Examples - 17.5 | 17. Independence of Random Variables | Mathematics - iii (Differential Calculus) - Vol 3
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Understanding Independence in Discrete Random Variables

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0:00
Teacher
Teacher

Today, we will explore the independence of random variables using an example with discrete random variables X and Y. Can anyone remind me what it means for two variables to be independent?

Student 1
Student 1

It means that knowing the value of one variable does not change the probability of the other.

Teacher
Teacher

Exactly! If we denote this formally, for discrete variables, we check if this equation holds: X and Y are independent if P(X=x, Y=y) = P(X=x) * P(Y=y). Let's look at the example now.

Student 2
Student 2

What's given in our example?

Teacher
Teacher

We have the joint probability mass function given in a table. Let's first calculate the marginal probabilities. Can someone help calculate P(X=1)?

Student 3
Student 3

It's 0.1 + 0.2, which gives us 0.3.

Teacher
Teacher

Correct! Now let's check if the independence condition holds!

Student 4
Student 4

P(X=1) times P(Y=1) would be 0.3 times 0.3, which is 0.09.

Teacher
Teacher

Right, but we found P(X=1, Y=1) = 0.1. Therefore, they are not independent since 0.1 does not equal 0.09.

Teacher
Teacher

So in summary, for discrete random variables, if the joint event probability does not equal the product of the marginals, they are dependent.

Exploring Independence in Continuous Random Variables

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0:00
Teacher
Teacher

Now, let’s shift our focus to continuous random variables. Who can remember how we determine independence using PDFs?

Student 1
Student 1

We check if the joint PDF equals the product of the marginal PDFs.

Teacher
Teacher

Exactly! Let's examine our example: we have the joint PDF given as f(x, y) = e^(-x)e^(-y) for x, y > 0. What can we say about the marginals?

Student 2
Student 2

The marginals would be f(X) = e^(-x) and f(Y) = e^(-y).

Teacher
Teacher

Good! Now, we should verify if the joint PDF equals the product of these marginals. What do we find?

Student 3
Student 3

When we multiply, it gives us e^(-x)e^(-y), which is the same as our joint PDF.

Teacher
Teacher

Great observation! Therefore, we conclude that X and Y are independent.

Student 4
Student 4

Why is knowing independence important again?

Teacher
Teacher

Independence simplifies calculations and analysis in various applications, particularly in fields like engineering and statistics where multiple random variables are involved.

Teacher
Teacher

To summarize, checking independence in continuous random variables involves confirming that the joint PDF equals the product of the marginal PDFs.

Introduction & Overview

Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.

Quick Overview

This section provides practical examples of testing the independence of random variables in both discrete and continuous cases.

Standard

In this section, we explore two examples demonstrating how to determine whether two random variables are independent, using joint distributions for discrete and continuous cases. The examples illustrate the application of mathematical definitions and calculations to assess independence.

Detailed

In-Depth Summary

This section focuses on practical examples to illustrate the concept of independence among random variables, as discussed in the preceding sections. Independence is a fundamental aspect in probability and statistics, which allows simplification in models involving multiple random variables.

Example 1: Discrete Case

This example involves discrete random variables X and Y with a joint probability table. We calculate the marginal probabilities and then check whether the condition for independence holds:
- Joint Distribution of X and Y is given in a tabular form.
- Marginal probabilities are calculated for each variable.
- We verify the independence condition by checking:

$$ P(X = x, Y = y) = P(X = x) imes P(Y = y) $$
for specific values. In this case, it is found that $X$ and $Y$ are not independent.

Example 2: Continuous Case

This example deals with continuous random variables and uses a joint probability density function (PDF).
- We define a joint PDF for random variables X and Y.
- Using the marginal PDFs derived from the joint PDF, we check the independence condition based on:

$$ f(x, y) = f(x) imes f(y) $$.
- The calculations show that X and Y are indeed independent for the specified joint PDF.

Understanding these examples is crucial in modeling complex systems in engineering and other fields, as they demonstrate how independence can simplify the analysis of random variables.

