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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]=∫0∞∫0∞x·1ye-y+xydydx

=∫0∞e-y∫0∞xye-xydxdy

=∫0∞ye-ydy

=ye-y-10∞-∫0∞e-y-1dy

=0+1

=1

03

Explanation

Calculate the value of E[Y],

E[Y]=∫0∞∫0∞y·1ye-y+xydxdy

=∫0∞∫0∞e-ye-xydxdy

=∫0∞e-y∫0∞e-xydxdy

=∫0∞e-yydy

=1

04

Explanation

Calculate the value of Cov(X,Y),

E(XY)=∫0∞∫0∞xy·1ye-y+xydxdy

=∫0∞e-y∫0∞xe-xydxdy

=∫0∞e-yxe-xy-1y0∞-∫0∞e-xy-1ydxdy

=∫0∞e-y0+y2dy

=∫0∞y2e-ydy

=-y2e-y0∞+∫0∞2ye-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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