chapter 1

Chapter 5: Joint Probability Distributions and Random Samples87.2 22 2 22a. Var(aX + Y)= a σ + 2aCov(X , Y)+ σ = a σ + 2aσσ ρ + σ .SubstitutingaσxY2 2 2 2= yields σ + 2σYρ + σY= 2σY( 1−ρ ) ≥ 0σXyxY, so ρ ≥ −1XYyb. Same argument as in ac. Suppose ρ = 1. Then Var ( aX −Y) = 2σ 2 Y( 1−ρ ) = 0 , which implies thataX − Y = k (a constant), so aX − Y = aX − k , which is of the form aX + b .2288. X + Y − t)= ∫ ∫ ( x + y −t)⋅101E ( f ( x,y)dxdy.To find the minimizing value of t,0take the derivative with respect to t and equate it to 0:011= ∫ ∫∫ ∫0 01 12( x + y −t)(−1)f ( x,y)= 0 ⇒ tf ( x,y)dxdy = t= ∫ ∫ ( x + y)⋅ f ( x,y)dxdy = E(X + Y), so the best prediction is the individual’s00expected score ( = 1.167).101089.a. With Y = X 1 + X 2 ,FY( y)=y⎧⎨⎩∫ ∫0 0y−x121⋅ν 21/2( ν / 2) 2 Γ( ν / )ν1 / 2Γ122But the inner integral can be shown to be equal to21 [y( ν 2 )/2)1 + νΓ ( ν + ν ) / 212y1⋅ xν1−121xν 2−122ex1+ x2−2( ν1+ν 2 )/2] −1− / 2, from which the result follows.e⎫dx2⎬dx1⎭.b. By a,ν= 3Z + is chi-squared with = 22 21Z2, etc, until2 2 2ν , so ( Z Z ) +1 2Z322+ 9s chi-squared withν= nZ1... + Zn+ is chi-squared withc.X iσ− µis standard normal, sois chi-squared with ν = n .⎡⎢⎣X i2− µ ⎤σ ⎥⎦is chi-squared with ν = 1 , so the sum202