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Show that can express the residuals from a multiple regression model as \(e=(\mathbf{I}-\mathbf{H}) \mathbf{y}\) where \(\mathbf{H}=\mathbf{X}(\mathbf{X} \mathbf{X})^{-1} \mathbf{X}^{\prime}\)

Short Answer

Expert verified
Residuals can be expressed as \( \textbf{e} = (\textbf{I} - \textbf{H})\textbf{y} \).

Step by step solution

01

Understanding the Residuals

Residuals in a multiple regression model are the differences between the observed values and the fitted values. If \( extbf{y} \) is the vector of observed responses, and \( extbf{X} \) is the design matrix, the fitted values are given by \( extbf{X} \hat{\beta} \), where \( \hat{\beta} \) is the vector of estimated coefficients.
02

Formulating the Estimator

The estimator \( \hat{\beta} \) for the coefficients in a multiple regression model is calculated using the formula \( \hat{\beta} = (\textbf{X}'\textbf{X})^{-1}\textbf{X}'\textbf{y} \). This formula comes from minimizing the sum of squared residuals.
03

Expressing Fitted Values

Substitute \( \hat{\beta} \) back into the fitted values equation to obtain \( \hat{\textbf{y}} = \textbf{X}\hat{\beta} = \textbf{X}(\textbf{X}'\textbf{X})^{-1}\textbf{X}'\textbf{y} \). The matrix inside the expression, \( \textbf{H} = \textbf{X}(\textbf{X}'\textbf{X})^{-1}\textbf{X}' \), is called the Hat matrix because it maps \( \textbf{y} \) to \( \hat{\textbf{y}} \).
04

Defining Residuals in Terms of Identity Matrix

The residual vector \( \textbf{e} \) is defined as \( \textbf{y} - \hat{\textbf{y}} \). Substituting the expression for fitted values, we get \( \textbf{e} = \textbf{y} - \textbf{H}\textbf{y} \), which can be re-arranged as \( \textbf{e} = (\textbf{I} - \textbf{H})\textbf{y} \), where \( \textbf{I} \) is the identity matrix.

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Key Concepts

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

Residuals
In a multiple regression model, understanding residuals is crucial because they tell us how well our model is performing. Residuals are the differences between the observed data points and the values predicted by the model. When we have observed values denoted as \( \mathbf{y} \), these are the actual measurements we collect.
The fitted values are the ones our regression model predicts, calculated as \( \mathbf{X} \hat{\beta} \), where \( \hat{\beta} \) represents the estimated coefficients. The formula for the residuals is simple: - Residuals = Observed values - Fitted values.
Mathematically, this is expressed as \( \mathbf{e} = \mathbf{y} - \hat{\mathbf{y}} \). By substituting the expression for \( \hat{\mathbf{y}} \), the residuals can be written as \( \mathbf{e} = \mathbf{y} - \mathbf{H}\mathbf{y} \). Here, the vector \( \mathbf{e} \) is crucial in diagnosing how well the model fits the data, and any odd patterns in these residuals might indicate that the model needs improvement.
Hat Matrix
The Hat matrix, often denoted as \( \mathbf{H} \), plays an essential role in the regression analysis as it helps in converting the observed data into the predicted values. The Hat matrix is derived as: - \( \mathbf{H} = \mathbf{X}(\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}' \)Let's break this down:
- \( \mathbf{X} \) is the design matrix, representing the data for the predictors.- \( \mathbf{X}' \) is the transpose of the design matrix.- \( (\mathbf{X}'\mathbf{X})^{-1} \) is the inverse of the product of the design matrix and its transpose.
This matrix \( \mathbf{H} \) is called the "Hat" matrix because it maps the observed values \( \mathbf{y} \) to the fitted values \( \hat{\mathbf{y}} \) by essentially putting a "hat" on \( \mathbf{y} \). This is why we use the term \( \hat{\mathbf{y}} \) for predicted values. The equation connecting them is \( \hat{\mathbf{y}} = \mathbf{H}\mathbf{y} \), which showcases how the Hat matrix influences our predictions.
Understanding the Hat matrix is key to analyzing how each point in the data contributes to the predicted values, and it helps identify leverage points that might disproportionately affect the regression results.
Design Matrix
The design matrix \( \mathbf{X} \) is fundamental in creating a multiple regression model. It's a structured way of organizing our independent variables or predictors, where each row represents an observation and each column corresponds to a predictor.
Imagine the simplest case where a single predictor is used. The design matrix would have two columns: one for the constant term (often filled with ones for the intercept) and one for the predictor values themselves. For multiple predictors, the design matrix grows in size, with additional columns representing each one.
For instance:- A simple design matrix for a regression with two predictors might look like: \[ \mathbf{X} = \begin{bmatrix} 1 & x_{11} & x_{12} \ 1 & x_{21} & x_{22} \ \vdots & \vdots & \vdots \ 1 & x_{n1} & x_{n2} \end{bmatrix} \]Here, the rows correspond to different data points, and the columns represent the different predictors, including a constant term for the intercept. The setup of the design matrix directly influences the calculation of coefficients \( \hat{\beta} \), as it is part of the formula \( \hat{\beta} = (\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'\mathbf{y} \). So, the way we structure our design matrix has a direct impact on the model's performance and interpretability.
A well-constructed design matrix is essential for the accuracy and functionality of the regression model, providing the framework for understanding the relationships between predictors and the response variable.

