Probability

Mathematical Expectation

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 (x1p1, x2p2, …, xnpn),

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.

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Properties of Expectation

  1. 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)

  1. 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)

  1. 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.

Mathematical Expectation

Source: freepik.com

  1. 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.

  1. 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(∑aiXi)=∑ aE(Xi)

Where, ai, (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.

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Mandeep P ShettyMandeepKumar Recent comment authors
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Kumar
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Kumar

There are so many errors in two of the lectures that I have watched. The flow of the lectures are also inappropriate. Firstly you never defined what an event is. For this lecture you can just say that an event is a subset of sample space. Therefore it can be any subset of sample space, even phi(empty set) or the whole sample space itself. You are confusing events with elements of sample space. There is a fundamental errors on tis page too. Like P(A|B) is probability of event A given that event B has (already) occurred. However the text in… Read more »

Mandeep P Shetty
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Mandeep P Shetty

@Kumar Thank you! I thought I was somehow wrong in my understanding. Another error is: “Even compound events (two events occurring at the same time) can be independent events. Ex. Tossing a coin and rolling a die. Sample space S = {(1,H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T) (5, T) (6, T)}.” The example is still simple even and not a compound event. An actual example for a compound event will be ex: Tossing a coin and rolling a die and getting a 1 every time. The sample… Read more »

Mandeep
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Mandeep

@Kumar Thank you! I thought I was somehow wrong in my understanding. Another error is: “Even compound events (two events occurring at the same time) can be independent events. Ex. Tossing a coin and rolling a die. Sample space S = {(1,H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T) (5, T) (6, T)}.” The example is still simple even and not a compound event. An actual example for a compound event will be ex: Tossing a coin and rolling a die and getting a 1 every time. The sample… Read more »

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