Index: The Book of Statistical ProofsModel Selection ▷ Bayesian model selection ▷ Bayes factor ▷ Computation using Savage-Dickey Density Ratio

Theorem: Consider two competing models on data $y$ containing parameters $\delta$ and $\varphi$, namely $m_0:\delta=\delta_0,\varphi$ and $m_1:\delta,\varphi$. In this context, we say that $\delta$ is a parameter of interest, $\varphi$ is a nuisance parameter (i.e., common to both models), and $m_0$ is a sharp point hypothesis nested within $m_1$. Suppose further that the prior for the nuisance parameter $\varphi$ in $m_0$ is equal to the prior for $\varphi$ in $m_1$ after conditioning on the restriction – that is, $p(\varphi\mid m_0) = p(\varphi\mid \delta=\delta_0,m_1)$. Then the Bayes factor for $m_0$ over $m_1$ can be computed as:

\[\label{eq:sd} \text{BF}_{01} = \frac{p(\delta=\delta_0\mid y,m_1)}{p(\delta=\delta_0\mid m_1)}.\]

Proof: By definition, the Bayes factor $\text{BF}_{01}$ is the ratio of marginal likelihoods of data $y$ over $m_0$ and $m_1$, respectively. That is,

\[\label{eq:bf} \text{BF}_{01}=\frac{p(y \mid m_0)}{p(y \mid m_1)}.\]

The key idea in the proof is that we can use a “change of variables” technique to express $\text{BF}_{01}$ entirely in terms of the “encompassing” model $m_1$. This proceeds by first unpacking the marginal likelihood for $m_0$ over the nuisance parameter $\varphi$ and then using the fact that $m_0$ is a sharp hypothesis nested within $m_1$ to rewrite everything in terms of $m_1$. Specifically,

\[\label{eq:ml-m0} \begin{split} p(y \mid m_0) &= \int p(y \mid \varphi,m_0) \, p(\varphi\mid m_0) \, \mathrm{d} \varphi \\ &= \int p(y \mid \varphi,\delta=\delta_0,m_1) \, p(\varphi\mid \delta=\delta_0,m_1) \, \mathrm{d} \varphi \\ &= p(y \mid \delta=\delta_0,m_1).\\ \end{split}\]

By Bayes’ theorem, we can rewrite this last line as

\[\label{eq:ml-m0-bt} p(y \mid \delta=\delta_0,m_1) = \frac{p(\delta=\delta_0\mid y,m_1) \, p(y \mid m_1)}{p(\delta=\delta_0\mid m_1)}.\]

Thus we have

\[\begin{split} \text{BF}_{01} &\overset{\eqref{eq:bf}}{=} \frac{p(y \mid m_0)}{p(y \mid m_1)}\\ &= p(y \mid m_0) \cdot \frac{1}{p(y \mid m_1)}\\ &\overset{\eqref{eq:ml-m0}}{=} p(y \mid \delta=\delta_0,m_1) \cdot \frac{1}{p(y \mid m_1)}\\ &\overset{\eqref{eq:ml-m0-bt}}{=} \frac{p(\delta=\delta_0\mid y,m_1) \, p(y \mid m_1)}{p(\delta=\delta_0\mid m_1)} \cdot \frac{1}{p(y \mid m_1)}\\ &= \frac{p(\delta=\delta_0 \mid y,m_1)}{p(\delta=\delta_0\mid m_1)}, \end{split}\]

which completes the proof of \eqref{eq:sd}.

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Metadata: ID: P156 | shortcut: bf-sddr | author: tomfaulkenberry | date: 2020-08-26, 12:00.