chapter 1

Chapter 5: Joint Probability Distributions and Random Samples1 11 x227. E[h(X,Y)] = x − y ⋅6xydxdy = 2 ( x − y)∫ ∫00∫ ∫0 0⋅6x2ydydx121 x3 2 2∫ ∫ ( x y − x y )0 0dydx = 12∫105x6dx =1328. E(XY) = ∑∑ ⋅ p(x,y)= ∑∑ xy⋅px(x)⋅ py( y)= ∑ xpx(x)⋅∑xy yp ( y)x y x yx= E(X) ⋅ E(Y). (replace Σ with ∫in the continuous case)yy22µ x= µ y= . E(X 2 12) =755∫ x ⋅ fx( x)dx012 112 1 3 21 4 1= ∫ x (1 − x dx)= = , so Var (X) = − =060 55 25 251−75 50Similarly, Var(Y) = , so ρ2X , Y= = − = − . 667251 1⋅ 7529. Cov(X,Y) = − and252530.a. E(X) = 5.55, E(Y) = 8.55, E(XY) = (0)(.02) + (0)(.06) + … + (150)(.01) = 44.25, soCov(X,Y) = 44.25 – (5.55)(8.55) = -3.20− 3.20σX=Y, so ρX , Y== −.207(12.45)(19.15)22b. 12.45, σ = 19. 1531.30302a. E(X) =∫ xf x( x)dx = x[ 10Kx+ .05] dx = 25.329 = E(Y)20 ∫2030 302 2E(XY) =∫ ∫ xy ⋅ K(x + y ) dxdy = 641. 4472020⇒ Cov ( X , Y ) = 641.447 −(25.329)2= −.11130∫,20so Var (X) = Var(Y) = 649.8246 – (25.329) 2 = 8.2664b. E(X 2 2 22) = x [ 10Kx+ .05] dx = 649.8246 = E(Y )⇒ ρ =− .111= −.0134(8.2664)(8.2664)185