A project report
On
Expectation
Submitted in partial fulfilment of
The requirement for 2nd semester
examination of MCA
NAME ROLL NUMBER
Satish Kumar Mahapatra 202468083
Debanand Das 202468092
Subject :- Mathematices
Submission date :- 17.5.2025
Table of Contents
1. Acknowledgment
2. Abstract
3. Introduction
4. Content
5. Objective
6. Examples
7. Application
8. Conclusion
9. References
Acknowledgment
I would like to express my heartfelt gratitude to my teacher, DR BHASKAR BHAULA, for
their constant support and guidance throughout this project. I also wish to thank my classmates
and family for their encouragement.
This project has been a valuable learning experience, and I am grateful to all those who provided
guidance, resources, and motivation, directly or indirectly.
I extend my sincere NIST UNIVERSITY for providing the resources, infrastructure, and
supportive academic environment for the project development.
Abstract
This project explores the concept of expectation in probability theory, a fundamental concept
with applications in various fields like statistics, finance, and artificial intelligence. Expectation,
often referred to as the expected value or mean, is a measure of the central tendency of a random
variable, providing insight into the average outcome of a random process. The project covers the
mathematical definition of expectation, including its calculation for both discrete and continuous
random variables. Various types of expectation, such as conditional expectation and joint
expectation, are discussed with detailed explanations. Difficult examples of both discrete and
continuous random variables are solved, demonstrating practical applications of expectation.
This project aims to provide a comprehensive understanding of expectation and its significance
in mathematical analysis and real-world scenarios.
Introduction
Expectation, also known as the expected value or mean, is a measure of the central tendency of a
random variable in probability theory. It gives an idea of the average value of a random variable
over many trials. This project will explore the concept of expectation in depth, its types, and
practical examples.
Expectation, also known as the expected value or mean, is a fundamental concept in probability
theory that represents the average or central tendency of a random variable. It is a mathematical
tool that helps in predicting the average outcome of a random process over a large number of
trials. The concept of expectation is essential in various fields, including statistics, finance,
artificial intelligence, economics, and decision theory. By understanding expectation, we can
analyze the long-term behavior of random phenomena and make informed predictions.
This project provides a detailed exploration of expectation, covering its definition, mathematical
representation, and various types. We will discuss the difference between the expectation of
discrete and continuous random variables, along with advanced concepts like conditional
expectation and joint expectation. Practical examples, ranging from simple calculations to
complex real-world scenarios, are provided to enhance understanding. The project also
highlights the importance of expectation in solving real-world problems and making strategic
decisions.
Content
What is Expectation?
Expectation is the weighted average of all possible values of a random variable, where the
weights are the probabilities of each value occurring.
Mathematical Definition:
For a discrete random variable X with possible values x1, x2, ..., xn and corresponding
probabilities p1, p2, ..., pn, the expectation (E[X]) is defined as:
E[X] = Σ (xi * pi)
For a continuous random variable, the expectation is defined as:
E[X] = ∫ x * f(x) dx
Types of Expectation
1. Expectation of Discrete Random Variables
2. Expectation of Continuous Random Variables
3. Conditional Expectation
4. Joint Expectation
5. Geometric Expectation
6. Exponential Expectation
Objective
The objective of this project is to understand the concept of expectation, its various types, and
how it can be applied in different mathematical and real-world scenarios.
Expectation of Discrete Random
Variables: Examples:
Example 1: Drawing Balls from a Bag
A bag contains 10 red balls, 15 blue balls, and 5 green balls. The random
variable X represents the color of a ball drawn at random. Calculate the
expected value of drawing each color.
Solution:
Total balls = 10 + 15 + 5 = 30
Probability of red = 10/30, blue = 15/30,
green = 5/30 Assign values: Red = 1, Blue =
2, Green = 3
E[X] = (1 * 10/30) + (2 * 15/30) + (3 * 5/30)
E[X] = (10/30) + (30/30) + (15/30) = 55/30 = 1.83
Example 2: Spinner in a Board Game
A spinner has 8 equal sections numbered from 1 to 8. Calculate the expected
value of a spin.
Solution:
Each section has equal probability: P(X = x) = 1/8 for x =
1, 2, …, 8 E[X] = (1/8) * (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)
E[X] = (1/8) * 36 = 4.5
Example 3: Defective Items in Manufacturing
A manufacturing machine has a 95% chance of producing defect-free items.
Let X represent the number of defective items in a batch of 100. Calculate the
expected number of defective items.
Solution:
Probability of defective = 0.05
Expected number of defective items = E[X] =
n * p E[X] = 100 * 0.05 = 5
Example 4: Survey Response
A survey of 100 people shows that 60% visited a website and 40% did not.
Let X be the number of people who visited the website. Calculate the
expected number.
Solutions:
E[X] = n * p = 100 * 0.6 = 60
Example 5: Sum of Two Dice
A game involves rolling two fair dice. Let X be the sum of the two dice.
Calculate the expected value of this sum.
