add [bf-tran]

Author: tomfaulkenberry

P/bf-tran.md

@@ -0,0 +1,49 @@
+---
+layout: proof
+mathjax: true
+
+author: "Thomas J. Faulkenberry"
+affiliation: "Tarleton State University"
+e_mail: "faulkenberry@tarleton.edu"
+date: 2020-09-07 12:00:00
+
+title: "Transitivity of Bayes Factors"
+chapter: "Model Selection"
+section: "Bayesian model selection"
+topic: "Bayes factor"
+theorem: "Transitivity of Bayes Factors"
+
+sources:
+
+proof_id: "P163"
+shortcut: "bf-tran"
+username: "tomfaulkenberry"
+---
+
+
+**Theorem:** Consider three competing [models](/D/gm) $m_1$, $m_2$, and $m3$ for observed data $y$. Then the [Bayes factor](/D/bf) for $m_1$ over $m_3$ can be written as:
+
+$$ \label{eq:bf-tran}
+\text{BF}_{13} = \text{BF}_{12}\cdot \text{BF}_{23}.
+$$
+
+**Proof:** By [definition](/D/bf), 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 equivlently write 
+
+$$
+\begin{split}
+  \text{BF}_{13} &\overset{\eqref{eq:bf}}{=} \frac{p(y \mid m_1)}{p(y \mid m_3)}\\
+  &= \frac{p(y \mid m_1)}{p(y \mid m_3)} \cdot \frac{p(y \mid m_2)}{p(y \mid m_2)}\\
+  &= \frac{p(y \mid m_1) \cdot p(y \mid m_2)}{p(y \mid m_3)\cdot p(y \mid m_2)}\\
+  &= \frac{p(y \mid m_1) \cdot p(y \mid m_2)}{p(y \mid m_2)\cdot p(y \mid m_3)}\\
+  &=\frac{p(y \mid m_1)}{p(y \mid m_2)} \cdot \frac{p(y \mid m_2)}{p(y \mid m_3)}\\
+  &\overset{\eqref{eq:bf}}{=}\text{BF}_{12} \cdot \text{BF}_{23},
+\end{split}
+$$
+
+which completes the proof of \eqref{eq:tran}.