/*! 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 102 Suppose that we are fitting the ... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that we are fitting the line \(Y=\beta_{0}+\beta_{1} x+\epsilon,\) but the variance of \(Y\) depends on the level of \(x\); that is, $$V\left(Y_{i} \mid x_{i}\right)=\sigma_{i}^{2}=\frac{\sigma^{2}}{w_{i}} \quad i=1,2, \ldots, n$$ where the \(w_{i}\) are constants, often called weights. Show that for an objective function in which each squared residual is multiplied by the reciprocal of the variance of the corresponding observation, the resulting weighted least squares normal equations are $$\begin{aligned}\hat{\beta}_{0} \sum_{i=1}^{n} w_{i}+\hat{\beta}_{1} \sum_{i=1}^{n} w_{i} x_{i} &=\sum_{i=1}^{n} w_{i} y_{i} \\\\\hat{\beta}_{0} \sum_{i=1}^{n} w_{i} x_{i}+\hat{\beta}_{1} \sum_{i=1}^{n} w_{i} x_{i}^{2} &=\sum_{i=1}^{n} w_{i} x_{i} y_{i}\end{aligned}$$ Find the solution to these normal equations. The solutions are weighted least squares estimators of \(\beta_{0}\) and \(\beta_{1}\).

Short Answer

Expert verified
The weighted least squares estimators are \(\hat{\beta}_1 = \frac{\sum_{i=1}^{n} w_i (x_i - \overline{x}_w) y_i}{\sum_{i=1}^{n} w_i (x_i - \overline{x}_w)^2}\) and \(\hat{\beta}_0 = \overline{y}_w - \hat{\beta}_1 \overline{x}_w\).

Step by step solution

01

Understand the Objective Function

We aim to minimize the weighted sum of squared residuals, which can be written as \( \sum_{i=1}^{n} w_i (y_i - \beta_0 - \beta_1 x_i)^2 \). Each squared residual is multiplied by the weight, which is the reciprocal of the variance.
02

Set Up the Normal Equations

To derive the normal equations, we take partial derivatives of the objective function with respect to \(\beta_0\) and \(\beta_1\), and then set them to zero. This results in the system of equations: \( \hat{\beta}_0 \sum_{i=1}^{n} w_i + \hat{\beta}_1 \sum_{i=1}^{n} w_i x_i = \sum_{i=1}^{n} w_i y_i \) and \( \hat{\beta}_0 \sum_{i=1}^{n} w_i x_i + \hat{\beta}_1 \sum_{i=1}^{n} w_i x_i^2 = \sum_{i=1}^{n} w_i x_i y_i \).
03

Express the System in Matrix Form

The normal equations can be expressed in matrix form as: \[ \begin{bmatrix} \sum_{i=1}^{n} w_i & \sum_{i=1}^{n} w_i x_i \ \sum_{i=1}^{n} w_i x_i & \sum_{i=1}^{n} w_i x_i^2 \end{bmatrix} \begin{bmatrix} \hat{\beta}_0 \ \hat{\beta}_1 \end{bmatrix} = \begin{bmatrix} \sum_{i=1}^{n} w_i y_i \ \sum_{i=1}^{n} w_i x_i y_i \end{bmatrix} \]
04

Solve the System of Equations

To solve the system in matrix form, we multiply both sides by the inverse of the coefficient matrix. This gives us:\[ \begin{bmatrix} \hat{\beta}_0 \ \hat{\beta}_1 \end{bmatrix} = \begin{bmatrix} \sum_{i=1}^{n} w_i & \sum_{i=1}^{n} w_i x_i \ \sum_{i=1}^{n} w_i x_i & \sum_{i=1}^{n} w_i x_i^2 \end{bmatrix}^{-1} \begin{bmatrix} \sum_{i=1}^{n} w_i y_i \ \sum_{i=1}^{n} w_i x_i y_i \end{bmatrix} \]
05

Compute the Weighted Estimators

The weighted least squares estimators are found as follows: \[ \hat{\beta}_1 = \frac{\sum_{i=1}^{n} w_i (x_i - \overline{x}_w) y_i}{\sum_{i=1}^{n} w_i (x_i - \overline{x}_w)^2} \] \[ \hat{\beta}_0 = \overline{y}_w - \hat{\beta}_1 \overline{x}_w \] where \( \overline{x}_w = \frac{\sum_{i=1}^{n} w_i x_i}{\sum_{i=1}^{n} w_i} \) and \( \overline{y}_w = \frac{\sum_{i=1}^{n} w_i y_i}{\sum_{i=1}^{n} w_i} \).

