Index: The Book of Statistical ProofsModel Selection ▷ Bayesian model selection ▷ Bayes factor ▷ Transitivity

Theorem: Consider three competing models $m_1$, $m_2$, and $m3$ for observed data $y$. Then the Bayes factor for $m_1$ over $m_3$ can be written as:

$\label{eq:bf-trans} \text{BF}_{13} = \text{BF}_{12}\cdot \text{BF}_{23}.$

Proof: By definition, the Bayes factor $\text{BF}_{13}$ is the ratio of marginal likelihoods of data $y$ over $m_1$ and $m_3$, respectively. That is,

$\label{eq:bf} \text{BF}_{13}=\frac{p(y \mid m_1)}{p(y \mid m_3)}.$

We can equivalently write

$\begin{split} \text{BF}_{13} &\overset{\eqref{eq:bf}}{=} \frac{p(y \mid m_1)}{p(y \mid m_3)}\\ &= \frac{p(y \mid m_1)}{p(y \mid m_3)} \cdot \frac{p(y \mid m_2)}{p(y \mid m_2)}\\ &=\frac{p(y \mid m_1)}{p(y \mid m_2)} \cdot \frac{p(y \mid m_2)}{p(y \mid m_3)}\\ &\overset{\eqref{eq:bf}}{=}\text{BF}_{12} \cdot \text{BF}_{23}, \end{split}$

which completes the proof of \eqref{eq:bf-trans}.

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Metadata: ID: P163 | shortcut: bf-trans | author: tomfaulkenberry | date: 2020-09-07, 12:00.