uniform pareto system error

Question

Let $X\sim U(0,\theta)$. Given a sample of size n, the likeliohood function is $l(\theta \mid x)=\frac{1}{\theta^n}$ Consider a pareto prior distribution $\theta\sim pareto(k,a)$ with density $\frac{ak^a}{\theta^{a+1}}$ with $\theta >k > 0$. By bayes theorem the posteori density is given by above, but this does not coincide with the correct value $\operatorname{pareto(a+n,max\{k,x_{max}\})}$.

$p(\theta\mid x)=\frac{\frac{1}{\theta^n}\frac{ak^a}{\theta^{a+1}}}{\int_k^\infty\frac{1}{\vartheta^n}\frac{ak^a}{\vartheta^{a+1}}}\ne \operatorname{pareto(a+n,max\{k,x_{max}\})}$

Could someone tell me what I am doing wrong, i.e. what is the integration border?

Answer 1

When using Bayes' theorem, it is often simpler to work using proportionality, ignoring the constant-of-integration completely. This lets you establish the kernel of the posterior density, and this usually lets you identify the posterior distribution and add in the appropriate constant-of-integration at the end. Now, in your case you have $X_1, ..., X_n \sim \text{IID U}(0, \theta)$ with prior $\theta \sim \text{Pareto}(k, a)$. If we define $k_n \equiv \max \{k, x_1, ..., x_n \}$ we have:

$$\begin{equation} \begin{aligned} p (\theta | \boldsymbol{x}) \propto L_\boldsymbol{x}(\theta) p(\theta) &= \frac{1}{\theta^n} \mathbb{I}(\theta \geqslant \max \{x_1, ..., x_n \}) \cdot \frac{a k^a}{\theta^{a+1}} \mathbb{I}(\theta \geqslant k) \\[6pt] &\propto \frac{1}{\theta^{n+a+1}} \mathbb{I}(\theta \geqslant k_n) \\[8pt] &\propto \text{Pareto}(\theta | k_n, n+a). \\[8pt] \end{aligned} \end{equation}$$

This shows that your posterior belief is $\theta | \boldsymbol{x} \sim \text{Pareto}(k_n, n+a)$. Hence, the full density function (including the relevant constant-of-integration) is:

$$p(\theta | \boldsymbol{x}) = \frac{(n+a)k_n^{n+a}}{\theta^{n+a+1}} \mathbb{I}(\theta \geqslant k_n).$$

This posterior is consistent with the list of conjugate posteriors listed here. It is possible to obtain this same result by doing the derivation without using proportionality, so it is a useful exercise to see if you can get this.