Index: The Book of Statistical ProofsGeneral Theorems ▷ Bayesian statistics ▷ Probabilistic modeling ▷ Combined posterior distribution from independent data

Theorem: Let $p(\theta \vert y_1)$ and $p(\theta \vert y_2)$ be posterior distributions, obtained using the same prior distribution from conditionally independent data sets $y_1$ and $y_2$:

$\label{eq:ind-cond} p(y_1,y_2|\theta) = p(y_1|\theta) \cdot p(y_2|\theta) \; .$

Then, the combined posterior distribution is proportional to the product of the individual posterior densities, divided by the prior density:

$\label{eq:post-ind} p(\theta|y_1,y_2) \propto \frac{p(\theta|y_1) \cdot p(\theta|y_2)}{p(\theta)} \; .$

Proof: Since $p(\theta \vert y_1)$ and $p(\theta \vert y_2)$ are posterior distributions, Bayes’ theorem holds for them:

$\label{eq:bayes-th} \begin{split} p(\theta|y_1) &= \frac{p(y_1|\theta) \cdot p(\theta)}{p(y_1)} \\ p(\theta|y_2) &= \frac{p(y_2|\theta) \cdot p(\theta)}{p(y_2)} \; . \end{split}$ $\label{eq:post-ind-s1} p(\theta|y_1,y_2) = \frac{p(y_1,y_2|\theta) \cdot p(\theta)}{p(y_1,y_2)} \; .$

With that, we can express the combined posterior distribution as follows:

$\label{eq:post-ind-s2} \begin{split} p(\theta|y_1,y_2) &\overset{\eqref{eq:post-ind-s1}}{=} \frac{p(y_1,y_2|\theta) \cdot p(\theta)}{p(y_1,y_2)} \\ &\overset{\eqref{eq:ind-cond}}{=} p(y_1|\theta) \cdot p(y_2|\theta) \cdot \frac{p(\theta)}{p(y_1,y_2)} \\ &\overset{\eqref{eq:bayes-th}}{=} \frac{p(\theta|y_1) \cdot p(y_1)}{p(\theta)} \cdot \frac{p(\theta|y_2) \cdot p(y_2)}{p(\theta)} \cdot \frac{p(\theta)}{p(y_1,y_2)} \\ &= \frac{p(\theta|y_1) \cdot p(\theta|y_2)}{p(\theta)} \cdot \frac{p(y_1) \cdot p(y_2)}{p(y_1,y_2)} \; . \end{split}$

Note that the second fraction does not depend on $\theta$ and thus, the posterior distribution over $\theta$ is proportional to the first fraction:

$\label{eq:post-ind-s3} p(\theta|y_1,y_2) \propto \frac{p(\theta|y_1) \cdot p(\theta|y_2)}{p(\theta)} \; .$
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Metadata: ID: P413 | shortcut: post-ind | author: JoramSoch | date: 2023-09-01, 13:46.