Unformatted text preview: 24 Solutions Manual for Statistical Inference 2.9 From the probability integral transformation, Theorem 2.1.10, we know that if u(x) = Fx (x), then Fx (X) uniform(0, 1). Therefore, for the given pdf, calculate u(x) = Fx (x) = . 2.10 a. We prove part b), which is equivalent to part a). b. Let Ay = {x : Fx (x) y}. Since Fx is nondecreasing, Ay is a half infinite interval, either open, say (, xy ), or closed, say (, xy ]. If Ay is closed, then FY (y) = P (Y y) = P (Fx (X) y) = P (X Ay ) = Fx (xy ) y. The last inequality is true because xy Ay , and Fx (x) y for every x Ay . If Ay is open, then FY (y) = P (Y y) = P (Fx (X) y) = P (X Ay ), as before. But now we have P (X Ay ) = P (X (  ,xy )) = lim P (X (, x]),
xy 0 if x 1 (x  1)2 /4 if 1 < x < 3 1 if 3 x Use the Axiom of Continuity, Exercise 1.12, and this equals limxy FX (x) y. The last inequality is true since Fx (x) y for every x Ay , that is, for every x < xy . Thus, FY (y) y for every y. To get strict inequality for some y, let y be a value that is "jumped over" by Fx . That is, let y be such that, for some xy , lim FX (x) < y < FX (xy ).
xy For such a y, Ay = (, xy ), and FY (y) = limxy FX (x) < y. 2.11 a. Using integration by parts with u = x and dv = xe x2 2 dx then EX 2 =
 x2 1 x2 1 e 2 dx = 2 2 xe x2 2 +
  e x2 2 dx = 1 (2) = 1. 2 Using example 2.1.7 let Y = X 2 . Then
y 1 y 1 1 1 y e 2 . fY (y) = e 2 + e 2 = 2 y 2y 2 2 Therefore, EY =
0 y y 1 y 1 e 2 dy = 2y 2 e 2 2y 2 +
0 0 y
1 1 2 e y 2 1 dy = ( 2) = 1. 2
y 2 This was obtained using integration by parts with u = 2y 2 and dv = 1 e 2 fY (y) integrates to 1. b. Y = X where  < x < . Therefore 0 < y < . Then FY (y) and the fact the = P (Y y) = P (X y) = P (y X y) = P (x y)  P (X y) = FX (y)  FX (y). ...
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics, Probability

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