# 11.3.1: Mean and Variance


Suppose you have a set of values $$a_{j}$$. By saying that this is a set, we mean that we have several values $$a_{1}$$, $$a_{2}$$, $$a_{3}$$, and so forth. The notation $$a_{j}$$, in this context, means that $$j$$ can be replaced by any integer between 1 and the total number of values that you have in order to refer to that specific value. Suppose that we have $$N$$ total values. The average of all of our values can be written as:

$\langle a\rangle=\frac{1}{N} \sum_{j} a_{j}\tag{11.10}$

The letter $$\Sigma$$ is the capital Greek letter “sigma”. This notation means that you sum together all of the values of $$a_{j}$$ that you have. For instance, suppose you had just four values, $$a_{1}, a_{2}, a_{3}$$, and $$a_{4}$$, then:

$\sum_{j} a_{j}=a_{1}+a_{2}+a_{3}+a_{4}\tag{11.11}$

Therefore, the mean (or average) value of $$a$$ in this context is:

$\langle a\rangle=\frac{1}{N} \sum_{j} a_{j}=\frac{1}{N}\left(a_{1}+a_{2}+a_{3}+a_{4}\right)\tag{11.12}$

To quantify the uncertainty on a set of values, we want to say something about how far, on average, a given value is from the mean of all the values. Thus, it’s tempting to try to define the uncertainty as follows:

$\frac{1}{N} \sum_{j}\left(a_{j}-\langle a\rangle\right)\tag{11.13}$

Remember that addition is commutative. Realizing that the $$\sum$$ symbol just indicates a sum, i.e. a whole lot of addition, we can rewrite this as:

$\frac{1}{N}\left(\sum_{j} a_{j}-\sum_{j}\langle a\rangle\right)\tag{11.14}$

The second term in the subtraction is a sum over $$j$$ of the average value. The average value doesn’t depend on which $$a_{j}$$ we’re talking about; it’s a constant, it’s the same for all of them. Therefore, the sum of that number $$N$$ times is just going to be equal to $$N\langle a\rangle$$. Making this substitution and distributing the 1/$$N$$ into the parentheses:

$\frac{1}{N} \sum_{j} a_{j}-\frac{1}{N} N\langle a\rangle\tag{11.15}$

But we recognize the first term in this subtraction as just $$\langle a\rangle$$. So, the total result of this is zero. Clearly, this is not a good expression for the uncertainty in $$a$$. If you think about it, the average deviation of $$a_{j}$$ from $$\langle a\rangle$$ ought to be zero. If $$\langle a\rangle$$ is the average value of $$a$$, then $$a_{j}$$ should be below $$\langle a\rangle$$ about as often as it is above, so your sum will have a mix of positive and negative terms. The very definition of the average insures that this sum will be zero.

Instead, we shall define the variance as:

$\Delta a^{2}=\frac{1}{N} \sum_{j}\left(a_{j}-\langle a\rangle\right)^{2}\tag{11.16}$

Here, we’re using $$\Delta a$$ to indicate the uncertainty in $$a$$. The variance is defined as the uncertainty squared.1 The advantage of this expression is that because we’re squaring the difference between each value $$a_{j}$$ and the average value, we’re always going to be summing together positive terms; there will be no negative terms to cancel out the positive terms. Therefore, this should be a reasonable estimate of how far, typically, the measurements $$a_{j}$$ are from their average.

We can unpack this sum a bit, first by multiplying out the squared polynomial:

$\Delta^{2}=\frac{1}{N} \sum_{j}\left(a_{j}^{2}-2\langle a\rangle a_{j}+\langle a\rangle^{2}\right)\tag{11.17}$

In order to clean this expression up, inside the parentheses both add and subtract $$\langle a\rangle^{2}$$:

\begin{aligned} \Delta a^{2} &=\frac{1}{N} \sum_{j}\left(a_{j}^{2}-2\langle a\rangle a_{j}+2\langle a\rangle^{2}-\langle a\rangle^{2}\right) \\ &=\frac{1}{N} \sum_{j}\left(a_{j}^{2}-\langle a\rangle^{2}+2\langle a\rangle\left(\langle a\rangle-a_{j}\right)\right) \\ &=\frac{1}{N} \sum_{j} a_{j}^{2}-\frac{1}{N} \sum_{j}\langle a\rangle^{2}+\frac{1}{N} 2\langle a\rangle \sum_{j}\left(\langle a\rangle-a_{j}\right) \end{aligned}\tag{11.18}

Notice that the last term is going to be zero, as it includes the average difference between the mean and each observation. The second term is just going to be $$\langle a\rangle^{2}$$, because once again $$\langle a\rangle$$ is the same for all terms of the sum; the sum will yield $$N\langle a\rangle^{2}$$, canceling the $$N$$ in the denominator. So, we have:

$\Delta a^{2}=\left\langle a^{2}\right\rangle-\langle a\rangle^{2}\tag{11.19}$

1If you know statistics, you may recognizing this as being very similar to how variance is defined there— only in statistics, we divide by $$N −1$$ rather than by $$N$$. The difference becomes unimportant as $$N$$ gets large.

This page titled 11.3.1: Mean and Variance is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pieter Kok via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.