/*! 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 45 In Exercise 5.1 we determined th... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 45

In Exercise 5.1 we determined that the joint distribution of \(Y_{1}\), the number of contracts awarded to firm \(\mathrm{A},\) and \(Y_{2},\) the number of contracts awarded to firm \(\mathrm{B}\), is given by the entries in the following table. $$\begin{array}{cccc}\hline & \multicolumn{3}{c} {y_{1}} \\\\\\)\cline { 2 - 4 }\\( y_{2} & 0 & 1 & 2 \\\\\hline 0 & 1 / 9 & 2 / 9 & 1 / 9 \\\1 & 2 / 9 & 2 / 9 & 0 \\\2 & 1 / 9 & 0 & 0 \\\\\hline\end{array}$$ The marginal probability function of \(Y_{1}\) was derived in Exercise 5.19 to be binomial with \(n=2\) and \(p=1 / 3 .\) Are \(Y_{1}\) and \(Y_{2}\) independent? Why?

Short Answer

Expert verified
No, \(Y_1\) and \(Y_2\) are not independent because their joint probabilities do not equal the product of their marginal probabilities.

Step by step solution

01

Define independence condition

Two random variables, say \(X\) and \(Y\), are independent if the joint probability \(P(X = x, Y = y)\) equals the product of their marginal probabilities \(P(X = x)\) and \(P(Y = y)\) for all values of \(x\) and \(y\).
02

Calculate marginal probabilities for Y2

To determine the marginal probability distribution of \(Y_2\), we need to sum the probabilities in the table over all values of \(Y_1\) for each fixed value of \(Y_2\). \(P(Y_2 = 0) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9}\)\(P(Y_2 = 1) = \frac{2}{9} + \frac{2}{9} + 0 = \frac{4}{9}\)\(P(Y_2 = 2) = \frac{1}{9} + 0 + 0 = \frac{1}{9}\)
03

Check independence condition for each combination

Compare the joint probability \(P(Y_1 = y_1, Y_2 = y_2)\) with the product \(P(Y_1 = y_1) \cdot P(Y_2 = y_2)\) for each combination of \(y_1\) and \(y_2\):- For \(Y_1 = 0, Y_2 = 0\): \(P(Y_1 = 0) \cdot P(Y_2 = 0) = \left(\frac{4}{9}\right) \left(\frac{4}{9}\right) = \frac{16}{81}\), but \(P(Y_1 = 0, Y_2 = 0) = \frac{1}{9} = \frac{9}{81}\).Since \(\frac{16}{81} eq \frac{9}{81}\), \(Y_1\) 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 Probability Distribution
Joint probability distribution is a way to express the likelihood of two or more random variables occurring at the same time. Think of it as a table or grid that shows the probability of different outcomes happening together. In our example, this is represented in the table showing the contracts awarded to firms A and B.
The table allows us to look up any pair of outcomes, such as the number of contracts for Firm A and Firm B, and see their associated probability.
  • Each cell shows the joint probability of both events happening. For example, the value in the first cell, \( \frac{1}{9} \), indicates the probability that both \( Y_1 \) and \( Y_2 \) are 0, meaning neither firm gets a contract.
  • This distribution captures all possible combinations of outcomes for the random variables involved.
Understanding joint probability helps in analyzing how two variables are correlated or interact. If you can predict one variable using the other, it says something about their statistical relationship.
Marginal Probability
Marginal probability refers to the probability of a single event or variable occurring, regardless of the outcome of another variable. It鈥檚 like adding up all the probabilities in a row or column to focus on one area of interest.
For example, in our exercise table, if we want to find out the marginal probability of \( Y_2 \), the number of contracts for Firm B, we sum up all the probabilities for each possible value of \( Y_2 \). This gives us a clearer picture of how likely each outcome is for Firm B individually.

Calculating Marginal Probability

  • To calculate the marginal probability of \( Y_2 = 0 \), add all probabilities from the row where \( Y_2 = 0 \): \( \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9} \).
  • Repeat this for each value of \( Y_2 \) to complete the marginal distribution.
Marginal probabilities are an important tool to understand each individual random variable's behavior without interference from others.
Random Variables
Random variables are a fundamental concept in statistics representing outcomes of random phenomena. They take on values based on the result of a random event or experiment. In our example, we have two random variables, \( Y_1 \) and \( Y_2 \), for the contracts awarded to firms A and B respectively.
Each random variable can assume various values, each with a certain probability.

Types of Random Variables

  • Discrete Random Variables: These take on distinct, separate values, like the number of contracts (0, 1, 2) a firm could win.
  • Continuous Random Variables: These can take on any value within a range and are not applicable in this discrete contracts example.
Random variables help in translating real-world scenarios into mathematical models so we can predict outcomes and make data-driven decisions.

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Most popular questions from this chapter

How big or small can \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)\) be? Use the fact that \(\rho^{2} \leq 1\) to show that $$-\sqrt{V\left(Y_{1}\right) \times V\left(Y_{2}\right)} \leq \operatorname{Cov}\left(Y_{1}, Y_{2}\right) \leq \sqrt{V\left(Y_{1}\right) \times V\left(Y_{2}\right)}$$

Contracts for two construction jobs are randomly assigned to one or more of three firms, A. B. and C. Let Y 1 denote the number of contracts assigned to firm A and Y 2 the number of contracts assigned to firm B. Recall that each firm can receive \(0.1 .\) or 2 contracts. a. Find the joint probability function for \(Y_{1}\) and \(Y_{2}\). b. Find \(F(1,0)\).

In Section 5.2 , we argued that if \(Y_{1}\) and \(Y_{2}\) have joint cumulative distribution function \(F\left(y_{1}, y_{2}\right)\) then for any \(a

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 $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} y_{1}+y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ Employee I has a higher productivity rating than employee II and a measure of the total productivity of the pair of employees is \(30 Y_{1}+25 Y_{2}\). Find the expected value of this measure of productivity.

In Exercise 5.6 , we assumed that if a radioactive particle is randomly located in a square with sides of unit length, a reasonable model for the joint density function for \(Y_{1}\) and \(Y_{2}\) is $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 1, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the marginal density functions for \(Y_{1}\) and \(Y_{2}\) b. What is \(P\left(.3 < Y_{1} < .5\right) ? P\left(.3 < Y_{2} < .5\right) ?\) c. For what values of \(y_{2}\) is the conditional density \(f\left(y_{1} | y_{2}\right)\) defined? d. For any \(y_{2}, 0 \leq y_{2} \leq 1\) what is the conditional density function of \(Y_{1}\) given that \(Y_{2}=y_{2} ?\) e. Find \(P\left(.3 < Y_{1} < .5 | Y_{2}=.3\right)\) f. Find \(P\left(.3 < Y_{1} < .5 | Y_{2}=.5\right)\) g. Compare the answers that you obtained in parts \((\mathrm{a}),(\mathrm{d}),\) and \((\mathrm{e}) .\) For any \(y_{2}, 0
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