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solution2

# solution2 - NTHU MATH 2820 2008 3-70 Fx(x y = Pr X(1 x X(n...

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c r r P ( A B ) = P ( B ) A B －P( ) 3-70 { } { } { } { } { } { } { } x r (1) (n) r (1) (n) r (n) r (1) (n) r 1 2 n r 1 2 n F (x, y) = P ( X x, X y ) = P ( X x X y ) = P ( X y ) P ( X x X y ) = P ( X ,X , ,X y ) P ( x < X ,X , ,X y ) > " " [ ] [ ] = F(y) F(y) F(x) n n 3-77 1 2 n ( k ) ( k) ( k 1) ( k) ( k ) ( k) ( k ) ( k ) k n k U , U , , U ~ U( 0, 1), n! the joint p.d.f. of U and U is f (x, y) = x (1 y) , (k 2)! (n k)! U W = U U U = W Z let | J | = Z = U U = Z " －１ －１ －1 －1 －2 ( k 1) ( k) ( k) ( k ) ( k) k n k U 0 1 W Z = 1 U U 1 1 W Z n! then the joint p.d.f. of W and Z is g (w , z) = f (z , w + z) | J | = z (1 z w) (k 2)! (n k)! since 0 < U < U 1 0 < < w + z 1 0 < z 1 w thus, the ma z = < < < －１ －2 － － 1 w 1 w 0 0 k n k k rginal p.d.f. of W is n! 1 g(w) = g (w , z) dz z (1 z w) dz , let u = , then du = dz (k 2)! (n k)! 1 w 1 w n! [ u (1 w) ] [ ( 1 u ) (1 (k 2)! (n k)! z = = －2 －2 － － 1 0 1 0 n k k n k k n w) ] (1 w) du n! (1 w) u ( 1 u ) du (k 2)! (n k)! n! ( ) ( ) (1 w) (k 2)! (n k)! ( ) = Γ Γ = Γ n－1 －2 n－1 －1 －k＋1 n n (1 w) , 0 w 1 = n－1 4-13 [ ] ( ) r 0 0 0 x 0 0 t suppose that f(x) is the p.d.f. of the random variable X, then 1 F(x) dx P ( X x ) dx f(t) dt dx 0 < x < t < f(t) dx dt = = = ∫ ∫ ∫ ∫ 0 x x t f(t) dt E [ X ] suppose that X ~ Exp( ), then f(x) = e , F(x) = 1 e , 0 < x < , then E λ λ λ λ = = [ ] x x 0 0 0 1 1 [ X ] = 1 F(x) dx e dx [ e ] = λ λ λ λ = = A B x t x < t ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~ beta function ~~~~~~~~~~ ~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~ NTHU MATH 2820, 2008 Solution to Homework 2 made by 陳陳陳陳陳

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4-36 2 r r r 0 2 X ~ U( 0, 1), f(x) = 1, 0 x 1, Y = X (a) let G(y) be the d.f. of Y, then G(y) = P ( Y y ) = P ( X y ) = P ( X y ) = 1 dx thus, the p.d.f. of Y is g(y) = G (y) = 2 y, 0 y 1 E [ Y ] = y g(y) y 2 3 1 0 0 0 3 1 2 0 0 1 1 1 2 2 dy = 2 y dy = [ y ] = 3 3 (b) by Theorem 4.1.1. A. 2 2 E [ Y ] = x dx = [ x ] = 3 3 4-49 x y x y X ~ ( , ) , Y ~ ( , ), , X and Y are indep. Z = X + (1 )Y, (a) E [ Z ] = E [ X + (1 )Y ] = E [ X ] + (1 ) E [Y ] = + (1 ) = μ σ μ σ σ σ α α α α α α α μ α μ μ ( ) 2 2 2 2 2 2 x y 2 2 x (b) Var( Z ) = Var( X + (1 )Y ) = Var( X ) + (1 ) Var( Y ) X and Y are indep. = + (1 ) f( ) = Var( Z ) = + (1 ) α α α α α σ α σ α α σ α 2 2 y 2 2 2 2 y x y y 2 2 x y 2 2 2 2 2 2 x y x y x y 2 2 x y x f ( ) = 2 2 (1 ) = 0 when = Var( Z ) has the minimum when = f ( ) = 2 + 2 > 0 1 X + Y (c) when = , we have Var( Z ) = Var( ) = 2 2 σ σ σ σ σ α α σ α σ α α σ σ σ σ σ σ
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