I am trying to calculate a posterior density given distribution and a prior. And I am a bit confused about how I should act as the domain of the distribution depends on the parameter. I am talking about the distribution with density:

$g(x) = \frac{\alpha p^{a}}{x^{\alpha+1}}\mathbb{1}_{[x\geq p]}$

and prior

$\pi(p)\propto p^{\gamma-1}e^{-\beta p}$, where $\alpha,\beta,\gamma>0$.

I am trying to use the basic approach for $iid$ $x_{i}$ with density $g$. Posterior is likelihood times prior. Then i get:

\begin{align}\pi(p|x_{1},...,x_{n})&\propto f(p|x_{1},...,x_{n}) \pi(p)\\ &\propto \prod_{i=1}^{n} g(x_{i}) p^{\gamma-1}e^{-\beta p}\\ &= \prod_{i=1}^{n} \frac{\alpha p^{a}}{x_{i}^{\alpha+1}}\mathbb{1}_{[x_{i}\geq p]} p^{\gamma-1}e^{-\beta p}\\ &\propto \alpha^{n} p^{\alpha n}\prod_{i=1}^{n}\mathbb{1}_{[x_{i}\geq p]} p^{\gamma-1}e^{-\beta p}\\ &\propto\mathbb{1}_{[\min_{i} x_{i}\geq p]} p^{n\alpha + \gamma -1}e^{-\beta p}\end{align}

Is this prior conjugate to the data model? It is if we ignore the indicator but I don't see how I can easily drop it. Or should I think of the domains in a different way? The prior and the model are conjugate in the support of the posterior? I am a little stuck and don't know how to proceed.

  • $\begingroup$ Thanks for the reply, but I am not sure I understand. On which set would I define that indicator, and wouldn't that alter the qualitative meaning of the prior if I assume further information? $\endgroup$
    – SebastianP
    Aug 10, 2022 at 19:53
  • $\begingroup$ To expand on @Xi'an 's comment, if you introduce the indicator $1_{[p < \infty]}$ into the prior, and rewrite the indicator in the posterior as $1_{[p < \min_i x_i]}$, I believe you're there. You can see how sequential updates of the posterior would work after that. $\endgroup$
    – jbowman
    Aug 10, 2022 at 20:05
  • $\begingroup$ @jbowman: I was thinking more of a generic $1_{[p<p^+]}$, not necessarily the case when $p^+=\infty$. $\endgroup$
    – Xi'an
    Aug 10, 2022 at 20:29

1 Answer 1


If$$\pi(p)\propto p^{\gamma-1}e^{-\beta p}\mathbb I_{(0,p^+)}(p)\qquad\gamma,\beta,p^+>0\tag{1}$$ then $$\pi(p|x_{1},...,x_{n})\propto p^{n\alpha+\gamma-1}e^{-\beta p}\mathbb I_{(0,p^+\wedge\min\{x_i\})}(p)$$ which is from the same family as (1) with \begin{align} \gamma &\mapsto n\alpha+\gamma\\ \beta &\mapsto \beta\\ p^+ &\mapsto p^+\wedge\min\{x_i\} \end{align} hence conjugate.

Note that the fact that $\beta$ is not actualised after observing the sample means that the conjugate family is over-parameterised. One could then argue that there is one conjugate family for each value of $\beta$, which can also be seen as part of the dominating measure. Or, else, that this part is altogether superfluous and that $\beta=0$ also leads to a conjugate (Beta) family.

  • $\begingroup$ Thanks a lot for the detailed response. That helps me a lot. I bow what you were trying to say. One quick question. Is there a small typo in the indicator for the posterior? I woukd assume it should be this indicator: $$\mathbb{1}_{(0,\min_{i} x_{i} \wedge p^{+})} (p).$$ Or am I misunderstanding something again? $\endgroup$
    – SebastianP
    Aug 12, 2022 at 9:54
  • $\begingroup$ Yes, this is a typo! Well-spotted. $\endgroup$
    – Xi'an
    Aug 12, 2022 at 11:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.