In order to motivate our definition of the notion of expectation, let us first discuss the meaning of the word “average”. Given a set of \(n\) quantities, which we denote by \(x_{1}, x_{2}, \ldots, x_{n}\) , we define their average, often denoted by \(\bar{x}\) , as the sum of the quantities divided by \(n\) ; in symbols

\[\bar{x}=\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}=\frac{1}{n} \sum_{i=1}^{n} x_{i} . \tag{1.1}\] 

The quantity \(\bar{x}\) is also called the arithmetic mean of the numbers \(x_{1}, x_{2}, \ldots\) , \(x_{n}\) .

For example, consider the scores on an examination of a class of 20 students:

\[\{10, 10, 10, 10, 9, 9, 9, 9, 9, 8, 8, 8, 8, 8, 7, 7, 6, 5, 5, 5\} \tag{1.2}\] 

The average of these scores is \(160 / 20=8\) .

Very often, a set of \(n\) numbers, \(x_{1}, x_{2}, \ldots, x_{n}\) , which is to be averaged, may be described in the following way. There are \(k\) real numbers, which we may denote by \(x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{k}^{\prime}\) , and \(k\) integers, \(n_{1}, n_{2}, \ldots, n_{k}\) (whose sum is \(n\) ), such that the set of numbers \(\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}\) consists of \(n_{1}\) repetitions of the number \(x_{1}^{\prime}, n_{2}\) repetitions of the number \(x_{2}^{\prime}\) , and so on, up to \(n_{k}\) repetitions of the number \(x_{k}^{\prime}\) . Thus the set of scores in (1.2) may be described by the following table:

\[\begin{array}{c|cccccc} \hline \text{Possible values } x'_i \text{ in the set} & 10 & 9 & 8 & 7 & 6 & 5 \\ \hline \text{Number } n_i \text{ of occurrences of } x'_i \text{ in the set} & 4 & 5 & 5 & 2 & 1 & 3 \\ \hline \end{array}\tag{1.3}\] 

In terms of this notation, the average \(\bar{x}\) defined by (1.1) may be written

\[\bar{x}=\frac{1}{n} \sum_{i=1}^{k} x_{i}^{\prime} n_{i} \tag{1.4}\] 

We may go one step further. Let us define the quantity

\[f\left(x_{i}^{\prime}\right)=\frac{n_{i}}{n} \tag{1.5}\] 

that represents the fraction of the set of numbers \(\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}\) , which is equal to the number \(x_{i}^{\prime}\) . Then (1.4) becomes

\[\bar{x}=\sum_{i=1}^{k} x_{i}^{\prime} f\left(x_{i}^{\prime}\right) \tag{1.6}\] 

In words, we may read (1.6) as follows: the average \(\bar{x}\) of a set of numbers, \(x_{1}, x_{2}, \ldots, x_{n}\) , is equal to the sum, over the set of numbers, \(x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{k}^{\prime}\) , which occur in the set \(\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}\) , of the product of the value of \(x_{i}^{\prime}\) and the fraction \(f\left(x_{i}^{\prime}\right) ; f^{\prime}\left(x_{i}^{\prime}\right)\) is the fraction of numbers in the set \(\left\{x_{1}, x_{2}, \ldots\right.\) , \(\left.x_{n}\right\}\) which are equal to \(x_{i}^{\prime}\) .

The question naturally arises as to the meaning to be assigned to the average of a set of numbers. It seems clear that the average of a set of numbers is computed for the purpose of summarizing the data represented by the set of numbers, so as to better comprehend it. Given the examination scores of a large number of students, it is difficult to form an opinion as to how well the students performed, except perhaps by forming averages.

However, it is also clear that the average of a set of numbers, as defined by (1.1) or (1.6) , does not serve to summarize the data completely. Consider a second group of twenty students who, in the same examination on which the scores in (1.2) were obtained, gave the following performance: \[\begin{array}{c|cccccc} \hline \text{ Scores } x_{i}^{\prime} & 10 & 9 & 8 & 7 & 6 & 5 \\ \hline \text{Number } n_{i} \text{ of students scoring the score }x_{i}^{\prime} & 3 & 6 & 6 & 2 & 3 & 1 \\ \hline \end{array}\tag{1.7}\] The average of this set of scores is 8, as it would have been if the scores had been \[\begin{array}{c|cccccc} \hline \text{ Scores } x_{i}^{\prime} & 10 & 9 & 8 & 7 & 6 & 5 \\ \hline \text{Number } n_{i} \text{ of students scoring the score }x_{i}^{\prime} & 3 & 3 & 8 & 3 & 3 & 0 \\ \hline \end{array}\tag{1.8}\] Consequently, if we are to summarize these collections of data, we shall require more than the average, in the sense of (1.6) , to do it.

