## What is meant by Expected Value?

When a large number of trials are performed for any random variable ‘X’, the predicted result is most likely the mean of all the outcomes for the random variable and it is known as expected value also known as expectation. The expected value, also known as the expectation, is denoted by: E(X).

In the middle of the 17th century, the so-called ‘problem of points’, which divides the stakes in a fair way between two players who must end their game before it is properly completed, gave rise to the idea of expected value.

The problem had been debated for centuries before French writer and amateur mathematician Chevalier de Méré introduced it to Blaise Pascal in 1654, and many conflicting ideas and solutions had been suggested over the years.

Méré said that this problem could not be solved and that it showed the flaws in mathematics when applied to real-world situations. Pascal, a mathematician, felt compelled to find a permanent solution to the problem.

## Pros and Cons of Expected Value

The pros and cons of the expected value are given below:

### Pros

• In case of uncertainty, the likelihood of each potential outcome is considered, and the information is used to determine an expected value.
• Information is reduced to a single number, which simplifies decision-making.
• Many of the equations are fairly straightforward.

### Cons

• The probabilities used in most cases are extremely subjective.
• The expected value has no meaning for a one­time project since it is simply a weighted average.
• The expected value offers no hint of the risk associated with the dispersion of possible expected value outcomes.
• It’s possible that the expected value has nothing to do with any of the real outcomes.

## Calculating Expected Value

Essentially, the expected value is the variable’s long-term average value. As the number of repetitions reaches infinity, the average value of the variable converges to the expectation value due to the law of large numbers.

Then according to the definition, the equation obtained for expectation can be expressed as follows:

$E\left(X\right)={\mu }_{n\to \infty }$ ……(1)

Where,

$\mu$ = Mean.

$n$= Number of Trails.

Thus, the value formula for any discrete random variable X is expressed as follows:

$E\left(X\right)=\sum x\cdot P\left(X=x\right)$ ……(2)

Where,

$x$ = Each possible value of X.

$P\left(X=x\right)$= Probability of each outcome that is occurring.

We have the value formula for any discrete random variable ‘X’, which can be used to solve examples.

To calculate the value, you can use an online expected value calculator.

### Case 1: Calculate the Expectation for Random Variable

Now, we will start solving an example in which the probability distribution of a continuous random variable is given in a table in the example below.

Example No. 1:

Question:

An experiment produces the discrete random variable X that has the probability distribution as shown in the table below. If a very high number of trials are carried out, what would be the expectation of all the outcomes for the random variable X?

We can recall that the law of large numbers states as follows:

As the number of trials approaches infinity, the mean of the results will approach the value of $E\left(X\right)$.

That is,

$E\left(X\right)={\mu }_{n\to \infty }$

Where,

$\mu$ = Mean.

$n$= Number of Trails.

The expectation is expressed as follows:

$E\left(X\right)=\sum x\cdot P\left(X=x\right)$

Where,

$x$ = Each possible value of X.

$P\left(X=x\right)$= Probability of each outcome that is occurring.

Thus, to find the mean, we will substitute the values given in the table in the equation which is for the expectation for the discrete random variables.

That is, we will multiply each of the possible outcomes with its probability and then find the sum of the products to get the expectation:

Thus,

$\begin{array}{l}E\left(X\right)=2\left(0.1\right)+3\left(0.3\right)+4\left(0.2\right)+5\left(0.4\right)\\ E\left(X\right)=0.2+0.9+0.8+2.0\\ E\left(X\right)=3.9\end{array}$

Thus, the expected value of random variable X is 3.9.

Hence, if a high number of trials are carried out on any case, the mean of all the outcomes would be 3.9.

### Case 2: Calculate the Expected Value from the Probabilities of a Discrete Uniform Random Variable

Calculate the expected value of a discrete random variable that has a uniform probability distribution with the help of the following example.

Example No. 2:

Question:

The probability distribution for a six-sided die is shown in the following table. Determine the $E\left(X\right)$.

Now,

The expected value formula of a random variable is expressed as follows:

$E\left(X\right)=\sum x\cdot P\left(X=x\right)$

Substitute the given values in the above equation,

Thus,

$\begin{array}{l}E\left(X\right)=1\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)+3\left(\frac{1}{6}\right)+4\left(\frac{1}{6}\right)+5\left(\frac{1}{6}\right)+6\left(\frac{1}{6}\right)\\ E\left(X\right)=\left(\frac{1}{6}\right)+\left(\frac{2}{6}\right)+\left(\frac{3}{6}\right)+\left(\frac{4}{6}\right)+\left(\frac{5}{6}\right)+\left(\frac{6}{6}\right)\\ E\left(X\right)=\left(\frac{21}{6}\right)\\ E\left(X\right)=3.5\end{array}$

Thus, the expected value of discrete random variable X is 3.5.

Now to cross-validate the above value of expected value, the expected value calculator can be used in the following way:

The expected value formula of a random variable with a uniform probability distribution can be given by:

$E\left(X\right)=\frac{n+1}{2}$ ……(3)

Where,

X = {1, 2, 3, 4, …., n}

n = Last consecutive integer in the set of all possible values of random variable X.

