贝叶斯估计习题

x x x 是来自如下指数分布的一个观察值

p ( x ∣ θ ) = e − ( x − θ ) , x ≥ θ p(x|\theta)=e^{-(x-\theta)},x\geq\theta p(x∣θ)=e−(x−θ),x≥θ

取柯西分布作为 θ \theta θ 的先验分布

π ( θ ) = 1 π ( 1 + θ 2 ) \pi(\theta)=\frac{1}{\pi(1+\theta^2)} π(θ)=π(1+θ2)1​

求 θ \theta θ 的最大后验估计 θ ^ M D \hat{\theta}_{MD} θ^MD​

π ( θ ∣ x ) = p ( x ∣ θ ) π ( θ ) m ( x ) = e − ( x − θ ) m ( x ) ( 1 + θ 2 ) π \pi(\theta|x) =\frac{p(x|\theta)\pi(\theta)}{m(x)} =\frac{e^{-(x-\theta)}}{m(x)(1+\theta^2)\pi} π(θ∣x)=m(x)p(x∣θ)π(θ)​=m(x)(1+θ2)πe−(x−θ)​

d d θ π ( θ ∣ x ) = e − x m ( x ) π [ e θ 1 + θ 2 − 2 θ e θ ( 1 + θ 2 ) 2 ] = e − x e θ ( θ − 1 ) 2 m ( x ) ( 1 + θ 2 ) 2 π ≥ 0 \begin{align*} \frac{\mathcal{d}}{\mathcal{d}\theta}\pi(\theta|x) &=\frac{e^{-x}}{m(x)\pi}\left[\frac{e^\theta}{1+\theta^2}- \frac{2\theta e^\theta}{(1+\theta^2)^2}\right]\\&= \frac{e^{-x}e^\theta(\theta-1)^2}{m(x)(1+\theta^2)^2\pi}\geq0 \end{align*} dθd​π(θ∣x)​=m(x)πe−x​[1+θ2eθ​−(1+θ2)22θeθ​]=m(x)(1+θ2)2πe−xeθ(θ−1)2​≥0​

又因为 x ≥ θ x\geq\theta x≥θ

所以 θ ^ M D = x \hat{\theta}_{MD}=x θ^MD​=x