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

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  • $\begingroup$ A general comment: You have many questions which you haven´t accepted although the corresponding answers are nice answers. $\endgroup$ – callculus Oct 2 '17 at 14:01
  • $\begingroup$ @callculus Thanks for reminding. Just checked them. $\endgroup$ – CoolKid Oct 2 '17 at 14:58
  • $\begingroup$ The method of moments can sometimes give nonsense results (try a binomial distribution with sample variance greater than sample mean as an obvious example), and in any case having sample moments which are finite while the population moments are infinite poses an intractable issue. If you are worried about this, try an alternative method $\endgroup$ – Henry Oct 2 '17 at 17:45
  • $\begingroup$ @Henry Suppose I am going for the method of moment, the concern I am having is that I don't know what to do when, say, first moment is infinite (i.e. $\alpha<=1$). I am thinking of dividing it into cases, but not sure if it's a standard way to deal with this kind of situation. $\endgroup$ – CoolKid Oct 2 '17 at 17:58
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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.

enter image description here

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