Exercise :
Using the method of moments, find the estimator for $\theta$ for a random sample $X_1, \dots, X_n$ that follows the distribution with pdf $f(x) = \theta x^{-2}, \; 0 < \theta \leq x < \infty$, where $\theta$ is the unknown parameter.
Attempt :
$$m_1' = \bar{X}$$
$$μ_1' = \int_\theta^xu\theta u^{-2}\mathrm{d}u=\int_\theta^x\theta u^{-1}\mathrm{d}u=\big[\theta \ln u\big]_\theta^x =\theta\ln x - \theta\ln\theta$$
Thus, the estimator is given by :
$$m_1' = μ_1' \Rightarrow \theta\ln x - \theta\ln\theta = \bar{X}$$
But we cannot solve that equation with respect to $\theta$, so we'll take the moments of bigger order :
$$m_2' = \frac{1}{n}\sum_{i=1}^nX_i^2$$ $$μ_2' = \int_\theta^x u^2 \theta u^{-2} \mathrm{d}u = \int_\theta^x \theta \mathrm{d}u = \big[\theta u\big]_\theta^x =\theta x - \theta ^2 $$
Thus the solution of the following quadratic equation for $\theta >0$, gives as the estimator of $\theta$ :
$$\theta^2-x\theta +\frac{1}{n}\sum_{i=1}^nX_i^2$$
Question : Is my solution correct ? Shall the upper bound be $x$ or $\infty$ ?
The upper bound must be $\infty$. That is a real problem since we have :
$$\int_{\theta}^A xf(x)dx\underset{A\to \infty}\to \infty.$$
Then every moment of order greater than $1$ does not exists. One way to estimate $\theta$ is to proceed as follows :
$$E(1/X)=\int_\theta^{\infty} \theta x^{-3} dx=\frac 1 {2 \theta}.$$
Since, by LLN :
$$\frac 1 n \sum_{i=1}^n \frac 1 {X_i} \to E(1/X)=\frac1{2 \theta}.$$
We deduce :
$$\frac 2 n \sum_{i=1}^n \frac 1 {X_i} \to \frac 1 \theta.$$