Chapter 10: Q. 10.11 (page 431)

Use the rejection method with g(x) = 1, 0 < x < 1, to determine an algorithm for simulating a random variable having density function
$f (x) \{\begin{cases} 60 x^{3} {(1 - x)}^{2} & 0 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$

Short Answer

Expert verified

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

$f (x) \{\begin{cases} 60 x^{3} {(1 - x)}^{2} & 0 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$

Simplify

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

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

which implies the equality

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

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

$U \leq \frac{f (Y)}{c g (Y)}$

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

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Use the rejection method with g(x) = 1, 0 < x < 1, to determine an algorithm for simulating a random variable having density function
$f (x) \{\begin{cases} 60 x^{3} {(1 - x)}^{2} & 0 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$

Short Answer

Step by step solution

Given Information

Simplify

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Use the rejection method with g(x) = 1, 0 < x < 1, to determine an algorithm for simulating a random variable having density functionf(x)60x3(1-x)20<x<10otherwise

Short Answer

Step by step solution

Given Information

Simplify

One App. One Place for Learning.

Most popular questions from this chapter

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Study anywhere. Anytime. Across all devices.

Use the rejection method with g(x) = 1, 0 < x < 1, to determine an algorithm for simulating a random variable having density function
$f (x) \{\begin{cases} 60 x^{3} {(1 - x)}^{2} & 0 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$