Unformatted text preview: 422 Solutions Manual for Statistical Inference 4.63 Since X = eZ and g(z) = ez is convex, by Jensen's Inequality EX = Eg(Z) g(EZ) = e0 = 1. In fact, there is equality in Jensen's Inequality if and only if there is an interval I with P (Z I) = 1 and g(z) is linear on I. But ez is linear on an interval only if the interval is a single point. So EX > 1, unless P (Z = EZ = 0) = 1. 4.64 a. Let a and b be real numbers. Then, a + b2 = (a + b)(a + b) = a2 + 2ab + b2 a2 + 2ab + b2 = (a + b)2 . Take the square root of both sides to get a + b a + b. b. X + Y  X + Y  EX + Y  E(X + Y ) = EX + EY . 4.65 Without loss of generality let us assume that Eg(X) = Eh(X) = 0. For part (a) E(g(X)h(X)) =
 g(x)h(x)fX (x)dx g(x)h(x)fX (x)dx +
{x:h(x)0} {x:h(x)0} = g(x0 ) g(x)h(x)fX (x)dx h(x)fX (x)dx
{x:h(x)0} h(x)fX (x)dx + g(x0 )
{x:h(x)0} =
 h(x)fX (x)dx = g(x0 )Eh(X) = 0. where x0 is the number such that h(x0 ) = 0. Note that g(x0 ) is a maximum in {x : h(x) 0} and a minimum in {x : h(x) 0} since g(x) is nondecreasing. For part (b) where g(x) and h(x) are both nondecreasing E(g(X)h(X)) =
 g(x)h(x)fX (x)dx g(x)h(x)fX (x)dx +
{x:h(x)0} {x:h(x)0} = g(x0 ) g(x)h(x)fX (x)dx h(x)fX (x)dx
{x:h(x)0} h(x)fX (x)dx + g(x0 )
{x:h(x)0} =
 h(x)fX (x)dx = g(x0 )Eh(X) = 0. The case when g(x) and h(x) are both nonincreasing can be proved similarly. ...
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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

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