Expected value (EV) is the average value of a random variable, calculated by multiplying each possible outcome by its probability and adding the results. It represents the average outcome expected from a random experiment over many trials.
Properties of Expected Value
- Linearity Principle: E(X + Y) = E(X) + E(Y)
- Addition of a Constant: E(X + C) = E(X) + C
- Multiplication by a Constant: E(CX) = C × E(X)
Where E(X) and E(Y) denote the expected values of the random variables X and Y, and C is a constant.
Random Variables and Expectations
A random variable is a function that assigns a real numerical value to each outcome of a random experiment. It helps us represent the results of random events in numerical form so they can be analysed using probability and statistics.
Random variables can be divided into two broad categories depending on the type of data they represent: discrete random variables and continuous random variables.
Example:
- Discrete Random Variable: When a die is rolled, the possible outcomes are 1, 2, 3, 4, 5, and 6.
- Continuous Random Variable: The height of people in a population can take any value within a range.
Expectation
For a random variable X, expectation represents the average value that X is expected to take when a random experiment is repeated many times. It gives the long-term average outcome associated with the values of the random variable.
For random variable X which assumes values x1, x2, x3,...xn with probability P(x1), P(x2), P(x3), ... P(xn)
Expectation of X is defined as,
E(x) =
\sum P(x_{i})x_{i}
Example: Aman's dad gave him a die. To check if it was a fair die, he rolled the die 500 times and noted down the frequencies of the values obtained in the table on a piece of paper. Due to rain, some values on the table were washed away. So, instead of doing all the experiments again, he rolled the die 20 times and concluded that the expected value in 20 rolls was 3.37. The washed-away table is given below. Find the values of the missing frequencies.
| Die Value | Absolute Frequency |
|---|---|
| 1 | A |
| 2 | 110 |
| 3 | 95 |
| 4 | 70 |
| 5 | 75 |
| 6 | B |
| Total | 500 |
Solution:
Assuming that the expected value remains the same in both 20 and 500 trials.
Let's say the random variable "X" is defined as the value obtained on the die.
X can take values like 1, 2, 3, 4, 5 and 6
The expected value of a random variable is given by,
E(x) =
\sum P(x_{i})x_{i}
E(X) = \sum P(x_{i})x_{i} \\ E(X) = P(X = 1)(1) + P(X = 2)2 + P(X = 3)3 + P(X = 4)4 + P(X = 5)5 + P(X = 6)6 \\ E(X) = \frac{A}{500} + \frac{110}{500}.2 + \frac{95}{500}.3 + \frac{70}{500}.4 + \frac{75}{500}.5 + \frac{B}{500}.6 \\ Substituting the expected value,
3.67 = \frac{A}{500} + \frac{110}{500}.2 + \frac{95}{500}.3 + \frac{70}{500}.4 + \frac{75}{500}.5 + \frac{B}{500}.6 \\ 1685 = A + 220 + 285 + 280 + 375 + 6B \\ 525 = A + 6B Now this equation has two variables, one more equation is required to solve this.
A + 110 + 95 + 70 + 75 + B = 500
⇒ A+ B = 150
So, the two equations are,
A + B = 150
A + 6B = 525
Solving these equations, the values of variables come out to be.
A =75 and B = 75.
