Expectation and Variance - Discrete Random Variable

Introduction

Before moving onto the main topic, let us discuss random variables and their types. A random variable is a term where the output depends on the random phenomenon. The three types of random variables are singular, continuous, discrete.

A random variable with a countable number of possible values is called a discrete random variable. The discrete random variables can have either a finite set or a countable number of discrete variables. The probability of every value in the discrete random variable is between 0 to 1. The sum of all the probabilities is equal to 1. For example, the number of children in a family, the number of chairs in a room, etc.

Expectation

The expectation is also an expected value of a random variable. It tells you about the behavior of a random variable. We can say expectation as an average value of the random variables, where each value is weighted according to its probability. 

Suppose X is a discrete random variable with values x₁, x₂, x₃, x₄, … x_n. The probability of every x_i is P(x_i) (1 ≤ i ≤ n). The expected value of the random variable X is E(X), defined by

Example 1. Find E(X) for the random variable X given in the table below:

Solution: E(X) = 1*(3/10) + 5*(1/2) + 8*(2/10) = 4.4

Example 2. Roll two unbiased 6-sided dice. You win $800 if the sum is 3 and lose $50 otherwise. How much do you expect to win on average per trial?

Solution: The probability of getting sum = 3 is 2/36 because you can either get (1, 2) or (2, 1) to get a sum equal to 3.

                 Then the probability of not getting the sum = 3 will be 34/36.

                 If you play N times, you can expect N*(2/36) of getting sum equal to 3 and N*(34/36) of getting sum not equal to 3.

                 Thus your expected winnings will be:

                 800*(2/36)N - 50*(34/36)N = -$2.78N

                  Now divide it by N to get the expected average per trial.

                  -$2.78 will be the expected average per trial.

Example 3. Suppose we have a die with six faces marked with four 2’s, one 3, and one 6. What would you expect the average of 12000 rolls to be?

Solution: The probability of getting 2 is 4/6.

                 The probability of getting 3 is 1/6.

                 The probability of getting 6 is 1/6.

Expected average of these 12000 rolls will be

(8000*2 + 2000*3 + 2000*6) / 12000 = 2.83

Expected average of 12000 rolls = 2.83