Mathematical Expectation
Data Science and A.I. Lecture Series
By Bindeshwar Singh Kushwaha – PostNetwork Academy
Introduction
- This unit explores the expectation of a random variable.
- Expectation provides a measure of central tendency in probability distributions.
- Expectation is useful in both discrete and continuous probability distributions.
- Problems and examples help in understanding practical applications.
Objectives
- Define expectation of a discrete and continuous random variable.
- Explore properties of expectation.
- Compute moments and other statistical measures.
- Apply addition and multiplication theorems of expectation.
- Solve problems on mathematical expectation.
Expectation of a Discrete Random Variable
If \( X \) is a discrete random variable with probability mass function \( p(x) \), the expectation is given by:
$$ E(X) = \sum_{i} x_i p(x_i) $$
Expectation represents the long-run average value of the random variable.
Expectation of a Continuous Random Variable
If \( X \) is a continuous random variable with probability density function \( f(x) \), then its expectation is given by:
$$ E(X) = \int_{-\infty}^{\infty} x f(x) dx $$
Expectation is also known as the mean of the probability distribution.
Example 1: Expected Value in a Game (Discrete)
A game involves drawing a card from a box containing cards numbered 1, 3, and 5. If a card numbered \( x \) is drawn, the player wins \( 2x \) rupees. Find the expected winnings.
Solution:
$$ E(X) = 2(1) \times \frac{1}{3} + 2(3) \times \frac{1}{3} + 2(5) \times \frac{1}{3} $$
$$ E(X) = \frac{2}{3} + \frac{6}{3} + \frac{10}{3} = 6 $$
Example 2: Expected Revenue (Discrete)
A fruit vendor sells mangoes for Rs 20 each. If the daily demand follows the probability distribution \( P(X=10) = 0.3 \), \( P(X=15) = 0.5 \), \( P(X=20) = 0.2 \), find the expected daily revenue.
Solution:
$$ E(X) = 20(10) \times 0.3 + 20(15) \times 0.5 + 20(20) \times 0.2 $$
$$ E(X) = 60 + 150 + 80 = 290 $$
Properties of Expectation
- \( E(k) = k \), where \( k \) is a constant.
- \( E(kX) = kE(X) \), where \( k \) is a constant.
- \( E(aX + b) = aE(X) + b \), where \( a \) and \( b \) are constants.
Expectation Calculation Example
Given the probability distribution:
| \( X \) |
\( p(X) \) |
| -2 |
0.15 |
| -1 |
0.30 |
| 0 |
0 |
| 1 |
0.30 |
| 2 |
0.25 |
Find:
- \( E(X) \)
- \( E(2X + 3) \)
- \( E(X^2) \)
- \( E(4X – 5) \)
Solution
\( E(X) = (-2)(0.15) + (-1)(0.30) + (0)(0) + (1)(0.30) + (2)(0.25) \)
$$ E(X) = -0.3 – 0.3 + 0 + 0.3 + 0.5 = 0.2 $$
\( E(2X + 3) = 2E(X) + 3 \)
$$ E(2X + 3) = 2(0.2) + 3 = 3.4 $$
\( E(X^2) = \sum_i x_i^2 p_i \)
$$ E(X^2) = (4)(0.15) + (1)(0.30) + (0)(0) + (1)(0.30) + (4)(0.25) $$
$$ E(X^2) = 0.6 + 0.3 + 0 + 0.3 + 1 = 2.2 $$
\( E(4X – 5) = 4E(X) – 5 \)
$$ E(4X – 5) = 4(0.2) – 5 = -4.2 $$
Summary
- Defined expectation for discrete and continuous random variables.
- Explained properties of expectation.
- Stated and proved addition and multiplication theorems.
- Solved problems for both discrete and continuous cases.
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