Mean and Variance of Random Variables

A unique numerical value is associated with every outcome of an experiment. The value varies with every trial of the experiment. A random experiment is one whose outcome is unknown. The experiment is usually repeated under identical conditions. In an example of a coin being tossed, the outcome is a head or a tail. Every time the coin is tossed, the outcome is either of the two. The outcome is not known prior and is unpredictable. A random variable with values being finite is a discrete random variable. A variable which can take any value between the given limits is a continuous random variable.

How to Define the Mean and Variance of Random Variables

The expectation of a random variable is a weighted average of all the values of a random variable. Here, the weights correspond to their respective probabilities. The formula for finding the mean of a random variable is as follows:

E (X) = μ = Σ_i x_i p_i, where i = 1, 2, …, n

E (X) = x₁p₁ + x₂p₂ + … + x_np_n, where p refers to the probabilities.

Variance gives the distance of a random variable from the mean. The smaller the variance, the random variable is closer to the mean.

The formula for finding the variance of a random variable is

σ_x² = Var (X) = ∑_i (x_i − μ)² p(x_i) = E (X − μ)² or, Var(X) = E(X²) − [E(X)]².

E(X²) = ∑_i x_i² p(x_i), and [E(X)]² = [∑_i x_i p(x_i)]² = μ².

Mean and Variance of Random Variables Examples

Examples of the expectation of random variables

1) Let X represent the outcome of a roll of a fair six-sided die. More specifically, X will be the number of pips showing on the top face of the die after the toss. The possible values for X are 1, 2, 3, 4, 5, and 6, all of which are equally likely with a probability of 1/6. The expectation of X is

If one rolls the die n times and computes the average (arithmetic mean) of the results, then as n grows, the average will almost surely converge to the expected value, a fact known as the strong law of large numbers.

That is, the bet of $1 stands to lose

Also Read

• How to Find the Variance
• Bayes Theorem of Probability
• Find Standard Deviation from Variance

Mean and Variance of Random Variables Problems

Example 1: If X and Y be the random variables with the distributions,

Then find the expected value of X and Y.

Solution:

E (x) = ∑ X_i P_i

= 2 × (0.2) + 3 × 0.2 + 6 × 0.5 + 10 × 0.1

= 0.4 + 0.6 + 3.0 + 1.0 = 5

E (Y) = ∑ Y_i P_i

= (0.8) × (0.2) + (0.2) (0.3) + 0 × 0.1 + 3 × 0.3 + 7 × 0.1

= 0.16 + 0.06 + 0 + 0.9 + 0.7 = 0.38

Example 2: There are 4 numbered cards beginning with 1. Two cards need to be drawn without replacement. If x denotes the sum of the numbers on the 2 cards drawn, find the mean and variance of x.

Solution:

The possible sums are 3, 4, 5, 6 and 7.

x 3 4 5 6 7

P(x) 2/12 2/12 4/12 2/12 2/12

x * P(x) 3/6 4/6 5/3 6/6 7/6

x² * P(x) 9/6 16/6 25/3 36/6 49/6

Mean = ∑ x * P(x) = 30/6 = 5

Variance = ∑ x² * P(x) – [∑ x * P(x)]² = [160/6] – 25 = 5/3

Example 3: Consider a meeting where a proposal is announced. Out of them, 70% of the members are in favour and 30% are against it. If a member is selected at random, taking X = 0 if he opposed and X = 1 if he is in favour, find the mean and variance of X.

Solution:
Here, the distribution of random variable X is
P (X = 0) = 30/100 = 0.3 and P(X = 1) = 70/100 = 0.7

Mean = E (X) = 0 * 0.3 + 1 * 0.7 = 0.7

E (X²) = 0² * 0.3 + 1² * 0.7 = 0.7

Var (X) = E (X²) – [E (X)]²

= 0.7 – [0.7]²

= 0.7 – 0.49

= 0.21

Example 4: A pair of dice is thrown, and let X be the random variable which represents the sum of the numbers that appear on the two dice. Find the mean of X.

Solution:
The S be the sample space when a pair of dice is thrown.
; n(S) = 36

The random variable X, i.e., the sum of the numbers on the two dice, takes the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12.

To practise past years’ problems on statistics, click here.

Statistics – Important Topics

Mean, Median and Mode