5.3: Expected value

Definition:

For a probability distribution defined by $P(X =x)$, we define the expectation of the random variable $X$ as

$$\begin{align*} E(X) &= \sum_{i=1}^{i=n} x_i P(X=x_i) \\[4pt] &= x_1 P(X=x_2) + x_2 P(X=x_2)+ \cdots+ x_n P(X=x_n) \end{align*}$$

where $x_i$ represents the observed outcome and $P(X=x_i)$ is the probability of the outcome occurring.

The “expected value of $X$” can be interpreted as the mean value of $X.$

The expectation values can be considered in two ways.

1. Long-run average

This is the measure one would see if the experiment was repeated a large number of times, namely $E(X)= np$, where $n$ is the number of times the experiment occurred and $p$ is the probability for the event to occur.

Example $\PageIndex{1}$

If we tossed a coin $1500$ times, and the random variable $X,$ represents the number of heads observed, we would expect $750$ heads, that is $E(X) = 750$.

2. Probability weighted average

This is the measure that takes into account the relative probabilities of each observed outcome.

Example $\PageIndex{2}$

For the probability distribution with random variable X defined by

Table $\PageIndex{2}$: Probability distribution

$x$	$2$	$3$	$4$	$5$
$P(X=x)$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$

$E(X)= \sum_{i=1}^{i=4} x_i P(X=x_i) = 2 \dfrac{1}{6} +3 \dfrac{1}{6}+4 \dfrac{1}{6} +5 \dfrac{1}{6}= \dfrac{15}{6}$.

Thus $X$ has a mean value of $\dfrac{15}{6}$.

Example $\PageIndex{3}$:

We toss four coins at the same time, then the probability of getting $X$ number of tails:

Table $\PageIndex{3}$: Tossing four coins at the same time

$x$	$0$	$1$	$2$	$3$	$4$	Total
$P(X=x)$	$\dfrac{1}{16}$	$\dfrac{4}{16}$	$\dfrac{6}{16}$	$\dfrac{4}{16}$	$\dfrac{1}{16}$

Then, the expected value is

$E(X)= \sum_{i=0}^{i=4} x_i P(X=x_i) = 0 \dfrac{1}{16} +1 \dfrac{4}{16}+2 \dfrac{6}{16}+3 \dfrac{4}{16}+ 4 \dfrac{1}{16}= \dfrac{32}{16}=2$.

Therefore, in the long run, we expect to get $2$ tails when tossing $94$ coins simultaneously.