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:

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

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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.