Let the sampling distribution be
$$\mathbf{Y}|\Theta\sim U(0,\theta)$$
and
$$f(\mathbf{y}|\theta)=\prod^n_{i=1}\frac{1}{\theta}\times \mathbb{1\{0<y_i\le\theta\}}=\frac{1}{\theta^n}\times \mathbb{1}\{0<y_i\}\times \mathbb{1}\{c'\le\theta\},$$
where $c'=\text{max}\{y_1,…,y_i\}.$
Let the prior distribution be Pareto distributed. That is
$$p(\theta)=\frac{Kb^K}{\theta^{K+1}}\mathbb\times{1\{b\le\theta\}}\propto \frac{1}{\theta^{K+1}}\mathbb\times{1\{b\le\theta\}}.$$
I'm trying to get the posterior distribution. That is,
$$p(\theta|\mathbf{y})=\frac{1}{\theta^n}\frac{1}{\theta^{K+1}}\times \mathbb{1}\{0<y_i\}\times \mathbb{1}\{\text{max}\{b,c'\}\le\theta\} \\
=\theta^{-(n+K+1)}\times \mathbb{1}\{0<y_i\}\times \mathbb{1}\{\text{max}\{b,c'\}\le\theta\}.$$
However, the correct answer is
$$p(\theta|\mathbf{y})=\theta^{-(n+K)}\times \mathbb{1}\{b\le\theta\}\times \mathbb{1}\{\text{max}\{b,c'\}\le\theta\}$$
but I don't know how to get there.
Best Answer
$$\Pi(\theta|y_1,\cdots ,y_n)\propto f(y_1,\cdots ,y_n |\theta) \Pi(\theta)$$
$$\propto \frac{1}{\theta^n}1_{(c\prime \leq \theta)} \frac{1}{\theta^{k+1}}1_{(b \leq \theta)}$$
$$\propto \frac{1}{\theta^{n+k+1}}1_{(c\prime \leq \theta)}1_{(b \leq \theta)}$$
$$\propto \frac{1}{\theta^{n+k+1}}1_{(\max\{c\prime,b\} \leq \theta)}$$
so $$(\theta|y_1,\cdots ,y_n)\sim Pareto(n+k,\max\{c\prime,b\} )$$