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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.

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  • $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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