Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

Q. 10.11

Expert-verified

A First Course in Probability

Found in: Page 431

Book edition 9th

Author(s) Sheldon M. Ross

Pages 432 pages

ISBN 9780321794772

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later

Short Answer

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

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

We have given the density function

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

Step 2: 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.

Most popular questions for Math Textbooks

In Example 4a, we showed that

E[(1 − V2) 1/2] = E[(1 − U2) 1/2] = π/4

when V is uniform (−1, 1) and U is uniform (0, 1). Now show that

Var[(1 − V2) 1/2] = Var[(1 − U2) 1/2]

and find their common value.

Use the inverse transformation method to present an approach for generating a random variable from the Weibull distribution

$F (t) = 1 - e^{- a t^{β}} t \geq 0$

Suppose we have a method for simulating random variables from the distributions F1 and F2. Explain how to simulate from the distribution

F(x) = pF1(x) + (1 − p)F2(x) 0 < p < 1

Give a method for simulating from

$F (x) \{\begin{cases} \frac{1}{3} (1 - e^{- 3 x}) + \frac{2}{3} x & 0 < x \leq 1 \\ \frac{1}{3} (1 - e^{- 3 x}) + \frac{2}{3} & x > 1 \end{cases}$

Let F be the distribution function

F(x) = xn 0 < x < 1

(a) Give a method for simulating a random variable having distribution F that uses only a single random number.

(b) Let U1, ... , Un be independent random numbers. Show that

P{max(U1, ... , Un) … x} = xn

(c) Use part (b) to give a second method of simulating a random variable having distribution F.

The following algorithm will generate a random permutation of the elements 1, 2, ... , n. It is somewhat faster than the one presented in Example 1a but is such that no position is fixed until the algorithm ends. In this algorithm, P(i) can be interpreted as the element in position i. Step 1. Set k = 1. Step 2. Set P(1) = 1. Step 3. If k = n, stop. Otherwise, let k = k + 1. Step 4. Generate a random number U and let P(k) = P([kU] + 1) P([kU] + 1) = k Go to step 3. (a) Explain in words what the algorithm is doing. (b) Show that at iteration k—that is, when the value of P(k) is initially set—P(1),P(2), ... ,P(k) is a random permutation of 1, 2, ... , k.

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free