# Properties Of Mean And Variance Of Random Variables

The meaning of random is uncertain. So, a random variable is the one whose value is unpredictable. A variable, whose possible values are the outcomes of a random experiment is a random variable. In this article, students will learn important properties of mean and variance of random variables with examples.

The mean or expected value or the expectation is called an average in statistics and probability. The expected value can be calculated if the probability distribution for a random variable is found. Mean of a random variable defines the location of a random variable whereas the variability of a random variable is given by the variance.

## 9 Important Properties of Mean and Variance of Random Variables

Property 1: E (X + Y) = E (X) + E (Y).

Property 2: E (X1 + X2 + … + Xn) = E (X1) + E (X2) + … + E (Xn) = Σi E (Xi).

Property 3: E (XY) = E (X) E (Y). Here, X and Y must be independent.

Property 4: E [aX] = a E [X] and E [X + a] = E [X] + a, where a is a constant

Property 5: For any random variable, X > 0, E(X) > 0.

Property 6: E(Y) ≥ E(X) if the random variables X and Y are such that Y ≥ X.

Property 7: The variance of a constant is 0.

Property 8: V [aX + b] = a2 Var (X), where a and b are constants.

Property 9: V (a1X1 + a2 X2 + … + anXn) = a12 V(X1) + a22 V(X2) + … + an2 V(Xn).

### Example on Variance of Random Variable

1] A fair six-sided die can be modeled as a discrete random variable, X, with outcomes 1 through 6, each with equal probability 1/6. The expected value of X is ${\displaystyle (1+2+3+4+5+6)/6=7/2.}{\displaystyle (1+2+3+4+5+6)/6=7/2.}$

Therefore, the variance of X is

${Var} (X)=\sum _{i=1}^{6}{\frac {1}{6}}\left(i-{\frac {7}{2}}\right)^{2}\\[5pt]\\={\frac {1}{6}}\left((-5/2)^{2}+(-3/2)^{2}+(-1/2)^{2}+(1/2)^{2}+(3/2)^{2}+(5/2)^{2}\right)\\[5pt]\\={\frac {35}{12}}\approx 2.92.$

The general formula for the variance of the outcome, X, of an n-sided die is

${Var} (X)= {E} (X^{2})- {E} (X)^{2}\\[5pt]\\={\frac {1}{n}}\sum _{i=1}^{n}i^{2}-\left({\frac {1}{n}}\sum _{i=1}^{n}i\right)^{2}\\[5pt]={\frac {(n+1)(2n+1)}{6}}-\left({\frac {n+1}{2}}\right)^{2}\\[4pt]={\frac {n^{2}-1}{12}}.$

Also, students are advised to solve JEE Past year solved Problems on Statistics and join BYJU’S online classes to crack the JEE Exam.