- Properties of mathematical expectation
- The mathematical expectation in betting
- Examples
- Example 1
- Example 2
- Exercise resolved
- Solution
- References
The mathematical expectation or expected value of the random variable X, is denoted as E (X) and is defined as the sum of the product between the probability of a random event occurring and the value of said event.
In mathematical form it is expressed as follows:
Figure 1. Mathematical expectation is widely used in the stock market and in insurance. Source: Pixabay.
Where x i is the value of the event and P (x i) its probability of occurrence. The summation extends over all the values that X admits. And if these are finite, the indicated sum converges to the value E (X), but if the sum does not converge, then the variable simply has no expected value.
When it is a continuous variable x, the variable can have infinite values and the integrals replace the summations:
Here f (x) represents the probability density function.
In general, the mathematical expectation (which is a weighted average) is not equal to the arithmetic mean or average, unless we are dealing with discrete distributions in which each event is equally probable. Then, and only then:
Where n is the number of possible values.
The concept is very useful in financial markets and insurance companies, where certainties are often lacking but probabilities exist.
Properties of mathematical expectation
Among the most important properties of mathematical expectation, the following stand out:
- Sign: if X is positive, then E (X) will also be positive.
- Expected value of a constant: the expected value of a real constant k is the constant.
- Linearity in the sum: the expectation of a random variable that is in turn the sum of two variables X and Y is the sum of the expectations.
E (X + Y) = E (X) + E (Y)
- Multiplication by a constant: if the random variable is of the form kX, where k is a constant (a real number), it comes out outside the expected value.
- Expected value of the product and independence between variables: if a random variable is the product of the random variables X and Y, which are independent, then the expected value of the product is the product of the expected values.
In general, if Y = g (X):
- Order in expected value: if X ≤ Y, then:
Since there are the expected values of each of them.
The mathematical expectation in betting
When the famous astronomer Christian Huygens (1629-1695) was not observing the skies, he devoted himself to studying, among other disciplines, probability in games of chance. It was he who introduced the concept of mathematical hope in his 1656 work entitled: Reasoning about games of chance.
Figure 2. Christiaan Huygens (1629-1625) was a brilliant and versatile scientist, to whom we owe the concept of expected value.
Huygens found that bets could be classified in three ways, based on expected value:
-Games with advantage: E (X)> 0
- Fair bets: E (X) = 0
-Game at a disadvantage: E (X) <0
The problem is that in a game of chance the mathematical expectation is not always easy to calculate. And when you can, the result is sometimes disappointing for those who wonder whether or not to bet.
Let's try a simple bet: heads or tails and the loser pays a $ 1 coffee. What is the expected value of this bet?
Well, the probability of a heads being rolled is ½, equal to a tails. The random variable is to win $ 1 or lose $ 1, the gain is denoted by + sign and loss by sign -
We organize the information in a table:
We multiply the values of the columns: 1. ½ = ½ and (-1). ½ = -½ and finally the results are added. The sum is 0 and it is a fair game, in which participants are expected to neither win nor lose.
French roulette and the lottery are handicap games in which most bettors lose. Later there is a slightly more complex bet in the solved exercises section.
Examples
Here are some simple examples where the concept of mathematical expectation is intuitive and clarifies the concept:
Example 1
We will start by rolling an honest die. What is the expected value of the launch? Well, if the die is honest and has 6 heads, the probability that any value (X = 1, 2, 3… 6) will roll is 1/6, like this:
E (X) = 1. (1/6) + 2. (1/6) + 3. (1/6) + 4. (1/6) + 5. (1/6) + 6. (1 / 6) = 21/6 = 3.5
Figure 3. In the roll of an honest die, the expected value is not a possible value. Source: Pixabay.
The expected value in this case is equal to the average, since each face has the same probability of coming out. But E (X) is not a possible value, since no heads are worth 3.5. This is perfectly possible in some distributions, although in this case the result does not help the bettor much.
Let's look at another example with the toss of two coins.
Example 2
Two honest coins are tossed into the air and we define the random variable X as the number of heads that are rolled. The events that can occur are the following:
-No heads come up: 0 heads which is equal to 2 tails.
-It comes out 1 head and 1 stamp or tails.
-Two faces come out.
Let C be a head and T a seal, the sample space that describes these events is the following:
S m = {Seal-Seal; Seal-Face; Face-Seal; Face-Face} = {TT, TC, CT, CC}
The probabilities of the events happening are:
P (X = 0) = P (T). P (T) = ½. ½ = ¼
P (X = 1) = P (TC) + P (CT) = P (T). P (C) + P (C). P (T) = ¼ + ¼ = ½
P (X = 2) = P (C). P (C) = ½. ½ = ¼
The table is built with the values obtained:
According to the definition given at the beginning, the mathematical expectation is calculated as:
Substituting values:
E (X) = 0. ¼ + 1. ½ + 2. ¼ = ½ + ½ = 1
This result is interpreted as follows: if a person has enough time to do a large number of experiments by tossing the two coins, he is expected to get a head on each toss.
However, we know that releases with 2 labels are perfectly possible.
Exercise resolved
In the toss of two honest coins, the following bet is made: if 2 heads come out you win $ 3, if 1 head comes out you win $ 1, but if two stamps come out you have to pay $ 5. Calculate the expected win of the bet.
Figure 4. Depending on the bet, the mathematical expectation changes when tossing two honest coins. Source: Pixabay.
Solution
The random variable X is the values that the money takes in the bet and the probabilities were calculated in the previous example, therefore the table of the bet is:
E (X) = 3. ¼ + 1. ½ + (-5). ¼ = 0
As the expected value is 0, this is fair game, so here the bettor is expected not to win and not to lose either. However, the bet amounts may be changed to make the bet a handicap game or a handicap game.
References
- Brase, C. 2009. Understandable Statistics. Houghton Mifflin.
- Olmedo, F. Introduction to the concept of expected value or mathematical expectation of a random variable. Recovered from: personal.us.es.
- Statistics LibreTexts. Expected Value of Discrete Random Variables. Recovered from: stats.libretexts.org.
- Triola, M. 2010. Elementary Statistics. 11th. Ed. Addison Wesley.
- Walpole, R. 2007. Probability and Statistics for Science and Engineering. 8th. Edition. Pearson Education.