Question

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

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

Step-by-step solution

Introduction

The joint density of X and Y is not independent.

Given Information

$$\begin{gathered} ThreepointsX1,X2,X3areselectedatrandomonalineL \\ Fromtheinformation, \\ observethatthejointdensityfunctionofXandYisasfollows: \\ f(x,y)=\left\{\begin{array}{ll}x{e}^{-(x+y)}& x>0,y>0\\ 0& Othervise\end{array}\right. \\ CheckwhetherXandYareindependentornot. \\ ThemarginaldensityofXis, \\ {f}_x\left(x\right)={\int }_0^{\infty }f(x,y)dy \\ {\int }_0^{\infty }x{e}^{-(x+y)}dy \\ {\int }_0^{\infty }x{e}^{-x}{e}^{-y}dy \\ x{e}^{-x}{\left(-e^{-y}\right)}_0^{\infty } \\ x{e}^{-x}\left(e^{-0}-e^{-0}\right) \\ x{e}^{-x} \\ CalculatethemarginaldensityofY \\ {f}_r\left(y\right)={\int }_0^{\infty }f(x,y)dx \\ ={\int }_0^{\infty }x{e}^{-(x+y)}dx \\ ={\int }_0^{\infty }x{e}^{-x}{e}^{-y}dx \\ =e^{-y}{\int }_0^{\infty }x{e}^{-x}dx \\ =e^{-y}\left[-x{\left(e^{-x}\right)}_0^{\infty }+{\int }_0^{\infty }{e}^{-x}dx\right]\text{(since integration by parts)} \\ ={\int }_0^{\infty }x{e}^{-(x+y)}dx \\ =e^{-y}\left[-x\left(e^{-x}-e^{-0}\right)-{\left(e^{-x}\right)}_0^{\infty }\right] \\ =e^{-y}\left[1\right] \\ =e^{-y} \\ Therefore, \\ {f}_x\left(x\right)f_Y\left(y\right)=x{e}^{-x}{e}^{-y} \\ =x{e}^{-x-y} \\ =x{e}^{-(x+y)} \\ =f(x,y) \\ Hence,XandYareindependent. \\ NowTheserandomvariablesarenotindependent.Iftheywereindependent,theirjointPDFwouldfactorize \\ f(x,y)=f\left(x\right)f\left(y\right) \\ Butforeverypoint(x,y)\in (0,1{)}^{2}suchthaty<xewouldhave{f}_x(x)>0and{f}_r(y)>0ontheotherhandf(x,y)=0 \\ Thatleadstothecontradiction. \\ Hence,theyarenotindependent. \end{gathered}$$