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Q. 10.11

Expert-verifiedFound in: Page 431

Book edition
9th

Author(s)
Sheldon M. Ross

Pages
432 pages

ISBN
9780321794772

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

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

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

We have given the density function

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

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

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

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.

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