麓山君陌2021-12-03 18:50:28.0
故成立 先验分布 π ( θ ) = α θ m i n α θ α + 1 \pi(\theta)=\frac{\alpha\theta_{min}^{\alpha}}{\theta^{\alpha+1}} π(θ)=θα+1αθminα 有样本的分布函数 p ( x ∣ θ ) = 1 θ n;θ)=n+ααxmax−nθn+α+1αxmaxα=θn+α+1(n+α)xmaxn+α 此时可以求后验期望估计 θ ^ E = ∫ x m a x ∞ θ π ( θ ∣ x ) d θ = ∫ x m a x ∞ ( n + α ) x m a x n + α θ n + α d θ = n + α n + α − 1 x m a x \widehat{\theta}_E=\int_{x_{max}}^{\infty}\theta\pi(\theta|x)d\theta=\int_{x_{max}}^{\infty}\frac{(n+\alpha)x_{max}^{n+\alpha}}{\theta^{n+\alpha}}d\theta=\frac{n+\alpha