What is the logic behind method of moments? Why in "Method of Moments", we equate sample moments to population moments for finding point estimator?
Where is the logic behind this?
 A: A sample consisting of $n$ realizations from identically and independently distributed random variables is ergodic. In a such a case, "sample moments" are consistent estimators of theoretical moments of the common distribution, if the theoretical moments exist and are finite.  
This means that
$$\hat \mu_k(n) = \mu_k(\theta) + e_k(n), \;\;\; e_k(n) \xrightarrow{p} 0 \tag{1}$$
So by equating the theoretical moment with the corresponding sample moment we have
$$\hat \mu_k(n) = \mu_k(\theta) \Rightarrow \hat \theta(n) = \mu_k^{-1}(\hat \mu_k(n)) = \mu_k^{-1}[\mu_k(\theta) + e_k(n)]$$
So ($\mu_k$ does not depend on $n$)
$$\text{plim} \hat \theta(n) = \text{plim}\big[\mu_k^{-1}(\mu_k(\theta) + e_k)\big] = \mu_k^{-1}\big(\mu_k(\theta) + \text{plim}e_k(n)\big)$$
$$=\mu_k^{-1}\big(\mu_k(\theta) + 0\big) = \mu_k^{-1}\mu_k(\theta)  = \theta$$
So we do that because we obtain consistent estimators for the unknown parameters.
A: Econometricians call this "the analogy principle". You compute the population mean as the expected value with respect to the population distribution; you compute the estimator as the expected value with respect to the sample distribution, and it turns out to be the sample mean. You have a unified expression
$$
T(F) = \int t(x) \, {\rm d}F(x)
$$
into which you plug either the population $F(x)$, say $F(x) = \int_{\infty}^x \frac1{\sqrt{2\pi\sigma^2}} \exp\bigl[ - \frac{(u-\mu)^2}{2\sigma^2} \bigr] \, {\rm d}u $ or the sample $F_n(x) = \frac 1n \sum_{i=1}^n 1\{ x_i \le x \}$, so that ${\rm d}F_n(x)$ is a bunch of delta-functions, and the (Lebesgue) integral with respect to ${\rm d}F_n(x)$ is the sample sum $\frac1n \sum_{i=1}^n t(x_i)$. If your functional $T(\cdot)$ is (weakly) differentiable, and $F_n(x)$ converges in the appropriate sense to $F(x)$, then it is easy to establish that the estimate is consistent, although of course more hoopla is needed to obtain say asymptotic normality.
A: I might be wrong but the way that I think about it is as follows:
Let's say you have samples $X_1, X_2, \dotsc, X_n$. Then, the method of the moments suggest that we should compare $m-th$ moment of sample with $m$th moment of population
$$(X_1 + X_2 + \dotsm + X_n) / n = μ$$
here, we are averaging all the samples out which seems like a good estimate of population mean
$$(X_1^2 + X_2^2 + \dotsm + X_n^2) / n = σ^2$$
here, we are averaging all the sample variances out which also seems like a good estimate of population variance
And so on,
That is what I think is a good explanation of how method of moment works!
