Index: The Book of Statistical Proofs ▷ Model Selection ▷ Bayesian model selection ▷ Bayes factor ▷ Derivation of the group Bayes factor

Theorem: Let $Y = \left\lbrace y_1, \ldots, y_N \right\rbrace$ be a set of $N$ conditionally independent data sets given two generate models $m_1$ and $m_2$ with model evidences $p(y_i \vert m_1)$ and $p(y_i \vert m_2)$ for $i = 1,\ldots,N$. Then, the group Bayes factor for $Y$ in favor of $m_1$ over $m_2$ can be expressed as

\[\label{eq:gbf-me} \mathrm{GBF}_{12} = \prod_{i=1}^N \frac{p(y_i \vert m_1)}{p(y_i \vert m_2)}\]

or

\[\label{eq:gbf-bf} \mathrm{GBF}_{12} = \prod_{i=1}^N \mathrm{BF}_{12}^{(i)} \; .\]

where $\mathrm{BF}_{12}^{(i)}$ is the Bayes factor for $y_i$ in favor of $m_1$ over $m_2$.

Proof:

1) The Bayes factor for the “group data set” $Y$ evaluates to

\[\label{eq:gbf-me-Y} \mathrm{GBF}_{12} = \frac{p(Y \vert m_1)}{p(Y \vert m_2)} \; .\]

Under conditional independence, we have

\[\label{eq:p-Y-yi} p(Y \vert m_1) = \prod_{i=1}^N p(y_i \vert m_1) \quad \mathrm{and} \quad p(Y \vert m_2) = \prod_{i=1}^N p(y_i \vert m_2)\]

such that substituting \eqref{eq:p-Y-yi} into \eqref{eq:gbf-me-Y} gives:

\[\label{eq:gbf-me-qed} \mathrm{GBF}_{12} = \prod_{i=1}^N \frac{p(y_i \vert m_1)}{p(y_i \vert m_2)} \; .\]

2) The Bayes factor for a “single data set” $y_i$ evaluates to

\[\label{eq:bf-me-yi} \mathrm{BF}_{12}^{(i)} = \frac{p(y_i \vert m_1)}{p(y_i \vert m_2)}\]

and substituting \eqref{eq:bf-me-yi} into \eqref{eq:gbf-me-qed}, we obtain:

\[\label{eq:gbf-bf-qed} \mathrm{GBF}_{12} = \prod_{i=1}^N \mathrm{BF}_{12}^{(i)} \; .\]
∎
Sources:

Metadata: ID: P543 | shortcut: gbf-der | author: JoramSoch | date: 2026-06-14, 15:58.