1-Bayesian Estimation （P317）
Suppose that x = θ + ν w h e r e ν i s a n N ( 0 , σ ) random variable and θ is the value of a n N ( θ 0 , σ 0 ) random variable θ (Fig. 8-7). Find the bayesian estimate θ o f θ . \begin{aligned}&\text{Suppose that x}=\theta+\nu\mathrm{~where~}\nu\mathrm{~is~an~}N(0,\sigma)\text{ random variable and }\theta\text{ is the value of}\\&\mathrm{an~}N(\theta_0,\sigma_0)\text{ random variable }\theta\text{ (Fig. 8-7). Find the bayesian estimate }\theta\mathrm{~of~}\theta.\end{aligned} Suppose that x=θ+ν where ν is an N(0,σ) random variable and θ is the value ofan N(θ0,σ0) random variable θ (Fig. 8-7). Find the bayesian estimate θ of θ.
条件：
f v ( v ) ∼ e − v 2 / 2 σ 2 f θ ( θ ) ∼ e − ( θ − θ 0 ) 2 / 2 σ 0 2 \mathrm{f}_{\mathbf{v}}(\mathbf{v})\sim e^{-\mathbf{v}^{2}/2\sigma^{2}}\quad\mathrm{f}_{\theta}(\theta)\sim e^{-(\theta-\theta_{0})^{2}/2\sigma_{0}^{2}} fv(v)∼e−v2/2σ2fθ(θ)∼e−(θ−θ0)2/2σ02
证明：
f θ ( θ ∣ x ) ∼ e − ( θ − θ 1 ) 2 / 2 σ 1 2 \mathrm{f}_{\theta}(\theta|\mathrm{x})\sim e^{-(\theta-\theta_{1})^{2}/2\sigma_{1}{}^{2}} fθ(θ∣x)∼e−(θ−θ1)2/2σ12，where 1 σ 1 2 ≡ 1 σ 0 2 + n σ 2 θ 1 ≡ σ 1 2 σ 0 2 θ 0 + n σ 1 2 σ 2 x ˉ \frac{1}{{\sigma_{1}}^{2}}\equiv\frac{1}{{\sigma_{0}}^{2}}+\frac{\mathrm{n}}{\sigma^{2}}\quad\theta_{1}\equiv\frac{{\sigma_{1}}^{2}}{{\sigma_{0}}^{2}}\:\theta_{0}+\frac{\mathrm{n}{\sigma_{1}}^{2}}{\sigma^{2}}\:\bar{\mathrm{x}} σ121≡σ021+σ2nθ1≡σ02σ12θ0+σ2nσ12xˉ
Proof
似然函数，观测值的条件分布为：
随机变量 v = x − θ v = x - \theta v=x−θ的分布与 ν \nu ν相同， f x ( x ∣ θ ) f_x(x|\theta) fx(x∣θ)可以等价地表示为 f v ( x − θ ) f_v(x-\theta) fv(x−θ)，这表明给定 θ \theta θ时
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