Second Edition7-97.21 a.E1niYixi=1niEYixi=1niβxixi=β.b.Var1niYixi=1n2iVarYix2i=σ2n2i1x2i.Using Example 4.7.8 withai= 1/x2iwe obtain1ni1x2i≥n∑ix2i.Thus,Varˆβ=σ2∑ix2i≤σ2n2i1x2i= Var1niYixi.Becauseg(u) = 1/u2is convex, using Jensen’s Inequality we have1¯x2≤1ni1x2i.Thus,Var∑iYi∑ixi=σ2n¯x2≤σ2n2i1x2i= Var1niYixi.7.22 a.f(¯x, θ) =f(¯x|θ)π(θ) =√n√2πσe-n(¯x-θ)2/(2σ2)1√2πτe-(θ-μ)2/2τ2.b. Factor the exponent in part (a) as-n2σ2(¯x-θ)2-12τ2(θ-μ)2=-12v2(θ-δ(x))2-1τ2+σ2/n(¯x-μ)2,whereδ(x) = (τ2¯x+(σ2/n)μ)/(τ2+σ2/n) andv= (σ2τ2/n)(τ+σ2/n). Let n(a, b) denotethe pdf of a normal distribution with meanaand varianceb. The above factorization showsthatf(x, θ) = n(θ, σ2/n)×n(μ, τ2) = n(δ(x), v2)×n(μ, τ2+σ2/n),where the marginal distribution of¯Xis n(μ, τ2+σ
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