/*! 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 17 This problem extends some of the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

This problem extends some of the material in Section \(14.2 .3 .\) Let \(X\) and \(Y\) be random variables with $$ \begin{array}{c} E(X)=\mu_{x} \quad E(Y)=\mu_{y} \\ \operatorname{Var}(X)=\sigma_{x}^{2} \quad \operatorname{Var}(Y)=\sigma_{y}^{2} \\ \operatorname{Cov}(X, Y)=\sigma_{x y} \end{array} $$ Consider predicting \(Y\) from \(X\) as \(\hat{Y}=\alpha+\beta X,\) where \(\alpha\) and \(\beta\) are chosen to minimize \(E(Y-\tilde{Y})^{2},\) the expected squared prediction error.

Short Answer

Expert verified
Calculate \( \beta = \frac{\sigma_{xy}}{\sigma_x^2} \) and \( \alpha = \mu_y - \beta \mu_x \).

Step by step solution

01

Understand the Problem Statement

The problem involves predicting the random variable \( Y \) from another random variable \( X \) using a linear function. The prediction function is given by \( \hat{Y} = \alpha + \beta X \), where the objective is to find the values of \( \alpha \) and \( \beta \) that minimize the expected squared prediction error \( E((Y - \hat{Y})^2) \).
02

Identify Required Formulas

To find \( \alpha \) and \( \beta \), we use the formulas from statistics for linear regression. Specifically: \( \beta = \frac{\operatorname{Cov}(X, Y)}{\operatorname{Var}(X)} \) and \( \alpha = E(Y) - \beta E(X) \).
03

Calculate \( \beta \)

Using the given covariance and variance values, we calculate \( \beta \):\[ \beta = \frac{\sigma_{xy}}{\sigma_x^2} \].
04

Calculate \( \alpha \)

Once \( \beta \) is known, we can find \( \alpha \) using the means of \( X \) and \( Y \):\[ \alpha = \mu_y - \beta \mu_x \].
05

Conclude the Solution

Thus, the optimal prediction equation is \( \hat{Y} = \alpha + \beta X \), where \( \alpha \) and \( \beta \) are determined by the formulas derived above to minimize the expected squared prediction error.

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 is a key concept in statistics that measures the degree to which two random variables change together. When we have two variables, say \( X \) and \( Y \), the covariance between them, denoted as \( \text{Cov}(X, Y) \, or \, \sigma_{xy} \), gives us insight into their relationship. If the covariance is positive, \( X \) and \( Y \) tend to increase together. Conversely, a negative covariance implies that as one variable increases, the other tends to decrease.
In the context of linear regression, covariance helps us determine the line of "best fit" for predicting one variable based on another. When calculating covariance, we rely on the means of the variables:
  • \( \mu_x = E(X) \)
  • \( \mu_y = E(Y) \)
The formula for covariance is expressed as:\[\text{Cov}(X, Y) = E((X - \mu_x)(Y - \mu_y))\]Understanding covariance is essential for determining the slope of the regression line, \( \beta \), which you can calculate as:\[\beta = \frac{\sigma_{xy}}{\sigma_x^2}\]
Variance
Variance is another fundamental concept in the realm of statistics and probability. It provides a measure of how much a set of random variables \( X \) or \( Y \) is spread out from its mean. The variance of a variable \( X \) is represented as \( \sigma_x^2 \), and similarly, for \( Y \), it is \( \sigma_y^2 \).
Variance is calculated by taking the average of the squared differences from the mean, and it allows us to quantify the variability in our data set. The formula to compute variance for a random variable \( X \) is:\[\text{Var}(X) = E((X - \mu_x)^2)\]In linear regression, variance is crucial for determining the fit of the model. A smaller variance means the data points are closely centered around the mean, which often translates into a more accurate predictive model. For calculating the regression coefficients, variance in \( X \) is particularly important in computing \( \beta \), as shown by the formula:\[\beta = \frac{\sigma_{xy}}{\sigma_x^2}\]
Expected Squared Prediction Error
The expected squared prediction error is a statistical measure used to assess the accuracy of a predictive model. In linear regression, it's vital to minimize this error to improve the reliability of predictions. The linear prediction of \( Y \) from \( X \) is given by:\[\hat{Y} = \alpha + \beta X\]Where \( \alpha \) and \( \beta \) are coefficients that you determine through minimizing this error. The expected squared prediction error is expressed as:\[E((Y - \hat{Y})^2)\]This equation represents the expected value of the square of the difference between the observed value \( Y \) and the predicted value \( \hat{Y} \). By choosing \( \alpha \) and \( \beta \) such that this error is minimized, we ensure that the regression model is the most accurate it can be, based on the given data.
To formally define the error, consider the variance and covariance between different elements in the dataset, as they will play an integral role in optimizing \( \alpha \) and \( \beta \), thus efficiently predicting \( Y \) based on \( X \). The formulas for \( \alpha \) and \( \beta \) are:
  • \( \beta = \frac{\sigma_{xy}}{\sigma_x^2} \)
  • \( \alpha = \mu_y - \beta \mu_x \)
Random Variables
Random variables are a fundamental aspect of probability and statistics. They are essentially variables that can take on different values, each with an associated probability. In the context of predicting outcomes using linear regression, both \( X \) and \( Y \) are treated as random variables.Random variables can be discrete or continuous:
  • Discrete random variables have a countable set of possible outcomes.
  • Continuous random variables have an infinite number of possible values that are measured over a range.
For instance, when predicting \( Y \) (a continuous outcome) from \( X \) (another continuous variable), we use the linear model where the coefficients \( \alpha \) and \( \beta \) are based on the statistical properties of these random variables.
One of the key aspects of working with random variables in linear regression is computing and utilizing their expectations, such as the expected means \( E(X) \) and \( E(Y) \), and dispersions like variance and covariance. These help in formulating predictions that minimize error, allowing us to develop a model that is both efficient and accurate.