The average, in the sense of (1.6) , is a measure of what might be called the mid-point, or mean, of the data, about which the numbers in the data are, loosely speaking, “centered.” More precisely, the mean \(\bar{x}\) represents the center of gravity of a long rod on which masses \(f\left(x_{1}^{\prime}\right), \ldots, f\left(x_{k}^{\prime}\right)\) have been placed at the points \(x_{1}^{\prime}, \ldots, x_{k}^{\prime}\) , respectively.

Perhaps another characteristic of the data for which one should have a measure is its spread or dispersion about the mean. Of course, it is not clear how this measure should be defined.

The dispersion might be defined as the average of the absolute value of the deviation of each number in the set from the mean \(\bar{x}\) ; in symbols,

\[\text { absolute dispersion }=\sum_{i=1}^{k}\left|x_{i}^{\prime}-\bar{x}\right| f\left(x_{i}^{\prime}\right). \tag{1.9}\] 

The value of the expression (1.9) for the data in (1.3) , (1.7) , and (1.8) is equal to 1.3, 1.1, and 0.9 , respectively, where in each case the mean \(\bar{x}=8\) .

Another possible measure of the spread of the data is the average of the squares of the deviation from the mean \(\bar{x}\) of each number \(x_{i}^{\prime}\) in the set; in symbols,

\[\text { square dispersion }=\sum_{i=1}^{k}\left(x_{i}^{\prime}-\bar{x}\right)^{2} f\left(x_{i}^{\prime}\right), \tag{1.10}\] 

which has the values \(2.7,2.0\) , and 1.5 for the data in (1.3), (1.7), and (1.8), respectively.

Next, one may desire a measure for the symmetry of the distribution of the scores about their mean, for which purpose one might take the average of the cubes of the deviation of each number in the set from the mid-point \(\bar{x}(=8)\) ; in symbols,

\[\sum_{i=1}^{k}\left(x_{i}^{\prime}-\bar{x}\right)^{3} f\left(x_{i}^{\prime}\right), \tag{1.11}\] 

which has the values \(-2.7,-1.2\) , and 0 for the data in (1.3) , (1.7) , and (1.8) , respectively.

From the foregoing discussion one conclusion emerges clearly. Given data \(\{x_{1}, x_{2}, \ldots, x_{n}\}\) , there are many kinds of averages one can define, depending on the particular aspect of the data in which one is interested. Consequently, we cannot speak of the average of a set of numbers. Rather, we must consider some function \(g(x)\) of a real variable \(x\) ; for example, \(g(x)=x, g(x)=(x-8)^{2}\) , or \(g(x)=(x-8)^{3}\) . We then define the average of the function \(g(x)\) with respect to a set of numbers \(\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}\) as \[\frac{1}{n} \sum_{j=1}^{n} g\left(x_{j}\right)=\sum_{i=1}^{k} g\left(x_{i}^{\prime}\right) f\left(x_{i}^{\prime}\right), \tag{1.12}\] in which the numbers \(x_{1}^{\prime}, \ldots, x_{k}^{\prime}\) occur in the proportions \(f\left(x_{1}^{\prime}\right), \ldots, f\left(x_{k}^{\prime}\right)\) in the set \(\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}\) .

Exercises

In each of the following exercises find the average with respect to the data given for these functions: (i) \(g(x)=x\) ; (ii) \(g(x)=(x-\bar{x})^{2}\) , in which \(\bar{x}\) is the answer obtained to question (i); (iii) \(g(x)=(x-\bar{x})^{3}\) ; (iv) \(g(x)=(x-\bar{x})\) ; (v) \(g(x)=|x-\bar{x}|\) .

Hint : First compute the number of times each number appears in the data.