Method of moments for symmetric mean zero distributions

Question

When using the method of moments to fit a symmetric mean zero distribution, does it make more sense to fit higher order moments or lower order absolute moments? I could not find any resources which discuss/compare the merits of these two approaches.

My intuition is that with fewer observations it would be preferable to fit the absolute lower order moments which will be a more stable estimation (just a guess). In practice (in industry not academia) I have seen both methods used (usually the method that is easier to solve analytically...).

Answer 1

Partial answer --- for the reader to complete.

We can simplify the question a bit by considering the relationship between the moments. Consider a random variable $X$ and $m_r \equiv \mathbb{E}(X^r)$ denote the $r$th raw moment and let $a_r \equiv \mathbb{E}(|X|^r)$ denote the $r$th absolute moment. For simplicity, we will assume that all the required moments exist. For a distribution that is symmetric around zero, we have:

$$\begin{matrix} m_r = 0 \ \ & & & \text{for odd } k, \ \\[6pt] m_r = a_r & & & \text{for even } k. \\ \end{matrix}$$

So, essentially, the question becomes, should we fit using the lowest order absolute moments or only every second absolute moment (i.e., those with even order). The lower absolute moments are usually going to be more robust, and may have lower variance, but estimators on the lower order absolute moments are also often biased. There may be a trade-off between bias and variance in such cases, and you will need to derive the bias and variance of your estimators in the case under consideration to be sure.

Ultimately, the relative merits of two competing estimators is assessed by their statistical properties (e.g., bias, variance, MSE, consistency, other asymptotic properties). Usually this process entails looking at the moments of the estimator (or even its full distribution in some cases) in a class of problems under consideration and seeing how the estimator performs at different sample sizes under different parametric inputs. I am not aware of any specific literature comparing the two estimators here (though there may well be some), but it is possible to make a comparison using ordinary techniques for the assessment of estimators.

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Example: Consider data $X_1,...,X_n \sim \text{N}(0, \sigma^2)$ from a normal distribution with zero mean. The standard MOM estimator and the alternative AMOM estimator for the variance are given by:

$$\hat{\sigma}_\text{MOM}^2 = \frac{1}{n} \sum_{i=0}^n X_i^2 \quad \quad \quad \hat{\sigma}_\text{AMOM}^2 = \frac{\pi}{2 n^2} \Big( \sum_{i=0}^n |X_i| \Big)^2.$$

The mean and variance of the MOM estimator are:

$$\begin{align} \mathbb{E}(\hat{\sigma}_\text{MOM}^2) &= \frac{1}{n} \sum_{i=0}^n \mathbb{E}(X_i^2) \quad \quad \quad \quad \ \\[6pt] &= \frac{1}{n} \sum_{i=0}^n \sigma^2 \\[6pt] &= \sigma^2, \\[6pt] \mathbb{V}(\hat{\sigma}_\text{MOM}^2) &= \frac{1}{n^2} \sum_{i=0}^n \mathbb{V}(X_i^2) \\[6pt] &= \frac{1}{n^2} \sum_{i=0}^n 2 \sigma^2 \\[6pt] &= \frac{\sigma^2}{n}. \\[6pt] \end{align}$$

The mean and variance of the AMOM estimator are:

$$\begin{align} \quad \quad \quad \mathbb{E}(\hat{\sigma}_\text{AMOM}^2) &= \frac{\pi}{2 n^2} \mathbb{E} \Bigg( \Big( \sum_{i=0}^n |X_i| \Big)^2 \Bigg) \\[6pt] &= \frac{\pi}{2 n^2} \mathbb{E} \Bigg( \sum_{i=0}^n X_i^2 + \sum_{i \neq j} |X_i| |X_j| \Bigg) \\[6pt] &= \frac{\pi}{2 n^2} \Bigg( n \sigma^2 + \frac{2n(n-1)}{\pi} \sigma^2 \Bigg) \\[6pt] &= \Bigg( \frac{\pi}{2n} + \frac{n-1}{n} \Bigg) \sigma^2, \\[6pt] &= \Bigg( 1 + \frac{\pi-2}{2n} \Bigg) \sigma^2, \\[6pt] \mathbb{E}(\hat{\sigma}_\text{AMOM}^4) &= \frac{\pi^2}{4 n^4} \mathbb{E} \Bigg( \Big( \sum_{i=0}^n |X_i| \Big)^4 \Bigg) \\[6pt] &= \frac{\pi^2}{4 n^4} \mathbb{E} \Bigg( \sum_{i=0}^n \sum_{j=0}^n \sum_{k=0}^n \sum_{l=0}^n |X_i| |X_j| |X_k| |X_l| \Bigg) \\[6pt] &= \frac{\pi^2}{4 n^4} \mathbb{E} \Bigg( \sum_{i=0}^n |X_i|^4 + 4 \sum_{i \neq j} |X_i|^3 |X_j| + 3 \sum_{i \neq j} |X_i|^2 |X_j|^2 \\[6pt] &\quad \quad \ \ \ + 6 \sum_{i \neq j \neq k} |X_i|^2 |X_j| |X_k| + 4 \sum_{i \neq j \neq k \neq l} |X_i| |X_j| |X_k| |X_l| \Bigg) \\[6pt] &= ... \\[6pt] \end{align}$$

Answer 2

I don't have much reputation for adding a comment. Hence I am adding an answer. Some reasons:

• Analytical answer is much comprehensible and straightforward, as suggested by @whuber
• Higher-order moments are usually less robust and more prone to outliers, especially when the sample size is small. Think about the median absolute deviation and standard deviation of a sample from the normal zero-mean distribution.
• Higher-order moments might not exist.
• For some distribution, lower-order moments might give the corresponding MLE, for some, higher-order moments might give the corresponding MLE. (Laplace vs Normal distribution). Those would give asymptotically the best solution under the 'usual' conditions.