I've got a gamma likelihood $\Gamma(\tau_c | \alpha_k, \frac{\alpha_k} {\tau_k})$ (parameterized with shape and rate)
with an InverseGamma prior $IG(\tau_k|a_0, b_0)$.
I know that the resulting posterior $P(\tau_k|\tau_c, \alpha_k, a_0, b_0)$ should have an InverseGamma ($IG(\tau|\alpha, \beta)$) form but I cannot manage to rewrite it as one. I got this far:

$ = \Gamma(\tau_c | \alpha_k, \frac{\alpha_k} {\tau_k}) IG(\tau_k|a_0, b_0)\\ = \frac{(\alpha_k/\tau_k)^{a_k}}{\Gamma(\alpha_k)}\tau_c^{\alpha_k - 1} \exp\left(-\tau_c \frac{\alpha_k}{\tau_k}\right) \cdot \frac{a_0^{b_0}}{\Gamma(a_0)}\tau_k^{-a_0-1}\exp\left(\frac{b_0}{\tau_c}\right)\\ \propto \tau_c^{\alpha_k - 1} \exp\left(-\tau_c \frac{\alpha_k}{\tau_k}\right) \cdot \tau_k^{-a_0-1}\exp\left(\frac{b_0}{\tau_c}\right)\\ = \tau_c^{\alpha_k - 1}\tau_k^{-a_0-1}\exp\left(-\tau_c\frac{\alpha_k}{\tau_k}+\frac{b_0}{\tau_c}\right) $

I'm not sure how to go from here (or if its even is correct so far). How to continue?

  • $\begingroup$ It should be $\exp\left(\frac{b_0}{\tau_k}\right)$, not $\exp\left(\frac{b_0}{\tau_c}\right)$. No? $\endgroup$ – Stéphane Laurent Nov 8 '17 at 15:03

Your Answer

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

Browse other questions tagged or ask your own question.