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Let \(Y_{1}\) and \(Y_{2}\) have joint density function \(f\left(y_{1}, y_{2}\right)\) and marginal densities \(f_{1}\left(y_{1}\right)\) and \(f_{2}\left(y_{2}\right)\) respectively. Show that \(Y_{1}\) and \(Y_{2}\) are independent if and only if \(f\left(y_{1} | y_{2}\right)=f_{1}\left(y_{1}\right)\) for all values of \(y_{1}\) and for all \(y_{2}\) such that \(f_{2}\left(y_{2}\right)>0 .\) A completely analogous argument establishes that \(Y_{1}\) and \(Y_{2}\) are independent if and only if \(f\left(y_{2} | y_{1}\right)=f_{2}\left(y_{2}\right)\) for all values of \(y_{2}\) and for all \(y_{1}\) such that \(f_{1}\left(y_{1}\right)>0\).

Short Answer

Expert verified
Two random variables are independent if their joint density is the product of their marginal densities, ensuring conditional density equals the marginal density.

Step by step solution

01

Definition of Conditional Density Function

The conditional density function of two random variables \( Y_1 \) and \( Y_2 \) is given by \( f(y_1|y_2) = \frac{f(y_1, y_2)}{f_2(y_2)} \). This describes the density of \( Y_1\) given \(Y_2 = y_2\), assuming that \(f_2(y_2) > 0\).
02

Condition for Independence

Two random variables \( Y_1 \) and \( Y_2 \) are independent if their joint density function is the product of their marginal densities, i.e., \( f(y_1, y_2) = f_1(y_1)f_2(y_2) \).
03

Show Independence Imposes 'If' Condition

Assume \( Y_1 \) and \(Y_2\) are independent. Then, by definition, \(f(y_1, y_2) = f_1(y_1)f_2(y_2)\). Substitute into the conditional density: \( f(y_1|y_2) = \frac{f_1(y_1)f_2(y_2)}{f_2(y_2)} = f_1(y_1)\). Thus, if \(Y_1\) and \(Y_2\) are independent, the conditional density \(f(y_1|y_2) \) equals \(f_1(y_1)\).
04

Show 'Only If' Condition

Assume \( f(y_1|y_2) = f_1(y_1) \) for all \(y_1\) and \(y_2\) such that \(f_2(y_2) > 0\). Therefore, we have \( \frac{f(y_1, y_2)}{f_2(y_2)} = f_1(y_1) \). Rearranging gives \( f(y_1, y_2) = f_1(y_1)f_2(y_2) \). Hence, the joint density is the product of marginal densities, indicating \(Y_1\) and \(Y_2\) are independent.
05

Analogous Argument for Opposite Condition

By symmetry, we show that the opposite condition \( f(y_2|y_1) = f_2(y_2) \) implies independence. Similarly, if this condition is satisfied for \( f(y_2|y_1) \), it follows that \( f(y_1, y_2) = f_1(y_1) f_2(y_2) \), confirming independence.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independence of Random Variables
In probability theory, two random variables are considered to be independent if the occurrence or value of one variable does not affect the occurrence or value of the other. For independent random variables, knowing the outcome of one variable does not provide any information about the other.
Mathematically, random variables \( Y_1 \) and \( Y_2 \) are said to be independent if their joint probability distribution \( f(y_1, y_2) \) can be expressed as the product of their individual marginal distributions, \( f_1(y_1) \) and \( f_2(y_2) \). This means the joint probability density is given by:
\[f(y_1, y_2) = f_1(y_1) f_2(y_2)\]
This equation confirms that each random variable behaves independently without any interaction with the other. In simple terms, the relationship between them is non-existent in a probabilistic sense, making them completely unrelated.
Joint Density Function
A joint density function is crucial when dealing with two or more continuous random variables. It essentially describes how these random variables behave together over their possible range of values. The joint density function \( f(y_1, y_2) \) captures the likelihood that the random variables \( Y_1 \) and \( Y_2 \) take on specific values simultaneously.
It is seen as an extension of the concept of a single-variable probability density function (PDF) to multiple dimensions. In mathematical terms, for any specific pair of values \( (y_1, y_2) \), the function \( f(y_1, y_2) \) represents the density of observing \( Y_1 = y_1 \) and \( Y_2 = y_2 \) together.
The joint density must satisfy two key conditions:
  • The function \( f(y_1, y_2) \) must be non-negative for all possible values of \( y_1 \) and \( y_2 \).
  • The integral over all possible values must equal 1, ensuring it represents a valid probability distribution.
When investigating relationships between random variables, the joint density plays an essential role in determining characteristics such as independence.
Marginal Density
The concept of marginal density refers to the probability distribution of a subset of a collection of random variables. When you look at the distribution of a single random variable, irrespective of others, you are considering its marginal density. This is derived from the joint density function by integrating out the other variables.
For two variables \( Y_1 \) and \( Y_2 \) with a joint density function \( f(y_1, y_2) \), the marginal density for \( Y_1 \) is determined by integrating over all possible values of \( Y_2 \). Mathematically, it is given as:
\[f_1(y_1) = \int f(y_1, y_2) \ dy_2\]
Similarly, the marginal density for \( Y_2 \) is computed by integrating over \( Y_1 \):
\[f_2(y_2) = \int f(y_1, y_2) \ dy_1\]
Marginal densities are valuable for understanding how each variable behaves individually when considered in the presence of others, highlighting the probability characteristics of each one independently.
Probability Theory
Probability theory is the mathematical foundation for analyzing random phenomena. It provides tools and concepts that allow us to understand and predict the behavior of random variables and assess the likelihood of different outcomes.
Key concepts include:
  • Random Variables: Functions that assign numerical values to each outcome of a random phenomenon.
  • Probability Distributions: Mathematical functions describing the probabilities of different possible outcomes of a random variable.
  • Expectation and Variance: Measures of central tendency and spread of a probability distribution.
Probability theory serves as the backbone for areas such as statistics, gambling, risk management, and many fields where predicting outcomes is essential. It combines mathematical rigor with real-world applications, offering powerful techniques to model uncertainty and make informed decisions based on probabilistic analysis.

