chapter 1

Chapter 5: Joint Probability Distributions and Random Samplesc. p(x,y) = 0 unless y = 0, 1, …, x; x = 0, 1, 2, 3, 4. For any such pair,⎛ x⎞⎝ y⎠y x−yp(x,y) = P(Y = y | X = x) ⋅ P(X = x) = ⎜ ⎟(. 6) (.4) ⋅ p ( x)p y (4) = p(y = 4) = p(x = 4, y = 4) = p(4,4) = (.6) 4 ⋅(.15) = .019443⎛ ⎞3p y (3) = p(3,3) + p(4,3) = (.6) (.25) + ⎜ ⎟(.6)(.4)(.15) = . 1058⎝3⎠32⎛ ⎞2p y (2) = p(2,2) + p(3,2) + p(4,2) = (.6) (.3) + ⎜ ⎟(.6)(.4)(.25)⎝2⎠⎛4⎞2 2+ ⎜ ⎟(.6)(.4) (.15) = .2678⎝2⎠⎛2⎞p y (1) = p(1,1) + p(2,1) + p(3,1) + p(4,1) = (. 6)(.2) + ⎜ ⎟(.6)(.4)(.3)⎝1⎠⎛3⎞42⎛ ⎞3⎜ ⎟(.6)(.4)(.25) + ⎜ ⎟(.6)(.4)(.15) = .3590⎝1⎠⎝1⎠p y (0) = 1 – [.3590+.2678+.1058+.0194] = .2480x7.a. p(1,1) = .030b. P(X ≤ 1 and Y ≤ 1 = p(0,0) + p(0,1) + p(1,0) + p(1,1) = .120c. P(X = 1) = p(1,0) + p(1,1) + p(1,2) = .100; P(Y = 1) = p(0,1) + … + p(5,1) = .300d. P(overflow) = P(X + 3Y > 5) = 1 – P(X + 3Y ≤ 5) = 1 – P[(X,Y)=(0,0) or …or (5,0) or(0,1) or (1,1) or (2,1)] = 1 - .620 = .380e. The marginal probabilities for X (row sums from the joint probability table) are p x (0) =.05, p x (1) = .10 , p x (2) = .25, p x (3) = .30, p x (4) = .20, p x (5) = .10; those for Y (columnsums) are p y (0) = .5, p y (1) = .3, p y (2) = .2. It is now easily verified that for every (x,y),p(x,y) = p x (x) ⋅ p y (y), so X and Y are independent.177