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Key Concepts

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

Variance
In the context of weighted least squares, variance plays a key role in understanding how data spreads around the mean. Variance is essentially a measure of how much the observed data can diverge from the expected value. In a standard least squares approach, each observation is assumed to have the same variance. However, in weighted least squares, the variance depends on the level of the independent variable, denoted as \(x\).

The formula given for variance is \( V\left(Y_{i} \mid x_{i}\right)=\sigma_{i}^{2}=\frac{\sigma^{2}}{w_{i}} \). Here, \(w_i\) are constants known as weights, highlighting the importance of variance that changes with \(x\). This variation implies that some observations have higher reliability due to lower variance, hence weighted more in the analysis. To manage this variance-dependent weighting, we incorporate these weights into the objective function to ensure that each observation contributes appropriately based on its variance.
Normal Equations
Normal equations are derived by setting the derivative of the objective function to zero, aiming to find the values of the coefficients, \(\beta_0\) and \(\beta_1\), that minimize the sum of squared errors. When working with weighted least squares, the objective function becomes:
  • \( \sum_{i=1}^{n} w_i (y_i - \beta_0 - \beta_1 x_i)^2 \)
This modified objective function when differentiated with respect to the parameters \(\beta_0\) and \(\beta_1\) provides us the normal equations:

\( \hat{\beta}_0 \sum_{i=1}^{n} w_i + \hat{\beta}_1 \sum_{i=1}^{n} w_i x_i = \sum_{i=1}^{n} w_i y_i \) and \( \hat{\beta}_0 \sum_{i=1}^{n} w_i x_i + \hat{\beta}_1 \sum_{i=1}^{n} w_i x_i^2 = \sum_{i=1}^{n} w_i x_i y_i \).

These equations are essential as they ensure the weighted residuals are as small as possible. Solution to these equations provides the linear model’s slope and intercept, adjusted for the variability within the data.
Matrix Form
The use of matrix representation for solving equations is very efficient, especially with multiple variables and large datasets. In weighted least squares, the normal equations can be expressed in a compact matrix form:

\[\begin{bmatrix}\sum_{i=1}^{n} w_i & \sum_{i=1}^{n} w_i x_i \\sum_{i=1}^{n} w_i x_i & \sum_{i=1}^{n} w_i x_i^2\end{bmatrix}\begin{bmatrix}\hat{\beta}_0 \\hat{\beta}_1\end{bmatrix}= \begin{bmatrix}\sum_{i=1}^{n} w_i y_i \\sum_{i=1}^{n} w_i x_i y_i \end{bmatrix}\]

This format is beneficial as it simplifies the computation by leveraging matrix operations. The system of equations becomes easier to handle when written in this matrix notation, allowing the use of linear algebra techniques. To solve for \(\hat{\beta}_0\) and \(\hat{\beta}_1\), we simply multiply both sides by the inverse of the coefficient matrix. This yields the parameter estimates that are optimized to reflect the variance and weight considerations in the data.
Estimators
The weighted least squares estimators \(\hat{\beta}_0\) and \(\hat{\beta}_1\) offer a refined method for estimating regression coefficients that account for variance variability. Unlike unweighted least squares, these estimators adjust to give observations with lower variance a larger influence.

The explicit formulas for these estimators are:
  • \(\hat{\beta}_1 = \frac{\sum_{i=1}^{n} w_i (x_i - \overline{x}_w) y_i}{\sum_{i=1}^{n} w_i (x_i - \overline{x}_w)^2}\)
  • \(\hat{\beta}_0 = \overline{y}_w - \hat{\beta}_1 \overline{x}_w \)
Where:
  • \( \overline{x}_w = \frac{\sum_{i=1}^{n} w_i x_i}{\sum_{i=1}^{n} w_i} \)
  • \( \overline{y}_w = \frac{\sum_{i=1}^{n} w_i y_i}{\sum_{i=1}^{n} w_i} \)
These weighted averages \(\overline{x}_w\) and \(\overline{y}_w\) reflect the central tendency of the data adjusted for variance. These calculations ensure that the resulting regression line is well adapted to the data’s variability, providing reliable predictions even when variance is not constant.

