11.02 Expected value

In the previous chapter, we considered the population of a maths class and their results on their latest pop quiz as a discrete random variable $X$X. In a class of ten students, two students scored $6$6, three scored $7$7, one scored $8$8 and four scored $9$9. We used these relative frequencies to create the discrete probability distribution shown in the table below.

$x$x	$6$6	$7$7	$8$8	$9$9
$P\left(X=x\right)$P(X=x)	$\frac{2}{10}$210​	$\frac{3}{10}$310​	$\frac{1}{10}$110​	$\frac{4}{10}$410​

Now that we have condensed all of the information into this probability distribution, what else can we do with it? Well, we could compute the average (mean) score on the maths test.

We could do this the old fashioned way, by adding up all the scores and dividing by the total number of scores, giving us a line of working something like this:

$\text{Mean score}=\frac{6\times2+7\times3+8\times1+9\times4}{10}=7.7$Mean score=6×2+7×3+8×1+9×410​=7.7

However, a probability distribution gives us another way of calculating the mean. We can multiply each outcome by its probability and add them together. These are simply the pairs within the probability distribution table. In this case, the probability represents the weighting for each outcome occurring.

$\text{Mean score}=6\times\frac{2}{10}+7\times\frac{3}{10}+8\times\frac{1}{10}+4\times\frac{4}{10}=7.7$Mean score=6×210​+7×310​+8×110​+4×410​=7.7

It's obvious that mathematically, this is the same process as all of the values that form the numerator and denominator of the overall calculation are the same. But calculating the mean from a probability distribution, where we won't necessarily know the relative frequencies of the scores, is a new idea! What exactly does the mean of a probability experiment, well, mean?

In the context of a probability experiment, the average (mean) represents how we expect that the results of our experiment will play out over a large number of trials. For this reason, we call this calculation the expected value of the experiment and denote it as $E\left(X\right)$E(X).

Expected value as a measure of central tendency

Having calculated a mean, it is clear that $E\left(X\right)$E(X) is more than just a value. If we consider this visually, in terms of the graph of a probability distribution, it defines the position that cuts the area of the probability function into two equal parts. You may also see the expected value written as $\mu$μ, which is the greek letter associated with the population mean.

The diagram below locates $\mu$μ on the discrete probability distribution of the class' test marks as discussed above.

 

Sample vs population mean

Up until now, we have used $\overline{x}$x, to represent the mean, not $\mu$μ. The notation $\overline{x}$x is the represents a sample mean. A sample is a smaller group taken from the entire set of data we have, which we call the population. In discrete probability distributions, whatever data set we are working with is considered to be our population. In the example we have been working with, the maths class we have is the population, and so, we will use $\mu$μ for the mean.

If we were to take a sample from our population, we could calculate its mean. This might give us the same value as $\mu$μ, but in all likelihood, it would be at least a little bit different. A sample mean therefore is an estimate of the population mean. Logically, as the sample size increases, the sample mean would become a better estimate of the population mean.

Remember

When we talk about expected value, we are talking about the result of multiple trials of an experiment, not a single occurrence. It's like when we say the average number of children per family is $2.3$2.3. No family will have $0.3$0.3 of a child! But the expected value is the mean value of many, many families, and this gives us an idea about how many children we can expect to see in a family; somewhere between two and three.

Similarly, the expected value is not to be confused with the most likely outcome. In our distribution above, the most likely outcome is $9$9 since $P\left(X=9\right)$P(X=9) has the highest probability of $0.4$0.4.

As well as calculating the expected value of a discrete probability question, we can be given $E\left(X\right)$E(X) and use this to find missing values in a probability distribution. Expected value also makes the financial outcome of a game very easy to find if you assign the possible amounts of a win or loss as the outcomes $X$X. Let's take a look at the worked examples below.

Worked examples

Example 1

Consider the probability distribution of a discrete random variable given in the table below, where $E(X)=3.05$E(X)=3.05

$x$x	$0$0	$1$1	$3$3	$5$5	$6$6
$P(X=x)$P(X=x)	$0.3$0.3	$a$a	$0.1$0.1	$b$b	$0.15$0.15

Determine the values of $a$a and $b$b.

Solution: Since we have two unknowns we need two equations to use in solving for $a$a and $b$b.

We are told that the table represents a discrete random variable, so we know all the probabilities must add to $1$1. So our first equation is:

$0.3+a+0.1+b+0.15$0.3+a+0.1+b+0.15

$=$=

$1$1

Simplifying gives us:

$a+b$a+b

$=$=

$0.45$0.45

Using the expected value we can form another equation:

$0\times0.3+1\times a+3\times0.1+5\times b+6\times0.15$0×0.3+1×a+3×0.1+5×b+6×0.15

$=$=

$3.05$3.05

Simplifying gives us:

$a+5b$a+5b

$=$=

$1.85$1.85

Solving these simultaneous equations using the elimination method gives us:

$4b$4b	$=$=	$1.4$1.4
$b$b	$=$=	$0.35$0.35
$a$a	$=$=	$0.1$0.1

Example 2

A card is drawn at random from a deck of playing cards. If it is red, the player wins $1$1 dollar; if it is black, the player loses $2$2 dollars. Find the expected value of the game.

Solution: The two outcomes, $+1$+1 and $-2$−2, are equally likely and so we have the following discrete probability distribution:

$x$x	$1$1	$-2$−2
$P\left(X=x\right)$P(X=x)	$0.5$0.5	$0.5$0.5

Now:

$E\left(X\right)$E(X)	$=$=	$1\times0.5+(-2)\times0.5$1×0.5+(−2)×0.5
$E\left(X\right)$E(X)	$=$=	$-0.5$−0.5

And so the expected value of this game is $-50$−50 c.

Practice questions

Question 1

Question 2

Question 3

Question 4