Solution:
Possible sums: 2 to 12
Probabilities:
P(2) = 1/36, P(3) = 2/36, P(4) = 3/36, …, P(12) = 1/36
E[X] = Σ (sum * probability)
E[X] = (2 * 1/36) + (3 * 2/36) + (4 * 3/36) + … +
(12 * 1/36) E[X] = 7
Example C:
X is a discrete random variable with the following probability mass function
(PMF):
(i) Determining the constant :
The sum of all probabilities in a PMF must equal 1:
0 + k + 2k + 2k + 3k + k + 2k^2 + (7k^2 + k) = 1
Combining like terms:
0 + k + 2k + 2k + 3k + k + 2k^2 + 7k^2 + k = 1
10k + 9k^2 = 1
This is a quadratic equation:
9k^2 + 10k - 1 = 0
Solving this quadratic using the quadratic formula:
Given:
X is a discrete random variable with the following probability mass function
(PMF):
(i) Determining the constant :
The sum of all probabilities in a PMF must equal 1:
0 + k + 2k + 2k + 3k + k + 2k^2 + (7k^2 + k) = 1
Combining like terms:
0 + k + 2k + 2k + 3k + k + 2k^2 + 7k^2 + k = 1
10k + 9k^2 = 1
This is a quadratic equation:
9k^2 + 10k - 1 = 0
Solving this quadratic using the quadratic formula:
Solving for the two solutions:
1. K = 1.cc/18=0.0S2
2. Therefore: K =0.0S2=1/10
(ii) Finding
p(x<6):
This means calculating the probability for and :
P(X < 6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)
= 0 + k + 2k + 2k + 3k + k = 9k
Substituting :
P(X < 6) = 9*1/10 = 0.9
(iii) Finding p(x>=6)
This means calculating
: P(X=6) + P(X=7):
P(x>=6)=P(X=6) + P(X=7)
= 2k^2 + (7k^2 + k)
= 2k^2 + 7k^2 + k = 9k^2 + k
Substituting k = 1/10:
P(X >= 6) = 9 * (1/10) ^ 2 + 1/10
= 9 * 1/100 + 1/10 = 9/100 + 10/100 = 19/100 = 0.19
Final Answers:
1. .
2. .
3. .
Solving for the two solutions:
1.
2. (Not valid as probability cannot be negative)
Therefore:
k = 0.092 \approx \frac{1}{10}
(ii) Finding :
This means calculating the probability for and :
P(X < 6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)
= 0 + k + 2k + 2k + 3k + k = 9k
Substituting :
P(X < 6) = 9 \times \frac{1}{10} = 0.9
(iii) Finding :
This means calculating :
P(X \geq 6) = P(X=6) + P(X=7)
= 2k^2 + (7k^2 + k)
= 2k^2 + 7k^2 + k = 9k^2 + k
Substituting :
P(X \geq 6) = 9 \left(\frac{1}{10}\right)^2 + \frac{1}{10}
= 9 \times \frac{1}{100} + \frac{1}{10} = \frac{9}{100} + \frac{10}{100} =
\frac{19}{100} = 0.19
Final Answers:
1. .
2. .
3. .
Expectation of Continuous Random Variables
Example 1:
The random variable X is uniformly distributed between 2 and 5. Calculate E[X].
Solution: E[X] = ∫ (x * (1 / (5 - 2))) dx from 2 to 5
= (1/3) ∫ x dx from 2 to 5
= (1/3) [(5^2/2) - (2^2/2)]
. = (1/3) [(25/2) - (4/2)] = (1/3) (21/2) = 7/2
Example 2:
X follows an exponential distribution with rate parameter λ = 0.5. Calculate E[X].
Solution: E[X] = 1 / λ = 1 / 0.5 = 2
Example 3:
X is a continuous random variable with probability density function f(x) = 3x^2 for x in [0, 1].
Calculate E[X].
Solution: E[X] = ∫ x * 3x^2 dx from 0 to 1
= ∫ 3x^3 dx from 0 to 1
= [3x^4 / 4] from 0 to 1
=3/4
Example 4:
The random variable X has a normal distribution N(μ = 4, σ² = 9). Calculate E[X].
Solution: For a normal distribution, E[X] = μ. Therefore, E[X] = 4.
Example 5:
X has a probability density function f(x) = (1 / 8) * (4 - x) for x in [0, 4]. Calculate E[X].
Solution: E[X] = ∫ x * (1/8) * (4 - x) dx from 0 to 4
= (1/8) ∫ (4x - x^2) dx from 0 to 4
= (1/8) [(4x^2 / 2) - (x^3 / 3)] from 0 to 4
= (1/8) [(8 * 4^2 / 2) - (4^3 / 3)] = (1/8) [(8 * 16 / 2) - (64 / 3)]
. = (1/8) [(64) - (64 / 3)]
= (64 / 8) - (8 / 3) = 8 - (8 / 3) = 16 / 3
Application
Expectation is widely used in various fields such as:
Statistics: Calculating mean values.
Finance: Determining expected returns.
Artificial Intelligence: Decision-making in probabilistic models.
Game Theory: Analyzing strategic outcomes
Conditional Expectation
Conditional expectation is the expected value (or mean) of a random variable
given that some condition is known to hold, usually related to another
variable. It provides a way to update our expectations based on new
information.
Example: -
Suppose we roll a fair six-sided die, and let:
XXX = the number on the die
YYY = 1 if the number is even, 0 otherwise
Example 2
Joint Expectation
Joint expectation refers to the expected value of the product of two
random variables. If XXX and YYY are two random variables, their joint
expectation is
E[XY]
This gives the average value of the product X⋅Y over their joint
distribution.
Example:-
Geometric Expectation
Geometric expectation refers to the expected value of a geometrically
distributed random variable.
The geometric distribution models the number of trials needed to get the
first success in a sequence of independent Bernoulli (yes/no) trials.
Example 1
Example 2 Rolling a Die
Exponential Expectation
The exponential distribution models the time between events in a Poisson
process, where events happen continuously and independently at a constant
average rate.
If a random variable )X∼Exponential(λ), where:
λ>0 is the rate parameter (average number of events per
unit time), Then the expected value (mean) of X is
Conclusion
Expectation is a fundamental concept in probability that helps in predicting the average outcome
of a random variable. Understanding its types and applications provides a solid foundation for
further study in probability and statistics.
References
1. Probability and Statistics Textbook by [Morris H. DeGroot C Mark J. Schervish ].
2. Online Lecture Notes on Probability Theory.
3. William Mendenhall C Robert J. Beaver