Thus, the uniform probability distribution in the above example is  X = {1, 2, 3, 4, 5, 6}.

Substitute n = 6 in the equation (3),

Thus,

$\begin{array}{l}E\left(X\right)=\frac{6+1}{2}\\ E\left(X\right)=\frac{7}{2}\\ E\left(X\right)=3.5\end{array}$

Thus, the expected value of discrete random variable X  is 3.5.

### Case 3: Calculate the Expected Value of a Random Variable Given a Distribution Function:

The probability distribution is given in the form of a function to calculate the $E\left(X\right)$ of a random variable with the help of the following example.

Example No. 3:

Question:

Let X denote a random variable that has the values –1, 𝑀, and 1. Given that 𝑋 has a probability distribution function $f\left(x\right)=\frac{x+2}{6}$, find the expectation of the random variable X.

Given is the probability distribution function for the random variable X and the three possible values that X can be substituted.

Evaluate the probability distribution of the variable function at each of the possible values of the variable in order to determine the probability associated with each value.

Now, calculate for x = -1,

Substitute the value in function given,

$\begin{array}{l}f\left(-1\right)=\left(\frac{-1+2}{6}\right)\\ f\left(-1\right)=\left(\frac{1}{6}\right)\end{array}$

Now, calculate for x = M,

Substitute the value in function given,

$f\left(M\right)=\left(\frac{M+2}{6}\right)$

Finally, calculate for x = 1,

Substitute the value in function given,

$\begin{array}{l}f\left(1\right)=\left(\frac{1+2}{6}\right)\\ f\left(1\right)=\left(\frac{3}{6}\right)\\ f\left(1\right)=\left(\frac{1}{2}\right)\end{array}$

We know that the sum of all of the probabilities in a probability distribution must be equal to 1.

Thus, we can write the equation as follows:

$\left(\frac{1}{6}\right)+\left(\frac{M+2}{6}\right)+\left(\frac{1}{2}\right)=1$

Now, find the common denominator for all the three fractions on the left-hand side,

$\left(\frac{1}{6}\right)+\left(\frac{M+2}{6}\right)+\left(\frac{3}{6}\right)=1$

Simplify further,

$\begin{array}{l}\left(\frac{M+6}{6}\right)=1\\ M+6=6\\ M=0\end{array}$

The probability of X=M is:

$P\left(X=M\right)=\left(\frac{M+2}{6}\right)$

The value of M = 0.

Thus,

The probability of X = 0 can be calculated as:

$\begin{array}{l}P\left(X=0\right)=\left(\frac{0+2}{6}\right)\\ P\left(X=0\right)=\frac{2}{6}\\ P\left(X=0\right)=\frac{1}{3}\end{array}$

To calculate the $E\left(X\right)$ of the random variable X, the table can be drawn as:

Now, use the formula for expected value as:

$E\left(X\right)=\sum x\cdot P\left(X=x\right)$

Thus, substitute the values in the above equation,

$\begin{array}{l}E\left(X\right)=-1\left(\frac{1}{6}\right)+0\left(\frac{1}{3}\right)+1\left(\frac{1}{2}\right)\\ E\left(X\right)=\left(\frac{-1}{6}\right)+\left(\frac{0}{3}\right)+\left(\frac{3}{6}\right)\\ E\left(X\right)=\left(\frac{2}{6}\right)\\ E\left(X\right)=\left(\frac{1}{3}\right)\end{array}$

Therefore, the expected value of random variable X is $\frac{1}{3}$.

## Common Mistakes

Below are some common mistakes that every student needs to take care of when solving the examples on $E\left(X\right)$.

• Look for the distribution used in the numerical i.e. normal distribution, probability distribution.
• For the requirement of risk-neutral agents, approximate values must be used.
• First, multiply the values of x and $P\left(X=x\right)$ and then add all the terms to get the expected value.

## Formulas

The equation obtained for expectation can be expressed as follows:

$E\left(X\right)={\mu }_{n\to \infty }$

Where,

$\mu$ = Mean.

$n$= Number of Trails.

The formula for any random variable X is expressed as follows:

$E\left(X\right)=\sum x\cdot P\left(X=x\right)$

Where,

$x$ = Each possible value of X.

$P\left(X=x\right)$= Probability of each outcome that is occurring.

## Context and Applications

Following are the applications when $E\left(X\right)$ comes into practice in real life:

• The estimated value is very important in many applications. We want a formula for the observed data that offers a good approximation of the parameter representing the effect of such explanatory variables on a dependent variable when conducting regression analysis.
• Based on different data sets, this formula often produces different estimates. As a result, a random variable is generated by the calculation. As a consequence, we test a formula to see whether it is an unbiased estimator or if the estimated value of the desired parameter equals the true value.
• The use of approximate values of uncertain quantities is a requirement of risk-neutral agents. However, it means optimizing the expected value of some objective function for risk-averse agents.

The below are the related concepts you should look at before going through the expected value as they may help you to get the concept clear.

• Random variables
• Discrete random variable
• Continuous random variable
• Probability

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