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The joint density function ofXandYis given by

f(x,y)=1ye-(y+x/y),x>0,y>0

Find E[X],E[Y]and show that Cov(X,Y)=1

Short Answer

Expert verified

The value of E[X]=1

The value of E[Y]=1

The value ofCov(X,Y)=1

Step by step solution

01

Given Information

The density of Xand Yis f(x,y)=1ye-(y+x/y),x>0,y>0

E[X]=?

E[Y]=?

Cov(X,Y)=?

02

Explanation

Calculate the value ofE[X],

E[X]=00x1ye-y+xydydx

=0e-y0xye-xydxdy

=0ye-ydy

=ye-y-10-0e-y-1dy

=0+1

=1

03

Explanation

Calculate the value of E[Y],

E[Y]=00y1ye-y+xydxdy

=00e-ye-xydxdy

=0e-y0e-xydxdy

=0e-yydy

=1

04

Explanation

Calculate the value of Cov(X,Y),

E(XY)=00xy1ye-y+xydxdy

=0e-y0xe-xydxdy

=0e-yxe-xy-1y0-0e-xy-1ydxdy

=0e-y0+y2dy

=0y2e-ydy

=-y2e-y0+02ye-ydy

=0+2=2

05

Final Answer

HenceCov(X,Y)=E(XY)-E(X)E(Y)=2-1.1

=2-1

=1

So, the value ofCov(X,Y)=1

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