Probability is used to denote the happening of a certain event, and the occurrence of that event, based on past experiences. The mathematical expectation is the events which are either impossible or a certain event in the experiment. Probability of an impossible event is zero, which is possible only if the numerator is 0. Probability of a certain event is 1 which is possible only if the numerator and denominator are equal.

**Mathematical ****Expectation**

Mathematical expectation, also known as the expected value**,** which is the summation of all possible values from a random variable.

It is also known as the product of the probability of an event occurring, denoted by P(x), and the value corresponding with the actually observed occurrence of the event.

For a random variable expected value is a useful property. E(X) is the expected value and can be computed by the summation of the overall distinct values that is the random variable. The mathematical expectation is denoted by the formula:

E(X)= Σ (x_{1}p_{1}, x_{2}p_{2}, …, x_{n}p_{n}),

where, x is a random variable with the probability function, *f*(x),

p is the probability of the occurrence,

and n is the number of all possible values.

The mathematical expectation of an indicator variable can be 0 if there is* *no occurrence of an event A, and the mathematical expectation of an indicator variable can be 1 if there is an occurrence of an event A.

For example, a dice is thrown, the set of possible outcomes is { 1,2,3,4,5,6} and each of this outcome has the same probability 1/6. Thus, the expected value of the experiment will be 1/6*(1+2+3+4+5+6) = 21/6 = 3.5. It is important to know that “expected value” is not the same as “most probable value” and, it is not necessary that it will be one of the probable values.

**Browse more Topics under Probability**

- Introduction to Probability
- Probability of an Event
- Events and its Types
- Events and Its Algebra
- Independent Events
- Conditional Probability
- Basic Theorems of Probability
- Multiplication Theorem on Probability
- Baye’s Theorem
- Random Variable and Its Probability Distribution
- Mean and Variance of Random Distribution
- Bernoulli Trials and Binomial Distribution

**Properties of Expectation**

- If X and Y are the two variables, then the mathematical expectation of the sum of the two variables is equal to the sum of the mathematical expectation of X and the mathematical expectation of Y.

Or

E(X+Y)=E(X)+E(Y)

- The mathematical expectation of the product of the two random variables will be the product of the mathematical expectation of those two variables, but the condition is that the two variables are independent in nature. In other words, the mathematical expectation of the product of the
*n*number of independent random variables is equal to the product of the mathematical expectation of the*n*independent random variables

Or

E(XY)=E(X)E(Y)

- The mathematical expectation of the sum of a constant and the function of a random variable is equal to the sum of the constant and the mathematical expectation of the function of that random variable.

Or,

E(a+*f*(X))=a+E(*f*(X)),

where, a is a constant and *f*(X) is the function.

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- The mathematical expectation of the sum of product between a constant and function of a random variable and the other constant is equal to the sum of the product of the constant and the mathematical expectation of the function of that random variable and the other constant.

Or,

E(aX+b)=aE(X)+b,

where, a and b are constants.

- The mathematical expectation of a linear combination of the random variables and constant is equal to the sum of the product of ‘n’ constant and the mathematical expectation of the ‘n’ number of variables.

Or

E(∑a_{i}X_{i})=∑ a_{i }E(X_{i})

Where, a_{i}, (i=1…n) are constants.

**Solved Example on Mathematical Expectation**

Q.What is the expected number of coin flips for getting two consecutive heads?

Sol: Let the expected number of coin flips be x.

If the first flip is a tail then the probability of the event is 1/2. Thus, the total number of flips required is x+1. If the first flip is a head and the second flip is a tail, then the probability of the event is 1/4 and the total number of flips we require is x+2. If the first flip is a head and the second flip is also heads, then the probability of the event is 1/4 and the total number of flips we require is 2.

By Adding, the equations we get

x = (1/2)(x+1) + (1/4)(x+2) + (1/4)2

By solving the equation, we get

x = 6.

So, the expected number of coin flips for getting two consecutive heads is 6.