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Audio Book

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Example 1: Discrete Case

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Example 1: Discrete Case
Let 𝑃(𝑋= π‘₯,π‘Œ = 𝑦) be given by:
X\Y 1 2
1 0.1 0.2
2 0.2 0.5
Find if X and Y are independent.
Solution:
Marginal:
β€’ 𝑃(𝑋 = 1)= 0.1+ 0.2 = 0.3
β€’ 𝑃(π‘Œ = 1) = 0.1 +0.2 = 0.3
Check:
$$P(X = 1, Y = 1) = 0.1 \ P(X = 1) \cdot P(Y = 1) = 0.3 \cdot 0.3 = 0.09 \neq 0.1$$
β‡’ X and Y are not independent

Detailed Explanation

In this example, we analyze the independence of two discrete random variables X and Y using their joint probability distribution. We are given the probabilities of the combinations of X and Y (formulated in a table). To assess independence, we start by calculating the marginal probabilities of X and Y separately. We find that P(X = 1) and P(Y = 1) both equal 0.3. Then, we check the condition for independence: P(X = 1, Y = 1) should equal the product of the marginal probabilities P(X = 1) and P(Y = 1). However, we find 0.1 (joint probability) β‰  0.09 (product of marginal probabilities), which indicates that X and Y are not independent.

Examples & Analogies

Consider rolling two dice. Let X and Y be the outcomes of each die. Knowing the result of one die affects the probabilities, as X and Y would show different outcomes. If rolling a die shows a 6 (X=6), it cannot influence the other die (Y), but the nature of dice makes specific combinations dependent on the overall rolling scenario. This scenario parallels how we analyzed the provided joint probabilities and their marginals.

Example 2: Continuous Case

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Example 2: Continuous Case
Suppose:
𝑓 (π‘₯,𝑦) = 𝑒^{βˆ’π‘₯}𝑒^{βˆ’π‘¦} = 𝑒^{βˆ’(π‘₯+𝑦)} for π‘₯,𝑦 > 0
This is:
𝑓 (π‘₯) = 𝑒^{βˆ’π‘₯}, 𝑓 (𝑦) = 𝑒^{βˆ’π‘¦}
So:
𝑓 (π‘₯,𝑦) = 𝑓 (π‘₯)⋅𝑓 (𝑦)
β‡’ X and Y are independent

Detailed Explanation

In this example, we examine the independence of two continuous random variables X and Y. We are provided with the joint probability density function f(x, y) which is the product of the individual density functions f(x) and f(y). The form e^(-x)*e^(-y) confirms that the distribution of X does not affect the distribution of Y. Thus, we conclude that X and Y are independent because the joint PDF can be expressed as the product of their individual PDFs, fulfilling the independence condition.

Examples & Analogies

Imagine a scenario where the weight of dogs and the height of trees in a park are measured. Even if both are influenced by environmental factors, the weight of a dog does not affect the height of a tree in any way. In this case, if we measure a dog (X) weighing 5 kg and note trees (Y) in the park can independently grow up to a certain height, the relationship remains independent - much like the continuous variables in this example.

Definitions & Key Concepts

Learn essential terms and foundational ideas that form the basis of the topic.

Key Concepts

  • Independence of Random Variables: Two random variables are independent if knowing the outcome of one does not affect the outcome of the other.

  • Joint Distribution: The combined distribution of two or more random variables.

  • Marginal Distribution: The distribution of a single random variable obtained from the joint distribution by summing or integrating over the other variables.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • Example 1: In a discrete probability table, it was found that P(X=1) * P(Y=1) does not equal P(X=1,Y=1), indicating that the random variables are dependent.

  • Example 2: For continuous variables, the joint PDF matched the product of individual PDFs, confirming that the variables are independent.

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

🎡 Rhymes Time

  • When variables don't interfere, their chance remains clear,
    Independent means that they're dear.

πŸ“– Fascinating Stories

  • Imagine two friends at a fair, one plays games and the other eats ice cream. The outcome of one friend's fun doesn’t change how much the other enjoysβ€”just like independent random variables!

🧠 Other Memory Gems

  • Remember 'PJI' – Product Just Indicates: For independence, we must check if P(X,Y) = P(X) * P(Y).

🎯 Super Acronyms

Use 'ID' for Independence Determination

  • P(X
  • Y) = P(X) * P(Y).

Flash Cards

Review key concepts with flashcards.

Glossary of Terms

Review the Definitions for terms.

  • Term: Random Variable

    Definition:

    A function that assigns real numbers to the outcomes of a random phenomenon.

  • Term: Joint Distribution

    Definition:

    The probability distribution of two or more random variables considered together.

  • Term: Independence

    Definition:

    Two random variables are independent if the occurrence of one does not affect the probability distribution of the other.

  • Term: Marginal Probability

    Definition:

    The probability distribution of a subset of a collection of random variables.

  • Term: Probability Mass Function (PMF)

    Definition:

    The function that gives the probability that a discrete random variable is equal to a particular value.

  • Term: Probability Density Function (PDF)

    Definition:

    The function that describes the likelihood of a continuous random variable taking on a particular value.