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Let $Y$ be a $\text{Gamma}(\alpha,\beta)$ random variable with known shape parameter $\alpha$ and unknown scale parameter $\beta$. Suppose we assign a $\Gamma(\alpha_0,\beta_0)$ prior to $\beta$. I am trying to show that the posterior distribution $\pi(\beta|\alpha_0,\beta_0)$ is again a Gamma random variable. I am confused because I got stuck when using the shape-scale parameterization for the Gamma distribution, but when I use the shape-rate distribution I was able to solve it. Here's my work:

Using the rule $\text{Posterior} \propto \text{Likelihood} \times \text{Prior}$, we have \begin{align*} f(\beta|\alpha_0,\beta_0) &\propto \frac{y^{\alpha - 1} e^{-y/\beta}}{\Gamma(\alpha) \beta^{\alpha}} \times \frac{\beta^{\alpha_0 - 1} e^{-\beta/\beta_0}}{\Gamma(\alpha_0) \beta_0^{\alpha_0}} \\[5pt] &\propto \beta^{\alpha_0 - \alpha -1} e^{-y/\beta - \beta/\beta_0} \end{align*} But this does not look like a Gamma random variable to me; the problem is that the $-y/\beta - \beta/\beta_0$ term contains $\beta$ in the denominator of the first fraction and the numerator of the second, so I don't see how this can be put into the form $-(\text{constant}) \beta$. Did I do something wrong? Then I attempted the same calculation using the shape-rate parameterization of the Gamma distribution: \begin{align*} f(y|r,v) = \frac{y^{r-1} e^{-vy} v^r}{\Gamma(r)}. \end{align*}

If $Y \sim \text{Gamma}(r,v)$ (with $r$ known, $v$ unknown) and $v \sim \text{Gamma}(r_0,v_0)$, then we have \begin{align*} f(v|r_0,v_0) &\propto \frac{y^{r-1} e^{-vy} v^r}{\Gamma(r)} \times \frac{v^{r_0-1} e^{-v_0 v} v_0^{r_0}}{\Gamma(r_0)} \\[5pt] &\propto v^{r + r_0 - 1} e^{-(y + v_0)v}, \end{align*} and from this we see that the posterior distribution is $\Gamma(r + r_0, y + v_0)$.

So my question is: Why does it work for the shape-rate parameterization, but not the shape-scale parameterization? Since the parameterizations are equivalent, I would expect that both should work...

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    $\begingroup$ The conjugate prior is different, that's all. The distribution of $\beta$ is not the same as that of $1/\beta$. $\endgroup$
    – jbowman
    Feb 17, 2023 at 21:31
  • $\begingroup$ @jbowman: Hmm, are you saying that the Gamma family is not the conjugate prior for Gamma when using the scale parameterization, but is conjugate when using the rate parameterization? $\endgroup$
    – Leonidas
    Feb 17, 2023 at 21:41
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    $\begingroup$ Well, yes, because the Inverse Gamma distributions and Gamma distributions are different. $\endgroup$
    – whuber
    Feb 17, 2023 at 22:53
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    $\begingroup$ The density$$f(\beta)\propto\beta^\omega\exp\{-a\beta-b\beta^{-1}\}$$is attached with a generalised inverse Gaussian distribution. $\endgroup$
    – Xi'an
    Feb 18, 2023 at 9:38
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    $\begingroup$ @jbowman After thinking about it some more, your point makes a lot of sense. I was under the false impression that two distributions with different but "equivalent" parameterizations should have the same properties w/respect to conjugacy, but now I see that's not the case. $\endgroup$
    – Leonidas
    Feb 18, 2023 at 12:43

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