Index: The Book of Statistical ProofsStatistical Models ▷ Count data ▷ Poisson distribution with exposure values ▷ Posterior distribution

Theorem: Consider data $y = \left\lbrace y_1, \ldots, y_n \right\rbrace$ following a Poisson distribution with exposure values:

$\label{eq:Poiss-exp} y_i \sim \mathrm{Poiss}(\lambda x_i), \quad i = 1, \ldots, n \; .$

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

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

Then, the posterior distribution is also a gamma distribution

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

and the posterior hyperparameters are given by

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

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

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

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

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

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

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

Resolving the product in the joint likelihood, we have

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

where $\bar{y}$ and $\bar{x}$ are the means of $y$ and $x$ respectively:

$\label{eq:xy-mean} \begin{split} \bar{y} &= \frac{1}{n} \sum_{i=1}^n y_i \\ \bar{x} &= \frac{1}{n} \sum_{i=1}^n x_i \; . \end{split}$ $\label{eq:Poiss-exp-post-s1} p(\lambda|y) \propto p(y,\lambda) \; .$

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

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

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

$\label{eq:Poiss-exp-post-s3} p(\lambda|y) = \frac{ {b_n}^{a_n}}{\Gamma(a_0)} \lambda^{a_n-1} \exp\left[-b_n \lambda\right] = \mathrm{Gam}(\lambda; a_n, b_n) \; .$
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: P42 | shortcut: poissexp-post | author: JoramSoch | date: 2020-02-04, 14:42.