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Most popular questions from this chapter

The following data gave \(X=\) the water content of snow on April 1 and \(Y=\) the yield from April to July (in inches) on the Snake River watershed in Wyoming for 1919 to 1935. (The data were taken from an article in Research Notes, Vol. \(61,1950,\) Pacific Northwest Forest Range Experiment Station, Oregon.) $$\begin{array}{cccc}\hline x & y & x & y \\\\\hline 23.1 & 10.5 & 37.9 & 22.8 \\\32.8 & 16.7 & 30.5 & 14.1 \\\31.8 & 18.2 & 25.1 & 12.9 \\ 32.0 & 17.0 & 12.4 & 8.8 \\\30.4 & 16.3 & 35.1 & 17.4 \\\24.0 & 10.5 & 31.5 & 14.9 \\\39.5 & 23.1 & 21.1 & 10.5 \\\24.2 & 12.4 & 27.6 & 16.1 \\\52.5 & 24.9 & & \\\\\hline\end{array}$$ (a) Estimate the correlation between \(Y\) and \(X\). (b) Test the hypothesis that \(\rho=0,\) using \(\alpha=0.05\). (c) Fit a simple linear regression model and test for significance of regression using \(\alpha=0.05 .\) What conclusions can you draw? How is the test for significance of regression related to the test on \(\rho\) in part (b)? (d) Analyze the residuals and comment on model adequacy.

The grams of solids removed from a material (y) is thought to be related to the drying time. Ten observations obtained from an experimental study follow: $$\begin{array}{c|c|c|c|c|c|c|c|c|c|c}y & 4.3 & 1.5 & 1.8 & 4.9 & 4.2 & 4.8 & 5.8 & 6.2 & 7.0 & 7.9 \\\\\hline x & 2.5 & 3.0 & 3.5 & 4.0 & 4.5 &5.0 & 5.5 & 6.0 & 6.5 & 7.0\end{array}$$ (a) Construct a scatter diagram for these data. (b) Fit a simple linear regression model. (c) Test for significance of regression. (d) Based on these data, what is your estimate of the mean grams of solids removed at 4.25 hours? Find a \(95 \%\) confidence interval on the mean. (e) Analyze the residuals and comment on model adequacy.

An article in Wear (Vol. \(152,1992,\) pp. \(171-181\) ) presents data on the fretting wear of mild steel and oil viscosity. Representative data follow, with \(x=\) oil viscosity and \(y=\) wear volume \(\left(10^{-4}\right.\) cubic millimeters) $$\begin{aligned}&\begin{array}{c|c|c|c|c|c}y & 240 & 181 & 193 & 155 & 172 \\\\\hline x & 1.6 & 9.4 & 15.5 & 20.0 & 22.0\end{array}\\\ &\begin{array}{l|c|c|c|c}y & 110 & 113 & 75 & 94 \\\\\hline x & 35.5 & 43.0 & 40.5 & 33.0\end{array}\end{aligned}$$ (a) Construct a scatter plot of the data. Does a simple linear regression model appear to be plausible? (b) Fit the simple linear regression model using least squares. Find an estimate of \(\sigma^{2}\) (c) Predict fretting wear when viscosity \(x=30\). (d) Obtain the fitted value of \(y\) when \(x=22.0\) and calculate the corresponding residual.

An article in the Journal of Sound and Vibration (Vol. \(151,1991,\) pp. \(383-394\) ) described a study investigating the relationship between noise exposure and hypertension. The following data are representative of those reported in the article. $$\begin{aligned}&\begin{array}{c|c|c|c|c|c|c|c|c|c|c}y & 1 & 0 & 1 & 2 & 5 & 1 & 4 & 6 & 2 & 3 \\\\\hline x & 60 & 63 & 65 & 70 & 70 & 70 & 80 & 90 & 80 & 80\end{array}\\\&\begin{array}{c|c|c|c|c|c|c|c|c|c|c}y & 5 & 4 & 6 & 8 & 4 & 5 & 7 & 9 & 7 & 6 \\\\\hline x & 85 & 89 & 90 & 90 & 90 & 90 & 94 & 100 & 100 & 100\end{array}\end{aligned}$$ (a) Draw a scatter diagram of \(y\) (blood pressure rise in millimeters of mercury) versus \(x\) (sound pressure level in decibels). Does a simple linear regression model seem reasonable in this situation? (b) Fit the simple linear regression model using least squares. Find an estimate of \(\sigma^{2}\) (c) Find the predicted mean rise in blood pressure level associated with a sound pressure level of 85 decibels.

Consider the following \((x, y)\) data. Calculate the correlation coefficient. Graph the data and comment on the relationship between \(x\) and \(y .\) Explain why the correlation coefficient does not detect the relationship between \(x\) and \(y\). $$\begin{array}{rrrr}\hline x & y & x & y \\\\\hline-4 & 0 & 0 & -4 \\\\-3 & -2.65 & 1 & 3.87 \\\\-3 & 2.65 & 1 & -3.87 \\\\-2 & -3.46 & 2 & 3.46 \\ -2 & 3.46 & 2 & -3.46 \\\\-1 & -3.87 & 3 & 2.65 \\\\-1 & 3.87 & 3 & -2.65 \\\0 & 4 & 4 & 0 \\\\\hline\end{array}$$

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