How to find estimator of Pareto distribution using method of mmoment with both parameters unknown?

Question

$f_{\alpha, \beta}(y)=\frac{\alpha}{\beta}(\frac{\beta}{y})^{\alpha +1}, y\ge\beta,\ \ \alpha,\beta\gt 0$. Both $\alpha, \beta$ unkown. 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.

How do we tackle such problem?

Answer 1

If you are given an exercise to find the MOM estimators of $\alpha$ and $\beta,$ then you can give the formal answers, and report the difficulty you encountered.

If you are actually trying to estimate parameters from Pareto data, then you should use ML estimators in Section 7 of this Wikipedia article, where the notation $\beta = x_{m}$ is used.

Addendum: Here is a demonstration of the MLEs for $n = 1000$ observations randomly sampled from $\mathsf{Pareto}(\alpha = 3/2, \beta = 1):$ $\hat \beta = 1.000166,\; \hat \alpha = 1.466662.$ Because $\hat\beta = \min(X_i),$ it is slightly biased upward from $\beta = 1.$ According to Wikipedia the standard error of $\hat \alpha$ is about $\hat\alpha/\sqrt{n} \approx 0.046.$

 set.seed(2017)
 alp = 3/2;  bta = 1;  n = 1000
 x = bta/runif(n)^(1/alp)                # random sample
 b = min(x);  a = n/(sum(log(x)-log(b))) # MLEs
 b; a
 ## 1.000166
 ## 1.466662

The Pareto distribution has such a heavy right tail that a histogram of the data doesn't show much. However $\log(X/\beta) \sim \mathsf{Exp}(\alpha),$ as illustrated below. Perhaps you could experiment with MOM estimates of an exponential distribution related to your Pareto distribution.