Index: The Book of Statistical ProofsStatistical Models ▷ Frequency data ▷ Beta-binomial data ▷ Method of moments

Theorem: Let $y = \left\lbrace y_1, \ldots, y_N \right\rbrace$ be a set of observed counts independent and identically distributed according to a beta-binomial distribution with number of trials $n$ as well as parameters $\alpha$ and $\beta$:

$\label{eq:binbeta} y_i \sim \mathrm{BetBin}(n, \alpha, \beta), \quad i = 1, \ldots, N \; .$

Then, the method-of-moments estimates for the parameters $\alpha$ and $\beta$ are given by

$\label{eq:binbeta-mome} \begin{split} \hat{\alpha} &= \frac{n m_1 - m_2}{n \left( \frac{m_2}{m_1} - m_1 - 1 \right) + m_1} \\ \hat{\beta} &= \frac{\left( n - m_1 \right)\left( n - \frac{m_2}{m_1} \right)}{n \left( \frac{m_2}{m_1} - m_1 - 1 \right) + m_1} \end{split}$

where $m_1$ and $m_2$ are the first two raw sample moments:

$\label{eq:y-m1-m2} \begin{split} m_1 &= \frac{1}{N} \sum_{i=1}^N y_i \\ m_2 &= \frac{1}{N} \sum_{i=1}^N y_i^2 \; . \end{split}$

Proof: The first two raw moments of the beta-binomial distribution in terms of the parameters $\alpha$ and $\beta$ are given by

$\label{eq:binbeta-mu1-mu2} \begin{split} \mu_1 &= \frac{n \alpha}{\alpha + \beta} \\ \mu_2 &= \frac{n \alpha (n \alpha + \beta + n)}{(\alpha + \beta)(n \alpha + \beta + 1)} \end{split}$

Thus, matching the moments requires us to solve the following equation system for $\alpha$ and $\beta$:

$\label{eq:binbeta-m1-m2} \begin{split} m_1 &= \frac{n \alpha}{\alpha + \beta} \\ m_2 &= \frac{n \alpha (n \alpha + \beta + n)}{(\alpha + \beta)(n \alpha + \beta + 1)} \; . \end{split}$

From the first equation, we can deduce:

$\label{eq:beta-as-alpha} \begin{split} m_1(\alpha+\beta) &= n \alpha \\ m_1 \alpha + m_1 \beta &= n \alpha \\ m_1 \beta &= n \alpha - m_1 \alpha \\ \beta &= \frac{n \alpha}{m_1} - \alpha \\ \beta &= \alpha \left( \frac{n}{m_1} - 1 \right) \; . \end{split}$

If we define $q = n/m_1 - 1$ and plug \eqref{eq:beta-as-alpha} into the second equation, we have:

$\label{eq:alpha-as-q} \begin{split} m_2 &= \frac{n \alpha (n \alpha + \alpha q + n)}{(\alpha + \alpha q)(\alpha + \alpha q + 1)} \\ &= \frac{n \alpha (\alpha (n + q) + n)}{\alpha (1 + q)(\alpha (1 + q) + 1)} \\ &= \frac{n (\alpha (n + q) + n)}{(1 + q)(\alpha (1 + q) + 1)} \\ &= \frac{n (\alpha (n + q) + n)}{\alpha (1 + q)^2 + (1 + q)} \; . \end{split}$

Noting that $1+q = n/m_1$ and expanding the fraction with $m_1$, one obtains:

$\label{eq:binbeta-mome-alpha} \begin{split} m_2 &= \frac{n \left(\alpha \left( \frac{n}{m_1} + n - 1 \right) + n \right)}{n \left( \alpha \frac{n}{m_1^2} + \frac{1}{m_1} \right)} \\ m_2 &= \frac{\alpha \left( n + n m_1 - m_1 \right) + n m_1}{\alpha \frac{n}{m_1} + 1} \\ m_2 \left( \frac{\alpha n}{m_1} + 1 \right) &= \alpha \left( n + n m_1 - m_1 \right) + n m_1 \\ \alpha \left( n \frac{m_2}{m_1} - (n + n m_1 - m_1) \right) &= n m_1 - m_2 \\ \alpha \left( n \left( \frac{m_2}{m_1} - m_1 - 1 \right) + m_1 \right) &= n m_1 - m_2 \\ \alpha &= \frac{n m_1 - m_2}{n \left( \frac{m_2}{m_1} - m_1 - 1 \right) + m_1} \; . \end{split}$

Plugging this into equation \eqref{eq:beta-as-alpha}, one obtains for $\beta$:

$\label{eq:binbeta-mome-beta} \begin{split} \beta &= \alpha \left( \frac{n}{m_1} - 1 \right) \\ \beta &= \left( \frac{n m_1 - m_2}{n \left( \frac{m_2}{m_1} - m_1 - 1 \right) + m_1} \right) \left( \frac{n}{m_1} - 1 \right) \\ \beta &= \frac{n^2 - n m_1 - n \frac{m_2}{m_1} + m_2}{n \left( \frac{m_2}{m_1} - m_1 - 1 \right) + m_1} \\ \hat{\beta} &= \frac{\left( n - m_1 \right)\left( n - \frac{m_2}{m_1} \right)}{n \left( \frac{m_2}{m_1} - m_1 - 1 \right) + m_1} \; . \end{split}$

Together, \eqref{eq:binbeta-mome-alpha} and \eqref{eq:binbeta-mome-beta} constitute the method-of-moment estimates of $\alpha$ and $\beta$.

Sources:

Metadata: ID: P357 | shortcut: betabin-mome | author: JoramSoch | date: 2022-10-07, 15:13.