# Posterior distribution of a $\text{Gamma}(\alpha,\beta)$ random variable given a Gamma prior for $\beta$

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...

• The conjugate prior is different, that's all. The distribution of $\beta$ is not the same as that of $1/\beta$. Feb 17, 2023 at 21:31
• @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? Feb 17, 2023 at 21:41
• Well, yes, because the Inverse Gamma distributions and Gamma distributions are different.
– whuber
Feb 17, 2023 at 22:53
• The density$$f(\beta)\propto\beta^\omega\exp\{-a\beta-b\beta^{-1}\}$$is attached with a generalised inverse Gaussian distribution. Feb 18, 2023 at 9:38
• @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. Feb 18, 2023 at 12:43