0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\alpha = \alpha^*$.


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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.