I have a problem to which I am trying to apply a Bayesian model. My data is generated as follows \begin{align} N_i \mid \mu &\sim \text{Poisson}(\mu) \\ Y_i \mid N_i, \theta_i &\sim \text{Binomial}({N_i, \theta_i}) \end{align}

In my problem domain $\theta_i$ is assume known, and $N_i$ is a hidden variable. So by taking the expectation over $N_i$ I get \begin{align} Y_i \mid \theta_i, \mu &\sim \text{Poisson}(\mu\theta_i) \end{align}

I am interested in the understanding some properties of the posterior on $\mu$ when $\mu$ has an uninformative prior. Specifically, I am trying to determine if when we know nothing about $\mu$ a priori, does using an uninformative prior give us a posterior on $\mu$ with properties (eg moments and mode) that match what we might expect in our problem domain (molecular biology)? Or does the posterior clash with our domain intuition and we need instead to change our methodology to measure $N_i$?

I found an improper prior on $\mu$ in the literature; \begin{align} \pi(\mu) \propto \mu^{-1} \end{align}

So I tried finding the posterior analytically; \begin{align} \pi(\mu|Y_1 ... Y_n, \theta_1, ... \theta_n) &\propto p(Y_1 ... Y_n \mid \theta_1, ... \theta_n, \mu) * \pi(\mu)\\ &\propto \frac{e^{-\mu\sum_i\theta_i}\mu^{\sum y_i}\prod_i \theta_i^{y_i}}{\prod_i y_i!} * \mu^{-1}\\ &\propto Ce^{-\mu \sum\theta_i} \mu^{(\sum_i y_i) - 1}\\ \mu|Y_1 ... Y_n, \theta_1, ... \theta_n &\sim \Gamma(\sum_iy_i, \sum_i \theta_i) \end{align} where $\sum_i \theta_i$ is a rate parameter.

So I am thinking I can interpret the moments and mode of this posterior in terms of my problem domain. I am assuming the subjective conclusions I make to my audience about the uninformative case with this improper prior will generalize to other uninformative or weakly informative priors.

Am I making sense? Is there something glaringly wrong that I missed? Is that posterior I found not really a gamma? Any comments or critiques welcome.

  • $\begingroup$ This is a correct derivation for the conditional posterior of μ given the $\theta_i$'s. $\endgroup$ – Xi'an Nov 7 '16 at 16:12

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