Second Moment of Beta distribution I am practicing the Method of Moments and in this problem, I am a little bit stuck on the algebra in my calculation of the second moment. I would sincerely appreciate any advice on what went wrong:
Let $X_1, \dots X_n$ be a random sample from $Beta(\alpha, \alpha)$, find a Method of Moments estimator of $\alpha$
My attempt:
$$ E(X) =\frac{\alpha}{\alpha + \alpha} = \frac{\alpha}{2\alpha} = \frac{1}{2}$$
and since the estimator cannot be a constant, we need to find the second moment (is this the correct logic?):
$$ E(X^2) = \frac{(\alpha + 1)\alpha}{(\alpha + \alpha + 1)(\alpha + \alpha)}= \frac{(\alpha + 1)\alpha}{(2\alpha + 1)(2\alpha)}=\frac{(\alpha + 1)}{(4\alpha + 2)}$$
Does the algebra seem correct or am I doing something wrong?
Another question is, would finding the second moment be enough here or would you recommend finding the variance as well?
 A: I guess you mean the right thing that the first moment does not allow you to apply the method-of-moments strategy of equating a population moment to a function of the parameter you aim to estimate, as the moment condition does not identify the parameter here. 
This is intuitive in case of the mean when the parameters of the beta distribution are equal to each other, as the mean (as you show) then always is equal to $1/2$, no matter what $\alpha$ generated the data, so the data will not be informative about $\alpha$. Or, from a more positive angle: if the parameters are identical, you do not need data to find the mean of the distribution.
You can utilize either $E(X^2)$ or $Var(X)$ to estimate $\alpha$ - this highlights the fact that method-of-moment estimators are not unique. Working with the variance seems convenient here, as
$$
Var(X)=\frac{1}{8\alpha+4},
$$
which can be easily solved for $\alpha$ to get
$$
\alpha=\frac{1-4Var(X)}{8Var(X)}
$$
Solving 
$$
E(X^2)=\frac{\alpha + 1}{4\alpha + 2}
$$
for $\alpha$ is of course not difficult, either, leading to 
$$
\alpha=\frac{1-2E(X^2)}{4E(X^2)-1}
$$
As method-of-moment estimators are generally consistent, estimators using these expressions will of course produce similar values in large samples, but are not numerically equivalent:
X <- rbeta(10000, shape1 = 2, shape2 = 2)
Xbarsq <- mean(X^2)
VX <- Xbarsq - mean(X)^2

> alpha1 <- (1-2*Xbarsq)/(4*Xbarsq-1)
[1] 2.033679

> alpha2 <- (1-4*VX)/(8*VX)
[1] 1.986139

Following up on @Glen_b's suggestion to also think about the sampling distribution, it does seem to be the case that the MoM estimator based on the variance is substantially more efficient.
Here are kernel density estimates from a little Monte Carlo study:

It ought to be possible to compute the asymptotic variances of the two estimators via the delta method, but I leave that for other interested readers :-).
A: Your approach (i.e. the logic by which you decided to look at the second moment) looks right.
Your algebra also looks right.
It's not necessary to calculate any further moments to estimate $\alpha$ but once you have an estimator it is generally of interest to compute its variance (and if possible, the sampling distribution of the estimator)
