/*! 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 32 Show that if \(Y=a X+b(a \neq 0)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that if \(Y=a X+b(a \neq 0)\), then \(\operatorname{Corr}(X, Y)=\) \(+1\) or \(-1\). Under what conditions will \(\rho=+1\) ?

Short Answer

Expert verified
\(\operatorname{Corr}(X, Y) = +1\) if \(a > 0\).

Step by step solution

01

Understand the Problem

We need to show that the correlation coefficient \( \rho \) between two random variables \(X\) and \(Y\), defined as \(Y = aX + b\), is either \(+1\) or \(-1\). Moreover, we need to find the condition for \(\rho = +1\).
02

Recall the Formula for Correlation Coefficient

The correlation coefficient between \(X\) and \(Y\) is given by the formula:\[ \operatorname{Corr}(X,Y) = \frac{\operatorname{Cov}(X,Y)}{\sigma_X \sigma_Y} \]where \(\operatorname{Cov}(X,Y)\) is the covariance of \(X\) and \(Y\), and \(\sigma_X, \sigma_Y\) are the standard deviations of \(X\) and \(Y\), respectively.
03

Express Covariance of X and Y

Since \(Y = aX + b\), we have: \[ \operatorname{Cov}(X,Y) = \operatorname{Cov}(X, aX + b) = a \operatorname{Cov}(X, X) + \operatorname{Cov}(X, b) \]Using \(\operatorname{Cov}(X, b) = 0\) because \(b\) is a constant, and \(\operatorname{Cov}(X, X) = \operatorname{Var}(X) = \sigma_X^2\), it follows that \( \operatorname{Cov}(X, Y) = a \sigma_X^2 \).
04

Relate Standard Deviations \( \sigma_Y\) and \( \sigma_X\)

From the equation \(Y = aX + b\), observe that the standard deviation \(\sigma_Y\) of \(Y\) is given by: \[ \sigma_Y = |a| \sigma_X \] because adding a constant \(b\) does not change the spread of the distribution.
05

Substitute Covariance and Standard Deviations into Correlation Formula

We substitute \( \operatorname{Cov}(X,Y) = a \sigma_X^2 \) and \( \sigma_Y = |a| \sigma_X \) into the correlation formula: \[ \operatorname{Corr}(X,Y) = \frac{a \sigma_X^2}{\sigma_X |a| \sigma_X} = \frac{a}{|a|} \] Simplifying, \(\frac{a}{|a|} = 1\) if \(a > 0\) and \(-1\) if \(a < 0\).
06

Conclusion and Deduce Conditions for \(\rho = +1\)

Thus, \(\operatorname{Corr}(X, Y) = 1\) when \(a > 0\) and \(\operatorname{Corr}(X, Y) = -1\) when \(a < 0\). For \(\rho = +1\), this occurs only when \(a > 0\) and \(Y\) is a positive linear transformation of \(X\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Covariance
Covariance helps us understand the relationship between two random variables, say, \(X\) and \(Y\). Picture it as a way of measuring how much the variables change together. When they increase or decrease together consistently, they have a positive covariance. If one increases while the other decreases, they have a negative covariance.

Mathematically, covariance is represented as \( \operatorname{Cov}(X, Y) \). In the problem, since \(Y = aX + b\), the formula for covariance becomes essential. We calculate it like this:
  • \(\operatorname{Cov}(X, Y) = \operatorname{Cov}(X, aX + b)\)
  • Using the formula \( \operatorname{Cov}(X, aX + b) = a \cdot \operatorname{Cov}(X, X) + \operatorname{Cov}(X, b)\)
  • Here, \(\operatorname{Cov}(X, b) = 0\) because \(b\) isn’t a variable, but a constant.
  • So, we simplify to \(a \cdot \operatorname{Var}(X) = a \cdot \sigma_X^2\), because \(\operatorname{Var}(X) = \sigma_X^2\).
This tells us that the covariance depends directly on the constant \(a\) and the variance of \(X\).
Standard Deviation
Standard deviation, denoted as \(\sigma\), shows how much variation or dispersion exists from the mean. In simpler terms, it's a measure of the spread of a set of values. Low standard deviation indicates that values tend to be close to the mean, while high standard deviation means they are spread out over a wider range.

