Index: The Book of Statistical ProofsStatistical Models ▷ Probability data ▷ Beta-distributed 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 distribution with shapes $\alpha$ and $\beta$:

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

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

$\label{eq:Beta-MoM} \begin{split} \hat{\alpha} &= \bar{y} \left( \frac{\bar{y} (1-\bar{y})}{\bar{v}} - 1 \right) \\ \hat{\beta} &= (1-\bar{y}) \left( \frac{\bar{y} (1-\bar{y})}{\bar{v}} - 1 \right) \end{split}$

where $\bar{y}$ is the sample mean and $\bar{v}$ is the unbiased sample variance:

$\label{eq:y-mean-var} \begin{split} \bar{y} &= \frac{1}{n} \sum_{i=1}^n y_i \\ \bar{v} &= \frac{1}{n-1} \sum_{i=1}^n (y_i - \bar{y})^2 \; . \end{split}$

Proof: Mean and variance of the beta distribution in terms of the parameters $\alpha$ and $\beta$ are given by

$\label{eq:Beta-E-Var} \begin{split} \mathrm{E}(X) &= \frac{\alpha}{\alpha+\beta} \\ \mathrm{Var}(X) &= \frac{\alpha\beta}{(\alpha+\beta)^2 (\alpha+\beta+1)} \; . \end{split}$

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

$\label{eq:Beta-mean-var} \begin{split} \bar{y} &= \frac{\alpha}{\alpha+\beta} \\ \bar{v} &= \frac{\alpha\beta}{(\alpha+\beta)^2 (\alpha+\beta+1)} \; . \end{split}$

From the first equation, we can deduce:

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

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

$\label{eq:alpha-as-q} \begin{split} \bar{v} &= \frac{\alpha \cdot \alpha q}{(\alpha + \alpha q)^2 (\alpha + \alpha q + 1)} \\ &= \frac{\alpha^2 q}{(\alpha (1+q))^2 (\alpha (1+q) + 1)} \\ &= \frac{q}{(1+q)^2 (\alpha (1+q) + 1)} \\ &= \frac{q}{\alpha (1+q)^3 + (1+q)^2} \\ q &= \bar{v} \left[ \alpha (1+q)^3 + (1+q)^2 \right] \; . \end{split}$

Noting that $1+q = 1/\bar{y}$ and $q = (1-\bar{y})/\bar{y}$, one obtains for $\alpha$:

$\label{eq:Beta-MoM-alpha} \begin{split} \frac{1-\bar{y}}{\bar{y}} &= \bar{v} \left[ \frac{\alpha}{\bar{y}^3} + \frac{1}{\bar{y}^2} \right] \\ \frac{1-\bar{y}}{\bar{y} \, \bar{v}} &= \frac{\alpha}{\bar{y}^3} + \frac{1}{\bar{y}^2} \\ \frac{\bar{y}^3(1-\bar{y})}{\bar{y} \, \bar{v}} &= \alpha + \bar{y} \\ \alpha &= \frac{\bar{y}^2(1-\bar{y})}{\bar{v}} - \bar{y} \\ &= \bar{y} \left( \frac{\bar{y} (1-\bar{y})}{\bar{v}} - 1 \right) \; . \end{split}$

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

$\label{eq:Beta-MoM-beta} \begin{split} \beta &= \bar{y} \left( \frac{\bar{y} (1-\bar{y})}{\bar{v}} - 1 \right) \cdot \left( \frac{1-\bar{y}}{\bar{y}} \right) \\ &= (1-\bar{y}) \left( \frac{\bar{y} (1-\bar{y})}{\bar{v}} - 1 \right) \; . \end{split}$

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

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Metadata: ID: P28 | shortcut: beta-mome | author: JoramSoch | date: 2020-01-22, 02:53.