chapter 1

Chapter 5: Joint Probability Distributions and Random Samples32. There is a difficulty here. Existence of r requires that both X and Y have finite means andvariances. Yet since the marginal pdf of Y is( )2∫( 1+y)∫1for y ≥ 0,1− y∞ y∞E ( y)= dy =dy0 2( 1+y −1)∞ 1∞ 1dy =( )∫ dy −0( + ) ∫1+y 1 y ( 1+y )0 2first integral is not finite. Thus r itself is undefined.0 2, and the33. Since E(XY) = E(X) ⋅ E(Y), Cov(X,Y) = E(XY) – E(X) ⋅ E(Y) = E(X) ⋅ E(Y) - E(X) ⋅ E(Y) =0, and since Corr(X,Y) =Cov(X,Y)σ σxy, then Corr(X,Y) = 034.a. In the discrete case, Var[h(X,Y)] = E{[h(X,Y) – E(h(X,Y))] 2 } =∑∑xy[ h(x,y)− E(h(X , Y ))]2p(x,y)=∑∑with∫∫ replacing ∑∑ in the continuous case.xy[ h(x,y)2p(x,y)]−[E(h(X , Y ))]2b. E[h(X,Y)] = E[max(X,Y)] = 9.60, and E[h 2 (X,Y)] = E[(max(X,Y)) 2 ] = (0) 2 (.02)+(5) 2 (.06) + …+ (15) 2 (.01) = 105.5, so Var[max(X,Y)] = 105.5 – (9.60) 2 = 13.3435.a. Cov(aX + b, cY + d) = E[(aX + b)(cY + d)] – E(aX + b) ⋅ E(cY + d)= E[acXY + adX + bcY + bd] – (aE(X) + b)(cE(Y) + d)= acE(XY) – acE(X)E(Y) = acCov(X,Y)b. Corr(aX + b, cY + d) =Cov(aX + b,cY + d)acCov(X , Y)=Var(aX + b)Var(cY + d)| a | ⋅ | c | Var(X ) ⋅Var(Y)= Corr(X,Y) when a and c have the same signs.c. When a and c differ in sign, Corr(aX + b, cY + d) = -Corr(X,Y).36. Cov(X,Y) = Cov(X, aX+b) = E[X⋅(aX+b)] – E(X) ⋅E(aX+b) = a Var(X),so Corr(X,Y) =aVar(X )=Var(X ) ⋅Var(Y )aVar(X )Var(X ) ⋅ aVar(X )2= 1 if a > 0, and –1 if a < 0186