I am new in statistical studies and on this site and I came across the "zero-sum property" in my book regarding the mean. It seems to be straight forward but I still can't understand the notion of it. The only information it gives with the formula is

the sum of the difference between each value of a variable $Y$, noted $Y_i$, and the mean value of $Y$, noted as $\bar Y$, is equal to zero.

Could someone explain the concept better?


4 Answers 4


You already got more formal answers. This answer is ought to give you some "intuition" behind the math.

Arithmetic mean is sensitive to your data (including outliers). Imagine a lever, like the one illustrated below. Your data are the orange balls that lie on a beam (imagine that it is an x-axis of some kind of plot and your data are values scattered around it on various positions). For the rod to be in horizontal position, the hinge needs to be placed in such place that balances the balls. You can recall from elementary physics (or just playground experiences from your childhood), that the placement of the balls plays a role in how much they influence the lever. The "outlying" balls, how we call them in statistics, have much greater influence then the balls cluttered around the "center". Mean is the value that places the hinge in the exact position that makes the lever balanced.

A lever

So we can say that mean lies in the center, between the values. The center is defined in terms of distances (i.e. differences) between the points and the mean. Since it is in the center, then we would expect that the distances are balanced, i.e. they zero-out each other, so the sum of distances needs to be zero and mean has this property (and only the mean).

Check also the related Arithmetic mean. Why does it work? thread on math.stackexchange.com.

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    $\begingroup$ This holds for the plain vanilla mean estimator, not for the population mean (in the limit), or more sophisticated mean estimators. In fact most shrinkage estimators yield a lower mean (in absolute terms) compared to the vanilla estimator, hence that level doesn't really balance. $\endgroup$ Jun 28, 2017 at 11:16
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    $\begingroup$ @CagdasOzgenc I am explicitly stating that this applies to arithmetic mean. $\endgroup$
    – Tim
    Jun 28, 2017 at 11:17
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    $\begingroup$ It is not a criticism. Additional remarks. The other answers were not comprehensive enough to comment on. $\endgroup$ Jun 28, 2017 at 11:19

let $y_1,y_2, \dots, y_n$ be $n$ observational values from a variable $Y$ and let $\overline{y} := \frac{1}{n}\sum_{i=1}^n y_i$ denote the arithmetic mean of the observations. The zero-sum property can be written mathematically as: $$0 = \sum_{i=1}^n (y_i - \overline{y}).$$ Proof: By definition of $\overline{y}$ we have $n\overline{y} = n\frac{1}{n}\sum_{i=1}^n y_i = \sum_{i=1}^n y_i$ and hence: $$\sum_{i=1}^n (y_i - \overline{y}) = \sum_{i=1}^n y_i - n \overline{y} =n \overline{y} - n \overline{y}= 0.$$ Interpretation: Note that $(y_i - \overline{y})$ is essentially the "distance" between the observation $y_i$ and the arithmetic mean $\overline{y}$ where the information wether the observation is smaller or greater than the arithmetic mean is still preserved through the sign of $(y_i - \overline{y})$ (of course, the distance itself would have to be nonnegative and would be $|y_i-\overline{y}|$).

The zero sum property can then be interpreted, that the arithmetic mean is the number $\overline{y}$ such that observational values of $Y$ which are smaller than $\overline{y}$ and the values of $Y$ which are larger than $\overline{y}$ keep in balance, i.e. they sum up to zero.

In fact it is easy to see from the proof that it is the only number for which this property holds.

You could obviously use this property to check if the calculations of the mean were correct.


Verba docent exempla trahunt.


Take three numbers: 1, 2 and 3.

Mean value is 2

Differences between values and a mean are:

1-2 = -1

2-2 = 0

3-2 = 1

Sum of these differences is

-1 + 0 + 1 = 0

Zero-sum property states that no matter what numbers you start with, a result (sum of diiferences between them and their mean) would be 0


Here's a simple handy little general proof of the result $\sum (x_i - \overline{x}) = 0$

Let's take the sequence of numbers: $$x_1,x_2,x_3,...,x_n$$ we acknowledge that the mean of this number set can be denoted by, $$\overline{x}=\frac{\sum x_i}{n}$$ Going back to the LHS of the original statement $\sum (x_i - \overline{x})$ we can write this out in full as follows: $$\sum (x_i - \overline{x}) = \Bigl(x_1-\frac{\sum x_i}{n}\Bigl) + \Bigl(x_2-\frac{\sum x_i}{n}\Bigl) + \Bigl(x_3-\frac{\sum x_i}{n}\Bigl) +...+\Bigl(x_n-\frac{\sum x_i}{n}\Bigl)$$ This can be simplified down to 0 in the following steps: $$x_1+x_2+x_3+...+x_n-\Bigl(\frac{n\sum x_i}{n}\Bigl)$$ $$\sum x_i-\sum x_i$$ $$=0$$


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