Dr. Hackney STA Solutions pg 66

Dr. Hackney STA Solutions pg 66 - 4-22 Solutions Manual for...

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Unformatted text preview: 4-22 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 + b|2 = (a + b)(a + b) = a2 + 2ab + b2 |a|2 + 2|ab| + |b|2 = (|a| + |b|)2 . Take the square root of both sides to get |a + b| |a| + |b|. b. |X + Y | |X| + |Y | E|X + Y | E(|X| + |Y |) = E|X| + E|Y |. 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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