solution3

# solution3 - 2.1 Solution (a) f X ( x) = 42*5(1 - x), 0 < x...

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2.1 Solution (a) () X f x = 42*5(1 x), 0 < x < 1; y = x^3= g(x), monotone, and y= (0, 1). Use Theorem 2.1.5. Y f y = 11 4 / 3 (( ) ) ( ) 1 41 4 , 0 1 X d f gy y y y dy −− =− < < =14y 14y4/3, 0 < y < 1. If one function f(x) is a p.d.f., it must satisfy two conditions, (1)f(x) is non negative (2) 1 X f xdx −∞ = Here the first condition is easily be checked. 1 4/3 2 7/3 0 1 ( 1 4 1 4 ) 7 6 1 0 Y f y dy y y dy y y −∞ = −= ∫∫ (b) 7 () 7 x X f xe = , 0 < x < 1, y = 4x + 3, monotone, and y = (3, ). Use Theorem 2.1.5. Y f y = 7 / 4 ( 3 ) ) ) ( ) 7 / 4 , 3 y X d fg y g y e y dy =< < Obviously, Y f y is a non negative function. To check the integral, 77 (3 ) ) 44 3 7 /4 1 3 yy Y fy d y e d y e −∞ == = (c) ( 0 ) ( ) YX Fy P X y F y =≤ = (( ) ) ) ) 1 2 X Y dF y dF y f yf y dy dy y = So, 1/2 2 1 ( ) 1 5 ( 1 ),0 1 2 f y y y y y < < Y f y is a non-negative function, then check the integral, 1 1/2 2 3/2 2 5/2 0 1 1 1 ) 1 0 1 5 6 1 0 Y f y dy y y dy y y y −∞ = + = 2.4 Solution (a)f(x) is a p.d.f. Because it’s a non negative function and 0 0 1 22 xx f xdx e dx λλ ∞∞ −∞ −∞ = += (b) If 0 t , 111 222 t t X t Ft ed x e e λ

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## This homework help was uploaded on 04/07/2008 for the course STAT 3659 taught by Professor Gerardohernandezdelvalle during the Spring '08 term at Columbia.

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solution3 - 2.1 Solution (a) f X ( x) = 42*5(1 - x), 0 < x...

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