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.