Deviation from average

Question

Say I have a finite list of numbers of some statistical experiment, what is the best way to represent the typical distance from the mean of the numbers based on this data? I have found the standard deviation, but there is also average absolute deviation. How do they differ?

Thanks in advance

Answer 1

Suppose a population is normally distributed with unknown mean $\mu$ and unknown variance $\sigma^2.$ Also suppose you have a random sample of size $n$ from this population. Then the 'best' estimator of the population mean $\mu$ is the sample mean $\bar X = \frac 1n \sum_{i=1}^n X_i.$ Also, the usual estimator of the population variance $\sigma^2$ is the sample variance $S^2 \ \frac{1}{n-1}\sum_{i=1}^n (X_i = \bar X)^2.$

By 'best' above, I mean that these estimators are unbiased: $E(\bar X) = \mu$ and $E(X^2) = \sigma^2.$ Moreover, among unbiased estimators, each of these has the smallest possible variance. There are other criteria for 'goodness' of estimators, and they lead to other estimators of $\sigma^2.$ But $\bar X$ and $S^2$ are the most commonly used ones.

In situations where the population distribution is not normal or is unknown in shape, other estimators of the center or dispersion may be used. For example, if the population is Laplace (double-exponential), then it may be best to estimate the population mean (which, by symmetry, is also the population median) may be the sample median. Moreover, there are situations in which the mean absolute deviation (from the mean) $\frac 1n \sum_{i=1}^n |X_i - \bar X|$ may be a better way to measure the dispersion of the population distribution. (Other way to describe the dispersion of a population or sample are also in use.)

In any discussion of alternative estimators, there must be a clear definition of what is meant by 'best' before meaningful choices can be made among estimators.

Answer 2

For a data set $D=(D_j)_{j\in\Bbb N, j\leq n}$ with mean $\mu=\frac{\sum_jD_j}{n}$, I denote the standard deviation $\sigma$ and absolute deviation $\alpha$. Then:

$$\alpha=\frac {\sum_j|\mu-D_j|}{n}$$ $$\sigma=\bigg(\frac{\sum_j|\mu-D_j|^2}{n}\bigg)^\frac12$$