For the exercise:
  • The standard deviation of \(X\) is \(\sigma_X\).
  • When we linearly transform \(X\) into \(Y = aX + b\), the standard deviation of \(Y\) doesn’t depend on \(b\), the constant that simply shifts the values.
  • Instead, \(\sigma_Y\) is scaled by the factor \(|a|\), leading to \( \sigma_Y = |a| \sigma_X \).
This relationship is key to solving problems involving linear transformations, as it shows us how the spread scales when the variable is transformed.
Linear Transformation
A linear transformation is an operation where we multiply by a constant and add another constant. Think of it as scaling and shifting the variable. In algebraic terms:
  • The formula \(Y = aX + b\) describes a linear transformation where \(a\) is the scale factor and \(b\) is the shift.
In statistical analysis, the idea of linear transformation is important when exploring how characteristics like correlation are affected. Within the context of covariance and standard deviation, any linear transformation affects the spread of data without altering certain relationships.

Let's break it down further:
  • Multiplying \(X\) by \(a\) scales \(X\)'s spread by \(|a|\). Therefore, the standard deviation of \(Y\) becomes \(\sigma_Y = |a| \sigma_X\).
  • Shifting the graph by \(b\) doesn't change the variation, but changes the location on the graph, leaving standard deviation unchanged.
This helps explain why the correlation remains unaffected by the constant \(b\) and highlights that correlation relies only on the sign and size of \(a\). This is why \( \operatorname{Corr}(X, Y) = +1 \) or \(-1\) depending on whether \(a\) is positive or negative, respectively, showing a perfectly linear relationship.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(X_{1}\) denote the time (hr) it takes to perform a first task and \(X_{2}\) denote the time it takes to perform a second one. The second take always takes at least as long to perform as the first task. The joint pdf of these variables is \(f\left(x_{1}, x_{2}\right)=\left\\{\begin{array}{cl}2\left(x_{1}+x_{2}\right) & 0 \leq x_{1} \leq x_{2} \leq 1 \\ 0 & \text { otherwise }\end{array}\right.\) a. Obtain the pdf of the total completion time for the two tasks. b. Obtain the pdf of the difference \(X_{2}-X_{1}\) between the longer completion time and the shorter time.

Let \(X\) and \(Y\) be the times for a randomly selected individual to complete two different tasks, and assume that \((X, Y)\) has a bivariate normal distribution with \(\mu_{X}=100, \sigma_{X}=50, \mu_{Y}=25, \sigma_{Y}=5, \rho=.5\). From statistical software we obtain \(P(X<100\), \(Y<25)=.3333, P(X<50, Y<20)=.0625\), \(P(X<50, Y<25)=.1274\), and \(P(X<100, Y<\) \(20)=.1274 .\) (a) Determine \(P(50

This week the number \(X\) of claims coming into an insurance office is Poisson with mean 100 . The probability that any particular claim relates to automobile insurance is \(.6\), independent of any other claim. If \(Y\) is the number of automobile claims, then \(Y\) is binomial with \(X\) trials, each with "success" probability .6. a. Determine \(E(Y \mid X=x)\) and \(V(Y \mid X=x)\). b. Use part (a) to find \(E(Y)\). c. Use part (a) to find \(V(Y)\).

When an automobile is stopped by a roving safety patrol, each tire is checked for tire wear, and each headlight is checked to see whether it is properly aimed. Let \(X\) denote the number of headlights that need adjustment, and let \(Y\) denote the number of defective tires. a. If \(X\) and \(Y\) are independent with \(p_{X}(0)=.5\), \(p_{X}(1)=.3, p_{X}(2)=.2\), and \(p_{Y}(0)=.6, p_{Y}(1)\) \(=.1, p_{Y}(2)=p_{Y}(3)=.05, p_{Y}(4)=.2\), display the joint pmf of \((X, Y)\) in a joint probability table. b. Compute \(P(X \leq 1\) and \(Y \leq 1)\) from the joint probability table, and verify that it equals the product \(P(X \leq 1) \cdot P(Y \leq 1)\) c. What is \(P(X+Y=0)\) (the probability of no violations)? d. Compute \(P(X+Y \leq 1)\)

The number of customers waiting for gift-wrap service at a department store is an rv \(X\) with possible values \(0,1,2,3,4\) and corresponding probabilities \(.1, .2, .3, .25, .15 .\) A randomly selected customer will have 1,2 , or 3 packages for wrapping with probabilities . 6, .3, and .1, respectively. Let \(Y=\) the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer). a. Determine \(P(X=3, Y=3)\), that is, \(p(3,3)\). b. Determine \(p(4,11)\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.