8. Poisson–Gamma Model

Lecture errata

At 1:39 in the video, the likelihood should be proportional to \(\lambda^{16} e^{-6\lambda}\). The lecture version is missing the 6.

Model

Here we have another conjugate model. The Poisson likelihood describes the number of marked cells on some plates.

\[\begin{split} \begin{align*} X_i | \lambda &\sim Poi(\lambda) \\ \lambda &\sim Ga(\alpha, \beta) \end{align*} \end{split}\]

This time there are two new wrinkles. One, we’re not given the gamma prior parameters directly. Instead we want a mean of 4 and a variance of \(1/4\).

We know that the gamma distribution’s mean is \(\alpha/\beta\) and the variance is \(\alpha/\beta^2\), so we use that knowledge to solve for the parameters \(\alpha=64, \beta=16\).

\(\sum_{i=1}^n X_i = 2 + 0 + 1 + 5 + 7 + 1 = 16, n=6\)

So,

\[\begin{split} \begin{align*} \lambda \mid X_i &\sim Ga(\alpha + \sum_{i=1}^n X_i, \beta + n) \\ &\sim Ga(64 + 16, 16 + 6) \\ &\sim Ga(80, 22) \end{align*} \end{split}\]

Our posterior mean is \(\alpha/\beta = 3.64\).