/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 10.9 Suppose we have a method for sim... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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)13(1-e-3x)+23x0<x≤113(1-e-3x)+23x>1

Short Answer

Expert verified

Observing flipping a coin. If, it falls Heads, declareX=X1otherwise, declareX=X2.

Step by step solution

01

Given Information

Consider

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

localid="1651486678603" F(x)=13(1-e-3x)+23x0<x≤113(1-e-3x)+23x>1

02

Simplify

Let us considerX1~Fi,i=1,2which we can generate. Now, generating random number in (0,1), call it U. If U<p,declare X=X1. If U≥p,declare X=X2. Because of the fact that Xassumes valueX1with probability pand value x2 with probability 1-p,we have that Xhas CDF

F(x)=pF1(x)+(1-p)F2(x)

To solve the second part of this problem

F1(x)=13(1-e-3x)+23xF2(x)=13(1-e-3x)+23

Now, we will write

F(x)=χ(0<x≤1)(x)F1(x)+χ(1<x)(x)F2(x)

Considering random variable G~Expo(1)and define P(G≤1).Now, we generate random variable X.

F(x)=pF1(x)+(1-p)F2(x)

and using the first part of this exercise, we haveXhas distribution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.