Empirical Bayes: method of moments The model for the data is $X_{i}$ ~ $Bin(n_{i},\theta_{i})$ (iid, $i=1,...,k$).
The prior distribution is $\theta_{i}$ ~ $Beta(\alpha,\beta)$.
How do we choose (and deduct) moment estimators for $\alpha$ and $\beta$ in this case ?
 A: First, note that the combination of a Binomial distribution for $X_i | n_i, \theta_i$ and a Beta distribution for $\theta_i | \alpha, \beta$ leads directly to a Beta-Binomial distribution for $X_i | n_i, \alpha, \beta$.
The first two moments of the Beta-Binomial distribution are:
$$\mathbb{E}X_i = n_i {\alpha \over \alpha+\beta} $$
$$\sigma^2(X_i) = n_i{(\alpha+\beta+n_i)\alpha \beta \over (\alpha+\beta)^2(\alpha+\beta+1)}$$
The variance can be rewritten as follows:
$$\sigma^2(X_i) = {n_i(\alpha+\beta)\alpha \beta \over (\alpha+\beta)^2(\alpha+\beta+1)} + {n_i^2 \alpha \beta \over (\alpha+\beta)^2(\alpha+\beta+1)}$$
$$ = {n_i\alpha \beta \over (\alpha+\beta)(\alpha+\beta+1)} + {n_i^2 \alpha \beta \over (\alpha+\beta)^2(\alpha+\beta+1)}$$
Let's define $X = \sum_{i=1}^kX_i$ and $n = \sum_{i=1}^k n_i$.  Now, since the $X_i$ are independent, we know that the first two moments of the sum of the $X_i$ are just the sum of the first two moments of the individual $X_i$:
$$\mathbb{E}X = n {\alpha \over \alpha+\beta} $$
$$\sigma^2(X) = {n\alpha \beta \over (\alpha+\beta)(\alpha+\beta+1)} + {\alpha \beta \sum n_i^2 \over (\alpha+\beta)^2(\alpha+\beta+1)}$$
Equating sample moments to the two moments above results in one equation that solves for an estimate $\hat{\theta}$ of the ratio  $\theta=\alpha / (\alpha+\beta)$ and another, messier, equation that can be partially written in terms of $\hat{\theta}$, $n$, and $\sum n_i^2$.
$$\hat{\theta} = {\sum X_i \over n}$$
$$s^2(X_i) = {n\hat{\theta}\hat{\beta} +\hat{\theta}(1-\hat{\theta})\sum n_i^2 \over \hat{\alpha}+\hat{\beta}+1}$$
Some more algebra, based upon the relationship $\hat{\beta} = (1-\hat{\theta})\hat{\alpha}/\hat{\theta}$ and substitution, gets us to:
$$s^2(X_i) = {n(1-\hat{\theta})\hat{\alpha} +\hat{\theta}(1-\hat{\theta})\sum n_i^2 \over \hat{\alpha}+(1-\hat{\theta})\hat{\alpha}/\hat{\theta}+1}={n(1-\hat{\theta})\hat{\alpha} +\hat{\theta}(1-\hat{\theta})\sum n_i^2 \over \hat{\alpha}/\hat{\theta}+1}$$
$$=\hat{\theta}(1-\hat{\theta}){n\hat{\alpha} +\hat{\theta}\sum n_i^2 \over \hat{\alpha} + \hat{\theta}}$$
$$=n\hat{\theta}(1-\hat{\theta}){\hat{\alpha} +\hat{\theta}\sum n_i^2/n \over \hat{\alpha} + \hat{\theta}}$$
And a few more steps leads to:
$${s^2(X_i) \over n\hat{\theta}(1-\hat{\theta})}\hat{\alpha} + \hat{\theta}=\hat{\alpha} +\hat{\theta}\sum n_i^2/n$$
$$\hat{\alpha} = \hat{\theta}\left(\sum n_i^2/n-1\right)\left({n\hat{\theta}(1-\hat{\theta})\over s^2(X_i)-n\hat{\theta}(1-\hat{\theta})}\right)$$
with some possibility I've messed up the algebra, despite a couple of checks.
Once you've gotten this far, finding $\hat{\beta}$ is straightforward.  Note that the MOM estimates won't exist if the sample variance is too small relative to the sample mean!  This is because there is a limit on how small the population variance can be; consider all the $\theta_i = \theta$, then the variance of the $X_i$ is just $n_i\theta(1-\theta)$, but the sample variance might be smaller by chance, which would contradict the math that shows that the population variance is larger for any values of $\alpha, \beta$.
