Now, suppose we play a game where we get $3$ dollars per heads and $4$ dollars per tails. Whether we get a heads or tails is random, so we can use random variable $X$ to denote the profit of a single toss. The probability mass function of $X$ , denoted as $p (x)$ , is as follows:

Outcome	$x$	$p (x)$
Heads	$3$	$0.8$
Tails	$4$	$0.2$

This table states that for a single toss:

the probability of getting $3$ dollars is $0.8$ .
the probability of getting $4$ dollars is $0.2$ .

Note that $x$ represents a specific value that a random variable $X$ could take on.

Now, the key question we wish to answer is:

If we toss the coin $n = 10$ times, how much do we expect to profit?

To answer this question, we should first find out how many heads and tails we expect from $10$ tosses. Intuitively, since the probability of heads is $0.8$ , we should expect the outcome to be $8$ heads and $2$ tails. Mathematically, we are performing the following calculations:

\begin{aligned} Expected number of heads & = n \times p (3) \\ = (10) (0.8) \\ = 8 \\ Expected number of tails & = n \times p (4) \\ = (10) (0.2) \\ = 2 \end{aligned}

Keep in mind that these are only the expected outcome since the outcome is random. Because we profit $3$ dollars for each heads and $4$ dollars for each tails, we can easily calculate the expected profit from heads and the expected profit from tails:

\begin{aligned} Expected profit from heads & = (Expected number of heads) \times (Profit per heads) \\ = (8) (3) \\ = 24 \\ Expected profit from tails & = (Expected number of tails) \times (Profit per tails) \\ = (2) (4) \\ = 8 \end{aligned}

The sum of the expected profits from heads and tails gives us the total expected profit:

\begin{aligned} Expected profit & = Expected profit from heads + Expected profit from tails \\ = 24 + 8 \\ = 32 \end{aligned}

Therefore, if we were to toss the coin $10$ times, then our expected profit is $32$ dollars! We can generalize the above calculations as follows:

\begin{aligned} Expected profit & = \overset{Expected profit of X=3}{\overset{⏞}{\underset{Expected number of X=3}{\underset{⏟}{[n \times p (3)]}} \times 3}} + \overset{Expected profit of X=4}{\overset{⏞}{\underset{Expected number of X=4}{\underset{⏟}{[n \times p (4)]}} \times 4}} \end{aligned}

Now, instead of tossing the coin $n = 10$ times, let's calculate the expected profit from a single coin toss, that is, $n = 1$ . Plugging in $n = 1$ in the above formula gives:

\begin{matrix} (1) & Expected profit of a single trial = 3 \cdot p (3) + 4 \cdot p (4) \end{matrix}

Recall that the random variable $X$ represents the profit of a single trial. We can mathematically denote the expected value of $X$ as $E (X)$ , which in this case means:

\begin{matrix} (2) & E (X) = Expected profit of a single trial \end{matrix}

Moreover, notice how the right-hand side of $(1)$ can be written as:

\begin{matrix} (3) & \sum_{x} x \cdot p (x) = 3 \cdot p (3) + 4 \cdot p (4) \end{matrix}

Here, the summation symbol $\sum_{x}$ means that we are summing over all possible values of $x$ .

Therefore, using $(2)$ and $(3)$ , we can express $(1)$ generally as:

E (X) = \sum_{x} x \cdot p (x)

For our example, we know that $p (3) = 0.8$ and $p (4) = 0.2$ , so the expected profit for a single toss is:

\begin{aligned} E (X) & = \sum_{x} x \cdot p (x) \\ = (3) \cdot p (3) + (4) \cdot p (4) \\ = (3) (0.8) + (4) (0.2) \\ = 3.2 \end{aligned}

Therefore, we should expect to profit $3.2$ dollars per toss on average! Note that this does not mean that we get $3.2$ dollars per toss - in fact, that's impossible because we either get $3$ or $4$ dollars per toss. An expected profit of $3.2$ dollars per toss means that if we were to toss the coin a large number of times, the average profit we make per toss is $3.2$ dollars.

Simulation to demonstrate expected values

Let's run a quick simulation to demonstrate this. Suppose we toss the coin $1000$ times - the simulated outcome is as follows:

As expected, we get around $800$ heads ( $X = 3$ ) and $200$ tails ( $X = 4$ ). Below is a graph showing the running average of the profit per toss:

We can see that the average profit per toss fluctuates a lot in the beginning. However, as we keep tossing the coin, the average profit per toss stabilizes to the theoretical expected value of $3.2$ that we calculated earlier! The more tosses we make, the closer the average profit per toss will be to $3.2$ .

In this example, the random variable $X$ represents the profit from a single toss and so $E (X)$ represents the expected profit from a toss. Of course, $X$ does not have to represent profit - it could for instance represent waiting time, in which case $E (X)$ would represent the expected waiting time.

We now state the formal definition of the expected value of random variables.

Definition.

Expected value of a random discrete variable

Let $X$ be a discrete random variable with probability mass function $p (x)$ . The expected value of $X$ is defined as follows:

E (X) = \sum_{x} x \cdot p (x)

In words, the expected value involves summing the products of all possible values of the random variable $X$ and their respective probability. Intuitively, the expected value of $X$ is a number that tells us the average value of $X$ we expect to see when we perform a large number of independent repetitions of an experiment.

Note that $E (X)$ is sometimes denoted as $μ$ .

Example.

Expected profit from a lottery

Consider a lottery game where the cost of a single ticket is $3$ dollars. Let the random variable $X$ denote the lottery prize. The following table describes the three possible outcomes and their corresponding probabilities:

$x$	$p (x)$
$0$	$0.90$
$10$	$0.08$
$100$	$0.02$

Compute and interpret the expected value of $X$ .

