Exercise 3.9 - Bayesian analysis of the uniform distribution

Answers

The conjugate prior for uniform distribution is the Pareto distribution, whose density function is defined by:

p (𝜃 | K, b) = Pa (𝜃 | K, b) = K \cdot b^{K} \cdot 𝜃^{- (K + 1)} \cdot 𝕀 [𝜃 \geq b] .

Let $m = \max {| x_{i} |}_{i = 1}^{n}$ , the joint distribution of $𝜃$ and $𝒟$ is:

\begin{aligned} p (𝜃, 𝒟 | K, b) = & p (𝜃 | K, b) \cdot p (𝒟 | 𝜃) \\ = & K \cdot b^{K} \cdot 𝜃^{- (K + 1)} \cdot 𝕀 [𝜃 \geq b] \cdot 𝕀 [𝜃 \geq m] \cdot {(𝜃)}^{- n} \\ = & K \cdot b^{K} \cdot 𝜃^{- (K + n + 1)} \cdot 𝕀 [𝜃 \geq \max (b, m)] . \end{aligned}

Now $p (𝜃, 𝒟 | K, b) = p (𝒟 | K, b) \cdot p (𝜃 | 𝒟, K, b)$ , hence the posterior distribution depends on $𝜃$ through:

𝜃^{- (K + n + 1)} \cdot 𝕀 [𝜃 \geq \max (b, m)] .

So the posterior distribution is another Pareto distribution with hyperparameters $K + n, \max (b, \max {| x_{i} |}_{i = 1}^{n})$ .

The evidence is computed from the Bayesian rule:

\begin{aligned} p (𝒟 | K, b) & = \int_{0}^{\infty} p (𝒟, 𝜃 | K, b) d 𝜃 \\ = \int_{\max (b, m)}^{\infty} \frac{K \cdot b^{K}}{𝜃^{K + n + 1}} d 𝜃 . \end{aligned}

The rest is trivial calculus.

solour_lfq

2021-03-24 13:42