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Q. 10.11

Expert-verified
Found in: Page 431

### A First Course in Probability

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}{\left(1-x\right)}^{2}& 0

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

See the step by step solution

## Step 1: Given Information

We have given the density function

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

## Step 2: Simplify

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

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

which implies the equality

$3\left(1-x\right)=2x⇒x=\frac{3}{5}$

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 $U\in \left(0,1\right).$ 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.