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3.6 Joint Distributions

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3.6 Joint Distributions Section 6. Joint Distributions (LECTURE NOTES 6) 101 3.6 Joint Distributions Properties of the joint (bivariate) discrete probability mass function pmf f(x; y) = P (X = x; Y = y) for random variables X and Y with ranges RX and RY where R = f(x; y)jx 2 RX ; y 2 RY g, are: • 0 ≤ f(x; y) ≤ 1, for all x 2 RX ; y 2 RY , • PPf(x; y) = 1, (x;y)2R • if S ⊂ R, P [(X; Y ) 2 S] = PPf(x; y), (x;y)2S with marginal pmfs of X and of Y , X X fX (x) = P (X = x) = f(x; y); fY (y) = P (Y = y) = f(x; y): y2RY x2RX Properties of the joint (bivariate) continuous probability density function pdf f(x; y) for continuous random variables X and Y , are: • f(x; y) ≥ 0, −∞ < x < 1; −∞ < y < 1, R 1 R 1 • −∞ −∞ f(x; y) dy dx = 1, RR • if S is a subset of two-dimensional plane, P [(X; Y ) 2 S] = S f(x; y) dy dx, with marginal pdfs of X and of Y , Z 1 Z 1 fX (x) = f(x; y) dy; fY (y) = f(x; y) dx: −∞ −∞ Random variables (discrete or continuous) X and Y are independent if and only if f(x; y) = fX (x) · fY (y): A set of n random variables are mutually independent if and only if f(x1; x2; : : : ; xn) = fX1 (x1) · fX2 (x3) ··· fXn (xn): Exercise 3.6 (Joint Distributions) 1. Discrete joint (bivariate) pmf: marbles drawn from an urn. Marbles chosen at random without replacement from an urn consist of 8 blue and 6 black marbles. Blue counts for 0 points and black counts for 1 point. Let X denote number of points from first marble chosen and Y denote number of points from second marble chosen. 102 Chapter 3. Continuous Random Variables (LECTURE NOTES 6) f(x, y) 0.60 0.40 0.20 0, blue 0, blue 1, black 1, black x, first draw y, second draw Figure 3.14: Discrete bivariate function: marbles (a) Chance of choosing two blue marbles is f(x; y) = f(0; 0) = 8·7 28 8·6 24 6·8 24 6·5 15 (i) 14·13 = 91 (ii) 14·13 = 91 (iii) 14·13 = 91 (iv) 14·13 = 91 . (b) Chance of a blue marble then black marble is f(x; y) = f(0; 1) = 8·7 28 8·6 24 6·8 24 6·5 15 (i) 14·13 = 91 (ii) 14·13 = 91 (iii) 14·13 = 91 (iv) 14·13 = 91 . (c) The joint density is first drawn, x blue, 0 blue, 0 black, 1 black, 1 second drawn, y blue, 0 black, 1 blue, 0 black, 1 8·7 28 8·6 24 6·8 24 6·5 15 f(x; y) 14·13 = 91 14·13 = 91 14·13 = 91 14·13 = 91 (i) True (ii) False (d) Chance of choosing a blue marble in first of two draws is fX (0) = P (X = 0) = f(0; 0) + f(0; 1) = 28 28 28 24 28 15 24 15 (i) 91 + 91 (ii) 91 + 91 (iii) 91 + 91 (iv) 91 + 91 . (e) Chance of choosing a black marble in first of two draws is fX (1) = P (X = 1) = f(1; 0) + f(1; 1) = 28 28 28 24 28 15 24 15 (i) 91 + 91 (ii) 91 + 91 (iii) 91 + 91 (iv) 91 + 91 . 24 24 (f) P (X + Y = 1) = f(0; 1) + f(1; 0) = 91 + 91 = 48 28 24 15 (i) 91 (ii) 91 (iii) 91 (iv) 91 . (g) The joint density, including the marginal probabilities, x f(x; y) blue, 0 black, 1 fY (y) = P (Y = y) 28 24 52 y blue, 0 91 91 91 24 15 39 black, 1 91 91 91 52 39 fX (x) = P (X = x) 91 91 1 (i) True (ii) False Section 6. Joint Distributions (LECTURE NOTES 6) 103 (h) Are first draw X and second draw Y independent? Since 28 52 52 f(0; 0) = ≈ 0:307 6= f (0) · f (0) = · ≈ 0:327 91 X Y 91 91 24 52 39 f(0; 1) = ≈ 0:264 6= f (0) · f (1) = · ≈ 0:245 91 X Y 91 91 24 39 52 f(1; 0) = ≈ 0:264 6= f (1) · f (0) = · ≈ 0:245 91 X Y 91 91 15 39 39 f(1; 1) = ≈ 0:165 6= f (1) · f (1) = · ≈ 0:184 91 X Y 91 91 X and Y are (i) independent (ii) dependent. Only necessary to show one of four equations unequal to one another to demonstrate dependence; on the other hand, must show all equations to be equal to one another to show independence. 