I need to specify priors for the parameters of a gamma distribution. Normally the gamma distribution is parametrized in either the "rate-form'':

$f\left(x;\alpha,\beta\right)=\frac{\beta^{\alpha}}{\Gamma\left(\alpha\right)}x^{\alpha-1}e^{-\beta x} $

or the "scale-form":


However, the only thing I know a priori is something about the mean ( "rate-form": $E\left[x\right]=\frac{\alpha}{\beta}$ or "scale-form" $E\left[x\right]=\kappa\theta$). Is it possible (and legal) to parametrize the gamma distribution as e.g.:

$f\left(x;\mu,\beta\right)=\frac{\beta^{\mu\beta}}{\Gamma\left(\mu\beta\right)}x^{\mu\beta-1}e^{-\beta x} $

where $\mu=E\left[x\right]$ and $\beta$ is the rate-parameter from the "rate-form". Using this form it would be much easier to specify the priors, since I know something about $\mu$.

Btw. I am doing everything simulation based, so I do not care about getting analytical expressions.


Certainly it's possible to write your gamma in shape-mean form and to place priors on the parameters written in that form.

If we start in 'shape-scale' form:

\begin{eqnarray*} f\left(x;\kappa,\theta\right)&=&\frac{1}{\Gamma\!\left(\kappa\right)\theta^{\kappa}}x^{\kappa-1}e^{-\frac{x}{\theta}}\\ &=&\frac{1}{\Gamma\!\left(\kappa\right)\theta}(\frac{_x}{^\theta})^{\kappa-1}e^{-\frac{x}{\theta}} \end{eqnarray*}

and then write $\mu=\theta\kappa$, equivalently $\theta=\mu/\kappa$, we get:

\begin{eqnarray*} f\left(x;\kappa,\mu\right)&=&\frac{\kappa}{\Gamma\!\left(\kappa\right)\mu}(\frac{_{x\kappa}}{^\mu})^{\kappa-1}e^{-\frac{x\kappa}{\mu}} \end{eqnarray*}

If it helps any, the conjugate prior for $\mu|\kappa$ is inverse-gamma, by inspection.

| cite | improve this answer | |

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.