/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 In Exercise \(5.16, Y_{1}\) and ... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercise \(5.16, Y_{1}\) and \(Y_{2}\) denoted the proportions of time that employees I and II actually spent working on their assigned tasks during a workday. The joint density of \(Y_{1}\) and \(Y_{2}\) is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll}y_{1}+y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\\0, & \text {elsewhere }\end{array}\right.$$ Are \(Y_{1}\) and \(Y_{2}\) independent?

Short Answer

Expert verified
\(Y_{1}\) and \(Y_{2}\) are not independent.

Step by step solution

01

Understand Independence

Two random variables \(Y_1\) and \(Y_2\) are independent if the joint probability density function (PDF) can be expressed as the product of their individual marginal densities. Mathematically, \(f(y_1, y_2) = f_{Y_1}(y_1) \cdot f_{Y_2}(y_2)\). We need to check this property for the given joint PDF.
02

Find Marginal Density of \(Y_1\)

The marginal density \(f_{Y_1}(y_1)\) is found by integrating the joint density function over all values of \(y_2\): \[f_{Y_1}(y_1) = \int_0^1 (y_1 + y_2) \, dy_2\]. Calculate this integral.
03

Evaluate the Integral for \(Y_1\)

Perform the integration: \[f_{Y_1}(y_1) = \int_0^1 y_1 + y_2 \, dy_2 = \left[ y_1y_2 + \frac{y_2^2}{2} \right]_0^1 = y_1 \cdot 1 + \frac{1^2}{2} = y_1 + \frac{1}{2}\].
04

Find Marginal Density of \(Y_2\)

Similarly, find the marginal density \(f_{Y_2}(y_2)\) by integrating over \(y_1\): \[f_{Y_2}(y_2) = \int_0^1 (y_1 + y_2) \, dy_1\]. Calculate this integral.
05

Evaluate the Integral for \(Y_2\)

Perform the integration: \[f_{Y_2}(y_2) = \int_0^1 y_1 + y_2 \, dy_1 = \left[ \frac{y_1^2}{2} + y_2 y_1 \right]_0^1 = \frac{1^2}{2} + y_2 \cdot 1 = \frac{1}{2} + y_2\].
06

Check for Independence

Now check if the joint PDF can be expressed as a product of the marginals; i.e., do we have \((y_1 + y_2) = (y_1 + \frac{1}{2})(\frac{1}{2} + y_2)\)? Expanding the product gives \((y_1 + \frac{1}{2})(\frac{1}{2} + y_2) = y_1 \cdot \frac{1}{2} + y_1y_2 + \frac{1}{2} \cdot y_2 + \frac{1}{4}\), which does not simplify to \(y_1 + y_2\). Thus, \(Y_1\) and \(Y_2\) are not independent.

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Key Concepts

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

Joint Density
The concept of a joint density function is central when dealing with two or more random variables. In our scenario with \( Y_1 \) and \( Y_2 \), which represent proportions of time spent by two employees on tasks, the joint density function \( f(y_1, y_2) \) essentially gives us a way to find probabilities for these random variables occurring together.
For \( 0 \leq y_1 \leq 1 \) and \( 0 \leq y_2 \leq 1 \), the function is defined as \( f(y_1, y_2) = y_1 + y_2 \). Outside this range, it becomes 0. This function allows us to understand how the two variables interact and exist together within the given range.
Knowing the joint density helps in determining the likelihood of different scenarios involving both random variables simultaneously. For example, you can calculate the probability that both employees will simultaneously spend certain proportions of time on their tasks by integrating over the specified range.
Marginal Density
To gain insights into each individual variable, we derive the marginal density functions from the joint density. This involves integrating the joint density with respect to one of the variables, effectively 'ignoring' the other.
For \( Y_1 \), the marginal density \( f_{Y_1}(y_1) \) is found by integrating \( f(y_1, y_2) \) over all values of \( y_2 \):
  • Integrate: \( f_{Y_1}(y_1) = \int_0^1 (y_1 + y_2) \, dy_2 \)
  • Result: \( y_1 + \frac{1}{2} \)
This marginal density tells us the probability distribution of \( Y_1 \) on its own.
Similarly, for \( Y_2 \), we follow the same procedure by integrating the joint density over \( y_1 \):
  • Integrate: \( f_{Y_2}(y_2) = \int_0^1 (y_1 + y_2) \, dy_1 \)
  • Result: \( \frac{1}{2} + y_2 \)
Marginal densities are crucial to understanding the individual behavior of random variables when they form part of a larger system.
Independence of Random Variables
Determining the independence of random variables \( Y_1 \) and \( Y_2 \) is essential when modeling related data points. Two random variables are independent if the joint density function can be decomposed into the product of their marginal densities; mathematically, \( f(y_1, y_2) = f_{Y_1}(y_1) \cdot f_{Y_2}(y_2) \).
The derived marginals were \( f_{Y_1}(y_1) = y_1 + \frac{1}{2} \) and \( f_{Y_2}(y_2) = \frac{1}{2} + y_2 \). Let’s compute \( (y_1 + \frac{1}{2})(\frac{1}{2} + y_2) \):
  • Result: \( \frac{y_1}{2} + y_1 y_2 + \frac{y_2}{2} + \frac{1}{4} \)
This expression does not match with the given \( y_1 + y_2 \), indicating the variables are not independent.
Understanding independence helps in predicting whether altering one variable would affect the other, which is critical in statistical analysis and decision-making. Thus, knowing \( Y_1 \) and \( Y_2 \) are dependent suggests that changes in one variable may influence the other.

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