2.5. Consider two continuous random variables X and Y with joint p.d.f. f X, Y ( x, y ) = <<<<otherwise00 ,10 23 xyxyxC a) Find the value of C so that f ( x, y ) is a valid joint p.d.f. 1 = dxdyyxxC 100 32 ∫ ∫ = dxyxyxyC 10042 4 ∫== = dxxC 109 4 ∫ = 0110 40 xC = 40C. ⇒ C = 40. b) Find P ( 5 Y > 4 X ). P ( 5 Y > 4 X ) = P ( Y > 0.80 X ) y = 0.80 x and y = x 2 ⇒ x = 0.80 = dxdyyxxx 180.080.0 32 40 ∫ ∫ = ( )dxyxxx 180.0 80.04 2 10 ∫ = ( )dxxx 180.059 4.096 10 ∫− = 80.01610 375256 −xx ≈ 0.3889. ORdydxyxyy 80.064.025.1 3 40 ∫ ∫ + dydxyxy 180.01 3 40 ∫ ∫ = … OR 1 – dxdyyxx 80.000 32 40 ∫ ∫ – dxdyyxx 180.080.00 3 40 ∫ ∫ = … OR 1 – dydxyxy 64.001 3 40 ∫ ∫ – dydxyxy 80.064.0125.1 3 40 ∫ ∫ = … c) Find P ( X + Y > 1 ). y = 1 – x and y = x 2 ⇒ x = 215 − P ( X + Y > 1 ) = dxdyyxxx 12151 3 2 40 ∫ ∫−− = . . . = 25756509 − ≈ 0.9808. d) Find the marginal probability density function of X, f X ( x ). f X ( x ) = ∫2 0 3 40 xdyyx = 10 x 9, 0 < x < 1.e) Find the marginal probability density function of Y, f Y ( y ). f Y ( y ) = ∫1 3 40 ydxyx = 20 y 3 – 20 y 4, 0 < y < 1. f) Find E ( X ), E ( Y ), E ( X ⋅ Y ). E ( X ) = ∫⋅10 9 10 dxxx = 1110. E ( Y ) = ( )∫−⋅10 43 20 20 dyyyy = 620520− = 32. OR E ( Y ) = dxdyyxyx 100 32 40 ∫ ∫⋅ = ∫10 11 8 dxx = 128 = 32. OR Y has Beta distribution with α = 4, β = 2. E ( Y ) = 244+ = 32. E ( X Y ) = dxdyyxyxx 100 32 40 ∫ ∫⋅ = ∫10 12 8 dxx = 138. Sneak preview of 4.2 : g) Find Cov ( X, Y ). Cov ( X, Y ) = E ( X Y ) – E ( X ) E ( Y ) = 321110138⋅− = 4294 ≈
View Full Document
Unlocking...