Tutorial 9_Solution

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1801/2901 Probability and Statistics: Foundations of Actuarial Science Example Class 9 (Suggested Solution) 1. Consider X 1 , X 2 , X 3 be independent integer-valued random variables such that p ij := P ( X i ≥ j ), p i 1 = 1 and p i 4 = 0 for i = 1 , 2 , 3. (a) Find the distribution of X min := min { X 1 , X 2 , X 3 } . (b) Find the distribution of X max := max { X 1 , X 2 , X 3 } . Solution: (a) Consider P ( X min ≥ x ) = P ( X 1 ≥ x, X 2 ≥ x, X 3 ≥ x ), P ( X min = 3) = P ( X min ≥ 3) = P ( X 1 ≥ 3 , X 2 ≥ 3 , X 3 ≥ 3) = p 13 p 23 p 33 P ( X min = 2) = P ( X min ≥ 2) - P ( X min ≥ 3) = P ( X 1 ≥ 2 , X 2 ≥ 2 , X 3 ≥ 2) - p 13 p 23 p 33 = p 12 p 22 p 32 - p 13 p 23 p 33 P ( X min = 1) = P ( X min ≥ 1) - P ( X min ≥ 2) = P ( X 1 ≥ 1 , X 2 ≥ 1 , X 3 ≥ 1) - p 12 p 22 p 32 = 1 - p 12 p 22 p 32 (b) Consider P ( X max ≤ x ) = P ( X 1 ≤ x, X 2 ≤ x, X 3 ≤ x ), P ( X max = 1) = P ( X max ≤ 1) = P ( X 1 ≤ 1 , X 2 ≤ 1 , X 3 ≤ 1) = P ( X 1 ≤ 1) P ( X 2 ≤ 1) P ( X 3 ≤ 1) = (1 - p 12 )(1 - p 22 )(1 - p 32 ) P ( X max = 2) = P ( X max ≤ 2) - P ( X max ≤ 1) = P ( X 1 ≤ 2 , X 2 ≤ 2 , X 3 ≤ 2) - (1 - p 12 )(1 - p 22 )(1 - p 32 ) = P ( X 1 ≤ 2) P ( X 2 ≤ 2) P ( X 3 ≤ 2) - (1 - p 12 )(1 - p 22 )(1 - p 32 ) = (1 - p 13 )(1 - p 23 )(1 - p 33 ) - (1 - p 12 )(1 - p 22 )(1 - p 32 ) P ( X max = 3) = P ( X max ≤ 3) - P ( X max ≤ 2) = 1 - (1 - p 13 )(1 - p 23 )(1 - p 33 ) 2. Let X 1 ∼ Gamma(3 . 9 , 1) and X 2 ∼ Gamma(6 . 1 , 1) be two independent random variables. (a) Determine the joint pdf of Y 1 = X 1 /X 2 and Y 2 = X 1 + X 2 . (b) Determine the marginal pdf of Y 2 . (c) Find P ( Y 2 ≥ 20 . 00) . Solution: 1

(a) From the transformation, we have X 1 = Y 1 Y 2 1 + Y 1 and X 2 = Y 2 1 + Y 1 . The inverse of Jacobian determinant is given by J - 1 ( y 1 , y 2 ) = [ J ( x 1 , x 2 )] - 1 = ∂x 1 ∂y 1 ∂x 1 ∂y 2 ∂x 2 ∂y 1 ∂x 2 ∂y 2 = y 2 (1 + y 1 ) 2 y 1 1 + y 1 - y 2 (1 + y 1 ) 2 1 1 + y 1 = y 2 (1 + y 1 ) 2 . The joint pdf of ( Y 1 , Y 2 ) is f ( y 1 , y 2 ) = f ( x 1 , x 2 ) × | J - 1 ( y 1 , y 2 ) | = f ( x 1 ) f ( x 2 ) × | J - 1 ( y 1 , y 2 ) | = 1 Γ(3 . 1) x 3 . 1 - 1 1 e - x 1 × 1 Γ(6 . 9) x 6 . 9 - 1 2 e - x 2 × | J - 1 ( y 1 , y 2 ) | = 1 Γ(3 . 1)Γ(6 . 9) x 2 . 1 1 x 5 . 9 2 e - ( x 1 + x 2 ) × y 2 (1 + y 1 ) 2 = 1 Γ(3 . 1)Γ(6 . 9) y 1 y 2 1 + y 1 2 . 1 y 2 1 + y 1 5 . 9 y 2 (1 + y 1 ) 2 e - y 2 = 1 Γ(3 . 1)Γ(6 . 9) 1 1 + y 1 10 y 2 . 1 1 y 9 2 e - y 2 , y 1 > 0 , y 2 > 0 . (b) Since the inner integral does not involve y 2 , 1 = Z ∞ 0 Z ∞ 0 1 Γ(3 . 1)Γ(6 . 9) 1 1 + y 1 10 y 2 . 1 1 y 9 2 e - y 2 dy 1 dy 2 = " Z ∞ 0 1 Γ(3 . 1)Γ(6 . 9) 1 1 + y 1 10 y 2 . 1 1 dy 1 # Z ∞ 0 y 9 2 e - y 2 dy 2 = " Z ∞ 0 Γ(10) Γ(3 . 1)Γ(6 . 9) 1 1 + y 1 10 y 2 . 1 1 dy 1 # Z ∞ 0 1 Γ(10) y 9 2 e - y 2 dy 2 . The second integrand is the probability density function of Gamma(10 , 1). Therefore, the second integral is 1. Hence, the first integral is also 1. Thus, the marginal density of Y 2 is f ( y 2 ) = Z ∞ 0 1 Γ(3 . 1)Γ(6 . 9) 1 1 + y 1 10 y 2 . 1 1 y 9 2 e - y 2 dy 1 = y 9 2 e - y 2 Z ∞ 0 1 Γ(3 . 1)Γ(6 . 9) 1 1 + y 1 10 y 2 . 1 1 dy 1 = 1 Γ(10) y 9 2 e - y 2 " Z ∞ 0 Γ(10) Γ(3 . 1)Γ(6 . 9) 1 1 + y 1 10 y 2 . 1 1 dy 1 # = 1 Γ(10) y 9 2 e - y 2 shows that Y 2 ∼ Gamma(10 , 1). (c) Consider 2 Y 2 ∼ Gamma(10 , 0 . 5), which is equivalent to χ 2 20 , P ( Y 2 ≥ 20 . 00) = P (2 Y 2 ≥ 40 . 00) = 0 . 005 2

