Bivariate or joint probability distributions

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(b) f(x, y) = f X ( x) f ( y) Y ; or 6 (c) f ( x y ) = function of x only or equivalently f ( y x ) = function of y only Example 3.5 The joint distribution function of X and Y is given by F x y y x 2 ⎛ ⎞ 2 ( , ) = 3 ⎜ + x⎟ ⎝ 2 ⎠ 0≤ x, y ≤1 = 0 otherwise (i) Find the marginal distribution and density functions. (ii) Find the joint density function. (iii) Are X and Y independent random variables? Example 3.6 X and Y have the joint probability density function 2 8x f ( x, y) = 1≤ x, y≤ 2 3 7y (a) Derive the marginal distribution function of X. (b) Derive the conditional density function of X given Y = y (c) Are X and Y independent? Given: Given: Joint density fn. f (x ,y) Joint distribution fn. F(x, y) ⏐ Integrate w.r.t ⏐ Differentiate (partially) ⏐ x and y ⏐ w.r.t. x and y ↓ ↓ Joint distribution fn. F(x, y) Joint density fn. f (x, y) v= y ∫ v=−∞ u= x ∫ u=−∞ ( , ) f u v du dv ∂ 2 F ∂x∂y .
Example 3.6(b) and (c) 7 Solution From 3.9 the conditional p.d.f of X given Y = y is denoted by f ( x y) and defined as f ( x y ) = f ( x Y y) = = ( , ) ( y) f x y and Y and fY ( y) is the marginal p.d.f. of Y. We know f ( x y) find f ( y) Y . There are two ways you can find f ( y) f Y where f ( x , y) is the joint p.d.f. of X x , = 8 7y 2 3 so we need to Y . The first way involves integration and the second way involves differentiation. I will do both ways to show you how to use the different results we have here but you should always choose the way you find easiest i.e you would not be expected to find fY ( y) both ways in any assessed work . Method 1 ∞ From 3.8 fY ( y) = ∫ f ( x , y) dx so f ( y) −∞ Y = 2 2 8x ∫ 3 7y dx = 1 8 y 2 7 3 2 ∫ x dx = 1 2 3 8 ⎡ x ⎤ 3 ⎢ 7y ⎣ 3 ⎥ = ⎦ 1 = 8 7y 3 3 ⎡2 ⎢ ⎣ 3 − 1⎤ ⎥ 3 = 8 ⎦ 7y 3 ⎡7 ⎣ ⎢3 ⎤ ⎦ ⎥ = 8 y 3 3 Method 2 From 3.11 f ( y) Y = dF Y ( y) dy From 3.10 FY ( y) = F( x y) F F where F ( y) MAX , where x MAX Y ( y) = F( 2, y) and from part (a), F( x y) ( y) Hence f ( y) 4 21 2 1 ⎛ 1 1 ⎞ ⎜ − 2 ⎟ = 4 ⎛ ⎝ y ⎠ 3 1 1 ⎞ ⎜ − 2 ⎟ ⎝ y ⎠ 3 2, = ( − ) Y = d ⎛ 4 ⎛ dy 3 1 1 ⎞⎞ ⎜ ⎜ − 2 ⎟⎟ = 4 ⎛ 2 ⎞ ⎜ 3 ⎟ = ⎝ ⎝ y ⎠⎠ 3 ⎝ y ⎠ Y is the marginal c.d.f of Y. is the largest value of x, so 4 21 3 , = ( x − ) 8 3y 3 hence F ( y) ⎛ 1 ⎜1− ⎝ 1 ⎞ 2 ⎟ so y ⎠ Y = 4 ⎛ 3 1 1 ⎞ ⎜ − 2 ⎟ . ⎝ y ⎠ as with method 1. Now therefore the conditional density function of X given Y = y , f ( x y) is given by So ( ) f f ( x y ) = ( x, y ) f ( y) Y = ⎛ 8x ⎜ ⎝ 7y ⎛ 8 ⎜ ⎝ 3y 2 3 3 ⎞ ⎟ ⎠ ⎞ ⎟ ⎠ = 3 x 7 f x y = 3 3 x 7 1 ó x ó 2 and 1 ó y ó 2 = 0 otherwise (c) Now f ( x y) is a function of x only, so using result 3.12(c), X and Y are independent. Notice also that f ( x y ) = f X ( x) which you would expect if X and Y are independent. 3
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Page 5: 3.9 Conditional probability density
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