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The Gamma distribution is the conjugate prior of Poisson distribution. What about the Truncated Gamma distribution? Is it still the conjugate prior of Poisson distribution?

Suki Hao
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  • Thanks for your answer. I thought it as still the conjugate prior but my lecturer told me it wouldn't be the conjugate prior. I am really confused. – Suki Hao Mar 08 '19 at 14:53

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A generic if rarely mentioned result about conjugate families is that they are defined in terms of an arbitrary dominating measure $\lambda$. This means that their density wrt this dominating measure is provided by the corresponding exponential family shape $$\exp\{A(\theta)\cdot S_0 -\lambda \psi(\theta)\}$$ but that the dominating measure $\lambda$ may include an arbitrary function $C(\theta)$ of $\theta$ wrt a standard dominating measure like the Lebesgue measure (or any other), in particular the indicator of a particular subset of the parameter space.

In the case of the Poisson example, with likelihood$$\theta^S\exp\{-n\theta\}$$the prior with density against the Lebesgue measure$$\pi(\theta)\propto \theta^{S_0}\exp\{-\lambda\theta\}\mathbb{I}_{(a,b)}(\theta)$$is associated with the posterior$$\pi(\theta)\propto \theta^{S+S_0}\exp\{-(n+\lambda)\theta\}\mathbb{I}_{(a,b)}(\theta)$$which is indeed of the same shape, $S_0$ being replaced with $S+S_0$ and $\lambda$ with $n+\lambda$.

Xi'an
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