/*! 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 90 In Exercise \(5.3,\) we determin... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercise \(5.3,\) we determined that the joint probability distribution of \(Y_{1}\), the number of married executives, and \(Y_{2},\) the number of never- married executives, is given by $$p\left(y_{1}, y_{2}\right)=\frac{\left(\begin{array}{c}4 \\\y_{1}\end{array}\right)\left(\begin{array}{c}3 \\ y_{2}\end{array}\right)\left(\begin{array}{c}2 \\\3-y_{1}-y_{2}\end{array}\right)}{\left(\begin{array}{l}9 \\\3 \end{array}\right)}$$ where \(y_{1}\) and \(y_{2}\) are integers, \(0 \leq y_{1} \leq 3,0 \leq y_{2} \leq 3,\) and \(1 \leq y_{1}+y_{2} \leq 3 .\) Find \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)\).

Short Answer

Expert verified
Covariance is calculated using the formula: \(\operatorname{Cov}(Y_1, Y_2) = E(Y_1Y_2) - E(Y_1)E(Y_2)\).

Step by step solution

01

Determine E(Y1) and E(Y2)

First, we need to calculate the expected values of \(Y_1\) and \(Y_2\). To find \(E(Y_1)\), multiply each \(y_1\) value by its probability \(p(y_1, y_2)\), and sum over all possible \(y_1\) and \(y_2\) pairs. Do the same for \(E(Y_2)\).
02

Determine E(Y1*Y2)

To find the joint expectation \(E(Y_1\cdot Y_2)\), multiply each \(y_1\) and \(y_2\) pair by their probability \(p(y_1, y_2)\), and by their products \(y_1 \cdot y_2\), then sum over all possible \(y_1\) and \(y_2\) combinations.
03

Calculate Cov(Y1, Y2)

Covariance is defined as \(\operatorname{Cov}(Y_1, Y_2) = E(Y_1Y_2) - E(Y_1)E(Y_2)\). Use the expected values from Steps 1 and 2 to calculate the covariance.

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Key Concepts

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

Joint Probability Distribution
A joint probability distribution gives us the probability of two random variables occurring simultaneously. In this exercise, we have two variables: the number of married executives, \(Y_1\), and the number of never-married executives, \(Y_2\). The joint probability distribution is represented by \(p(y_1, y_2)\), which provides probabilities for different combinations of \(y_1\) and \(y_2\).
The formula provided in the exercise is\[p\left(y_{1}, y_{2}\right)=\frac{\binom{4}{y_1}\binom{3}{y_2}\binom{2}{3-y_1-y_2}}{\binom{9}{3}}\]Here:
  • \(\binom{n}{k}\) is the binomial coefficient, equal to \(\frac{n!}{k!(n-k)!}\).
  • \(9\) is the total number of executives, while \(3\) is the sum of executives being considered for marriage status.
The probability is essentially calculated by considering how these individuals can be grouped based on their marital status. Each valid \(y_1, y_2\) pair has an associated probability that reflects the likely occurrence of that combination. This distribution is essential for understanding how the number of married and never-married executives relate to each other probabilistically.
Expected Value
The expected value is a fundamental concept in probability, often referred to as the mean or the average. When dealing with random variables, it tells us the average outcome we can expect over a large number of trials. For a random variable \(Y\), the expected value \(E(Y)\) can be calculated by summing up all possible values it can take, each multiplied by its probability.
In the case of our exercise:- To find \(E(Y_1)\), you compute \(\sum_{y_1, y_2} y_1 \cdot p(y_1, y_2)\). This means you multiply each value \(y_1\) by its associated joint probability and sum them up across all combinations of \(y_1\) and \(y_2\).- Similarly, \(E(Y_2) = \sum_{y_1, y_2} y_2 \cdot p(y_1, y_2)\) follows the same process, focusing on \(y_2\).
Ultimately, the expected values \(E(Y_1)\) and \(E(Y_2)\) give an average sense of how many married versus never-married executives we can expect, based on our joint probability distribution.
Covariance
Covariance is a measure of how much two random variables change together. If two variables tend to increase together, the covariance is positive. If one tends to decrease when the other increases, it's negative. The formula for covariance \(\operatorname{Cov}(Y_1, Y_2)\) is given by:\[\operatorname{Cov}(Y_1, Y_2) = E(Y_1 Y_2) - E(Y_1)E(Y_2)\]To find \(E(Y_1 Y_2)\), we calculate \(\sum_{y_1, y_2} y_1 y_2 \cdot p(y_1, y_2)\), where each product \(y_1 y_2\) is weighted by the probability \(p(y_1, y_2)\) and summed over all possibilities.
This outcome \(E(Y_1 Y_2)\) represents the expected value of the product of \(Y_1\) and \(Y_2\). By subtracting the product of \(E(Y_1)\) and \(E(Y_2)\), we can see how the actual joint expectation differs from what we would expect if \(Y_1\) and \(Y_2\) were independent.
Thus, covariance helps us understand the relationship between how often executives are married or never-married, revealing whether an increase in one correlates with a decrease or increase in the other.

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Most popular questions from this chapter

A quality control plan calls for randomly selecting three items from the daily production (assumed large) of a certain machine and observing the number of defectives. However, the proportion \(p\) of defectives produced by the machine varies from day to day and is assumed to have a uniform distribution on the interval (0,1) . For a randomly chosen day, find the unconditional probability that exactly two defectives are observed in the sample.

Suppose that the random variables \(Y_{1}\) and \(Y_{2}\) have joint probability density function, \(f\left(y_{1}, y_{2}\right)\) given by (see Exercises 5.14 and 5.32 ) $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll}6 y_{1}^{2} y_{2}, & 0 \leq y_{1} \leq y_{2}, y_{1}+y_{2} \leq 2 \\\0, & \text { elsewhere }\end{array}\right.$$ Show that \(Y_{1}\) and \(Y_{2}\) are dependent random variables.

Let the discrete random variables \(Y_{1}\) and \(Y_{2}\) have the joint probability function $$p\left(y_{1}, y_{2}\right)=1 / 3, \quad \text { for }\left(y_{1}, y_{2}\right)=(-1,0),(0,1),(1,0)$$ Find \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)\). Notice that \(Y_{1}\) and \(Y_{2}\) are dependent. (Why?) This is another example of uncorrelated random variables that are not independent.

Suppose that the random variables \(Y_{1}\) and \(Y_{2}\) have means \(\mu_{1}\) and \(\mu_{2}\) and variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\) respectively. Use the basic definition of the covariance of two random variables to establish that a. \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)=\operatorname{Cov}\left(Y_{2}, Y_{1}\right)\). b. \(\operatorname{Cov}\left(Y_{1}, Y_{1}\right)=V\left(Y_{1}\right)=\sigma_{1}^{2} .\) That is, the covariance of a random variable and itself is just the variance of the random variable.

In Exercise 5.42, we determined that the unconditional probability distribution for \(Y\), the number of defects per yard in a certain fabric, is $$p(y)=(1 / 2)^{y+1}, \quad y=0,1,2, \dots$$ Find the expected number of defects per yard.

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