Index: The Book of Statistical Proofs ▷ Statistical Models ▷ Count data ▷ Poisson-distributed data ▷ Posterior distribution

Theorem: Let there be a Poisson-distributed data set $y = \left\lbrace y_1, \ldots, y_n \right\rbrace$:

\[\label{eq:Poiss} y_i \sim \mathrm{Poiss}(\lambda), \quad i = 1, \ldots, n \; .\]

Moreover, assume a gamma prior distribution over the model parameter $\lambda$:

\[\label{eq:Poiss-prior} p(\lambda) = \mathrm{Gam}(\lambda; a_0, b_0) \; .\]

Then, the posterior distribution is also a gamma distribution

\[\label{eq:Poiss-post} p(\lambda|y) = \mathrm{Gam}(\lambda; a_n, b_n)\]

and the posterior hyperparameters are given by

\[\label{eq:Poiss-post-par} \begin{split} a_n &= a_0 + n \bar{y} \\ b_n &= b_0 + n \; . \end{split}\]

Proof: With the probability mass function of the Poisson distribution, the likelihood function for each observation implied by \eqref{eq:Poiss} is given by

\[\label{eq:Poiss-LF-s1} p(y_i|\lambda) = \mathrm{Poiss}(y_i; \lambda) = \frac{\lambda^{y_i} \cdot \exp\left[-\lambda\right]}{y_i !}\]

and because observations are independent, the likelihood function for all observations is the product of the individual ones:

\[\label{eq:Poiss-LF-s2} p(y|\lambda) = \prod_{i=1}^n p(y_i|\lambda) = \prod_{i=1}^n \frac{\lambda^{y_i} \cdot \exp\left[-\lambda\right]}{y_i !} \; .\]

Combining the likelihood function \eqref{eq:Poiss-LF-s2} with the prior distribution \eqref{eq:Poiss-prior}, the joint likelihood of the model is given by

\[\label{eq:Poiss-JL-s1} \begin{split} p(y,\lambda) &= p(y|\lambda) \, p(\lambda) \\ &= \prod_{i=1}^n \frac{\lambda^{y_i} \cdot \exp\left[-\lambda\right]}{y_i !} \cdot \frac{ {b_0}^{a_0}}{\Gamma(a_0)} \lambda^{a_0-1} \exp[-b_0 \lambda] \; . \end{split}\]

Resolving the product in the joint likelihood, we have

\[\label{eq:Poiss-JL-s2} \begin{split} p(y,\lambda) &= \prod_{i=1}^n \frac{1}{y_i !} \prod_{i=1}^n \lambda^{y_i} \prod_{i=1}^n \exp\left[-\lambda\right] \cdot \frac{ {b_0}^{a_0}}{\Gamma(a_0)} \lambda^{a_0-1} \exp[-b_0 \lambda] \\ &= \prod_{i=1}^n \left(\frac{1}{y_i !}\right) \lambda^{n \bar{y}} \exp\left[-n \lambda\right] \cdot \frac{ {b_0}^{a_0}}{\Gamma(a_0)} \lambda^{a_0-1} \exp[-b_0 \lambda] \\ &= \prod_{i=1}^n \left(\frac{1}{y_i !}\right) \cdot \frac{ {b_0}^{a_0}}{\Gamma(a_0)} \cdot \lambda^{a_0 + n \bar{y} - 1} \cdot \exp\left[-(b_0 + n \lambda)\right] \\ \end{split}\]

where $\bar{y}$ is the mean of $y$:

\[\label{eq:y-mean} \bar{y} = \frac{1}{n} \sum_{i=1}^n y_i \; .\]

Note that the posterior distribution is proportional to the joint likelihood:

\[\label{eq:Poiss-post-s1} p(\lambda|y) \propto p(y,\lambda) \; .\]

Setting $a_n = a_0 + n \bar{y}$ and $b_n = b_0 + n$, the posterior distribution is therefore proportional to

\[\label{eq:Poiss-post-s2} p(\lambda|y) \propto \lambda^{a_n-1} \cdot \exp\left[-b_n \lambda\right]\]

which, when normalized to one, results in the probability density function of the gamma distribution:

\[\label{eq:Poiss-post-s3} p(\lambda|y) = \frac{ {b_n}^{a_n}}{\Gamma(a_0)} \lambda^{a_n-1} \exp\left[-b_n \lambda\right] = \mathrm{Gam}(\lambda; a_n, b_n) \; .\]
∎
Sources:
  • Gelman A, Carlin JB, Stern HS, Dunson DB, Vehtari A, Rubin DB (2014): "Other standard single-parameter models"; in: Bayesian Data Analysis, 3rd edition, ch. 2.6, p. 45, eq. 2.15; URL: http://www.stat.columbia.edu/~gelman/book/.

Metadata: ID: P226 | shortcut: poiss-post | author: JoramSoch | date: 2020-04-21, 08:48.