
# 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

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

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

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

Example $$\PageIndex{3}$$

We toss 4 coins at the same time, then the probability of getting $$X$$ number of tails:

 $$x$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ 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 would expect to get $$2$$ tails, when we toss $$94$$ coins at the same time.