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

(Weighted Least Squares) Suppose that in the model \(y_{i}=\beta_{0}+\beta_{1} x_{i}+e_{i},\) the errors have mean zero and are independent, but \(\operatorname{Var}\left(e_{i}\right)=\rho_{i}^{2} \sigma^{2},\) where the \(\rho_{i}\) are known constants, so the errors do not have equal variance. This situation arises when the \(y_{i}\) are averages of several observations at \(x_{i}\); in this case, if \(y_{i}\) is an average of \(n_{i}\) independent observations, \(\rho_{i}^{2}=1 / n_{i}\) (why?). Because the variances are not equal, the theory developed in this chapter does not apply; intuitively, it seems that the observations with large variability should influence the estimates of \(\beta_{0}\) and \(\beta_{1}\) less than the observations with small variability. The problem may be transformed as follows: $$ \rho_{i}^{-1} y_{i}=\rho_{i}^{-1} \beta_{0}+\rho_{i}^{-1} \beta_{1} x_{i}+\rho_{i}^{-1} e_{i} $$ or $$ z_{i}=u_{i} \beta_{0}+v_{i} \beta_{1}+\delta_{i} $$ where $$ u_{i}=\rho_{i}^{-1} \quad v_{i}=\rho_{i}^{-1} x_{i} \quad \delta_{i}=\rho_{i}^{-1} e_{i} $$ a. Show that the new model satisfies the assumptions of the standard statistical model. b. Find the least squares estimates of \(\beta_{0}\) and \(\beta_{1}\) c. Show that performing a least squares analysis on the new model, as was done in part (b), is equivalent to minimizing $$ \sum_{i=1}^{n}\left(y_{i}-\beta_{0}-\beta_{1} x_{i}\right)^{2} \rho_{i}^{-2} $$ This is a weighted least squares criterion; the observations with large variances are weighted less. d. Find the variances of the estimates of part (b).

Chang (1945) studied the rate of sedimentation of amoebic cysts in water, in attempting to develop methods of water purification. The following table gives the diameters of the cysts and the times required for the cysts to settle through \(720 \mu \mathrm{m}\) of still water at three temperatures. Each entry of the table is an average of several observations, the number of which is given in parentheses. Does the time required appear to be a linear or a quadratic function of diameter? Can you find a model that fits? How do the settling rates at the three temperatures compare? (See Problem 7.) $$\begin{array}{c|c|c|c} \hline & \multicolumn{3}{|c} {\text { Setling Times of Cysts (sec) }} \\ \hline \text { Diameter }(\mu \mathrm{m}) & 10^{\circ} \mathrm{C} & 25^{\circ} \mathrm{C} & 28^{\circ} \mathrm{C} \\ \hline 11.5 & 217.1(2) & 138.2(1) & 128.4(2) \\ 13.1 & 168.3(3) & 109.3(3) & 103.1(4) \\ 14.4 & 136.6(11) & 89.1(13) & 82.7(11) \\ 15.8 & 114.6(17) & 73.0(11) & 70.5(18) \\ 17.3 & 96.4(8) & 61.3(6) & 59.7(6) \\ 18.7 & 80.8(5) & 56.2(4) & 50.0(4) \\ 20.2 & 70.4(2) & 46.3(1) & 41.4(2) \end{array}$$

An investigator wants to use multiple regression to predict a variable, \(Y,\) from two other variables, \(X_{1}\) and \(X_{2}\). She proposes forming a new variable \(X_{3}=X_{1}+X_{2}\) and using multiple regression to predict \(Y\) from the three \(X\) variables. Show that she will run into problems because the design matrix will not have full rank.

Show that the least squares estimates of the slope and intercept of a line may be expressed as $$ \beta_{0}=\bar{y}-\beta_{1} \bar{x} $$ and $$ \hat{\beta}_{1}=\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)}{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}} $$

Plot \(y\) versus \(x\) for the following pairs: $$\begin{array}{c|cccccccccc} x & 34 & 1.38 & -.65 & .68 & 1.40 & -.88 & -.30 & -1.18 & 50 & -1.75 \\ \hline y & .27 & 1.34 & -.53 & .35 & 1.28 & -.98 & -.72 & -.81 & .64 & -1.59 \end{array}$$ a. Fit a line \(y=a+b x\) by the method of least squares, and sketch it on the plot. b. Fit a line \(x=c+d y\) by the method of least squares, and sketch it on the plot. c. Are the lines in parts (a) and (b) the same? If not, why not?

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.