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Most popular questions from this chapter

Let \(Y_{1}\) and \(Y_{2}\) have a joint density function given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 3 y_{1}, & 0 \leq y_{2} \leq y_{1} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the marginal density functions of \(Y_{1}\) and \(Y_{2}\) b. Find \(P\left(Y_{1} \leq 3 / 4 | Y_{2} \leq 1 / 2\right)\) c. Find the conditional density function of \(Y_{1}\) given \(Y_{2}=y_{2}\) d. Find \(P\left(Y_{1} \leq 3 / 4 | Y_{2}=1 / 2\right)\)

Let \(Z\) be a standard normal random variable and let \(Y_{1}=Z\) and \(Y_{2}=Z^{2}\). a. What are \(E\left(Y_{1}\right)\) and \(E\left(Y_{2}\right) ?\) b. What is \(E\left(Y_{1} Y_{2}\right) ?\left[\text { Hint: } E\left(Y_{1} Y_{2}\right)=E\left(Z^{3}\right), \text { recall Exercise 4.199. }\right]\) c. What is \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right) ?\) d. Notice that \(P\left(Y_{2}>1 | Y_{1}>1\right)=1 .\) Are \(Y_{1}\) and \(Y_{2}\) independent?

Assume that \(Y_{1}, Y_{2},\) and \(Y_{3}\) are random variables, with $$E\left(Y_{1}\right)=2, \quad E\left(Y_{2}\right)=-1, \quad E\left(Y_{3}\right)=4 $$ $$V\left(Y_{1}\right)=4, \quad V\left(Y_{2}\right)=6, \quad V\left(Y_{3}\right)=8$$ $$\operatorname{Cov}\left(Y_{1}, Y_{2}\right)=1, \operatorname{Cov}\left(Y_{1}, Y_{3}\right)=-1, \operatorname{Cov}\left(Y_{2}, Y_{3}\right)=0$$ Find \(E\left(3 Y_{1}+4 Y_{2}-6 Y_{3}\right)\) and \(V\left(3 Y_{1}+4 Y_{2}-6 Y_{3}\right)\)

Two friends are to meet at the library. Each independently and randomly selects an arrival time within the same one-hour period. Each agrees to wait a maximum of ten minutes for the other to arrive. What is the probability that they will meet?

In Exercise 5.9 we determined that $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 6\left(1-y_{2}\right), & 0 \leq y_{1} \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ is a valid joint probability density function. Find $$\text { a. } E\left(Y_{1}\right) \text { and } E\left(Y_{2}\right)$$ $$\text { b. } V\left(Y_{1}\right) \text { and } V\left(Y_{2}\right)$$ $$\text { c. } E\left(Y_{1}-3 Y_{2}\right)$$

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