Averages: There is Still Something to Learn

Abstract

The common way to deal with outliers in empirical Economics and Finance is to delete them, either by trimming or winsorizing, or by computing statistics robust to outliers. However, due to their importance, there are situations where the exclusion of these observations is not reasonable and may even be counterproductive. For example, should we exclude the very high stock prices of Amazon and Google from an empirical analysis? Even if the purpose is to compute an average of tech stock prices, does it make economic and financial sense? Maybe not. A solution that would keep the two companies in the data set and yet not penalize the higher observations as much as the median, harmonic and geometric averages, might—were such a solution to be available—constitute an attractive alternative. In this paper we propose and analyze a modified measure, the adjusted median, where the influence of the outlying observations, while not as high as in the arithmetic average would, however, give more weight to the outlying observations than the median, harmonic and geometric averages. Monte Carlo simulations and bootstrapping real financial data confirm how useful the adjusted median could be.

Appendices

A Contributions of Each Data Point

See Eq. (5):

$$\begin{aligned} \bar{X}_{H}&= \sum _{i=1}^n x_i \frac{w_i}{\sum _{i=1}^n w_i} = \sum _{i=1}^n \left[ x_i \frac{ \frac{\sum _{i=1}^n x_i}{x_i} }{\frac{\sum _{i=1}^n x_i}{x_1}+\frac{\sum _{i=1}^n x_i}{x_2}+ \ldots + \frac{\sum _{i=1}^n x_i}{x_n}} \right] \\&= \sum _{i=1}^n \left[ \frac{x_i \sum _{i=1}^n x_i}{x_i \sum _{i=1}^n x_i {\sum _{i=1}^n \frac{1}{x_i}}} \right] =,\\&= \sum _{i=1}^n \left[ \frac{1}{\sum _{i=1}^n \frac{1}{x_i}}\right] = \frac{1}{\sum _{i=1}^n \frac{1}{x_i}}+\frac{1}{\sum _{i=1}^n \frac{1}{x_i}} + \ldots + \frac{1}{\sum _{i=1}^n \frac{1}{x_i}}=\frac{n}{\sum _{i=1}^n \frac{1}{x_i}}, \end{aligned}$$

if $x_i \sum _{i=1}^n x_i \ne 0.$

See Eq. (7):

$$\begin{aligned} x_i \frac{w_i}{\sum _{i=1}^n w_i}=x_i \frac{ \frac{\sum _{i=1}^n x_i}{x_i} }{\frac{\sum _{i=1}^n x_i}{x_1}+\frac{\sum _{i=1}^n x_i}{x_2}+ \ldots + \frac{\sum _{i=1}^n x_i}{x_n}}= \frac{x_i \sum _{i=1}^n x_i}{x_i \sum _{i=1}^n x_i {\sum _{i=1}^n \frac{1}{x_i}}}=\frac{1}{\sum _{i=1}^n \frac{1}{x_i}}, \end{aligned}$$

if $x_i \sum _{i=1}^n x_i \ne 0.$

B Averages Inequalities

According to the means definition, see Eqs. (1), (2) and (3), their logarithms are:

$$\begin{aligned} \ln \left( \bar{X} \right) =\ln \left( \frac{1}{n} \sum _{i=1}^{n}X_i \right) , \quad \ln \left( \bar{X}_G \right) =\frac{1}{n}\sum _{i=1}^{n} \ln \left( X_i\right) \text{ and } \ln \left( \bar{X}_H \right) =-\ln \left( \frac{1}{n} \sum _{i=1}^{n}\frac{1}{X_i}\right) . \end{aligned}$$

By Jensen’s inequality,

$$\begin{aligned} \ln \left( \frac{1}{n}\sum _{i=1}^nX_i \right) \ge \frac{1}{n} \sum _{i=1}^n \ln \left( X_i\right) , \end{aligned}$$

which can be exponentiated to give the arithmetic mean-geometric mean inequality:

$$\begin{aligned} \underbrace{\frac{1}{n} \sum _{i=1}^{n}X_i}_{\bar{X}} \ge \underbrace{\left( \prod _{i=1}^{n} X_i\right) ^{\frac{1}{n}}}_{\bar{X}_G}, \;\; \text{ thus } \bar{X} \ge \bar{X}_G. \end{aligned}$$