2. Another discrete joint pmf. Consider bivariate function xy f(x; y) = ; x = 1; 2; y = 1; 2; 3: 18 (a) This function is a pmf because 0 ≤ f(x; y) ≤ 1, x = 1; 2; y = 1; 2; 3 and 2 3 X X xy 1 · 1 1 · 2 1 · 3 2 · 1 2 · 2 2 · 3 = + + + + + = 18 18 18 18 18 18 18 x=1 y=1 (i) 0 (ii) 0:5 (iii) 0:75 (iv) 1. (b) Marginal pmf of X is 3 X xy x + 2x + 3x f (x) = = = X 18 18 y=1 x x x x (i) 3 (ii) 6 (iii) 9 (iv) 10 . (c) Marginal pmf of Y is 2 X xy y + 2y f (y) = = = Y 18 18 x=1 y y y y (i) 3 (ii) 6 (iii) 9 (iv) 10 . (d) Since xy x y f(x; y) = = f (x) · f (y) = · 18 X Y 3 6 X and Y are (i) independent (ii) dependent. 1·1 1·2 2·1 (e) P (X + Y ≤ 3) = f(1; 1) + f(1; 2) + f(2; 1) = 18 + 18 + 18 = 3 4 5 6 (i) 18 (ii) 18 (iii) 18 (iv) 18 . 104 Chapter 3. Continuous Random Variables (LECTURE NOTES 6) X 1·1 2·2 (f) P Y = 1 = f(1; 1) + f(2; 2) = 18 + 18 = 3 4 5 6 (i) 18 (ii) 18 (iii) 18 (iv) 18 . 3. Continuous pdf: weight and amount of salt in potato chips. Two machines fill potato chip bags. Although each bag should weigh 50 grams each and contain 5 milligrams of salt, in fact, because of differing machines, weight, X, and amount of salt, Y , placed in each bag varies according to two graphs below. Consider amount volume = 1 f(x ,y ) of salt y 8 7 6 5 1/12 4 3 2 49 51 x weight of bag Figure 3.15: Continuous joint bivariate function: potato chips following function for potato chip machine 1 ; 49 ≤ x ≤ 51; 2 ≤ y ≤ 8 f(x; y) = 12 0 elsewhere (a) This function is a pdf because 49 ≤ x ≤ 51; 2 ≤ y ≤ 8 and Z 8 Z 51 Z 8 Z 8 1 1 x=51 1 2 y=8 2 dx dy = (x)x=49 dy = 2 dy = (y)y=2 = · 6 = 2 49 12 12 2 12 2 12 12 (i) 0 (ii) 0:5 (iii) 0:75 (iv) 1. (b) Marginal for X Z 1 Z 8 1 1 y=8 fX (x) = f(x; y) dy = dy = (y)y=2 = −∞ 2 12 12 1 1 1 1 (i) 2 (ii) 3 (iii) 4 (iv) 5 , where 49 ≤ x ≤ 51. (c) Marginal for Y Z 1 Z 51 1 1 x=51 fY (y) = f(x; y) dx = dx = (x)x=49 = −∞ 49 12 12 1 1 1 1 (i) 3 (ii) 4 (iii) 5 (iv) 6 , where 2 ≤ y ≤ 8. Section 6. Joint Distributions (LECTURE NOTES 6) 105 (d) Independence? Since 1 1 1 f(x; y) = = f (x)f (y) = × ; 12 1 2 2 6 random variables X and Y are (i) dependent (ii) independent 4. Continuous Bivariate Function: What Is Constant k? f(x ,y ) y y Y > X y = x volume = 1 1 1 x x 0 1 2 0 1 2 Figure 3.16: Continuous bivariate density f(x; y) = (2 − x)(1 − y) (a) Determine k so that k(2 − x)(1 − y) 0 ≤ x ≤ 2; 0 ≤ y ≤ 1 f(x; y) = 0 elsewhere is a joint probability density function. Since Z 1 Z 2 Z 1 Z 2 k(2 − x)(1 − y) dx dy = k (2 − x − 2y + xy) dx dy 0 0 0 0 Z 1 1 1 x=2 = k 2x − x2 − 2xy + x2y dy 0 2 2 x=0 Z 1 = k (2 − 2y) dy 0 2y=1 = k 2y − y y=0 = k (2 − 1) = 1; 1 2 3 so k = (i) 4 (ii) 4 (iii) 4 (iv) 1. (b) Marginal fX (x). Z 1 Z 1 (2 − x)(1 − y) dy = (2 − x − 2y + xy) dy 0 0 1 y=1 = 2y − xy − y2 + xy2 = 2 y=0 106 Chapter 3. Continuous Random Variables (LECTURE NOTES 6) 1 1 1 3 1 (i) 2 − 2 x (ii) 1 − 2 x (iii) 2 − 2 x, where 0 ≤ x ≤ 2. (c) Marginal fY (y). Z 2 Z 2 (2 − x)(1 − y) dx = (2 − x − 2y + xy) dx 0 0 1 1 x=2 = 2x − x2 − 2xy + x2y = 2 2 x=0 1 3 1 (i) 2 − 2y (ii) 1 − 2 y (iii) 2 − 2 y, where 0 ≤ y ≤ 1. (d) Since 1 f(x; y) = (2 − x)(1 − y) = f (x)f (y) = 1 − x × (2 − 2y) ; X Y 2 random variables X and Y are (i) dependent (ii) independent (e) Determine P (Y > X) Z 1 Z y Z 1 Z y (2 − x)(1 − y) dx dy = (2 − x − 2y + xy) dx dy 0 0 0 0 Z 1 1 1 x=y = 2x − x2 − 2xy + x2y dy 0 2 2 x=0 Z 1 5 1 = 2y − y2 + y3 dy 0 2 2 5 1 y=1 = y2 − y3 + y4 = 6 8 y=0 6 7 8 (i) 24 (ii) 24 (iii) 24 (iv) 1.
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