According to Wikipedia (https://en.wikipedia.org/wiki/Conjugate_prior) the gamma distribution is a conjugate prior for the exponential distribution (with unknown rate-parameter, $\lambda$, and hyperparameters $\alpha$ and $\beta$). Moreover the posterior predictive is the Lomax (a.k.a. Pareto type II) distribution.

While I have no doubt that these results are correct I have not been able to find any proof leading to the Lomax distribution (the part concerning the gamma distribution is easy to find). I would appreciate if someone would share a reference.


It's easy to show if you have an integer value for $\alpha$ in the prior.

We have that the posterior is $\textrm{Gamma}(\alpha^* = \alpha+n,\beta^* = \beta+\sum_{i=1}^n x_i)$, which means that the posterior density is given by:

\begin{equation} p[\lambda | \alpha^*, \beta^*] = \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\lambda^{\alpha^*-1}\exp[-\beta^*\lambda] \end{equation}

The likelihood is an exponential density:

\begin{equation} p[x_{new} | \lambda] = \lambda\exp[-\lambda x_{new}] \end{equation}

Multiplying these two together and moving things around gives us:

\begin{equation} \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\lambda^{\alpha^*}\exp[-(\beta^*+x_{neq})\lambda] \end{equation}

Integrating out $\lambda$, we get:

\begin{equation} \begin{split} & \hspace{6mm }\frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\int_{\lambda \in (0,\infty)}\lambda^{\alpha^*}\exp[-(\beta^*+x_{neq})\lambda]\\ & = \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\int_{\lambda \in (0,\infty)}\frac{(\beta^*+x_{neq})}{(\beta^*+x_{neq})}\lambda^{\alpha^*}\exp[-(\beta^*+x_{neq})\lambda]\\& = \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\frac{1}{(\beta^*+x_{neq})}\int_{\lambda \in (0,\infty)}(\beta^*+x_{neq})\lambda^{\alpha^*}\exp[-(\beta^*+x_{neq})\lambda] \end{split} \end{equation}

Which is the $\alpha^*$ moment of an $\textrm{Exp}(\beta^*+x_{neq})$ distribution. This is equivalent to $\frac{\alpha^*!}{(\beta^*+x_{neq})^{\alpha^*}}$; giving us that the previous expression is: \begin{equation} \begin{split} \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\frac{1}{(\beta^*+x_{neq})}\int_{\lambda \in (0,\infty)}(\beta^*+x_{neq})\lambda^{\alpha^*}\exp[-(\beta^*+x_{neq})\lambda] & = \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\frac{\alpha^*!}{(\beta^*+x_{neq})^{\alpha^*+1}} \end{split} \end{equation}

The $\frac{\alpha^*!}{\Gamma(\alpha^*)}$ term reduces to $\alpha^*$, giving us that the expression reduces to:

\begin{equation} \begin{split} \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\frac{\alpha^*!}{(\beta^*+x_{new})^{\alpha^*+1}} & =\frac{\beta^{\alpha^*}\alpha^*}{(\beta^*+x_{new})^{\alpha^*+1}} \end{split} \end{equation}

Which matches the density given here with $\lambda = \beta^*$ and $\alpha = \alpha^*$.

  • $\begingroup$ This was exactly what I was looking for. Thanks for the proof. $\endgroup$ – user259047 Jan 18 at 8:42

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