Now comparing the harmonic with the geometric mean (and by Jensen’s inequality):

$$\begin{aligned} -\ln \left( \frac{1}{n} \sum _{i=1}^{n}\frac{1}{X_i}\right) \le -\frac{1}{n} \sum _{i=1}^{n}\ln \left( \frac{1}{X_i}\right) =\frac{1}{n}\sum _{i=1}^{n} \ln \left( X_i\right) , \end{aligned}$$

and by exponentiating both sides:

$$\begin{aligned} \underbrace{\frac{n}{\sum _{i=1}^{n} \frac{1}{X_i}}}_{\bar{X}_H} \le \underbrace{\left( \prod _{i=1}^{n} X_i\right) ^{\frac{1}{n}}}_{\bar{X}_G}, \; \; \text{ thus } \bar{X}_H \le \bar{X}_G. \end{aligned}$$

C Different Meanings of the Center

Median

Consider a first data set: 4, 6, 10, 100 ($n=4$, even). Thus, the median rank is $r_M=\frac{1+4}{2}=2.5$, the median is $\bar{X}_M= \frac{6+10}{2}=8$ and $\sum _{i=1}^4 \left( r_i-r_M\right) =(1-2.5)+(2-2.5)+(3-2.5)+(4-2.5)=0$.

For a second data set 4, 6, 10, 20, 100 ($n=5$, odd), the median rank is $r_M=\frac{1+5}{2}=3$, the median is $\bar{X}_M= 10$ and $\sum _{i=1}^5 \left( r_i-r_M \right) =(1-3)+(2-3)+(3-3)+(4-3)+(5-3)=0$. Thus, the median is the center of the distribution in terms of the counting observations: one half of the observations is on the left and one half is on the right of the median, no matter the value of the observations.

Arithmetic average

Consider again the second data set: $\bar{X}_A=\frac{4+6+10+20+100}{5}=28$ and

$\sum _{i=1}^5 \left( x_i-\bar{X}_A\right) =(4-28)+(6-28) + (10-28)+ (20-28) + (100-28)=0$.

Thus, the arithmetic average is the center of the distribution in terms of the deviations in absolute terms: the arithmetic mean is such that the absolute deviations on its right is compensated by the absolute deviations on its left. So, the center is defined in terms of the absolute deviations (or distances) between each value and the arithmetic average:

$$\begin{aligned} \underbrace{\underbrace{(4-28)}_{-24}+\underbrace{(6-28)}_{-22}+ \underbrace{(10-28)}_{-18}\underbrace{(20-28)}_{-8}}_{-72}+\underbrace{(100-28)}_{+72}. \end{aligned}$$

Geometric average

For the second data set: $\bar{X}_G=\root 5 \of {4 \times 6 \times 10 \times 20 \times 100}=13.69$ and

$x_i$	$\ln (x_i)$	$\ln (x_i)-\ln (\bar{X}_G)$
4	1.386	− 123.00%
6	1.792	− 82.45%
10	2.303	− 31.37%
20	2.996	37.94%
100	4.605	198.89%
	sum	0

Compounding percentage deviation means that:

$4=13.69 \times \exp (-123\%)$, $\ldots$, $100=13.69 \times \exp (198.89\%)$.

Thus, the geometric average is the value that balances the negative percentage deviations with the positive ones.

Harmonic average

$x_i$	$x_i-\bar{X}_H$	$w_i=1/x_i$	$\frac{w_i}{\sum _{i=1}^n w_i}$	$\left( x_i-\bar{X}_H\right) \left( \frac{w_i}{\sum _{i=1}^n w_i}\right)$
4	− 4.671	0.250	0.434	− 2.025
6	− 2.671	0.167	0.289	− 0.772
10	1.329	0.100	0.173	0.231
20	11.329	0.050	0.087	0.982
100	91.329	0.010	0.017	1.584
		0.577	1	0

Thus, the harmonic average defines the center of the distribution in order that the weighted deviations on its left compensate the weighted deviations on its right. The weights are inversely proportional to the original values.