Solution. By definition, the expected value of $X$ is:

\begin{aligned} E (X) & = \sum_{x} x \cdot p (x) \\ = (0) \cdot p (0) + (10) \cdot p (10) + (100) \cdot p (100) \\ = (0) (0.9) + 10 (0.08) + 100 (0.02) \\ = 2.8 \end{aligned}

This means that we should expect to win $2.8$ dollars per lottery game. However, since the cost of a single lottery ticket is $3$ dollars, we will lose $0.2$ dollars per game on average. If we play $100$ games, then we should expect to lose a total of $20$ dollars.

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Example.

Expected waiting time for food

The waiting time (in minutes) for food delivery is represented by the random variable $X$ . The probability mass function of $X$ is:

$x$	$p (x)$
$10$	$0.2$
$20$	$0.3$
$30$	$0.5$

On average, how long do we have to wait to get our food?

Solution. Let random variable $X$ be the waiting time for the food to arrive. The expected value of $X$ is:

\begin{aligned} E (X) & = \sum_{x} x \cdot p (x) \\ = (10) (0.2) + (20) (0.3) + (30) (0.5) \\ = 2 + 6 + 15 \\ = 23 \end{aligned}

Therefore, on average, we must wait 23 minutes for our food!

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Example.

Expected value of a symmetric distribution

Suppose we have a probability mass function like so:

$x$	$4$	$5$	$6$
$p (x)$	$1 / 4$	$2 / 4$	$1 / 4$

This is visualized below:

When we have symmetric distributions like this, we can find the expected value without having to calculate it. We know that the expected value is the mean value that a random variable takes in the long run. For a symmetric distribution, the mean is at the center, which means $E (X) = 5$ .

Example.

Expected value of rolling a dice

Suppose we roll a fair dice once. Let the random variable $X$ denote the face of the rolled dice. What is the expected value of $X$ ?

Solution. Since we have a fair dice, the probability mass function of $X$ is:

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$p (x)$	$1 / 6$	$1 / 6$	$1 / 6$	$1 / 6$	$1 / 6$	$1 / 6$

The expected value of $X$ is:

\begin{aligned} E (X) & = \sum_{x} x \cdot p (x) \\ = (1) (1 / 6) + (2) (1 / 6) + (3) (1 / 6) + (4) (1 / 6) + (5) (1 / 6) + (6) (1 / 6) \\ = 3.5 \end{aligned}

One way of interpreting this is to think of $X$ as points - for instance, if we roll a $6$ , we get $6$ points. On average, we would get $3.5$ points per roll.

As a side note, notice that $p (x)$ is a symmetric probability distribution. We know from earlier that the expected value of a symmetric distribution is the mean value of $x$ , that is:

\frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5

This way of computing the expected value is convenient because we don't have to deal with probabilities - we simply compute the average of all possible $x$ values!

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We will now discuss the expected values of continuous random variables. The underlying intuition is the same as that for the discrete case - the main difference is that instead of summing over all possible values of a random variable, we integrate over them!

Definition.

Expected value of a continuous random variable

Let $X$ be a continuous random variable with probability density function $f (x)$ . The expected value of $X$ is defined as follows:

E (X) = \int_{- \infty}^{\infty} x \cdot f (x) d x

Note the following:

this definition holds only if the integral exists.
the bounds of the integral are typically written as $- \infty$ and $\infty$ for the formal definition of expected values. In practice, we use the bounds of $X$ instead.

Example.

Computing the expected value of a continuous random variable (1)

Suppose random variable $X$ has the following probability density function:

f (x) = {\begin{cases} 2 x, & 0 \leq x \leq 1 \\ 0, & elsewhere \end{cases}

Compute the expected value of $X$ .

Solution. By definition, the expected value of $X$ is:

\begin{aligned} E (X) & = \int_{- \infty}^{\infty} x \cdot f (x) d x \\ = \int_{0}^{1} x (2 x) d x \\ = 2 \int_{0}^{1} x^{2} d x \\ = 2 [\frac{x^{3}}{3}]_{0}^{1} \\ = 2 (\frac{1}{3}) \\ = \frac{2}{3} \end{aligned}

Therefore, if we repeatedly sample from this distribution a large number of times, the average value that $X$ will take on is $2 / 3$ .

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Example.

Computing the expected value of a continuous random variable (2)

Suppose random variable $X$ has the following probability density function:

f (x) = {\begin{cases} 1 / 3, & 5 \leq x \leq 8 \\ 0, & elsewhere \end{cases}

Compute $E (X)$ .

Solution. By definition, the expected value of $X$ is:

\begin{aligned} E (X) & = \int_{- \infty}^{\infty} x \cdot f (x) d x \\ = \int_{5}^{8} x (\frac{1}{3}) d x \\ = \frac{1}{3} \int_{5}^{8} x d x \\ = \frac{1}{3} [\frac{x^{2}}{2}]_{5}^{8} \\ = \frac{1}{6} (64 - 25) \\ = 6.5 \end{aligned}

Therefore, the average value that $X$ takes on in the long run is $6.5$ .

Just as in the discrete case, we don't have to rely on the formula to compute the expected value when the distribution is symmetric. The probability density function of $X$ in this case is symmetric:

Because $E (X)$ represents the average value that $X$ takes on upon repeated sampling, we have that $E (X)$ must be at the center when the distribution of $X$ is symmetric. Therefore, we can easily conclude that $E (X) = 6.5$ in this case.

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In the next section, we will go over the mathematical properties of expected value of random variables.

Published by Isshin Inada

Edited by 0 others

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