Bayesian batting average prior I wanted to ask a question inspired by an excellent answer to the query about the intuition for the beta distribution. I wanted to get a better understanding of the derivation for the prior distribution for the batting average. It looks like David is backing out the parameters from the mean and the range.
Under the assumption that the mean is $0.27$ and the standard deviation is $0.18$, can you back out $\alpha$ and $\beta$ by solving these two equations:
\begin{equation}
\frac{\alpha}{\alpha+\beta}=0.27 \\
\frac{\alpha\cdot\beta}{(\alpha+\beta)^2\cdot(\alpha+\beta+1)}=0.18^2
\end{equation}
 A: I wanted to add this as a comment on the excellent answer but it ran long and will look better with answer formatting.
Something to keep in mind is that not all $(\mu, \sigma^2)$ are possible. It's clear $\mu \in [0,1]$, but not as clear are the limitations for $\sigma^2$.
Using the same reasoning as David, we can express
$$ \sigma^2(\alpha, \mu) = \frac{\mu^2 (1-\mu)}{\alpha + \mu} $$
This is decreasing with respect to $\alpha$, so the largest $\sigma^2$ can be for a given $\mu$ is:
$$\lim_{\alpha \rightarrow 0}\sigma^2(\alpha, \mu) = \mu(1-\mu)$$
This is only a supremum since the set of valid $\alpha$ is open (i.e., for Beta, we must have $\alpha > 0$); this limit is itself maximized at $\mu = \frac12$.
Notice the relationship to a corresponding Bernoulli RV. The Beta distribution with mean $\mu$, since it is forced to take all values between 0 and 1, must be less dispersed (i.e., have lower variance) than the Bernoulli RV with the same mean (which has all of its mass at the ends of the interval). In fact, sending $\alpha$ to 0 and fixing $\beta = \frac{1-\mu}{\mu} \alpha$ amounts to putting more and more of the mass of the PDF close to 0 and 1, i.e., getting closer to a Bernoulli distribution, which is why the supremum of the variance is exactly the corresponding Bernoulli variance.
Taken together, here is the set of valid means and variances for Beta:

(Indeed this is noted on the Wikipedia page for Beta)
A: Notice that:
\begin{equation}
\frac{\alpha\cdot\beta}{(\alpha+\beta)^2}=(\frac{\alpha}{\alpha+\beta})\cdot(1-\frac{\alpha}{\alpha+\beta})
\end{equation}
This means the variance can therefore be expressed in terms of the mean as
\begin{equation}
\sigma^2=\frac{\mu\cdot(1-\mu)}{\alpha+\beta+1} \\
\end{equation}
If you want a mean of $.27$ and a standard deviation of $.18$ (variance $.0324$), just calculate:
\begin{equation}
\alpha+\beta=\frac{\mu(1-\mu)}{\sigma^2}-1=\frac{.27\cdot(1-.27)}{.0324}-1=5.083333 \\
\end{equation}
Now that you know the total, $\alpha$ and $\beta$ are easy:
\begin{equation}
\alpha=\mu(\alpha+\beta)=.27 \cdot 5.083333=1.372499 \\
\beta=(1-\mu)(\alpha+\beta)=(1-.27) \cdot 5.083333=3.710831
\end{equation}
You can check this answer in R:
> mean(rbeta(10000000, 1.372499, 3.710831))
[1] 0.2700334
> var(rbeta(10000000, 1.372499, 3.710831))
[1] 0.03241907

