Exposure-adjusted Gamma-Poisson conjugate update

For fixed $e>0$ and $y\in {\mathbb{N}}_0$, the mass $q_{e,y}(\lambda )=e^{-e\lambda }(e\lambda {)}^y\mathrm{/}y!$ is continuous on $(0,\mathrm{\infty })$ by [F8]; for $e=0$ it is the constant $1_{\{y=0\}}$, also continuous. Conditional independence gives the data mass $\mathrm{\ell }(\lambda ,\mathbf{y})={\prod }_i{q}_{e_i,y_i}(\lambda )$, continuous in $\lambda$ for each fixed $\mathbf{y}$ (and equal to $1$ for the empty sample), hence Borel by [F9]. The count space is a countable union of the finite boxes $\{0,\dots ,m{\}}^n$ and has its discrete sigma-algebra, so $\mathrm{\ell }$ is jointly measurable on $(0,\mathrm{\infty })\times {\mathbb{N}}_0^n$: for each Borel $C\subseteq [0,\mathrm{\infty }]$, its inverse image is the countable union over $\mathbf{y}$ of the Borel section $\{\lambda :\mathrm{\ell }(\lambda ,\mathbf{y})\in C\}\times \{\mathbf{y}\}$. For any data event $D\subseteq {\mathbb{N}}_0^n$, its probability under parameter $\lambda$ is $K(\lambda ,D)={\sum }_{\mathbf{y}\in D}\mathrm{\ell }(\lambda ,\mathbf{y})$; the sums over $D\cap \{0,\dots ,m{\}}^n$ are finite and increase to this value, so $K(\cdot ,D)$ is measurable by [F8]. For each $\lambda$, conditional independence and the normalized marginal Poisson laws [F2] make $K(\lambda ,\cdot )$ a probability measure; the empty sample gives the point mass on its singleton data space. Thus $K$ is a probability kernel and $\mathrm{\ell }$ is the jointly measurable density required by [F7]. If some $e_i=0$ and $y_i>0$, [F2] makes the likelihood factor zero and its evidence zero by [F5]. Otherwise $y_i=0$ whenever $e_i=0$, and multiplying the mass factors in [F2] gives $L_{\mathbf{y}}(\lambda )=c(\mathbf{e},\mathbf{y}){\lambda }^S{e}^{-E\lambda }$. The constant $c$ is finite and positive, including when there are no positive exposures; the prior [F1] and kernel $K$ give the Bayesian model [F4].

Multiply the likelihood from step 1.1 by the prior density [F1]. The parameter-dependent part is $c(\mathbf{e},\mathbf{y})\frac{b^a}{\mathrm{\Gamma }(a)}{\lambda }^{a+S-1}{e}^{-(b+E)\lambda }\mathrm{.}$ Because $a+S>0$ and $b+E>0$, integrating this expression and applying [F3] with shape $a+S$ and rate $b+E$ yields the displayed evidence. Every factor is finite and strictly positive.

Dividing by that evidence leaves $(b+E{)}^{a+S}{\lambda }^{a+S-1}{e}^{-(b+E)\lambda }\mathrm{/}\mathrm{\Gamma }(a+S),$ the normalized Gamma$(a+S,b+E)$ density by [F3]. Its evidence is finite positive, so [F5] and [F6] permit the update and [F7] identifies it as the posterior kernel. When all exposures and counts vanish, $S=E=0$ and the formula is the original prior density.