How to use method of moment to find Pareto distribution estimator? I have $f_{\alpha, \beta}(y)=\frac{\alpha}{\beta}(\frac{\beta}{y})^{\alpha +1}, y\ge\beta,\ \ \alpha,\beta\gt 0$. Both $\alpha, \beta$ unknown. 
To find estimators using the method of moment, we equate $E(Y)=\frac{ \alpha\beta}{\alpha-1}=\frac1n\sum y_i$, 
$E(Y^2)=(E(Y))^2+Var(Y)=(\frac{ \alpha\beta}{\alpha-1})^2+\frac{\beta^2\alpha}{(\alpha-1)^2(\alpha-2)}=\frac1n\sum y_i^2$ 
The problem comes because the mean of Pareto is the $E(Y)$ above only when $\alpha \gt 1$, otherwise it's $\infty$. Same problem for the second moment since $Var(Y)$ is not $\infty$ only when $\alpha>2$, but in the question we only assume $\alpha, \beta>0$. I know that method of moments is not restricted to moments, as we can also equate $E_\theta(g_i(Y))=g_i(Y)$, but I don't see a way out.
 A: If I understand your concern, it's that the true $\alpha$ might be less than 1 (or 2). In this case, the logic behind MoM is that the true mean and the empirical mean should be close. However, the true mean would then be infinite yet the empirical mean is always finite. Setting the two to be equal and solving the MoM equations seems destined to yield an inaccurate estimate of $\alpha$. Generally speaking, this is a fair assessment of what will happen when $\alpha$ is too small.
You alluded to the fact that the method of moments does not necessarily require one to use the first two moments. This is simply the standard approach but as you've just noted, it's not always useful. You could instead use the fact that for any $\gamma \ge 0$, we have that 
$$E(Y^\gamma) = \dfrac {\alpha \beta^\gamma} {\alpha - \gamma} $$
If you suspect that $\alpha$ may be small, you can instead just pick two small values of $\gamma$ and then solve for the corresponding MoM estimates. Doing so should yield a more precise estimator of $\alpha$.
