Calculating the parameters of a Beta-Binomial distribution using the mean and variance

Question

I'm trying to do the same thing that was done in this question:

Calculating the parameters of a Beta distribution using the mean and variance

for the Beta-Binomial distribution for which the mean is

$\mu = n\frac{\alpha}{\alpha+\beta}$

and the variance is

$\sigma^2 = n\frac{\alpha\beta(\alpha+\beta+n)}{(\alpha+\beta)^2(\alpha+\beta+1)}$

How can I calculate $\alpha$ and $\beta$ in terms of $\mu$ and $\sigma^2$ for a given $n$?

Also some information regarding the bound of the mean and variance similar to the answer above would be appreciated, e.g. I know that $\mu \in (0,n)$.

Answer 1

$$\alpha = (n\mu-\sigma^2 -\mu^2)/T$$ $$\beta = (n-\mu)\left(n-\frac {\sigma^2 +\mu^2}{\mu}\right)/T$$

where $$T=\frac {n\sigma^2}{\mu} - n +\mu$$

$$0 \lt \mu \lt n$$ $$0 \lt \sigma^2 \lt n^2/4$$

$$\sigma^2 = n\frac{\alpha\beta(\alpha+\beta+n)}{(\alpha+\beta)^2(\alpha+\beta+1)} = n\frac{\alpha}{\alpha+\beta}\frac{\beta}{\alpha+\beta}\left (1+\frac {(n-1)}{(\alpha+\beta+1)}\right)$$

$max(\frac{\alpha}{\alpha+\beta}\frac{\beta}{\alpha+\beta}) = 1/4 $ when $\alpha = \beta$.

$max\left (1+\frac {(n-1)}{(\alpha+\beta+1)}\right) = n$ when $\alpha+\beta$ goes to zero.

$\alpha = \beta$ and $\alpha+\beta$ goes to zero can be true simultaneously, so got the results.