/*! 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.6.19 Show that f(x, y) = 1/x, 0 < ... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that f(x, y) = 1/x, 0 < y < x < 1, is a joint density function. Assuming that f is the joint density function of X, Y, find

(a) the marginal density of Y;

(b) the marginal density of X;

(c) E[X]; (d) E[Y].

Short Answer

Expert verified

The joint density of X and Y is given by f(X,Y) is not independent.

Step by step solution

01

Introduction

The joint density of X and Y is not independent.

02

Given Information

ThreepointsX1,X2,X3areselectedatrandomonalineLFromtheinformation,observethatthejointdensityfunctionofXandYisasfollows:f(x,y)=xe-(x+y)x>0,y>00OtherviseCheckwhetherXandYareindependentornot.ThemarginaldensityofXis,fx(x)=∫0∞f(x,y)dy∫0∞xe-(x+y)dy∫0∞xe-xe-ydyxe-x-e-y0∞xe-xe-0-e-0xe-xCalculatethemarginaldensityofYfr(y)=∫0∞f(x,y)dx=∫0∞xe-(x+y)dx=∫0∞xe-xe-ydx=e-y∫0∞xe-xdx=e-y-xe-x0∞+∫0∞e-xdx(since integration by parts)=∫0∞xe-(x+y)dx=e-y-xe-x-e-0-e-x0∞=e-y[1]=e-yTherefore,fx(x)fY(y)=xe-xe-y=xe-x-y=xe-(x+y)=f(x,y)Hence,XandYareindependent.NowTheserandomvariablesarenotindependent.Iftheywereindependent,theirjointPDFwouldfactorizef(x,y)=f(x)f(y)Butforeverypoint(x,y)∈(0,1)2suchthaty<xewouldhavefx(x)>0andfr(y)>0ontheotherhandf(x,y)=0Thatleadstothecontradiction.Hence,theyarenotindependent.

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