Expectation, Variance and Moment estimator of Beta Distribution I'm given a beta distributed random variable: $X \sim \text{Beta}_{(\theta, 1)} =: \mathbb{P}_\theta$. Where $\theta \geq q$ and
$$\mathbb{f}_\theta(x) = \theta \cdot x^{\theta-1} \cdot \mathbb{1}_{[0,1]}(x)$$
I was asked to compute the expectation and variance of $X$ and came up with the following solution:
$$\mathbb{E}X = \int_\mathbb{R} x \cdot \mathbb{f}_\theta(x) \, \text{dx} = \theta \int_0^{1} x^{\theta} \, \text{dx} = \frac{\theta}{\theta + 1}$$
$$\mathbb{E}X^2 = \int_\mathbb{R} x^2 \cdot \mathbb{f}_\theta(x) \, \text{dx} = \theta \int_0^{1} x^{\theta+1} \, \text{dx} = \frac{\theta}{\theta + 2}$$
$$\text{Var}X = \mathbb{E}X^2 - (\mathbb{E}X)^2 = \frac{\theta}{(\theta+1)^2(\theta+2)}$$
Are these computations correct or did I end up making a mistake? Now let $X_1, \dots, X_n \sim \text{Beta}_{(\theta,1)}$ iid. How can I justify that the moment estimator of $\theta$ is given by $\hat\theta:n = \frac{\bar X_n}{1 - \bar X_n}$? Is this estimator consistent?
I'd say that the estimator is consistent. The Expectation and Variance of $X_1$ is finite. Now after the strong law of large numbers we get 
$$\lim_{n \rightarrow \infty} \hat\theta_n = \frac{\mathbb{E}X_1}{1 - \mathbb{E}X_1} = \theta$$
Therefore the estimator should be consistent, but how can I show that the moment estimator of $\theta$ is given by $\hat\theta_n$? I'm familiar with the method of moments, but don't understand how to apply it in this case.

The idea behind the Method of Moments estimator is the following.
Let suppose we have $\underline{X}$ $=$ $(X_{1},...,X_{n})$ iid observations, distributed according to $f(\cdot \mid \boldsymbol{\theta})$, $\boldsymbol{\theta}$ $\in$ $\Theta$ $\subseteq$ $\mathbb{R}^{d}$.
Define:
$$\mu_{k} = E(X_{i}^{k}) = \mu_{k}(\theta_{1},...,\theta_{d})$$
the $k$-th moment of $X_{i}$ and 
$$m_{k} = \frac{1}{n}\sum_{i = 1}^{n}X_{i}^{k}$$
the $k$-th sample moment. Then we estimate the vector of parameters $\boldsymbol{\theta}$ = $(\theta_{1},...,\theta_{d})$ solving the following system of equations:
\begin{cases} m_{1} = \mu_{1}(\theta_{1},...,\theta_{d})\\
.\\
.\\
m_{d} = \mu_{d}(\theta_{1},...,\theta_{d})\end{cases}
leading to $(\hat{\theta_{1}},...,\hat{\theta_{d}})$ $=$ $\boldsymbol{\hat{\theta}}$, our method of moments estimator. 
Now, consider a generic beta distribution:
$$f(x_{i} \mid \theta, \alpha) = \frac{\Gamma(\theta + \alpha)}{\Gamma(\theta)\Gamma(\alpha)}x_{i}^{\theta - 1}(1 - x_{i})^{\alpha - 1}I(0 < x_{i} < 1)$$
In this case, $\boldsymbol{\theta}$ $=$ $(\theta, \alpha)$, hence we need to consider the first two moments.
The first moment of $X_{i}$ is:
$$E(X_{i}) = \frac{\theta}{\theta + \alpha}$$
while the second moment can be proved to be:
$$E(X_{i}^{2}) = \frac{\theta(\theta + 1)}{(\theta + \alpha)(\theta + \alpha + 1)}$$
Now, at this point, for the first moment in the sample, we have:
$$m_{1} = \frac{1}{n}\sum_{i = 1}^{n}X_{i} = \overline{X_{n}}$$
while for the second moment in the sample we exploit the sample variance:
$$s^{2} = m_{2} - m_{1}^{2} \Rightarrow m_{2} = s^{2} + m_{1}^{2}$$
and, at this point, we apply the system of equations described before to this case, to be then solved for both $\theta$ and $\alpha$:  

\begin{cases} m_{1} = \overline{X_{n}} = E(X_{i}) = \frac{\theta}{\theta + \alpha}\\

m_{2} = s^{2} + m_{1}^{2} = E(X_{i}^{2}) = \frac{\theta(\theta + 1)}{(\theta + \alpha)(\theta + \alpha + 1)}\end{cases}

which would yield the two Method of Moments estimators $\hat{\theta}$ and $\hat{\alpha}$, and this would be in the general case with both parameters of the beta unknown. In your case, the problem is simplified being $\alpha$ $=$ $1$ known, so that we have to estimate only $\theta$ using the first sample moment, that is we have only the first equation to solve with $\alpha$ $=$ $1$:
$$\overline{X_{n}} = \frac{\theta}{\theta + 1} \Rightarrow \hat{\theta} = \frac{\overline{X_{n}}}{1 - \overline{X_{n}}}$$
Hope it clarifies.
