Expectation

The expectation or expected value of a random variable $X$ is denoted by $E[X]$. It can be considered as a kind of weighted average for $X$, in which the weights are obtained from the probability distribution. If $S$ is discrete, then

$$E[X] = \sum_{s \in S} X(s) P(s) .$$ (9.10)

If $S$ is continuous, then^9.2

$$E[X] = \int_{S} X(s) p(s) ds.$$ (9.11)

One can then define conditional expectation, which applies a given condition to the probability distribution. For example, if $S$ is discrete and an event $F$ is given, then

$$E[X\vert F] = \sum_{s \in S} X(s) P(s\vert F) .$$ (9.12)

Example 9..5 (Tossing Dice)   Returning to Example 9.4, the elements of $S$ are already real numbers. Hence, a random variable $X$ can be defined by simply letting $X(s) = s$. Using (9.11), the expected value, $E[X]$, is $3.5$. Note that the expected value is not necessarily a value that is ``expected'' in practice. It is impossible to actually obtain $3.5$, even though it is not contained in $S$. Suppose that the expected value of $X$ is desired only over trials that result in numbers greater then $3$. This can be described by the event $F = \{4,5,6\}$. Using conditional expectation, (9.12), the expected value is $E[X\vert F] = 5$.

Now consider tossing two dice in succession. Each element $s \in S$ is expressed as $s = (i,j)$ in which $i,j \in \{1,2,3,4,5,6\}$. Since $S \not \subset {\mathbb{R}}$, the random variable needs to be slightly more interesting. One common approach is to count the sum of the dice, which yields $X(s) = i+j$ for any $s \in S$. In this case, $E[X] = 7$. $\blacksquare$