{5}$

So, the maximum value of $f$is assumed to be in point $x=\frac{3}{5}$and it is equal to localid="1648210382401" $f\left(\frac{3}{5}\right)\approx 2.0736:=c$. So, the algorithm is as follows: generate $Y~g$(which is uniform) and take random number $U\in (0,1).$Consider if

$U\le \frac{f\left(Y\right)}{cg\left(Y\right)}$

and in that case declare $X=Y.$Otherwise, go to the step $1$again.