统计推断答案(打印版)
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Solutions Manual for Statistical Inference, Second Edition
George Casella University of Florida
Roger L. Berger
North Carolina State University Damaris Santana
University of Florida
α {−∞ 2λ 2e θ c (α) = µ µ . 2 4 σ α − Second Edition 3-13 c. (i) h (x ) = 1 I (x ), c (α) = α α > 0, w (α) = α, w (α) = α, x {0 Γ(α) 1 2 t 1(x ) = log(x ), t 2(x ) = −x . (ii) A line. d. (i) h (x ) = C exp(x 4)I (x ), c (θ) = exp(θ4) − ∞ < θ < ∞, w 1 (θ) = θ, w 2(θ) = θ2, w 3(θ) = θ3, t 1(x ) = −4x 3, t 2(x ) = 6x 2, t 3(x ) = −4x . (ii) The curve is a spiral in 3-space. (iii) A good picture can be generated with the Mathematica statement ParametricPlot3D[{t, t^2, t^3}, {t, 0, 1}, ViewPoint -> {1, -2, 2.5}]. 3.35 a. In Exercise 3.34(a) w 1(λ) = 1 and for a n(e θ, e θ ), w 1(θ) = 1 . b. E X = µ = αβ, then β = µ . Therefore h (x ) = 1 I (x ), α x {0 α Γ(α)( α )α , α > 0, w 1 (α) = α, w 2 (α) = α , t 1(x ) = log(x ), t 2(x ) = −x . c. From (b) then (α1, . . . , αn , β1, . . . , βn ) = (α1, . . . , αn , α1 , . . . , αn ) µ µ The pdf ( 1 )f ( (x −µ) ) is symmetric about µ because, for any ǫ > 0, o σ 1 f .(µ+ǫ)−µ . = σ σ 1 f . ǫ . = σ σ 1 f . ǫ . = σ σ 1 f .(µ−ǫ)−µ . . σ σ Thus, by Exercise 2.26b, µ is the median. P (X > x α) = P (σZ + µ > σz α + µ) = P (Z > z α) by Theorem 3.5.6. First take µ = 0 and σ = 1. a. The pdf is symmetric about 0, so 0 must be the median. Verifying this, write ¸ ∞ 1 1 1 1 .∞ 1 . π . 1 P (Z ≥ 0) = 0 π 1+z 2 dz = tan − π (z ). = .0 π 2 −0 = 2 . b. P (Z ≥ 1) = 1 tan −1(z ).∞ = 1 π − π = 1 . By symmetry this is also equal to P (Z ≤ −1). π .1 π . 2 4 . 4 Writing z = (x − µ)/σ establishes P (X ≥ µ) = 1 and P (X ≥ µ + σ) = 1 . Let X ∼ f (x ) have mean µ and variance σ2. Let Z = X −µ . Then E Z = . 1 . σ E(X − µ ) = 0 and Var Z = Var . X − µ . σ . 1 . = σ2 Var(X − µ) = . 1 . σ2 σ2 Var X = σ2 = 1. Then compute the pdf of Z , f Z (z ) = f x (σz + µ)· σ = σf x (σz + µ) and use f Z (z ) as the standard pdf. a. This is a special case of Exercise 3.42a. b. This is a special case of Exercise 3.42b. a. Let θ1 > θ2. Let X 1 ∼ f (x − θ1) and X 2 ∼ f (x − θ2). Let F (z ) be the cdf corresponding to f (z ) and let Z ∼ f (z ).Then F (x | θ1) = P (X 1 ≤ x ) = P (Z + θ1 ≤ x ) = P (Z ≤ x − θ1) = F (x − θ1) ≤ F (x − θ2) = P (Z ≤ x − θ2) = P (Z + θ2 ≤ x ) = P (X 2 ≤ x ) = F (x | θ2).