Solution to HW 8

# Solution to HW 8 - Solutions to the EE351K homework...

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Solutions to the EE351K homework assignment 8 2. a. Since X and Y are independent, we can have their join PDF as 3 2 , 1 0, 0 ( , ) ( ) ( ) 6 0 x y X Y X Y e x y f x y f x f y otherwise - - = =

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3 2 , 0 0 3 3 2 2 0 0 0 0 5 3 6 3 2 0 0 1 ( ) ( , ) 6 1 1 ( 2 | ) 6 6 1 1 6 ( 1) ( 3 ) | 0.6 3 3 5 x x y X Y X Y x x x y y x x x x x P X Y f x y dxdy e e dydx e e dydx e e dx e e dx e e +∞ - - > +∞ +∞ - - - - +∞ - - - - +∞ > = = = = - = - - = - - + = ∫∫ ∫ ∫ b. , 0 0 0 0 ( ) ( , ) ( ) ( ) ( ) ( ) X Y X Y E XY xyf x y dxdy xf x dx yf y dy E X E Y +∞ +∞ +∞ +∞ = = = ∫ ∫ Let’s use Z to represent a variable whose PDF is 1 ( ) z Z f z e λ - = when 0 z , and ZOW, hence 0 0 0 0 0 0 0 1 ( ) ( ) ( ) | ( ) 0 0 ( ) | z z z z Z z z E Z zf z dz z e dz zde ze e d z e dz e +∞ +∞ +∞ +∞ - - - - +∞ +∞ - - +∞ = = = - = - - - = - + = - = So E(XY) = E(X) E(Y) = 2*3 =6 c. Cov(X,Y) = E(XY)-E(X)E(Y) = 6-2*3=0 3. a. , 2,0,2
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## Solution to HW 8 - Solutions to the EE351K homework...

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