By George Casella and Roger Berger

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**Additional info for Solutions Manual for Statistical Inference**

**Sample text**

J2 k ∂ 2 wi (θ) ti (x) i=1 ∂θj2 exp( −θ 2 )θ > 0, w1 (θ) = 1 2θ , (ii) The nonnegative real line. 1 b. (i) h(x) = I{−∞

1 b. (i) h(x) = I{−∞

The inequality is because x/σ2 > x/σ1 (because x > 0 and σ1 > σ2 > 0), and F is nondecreasing. For x ≤ 0, F (x | σ1 ) = P (X1 ≤ x) = 0 = P (X2 ≤ x) = F (x | σ2 ). To get strict inequality for some x, let (a, b] be an interval such that a > 0, b/a = σ1 /σ2 and P (a < Z ≤ b) = F (b) − F (a) > 0. Let x = aσ1 . Then F (x | σ1 ) = F (x/σ1 ) = F (aσ1 /σ1 ) = F (a) < F (b) = F (aσ1 /σ2 ) = F (x/σ2 ) = F (x | σ2 ). 43 a. 3. For θ1 > θ2 , FY (y|θ1 ) = 1 − FX 1 θ1 y ≤ 1 − FX 1 θ2 y = FY (y|θ2 ) for all y, since FX (x|θ) is stochastically increasing and if θ1 > θ2 , FX (x|θ2 ) ≤ FX (x|θ1 ) for all x.