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

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

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

Then, the conjugate prior for the model parameter $\lambda$ is a gamma distribution:

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

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:

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

Resolving the product in the likelihood function, we have

$\label{eq:Poiss-exp-LF-s3} \begin{split} p(y|\lambda) &= \prod_{i=1}^n \frac{ {x_i}^{y_i}}{y_i !} \cdot \prod_{i=1}^n \lambda^{y_i} \cdot \prod_{i=1}^n \exp\left[-\lambda x_i\right] \\ &= \prod_{i=1}^n \left(\frac{ {x_i}^{y_i}}{y_i !}\right) \cdot \lambda^{\sum_{i=1}^n y_i} \cdot \exp\left[-\lambda \sum_{i=1}^n x_i\right] \\ &= \prod_{i=1}^n \left(\frac{ {x_i}^{y_i}}{y_i !}\right) \cdot \lambda^{n \bar{y}} \cdot \exp\left[-n \bar{x} \lambda\right] \end{split}$

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

$\label{eq:xy-mean} \begin{split} \bar{y} &= \frac{1}{n} \sum_{i=1}^n y_i \\ \bar{x} &= \frac{1}{n} \sum_{i=1}^n x_i \; . \end{split}$

In other words, the likelihood function is proportional to a power of $\lambda$ times an exponential of $\lambda$:

$\label{eq:Poiss-exp-LF-prop} p(y|\lambda) \propto \lambda^{n \bar{y}} \cdot \exp\left[-n \bar{x} \lambda\right] \; .$

The same is true for a gamma distribution over $\lambda$

$\label{eq:Poiss-exp-prior-s1} p(\lambda) = \mathrm{Gam}(\lambda; a_0, b_0)$ $\label{eq:Poiss-exp-prior-s2} p(\lambda) = \frac{ {b_0}^{a_0}}{\Gamma(a_0)} \lambda^{a_0-1} \exp[-b_0 \lambda]$

exhibits the same proportionality

$\label{eq:Poiss-exp-prior-s3} p(\lambda) \propto \lambda^{a_0-1} \cdot \exp[-b_0 \lambda]$

and is therefore conjugate relative to the likelihood.

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. 2.14ff.; URL: http://www.stat.columbia.edu/~gelman/book/.

Metadata: ID: P41 | shortcut: poissexp-prior | author: JoramSoch | date: 2020-02-04, 14:11.