/*! 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} Problem 16 Let \(X\) and \(Y\) have the joi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 16

Let \(X\) and \(Y\) have the joint pdf \(f(x, y)=6(1-x-y), x+y<1,0

Short Answer

Expert verified
The probability \(P(2X + 3Y < 1)\) is 0.125 and the expectation \(E\left(XY + 2X^{2}\right)\) is 0.15.

Step by step solution

01

Identify Distribution and Boundaries

The joint pdf is defined as \(f(x, y)=6(1-x-y)\) for \(x>0\), \(y>0\), and \(x+y<1\), and zero elsewhere. First, we will compute \(P(2X + 3Y < 1)\). To make it simple, let's consider the inequality \(2X + 3Y = 1\). Obviously, the condition \(x+y<1\) has to be satisfied, after substituting \(x = 0.5\) and \(y = 0.5(1 - 2x)\). So the limits of the integral become: \(0 < x < 0.5\) and \(0 < y < 0.5(1 - 2x)\).
02

Compute the Probability

Now compute the integral over the region bounded by \(0 < x < 0.5\) and \(0 < y < 0.5(1 - 2x)\) to find the probability \(\int_{0}^{0.5} \int_{0}^{0.5(1 - 2x)} 6(1-x-y) dy\, dx\). Solving this integral gives a probability of 0.125.
03

Identify Boundaries for Expectation

Next, consider the expectation \(E\left(XY + 2X^{2}\right)\). For probability density function, the limits are \(0<x<1, 0<y<1-x\). We can calculate it using the definition of expectation.
04

Compute the Expectation

To calculate the expectation, compute the double integral over the region bounded by \(0<x<1, 0<y<1-x\), for \(x(y + 2x^{2}) \cdot f(x, y)\). The integral can be written as \(\int_{0}^{1} \int_{0}^{1-x} x(y + 2x^{2})6(1 - x - y) dy\, dx\). Solving this integral gives the expectation as 0.15

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 91影视!

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

Let \(f(x, y)=e^{-x-y}, 0

Find the probability of the union of the events \(a

Let \(X\) and \(Y\) have the joint pdf \(f(x, y)=1,-x

Let the random variables \(X\) and \(Y\) have the joint pmf (a) \(p(x, y)=\frac{1}{3},(x, y)=(0,0),(1,1),(2,2)\), zero elsewhere. (b) \(p(x, y)=\frac{1}{3},(x, y)=(0,2),(1,1),(2,0)\), zero elsewhere. (c) \(p(x, y)=\frac{1}{3},(x, y)=(0,0),(1,1),(2,0)\), zero elsewhere. In each case compute the correlation coefficient of \(X\) and \(Y\).

Let \(X_{1}, X_{2}, X_{3}\) be iid, each with the distribution having pdf \(f(x)=e^{-x}, 0<\) \(x<\infty\), zero elsewhere. Show that $$Y_{1}=\frac{X_{1}}{X_{1}+X_{2}}, \quad Y_{2}=\frac{X_{1}+X_{2}}{X_{1}+X_{2}+X_{3}}, \quad Y_{3}=X_{1}+X_{2}+X_{3}$$ are mutually independent.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.