since from the χ 2 table, we could find the upper 0 . 005 quantile to be 40 . 00. 3. Suppose that X and Y are random variables with the joint probability density function f ( x, y ) =    x e - x ( y +1) , for x > 0, y > 0 , 0 , otherwise . (a) Are X and Y independent? Why? (b) Let Z = XY . Determine the pdf of Z . (c) Let U = X ( Y + 1), V = 1 Y + 1 . Determine the joint pdf of U and V . Solution: (a) The marginal pdf of X and Y are given by f X ( x ) = Z ∞ 0 x e - x ( y +1) dy = h - e - x ( y +1) i | ∞ y =0 = e - x , x > 0 , f Y ( y ) = Z ∞ 0 x e - x ( y +1) dx = - Z ∞ 0 x y + 1 d e - x ( y +1) = 1 ( y + 1) 2 , y > 0 . Since f ( x, y ) 6 = f X ( x ) · f Y ( y ), X and Y are not independent. (b) f Z ( z ) = Z ∞ -∞ f ( x, z/x ) 1 x dx. Since x > 0 , y = z/x > 0 → x > 0 , z > 0, then f Z ( z ) = Z ∞ 0 f ( x, z/x ) 1 x dx = Z ∞ 0 x e - ( z + x ) · 1 x dx = e - z , z > 0 . (c) From U = X ( Y + 1), V = 1 Y + 1 , we have X = UV and Y = 1 V - 1 . The inverse of Jacobian determinant is given by J - 1 ( u, v ) = ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v = v u 0 - 1 v 2 = - 1 v . Then joint distribution of ( U, V ) is given by f ( u, v ) = f ( x, y ) × J - 1 ( u, v ) = uv e - u × 1 v = u e - u , u > 0 . 4. For n = 1 , 2 , . . . , let X n i . i . d ∼ N (0 , 0 . 5), W n = X 2 n - X 2 n - 1 and Y n = ∑ n j =1 W 2 j . 3

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(a) What is the distribution of W n ? (b) What is the distribution of Y n ? (c) Find P ( Y 3 > 9 . 35) (d) Find P ( Y 24 > 8 . 09) (e) Find P (24 . 43 ≤ Y 40 < 59 . 24) (f) Find the 95 th percentile of Y 11 . (g) Find the 1 st percentile of Y 15 . Solution: (a) From the result in Tutorial 8 Question 2 part (g), for a , b i are real number, Y i ∼ N ( μ i , σ 2 i ) independently ⇒ a + n X i =1 b i Y i ∼ N a + n X i =1 b i μ i , n X i =1 b 2 i σ 2 i ! , we have W n = X 2 n - X 2 n - 1 ∼ N (0 , 1). (b) Since W n i . i . d ∼ N (0 , 1), from the result in Tutorial 8 Question 2 part (h), we have Y n ∼ χ 2 n . (c) From the chi-square table, we found that P ( Y 3 > 9 . 35) = 0 . 025. (One tailed upper 0.025 quantile) (d) P ( Y 24 > 8 . 09) = 1 - P ( Y 24 ≤ 8 . 09) = 0 . 999 (One tailed lower 0.001 quantile). (e) P (24 . 43 ≤ Y 40 < 59 . 24) = 1 - 0 . 05 = 0 . 95 (Two tailed 0.05 quantile). (f) Since P ( Y 11 ≤ 19 . 68) = 0 . 95, the 95 th percentile of Y 11 is 19.68. (g) Since P ( Y 15 ≤ 5 . 23) = 0 . 01, the 1 th percentile of Y 11 is 5.23. 4

Tutorial 9_Solution

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