Posterior mean of $mu$ in Bayesian Hierarchical model (Poisson-Gamma)

Cross Validated Asked by Maverick Meerkat on November 17, 2020

Chapter 7 of Jim Albert’s book considers the case of using a hierarchical model, to estimate heart-transplant mortality rates ($lambda_i$) from 94 hospitals, each with it’s own exposure (# of operations, $e_i$), with:

y_i sim Poisson(e_ilambda_i) \
lambda_i sim Gamma(alpha, frac{alpha}{mu}) \

Under this model, we get that the posterior probability of $lambda_i$ is $Gamma(y_i + alpha, e_i + frac{alpha}{mu})$. It is also shown that the posterior mean of $lambda_i$ is equal to $mathbb E(lambda_i | y_i,alpha,mu) = (1-B_i)frac{y_i}{e_i} + B_i mu$, which demonstrates Shrinkage.

In addition, after assuming the following prior distribution for the hyperparameters $mu$ and $alpha$:

g(mu) proptofrac{1}{mu} \
h(alpha) = frac{z_0}{(alpha+z_0)^2}

We get also a Marginal Posterior density of the two. Which allows us to sample values for these hyperparameters, and then use these samples to sample plausible $lambda_i$ values and in turn also $y_i$.

Now at some point he states that $mathbb E(lambda_i | data)$ can be approximated by $(1-mathbb E(B_i|data))frac{y_i}{e_i} + mathbb E(B_i|data) frac{sum y_i}{sum e_i}$.

Now I think that he meant that we replace the parameters with our simulated averages of these parameters. This is what he does with $mathbb E(B_i|data)$ anyway. But I wonder why he replaced $mu$ with $frac{sum y_i}{sum e_i}$ ?

My hunch is that the posterior mean of $mu$ is equal to $frac{sum y_i}{sum e_i}$. At least they are very close numerically from the simulated samples. But I wonder if there is a way to show this analytically?

Add your own answers!

Ask a Question

Get help from others!

© 2024 All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP