Index: The Book of Statistical ProofsStatistical Models ▷ Count data ▷ Binomial observations ▷ Posterior probability

Theorem: Let $y$ be the number of successes resulting from $n$ independent trials with unknown success probability $p$, such that $y$ follows a binomial distribution:

$\label{eq:Bin} y \sim \mathrm{Bin}(n,p) \; .$

Moreover, assume two statistical models, one assuming that $p$ is 0.5 (null model), the other imposing a beta distribution as the prior distribution on the model parameter $p$ (alternative):

$\label{eq:Bin-m01} \begin{split} m_0&: \; y \sim \mathrm{Bin}(n,p), \; p = 0.5 \\ m_1&: \; y \sim \mathrm{Bin}(n,p), \; p \sim \mathrm{Bet}(\alpha_0, \beta_0) \; . \end{split}$

Then, the posterior probability of the alternative model is given by

$\label{eq:Bin-PP1} p(m_1|y) = \frac{1}{1 + 2^{-n} \left[ B(\alpha_0,\beta_0) / B(\alpha_n,\beta_n) \right]}$

where $B(x,y)$ is the beta function and $\alpha_n$ and $\beta_n$ are the posterior hyperparameters for binomial observations which are functions of the number of trials $n$ and the number of successes $y$.

$\label{eq:PP-LBF} p(m_1|y) = \frac{\exp(\mathrm{LBF}_{12})}{\exp(\mathrm{LBF}_{12}) + 1} \; .$ $\label{eq:Bin-LBF10} \mathrm{LBF}_{10} = \log B(\alpha_n,\beta_n) - \log B(\alpha_0,\beta_0) - n \log \left( \frac{1}{2} \right) \; .$

and the corresponding Bayes factor, i.e. exponentiated log Bayes factor, is equal to

$\label{eq:Bin-BF10} \mathrm{BF}_{10} = \exp(\mathrm{LBF}_{10}) = 2^n \cdot \frac{B(\alpha_n,\beta_n)}{B(\alpha_0,\beta_0)} \; .$

Thus, the posterior probability of the alternative, assuming a prior distribution over the probability $p$, compared to the null model, assuming a fixed probability $p = 0.5$, follows as

$\label{eq:Bin-PP1-qed} \begin{split} p(m_1|y) &\overset{\eqref{eq:PP-LBF}}{=} \frac{\exp(\mathrm{LBF}_{10})}{\exp(\mathrm{LBF}_{10}) + 1} \\ &\overset{\eqref{eq:Bin-BF10}}{=} \frac{2^n \cdot \frac{B(\alpha_n,\beta_n)}{B(\alpha_0,\beta_0)}}{2^n \cdot \frac{B(\alpha_n,\beta_n)}{B(\alpha_0,\beta_0)} + 1} \\ &= \frac{2^n \cdot \frac{B(\alpha_n,\beta_n)}{B(\alpha_0,\beta_0)}}{2^n \cdot \frac{B(\alpha_n,\beta_n)}{B(\alpha_0,\beta_0)} \left( 1 + 2^{-n} \frac{B(\alpha_0,\beta_0)}{B(\alpha_n,\beta_n)} \right)} \\ &= \frac{1}{1 + 2^{-n} \left[ B(\alpha_0,\beta_0) / B(\alpha_n,\beta_n) \right]} \end{split}$ $\label{eq:Bin-post-par} \begin{split} \alpha_n &= \alpha_0 + y \\ \beta_n &= \beta_0 + (n-y) \end{split}$

with the number of trials $n$ and the number of successes $y$.

Sources:

Metadata: ID: P384 | shortcut: bin-pp | author: JoramSoch | date: 2022-11-26, 11:42.