# Mathematical Expectation of Random Variables With Examples And Expected Value Formula

Probability / Tuesday, September 25th, 2018

# Mathematical Expectation of Random Variables

In the last three articles of probability we studied about Random Variables of single and double variables, in this article based on these types of random variables we will study their expected values using respective expected value formula.

1. Random Variable And Its Distribution
2. Distribution Function Of Random Variables
3. Introduction To Cumulative Distribution Function, Marginal Probability And Joint Density Function

Expectation is used almost in every sphere of life. For example, in a year what is the expectation of getting good rain, a gambler might be interested in his average winning at a game, for a business person he will be interested in his average profits on a product, all these answers can be well given with the help of expectation. Mathematical expectation is the average value of a random phenomenon.

## Mathematical expectation of one dimensional random variable

Let X be discrete random variable and f(x)be probability mass function (pmf). Then the mathematical expectation or expectation or expected value formula of f(x) is defined as:
$E(X)=\sum\limits_{x}{x.f(x)}$
If X is a continuous random variable and f(x) be probability density function (pdf), then the expectation is defined as:
$E(X)=\int\limits_{x}{x.f(x)}$
Provided that the integral and summation converges absolutely. It is also known as mean of random variable X.

 Example 01

Suppose a dice is tossed and let the random variable X denotes the numbers on the size i.e., X = 1, 2, 3, 4, 5, 6. The probability mass function (pmf) of X is (1/6). Find expectation.

Solution:
$Given,f(x)=\frac{1}{6},x=1,2,3,4,5,6$
$Then,E(X)=\sum\limits_{x}{x.f(x)}$
$\Rightarrow E(X)=1.\frac{1}{6}+2.\frac{1}{6}+3.\frac{1}{6}+4.\frac{1}{6}+5.\frac{1}{6}+6.\frac{1}{6}$
$\therefore E(X)=\frac{21}{6}$

 Example 02

Let X be a random variable with the following probability distribution function

 X f(X) 2 1/6 3 ½ 4 1/3

Find (a) E(X) (b) E(X2) (c) E(2X+1)2

Solution:
$(a)E(X)=\sum\limits_{x}{x.f(x)}=2.\frac{1}{6}+3.\frac{1}{2}+4.\frac{1}{3}=\frac{19}{6}$
$(b)E({{X}^{2}})=\sum\limits_{x}{{{x}^{2}}.f(x)}={{2}^{2}}.\frac{1}{6}+{{3}^{2}}.\frac{1}{2}+{{4}^{2}}.\frac{1}{3}=\frac{21}{3}$
$(c)E{{\left( 2X+1 \right)}^{2}}=\sum\limits_{x}{{{\left( 2x+1 \right)}^{2}}.f(x)}$
$\Rightarrow E{{\left( 2X+1 \right)}^{2}}={{\left( 2.2+1 \right)}^{2}}.\frac{1}{6}+{{\left( 2.3+1 \right)}^{2}}.\frac{1}{2}+{{\left( 2.4+1 \right)}^{2}}.\frac{1}{3}$
$\therefore E{{\left( 2X+1 \right)}^{2}}=\frac{25}{6}+\frac{49}{2}+\frac{81}{3}=\frac{167}{3}$

 Example 03

An urn contains 7 white and 3 red balls. Two balls are drawn together, at random from this urn. Compute the probability that (a) neither of them is white (b) one white one red. Hence compute the expected number of white balls drawn.

Solution:

Let X denotes the number of white balls drawn. Then the probability of X is

 X 0 1 2 f(X) $\frac{{}^{3}{{C}_{2}}}{{}^{10}{{C}_{2}}}=\frac{1}{15}$ $\frac{{}^{7}{{C}_{1}}\times {}^{3}{{C}_{1}}}{{}^{10}{{C}_{2}}}=\frac{7}{15}$ $\frac{{}^{7}{{C}_{2}}}{{}^{10}{{C}_{2}}}=\frac{7}{15}$

$(a)P(\text{neither of them is white})=\frac{{}^{3}{{C}_{2}}}{{}^{10}{{C}_{2}}}=\frac{1}{15}$
$(b)P(\text{one white one red})=\frac{{}^{7}{{C}_{1}}\times {}^{3}{{C}_{1}}}{{}^{10}{{C}_{2}}}=\frac{7}{15}$
Hence, expected number of white balls drawn is:
$E(X)=0.\frac{1}{15}+1.\frac{7}{15}+2.\frac{7}{15}=\frac{21}{15}$

 Example 04

From a bag containing 2 rupee-coins and 3 twenty paise coins, a person is asked to draw two coins at random. Find the value of his expectation.

Solution:

There are three following possibilities of drawing two coins at random.
$(1)\text{Both of these coins be rupee coins},\text{ its probability}=\frac{{}^{2}{{C}_{2}}}{{}^{5}{{C}_{2}}}=\frac{1}{10}$
$(2)\text{Both of these coins be twenty paise coins},\text{ and its probability}$
$\text{=}\frac{{}^{3}{{C}_{2}}}{{}^{5}{{C}_{2}}}=\frac{3}{10}$
$(3)\text{One of these coins be a rupee coin and the other a twenty paise coin}$
$\text{and its probability}=\text{=}\frac{{}^{2}{{C}_{1}}\times {}^{3}{{C}_{1}}}{{}^{5}{{C}_{2}}}=\frac{3}{5}$
Also 2 rupee coin = 10 twenty paise coins
1 rupee coin and 1 twenty paise coin = 6 twenty paise coins.
$\text{Required expectation}=\sum{{{p}_{i}}{{x}_{i}}}$
$=\left[ \left( \frac{1}{10}.10 \right)+\left( \frac{3}{10}.2 \right)+\left( \frac{3}{5}.6 \right) \right]\text{twenty paise coins}$
$\text{=}\left[ \text{1+}\frac{3}{5}+\frac{18}{5} \right]\text{twenty paise coins}$
$\text{=}\frac{26}{5}\times 20\text{ paise=104 paise}$

## Mathematical expectation of two dimensional random variable

Let X and Y be random variable with joint probability distribution function f(x, y). Then the mean or expected value of random variable g(X, Y) is given by
$\text{E}\left[ g(X,Y) \right]=\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{g(x,y)f(x,y)dxdy}}$
Where X, Y are continuous random variables
$\text{E}\left[ g(X,Y) \right]=\sum\limits_{x}{\sum\limits_{x}{g(x,y)f(x,y)}}$
$=\sum\limits_{i}{\sum\limits_{j}{{{x}_{i}}{{y}_{j}}P\left( X={{x}_{i}}\bigcap Y={{y}_{j}} \right)}}$
Where X, Y are discrete random variables.

### Properties of Expectation:

1. If X and Y are two random variables, then E(X+Y) = E(X)+E(Y)
2. If X and Y are independent random variables, then E(XY) = E(X).E(Y)
3. If X is a random variable and ‘a’ and ‘b’ are constants, then E(aX+b) = aE(X)+b
4. If X ≥ 0, E(X) ≥ 0
5. If X and Y are two random variables such that Y ≤ X, then E(X) ≤ E(Y)

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