chapter 1

Chapter 5: Joint Probability Distributions and Random Samplesd. f x (x) =∫ ∞ −∞302 22f ( x,y)dy = ∫ K ( x + y ) dy = 10Kx+20= 10Kx 2 + .05, 20 ≤ x ≤ 30yK333020e. f y (y) is obtained by substituting y for x in (d); clearly f(x,y) ≠ f x (x) ⋅ f y (y), so X and Y arenot independent.10.a. f(x,y) = ⎨ ⎧ 1⎩ 05 ≤ x ≤ 6,5 ≤ y ≤ 6otherwisesince f x (x) = 1, f y (y) = 1 for 5 ≤ x ≤ 6, 5 ≤ y ≤ 6b. P(5.25 ≤ X ≤ 5.75, 5.25 ≤ Y ≤ 5.75) = P(5.25 ≤ X ≤ 5.75) ⋅ P(5.25 ≤ Y ≤ 5.75) = (byindependence) (.5)(.5) = .25c.6Iy =x+1/6y =x−1/ 65IIP((X,Y) ∈ A) =∫∫A1 dxdy= area of A = 1 – (area of I + area of II )25 11= 1 − = = . 30636 365611.a. p(x,y) =−λ xe λ−µe µx! y!y⋅ for x = 0, 1, 2, …; y = 0, 1, 2, …−λ−µb. p(0,0) + p(0,1) + p(1,0) = e [ 1+λ + µ ]c. P( X+Y=m ) = ∑ P(X = k,Y = m − k)= ∑em!−(λ + µ )mmk m=k−λ−µek = k=0 k!( m − km−(λ + µ )m⎛ m⎞k m−ke ( λ + µ )∑k=00 )!⎜ ⎟λµ⎝ k ⎠=m!λ + .Poisson distribution with parameter µλµ, so the total # of errors X+Y also has a179