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Most popular questions from this chapter

An engineer at a semiconductor company wants to model the relationship between the device HFE \((y)\) and three parameters: Emitter-RS \(\left(x_{1}\right),\) Base-RS \(\left(x_{2}\right),\) and Emitter-toBase \(\operatorname{RS}\left(x_{3}\right)\). The data are shown in the Table \(\mathrm{E} 12-5\) (a) Fit a multiple linear regression model to the data. (b) Estimate \(\sigma^{2}\). (c) Find the standard errors \(\operatorname{se}\left(\hat{\beta}_{j}\right)\). Are all of the model parameters estimated with the same precision? Justify your answer. (d) Predict HFE when \(x_{1}=14.5, x_{2}=220,\) and \(x_{3}=5.0 .\) $$ \begin{array}{cccc} \hline \begin{array}{c} x_{1} \\ \text { Emitter-RS } \end{array} & \begin{array}{c} x_{2} \\ \text { Base-RS } \end{array} & \begin{array}{c} x_{3} \\ \text { E-B-RS } \end{array} & \begin{array}{c} y \\ \text { HFE-1M-5V } \end{array} \\ \hline 14.620 & 226.00 & 7.000 & 128.40 \\ 15.630 & 220.00 & 3.375 & 52.62 \\ 14.620 & 217.40 & 6.375 & 113.90 \\ 15.000 & 220.00 & 6.000 & 98.01 \\ 14.500 & 226.50 & 7.625 & 139.90 \\ 15.250 & 224.10 & 6.000 & 102.60 \\ 16.120 & 220.50 & 3.375 & 48.14 \\ 15.130 & 223.50 & 6.125 & 109.60 \\ 15.500 & 217.60 & 5.000 & 82.68 \mathrm{~b} \\ 15.130 & 228.50 & 6.625 & 112.60 \\ 15.500 & 230.20 & 5.750 & 97.52 \\ 16.120 & 226.50 & 3.750 & 59.06 \\ 15.130 & 226.60 & 6.125 & 111.80 \\ 15.630 & 225.60 & 5.375 & 89.09 \\ 15.380 & 229.70 & 5.875 & 101.00 \\ 14.380 & 234.00 & 8.875 & 171.90 \\ 15.500 & 230.00 & 4.000 & 66.80 \\ 14.250 & 224.30 & 8.000 & 157.10 \\ 14.500 & 240.50 & 10.870 & 208.40 \\ 14.620 & 223.70 & 7.375 & 133.40 \\ \hline \end{array} $$

An article in the Journal of Pharmaceuticals Sciences (1991, Vol. \(80,\) pp. \(971-977\) ) presents data on the observed mole fraction solubility of a solute at a constant temperature and the dispersion, dipolar, and hydrogen-bonding Hansen partial solubility parameters. The data are as shown in the Table E12-13, where \(y\) is the negative logarithm of the mole fraction solubility, \(x_{1}\) is the dispersion partial solubility, \(x_{2}\) is the dipolar partial solubility, and \(x_{3}\) is the hydrogen-bonding partial solubility. (a) Fit the model \(Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\beta_{3} x_{3}+\beta_{12} x_{1} x_{2}+\) \(\beta_{13} x_{1} x_{3}+\beta_{23} x_{2} x_{3}+\beta_{11} x_{1}^{2}+\beta_{22} x_{2}^{2}+\beta_{33} x_{3}^{2}+\epsilon\) (b) Test for significance of regression using \(\alpha=0.05\). (c) Plot the residuals and comment on model adequacy. (d) Use the extra sum of squares method to test the contribution of the second-order terms using \(\alpha=0.05\). $$ \begin{array}{ccccc} \hline \text { Observation } & & & & \\ \text { Number } & \boldsymbol{y} & \boldsymbol{x}_{\mathbf{1}} & \boldsymbol{x}_{2} & \boldsymbol{x}_{3} \\ \hline 1 & 0.22200 & 7.3 & 0.0 & 0.0 \\ 2 & 0.39500 & 8.7 & 0.0 & 0.3 \\ 3 & 0.42200 & 8.8 & 0.7 & 1.0 \\ 4 & 0.43700 & 8.1 & 4.0 & 0.2 \\ 5 & 0.42800 & 9.0 & 0.5 & 1.0 \\ 6 & 0.46700 & 8.7 & 1.5 & 2.8 \\ 7 & 0.44400 & 9.3 & 2.1 & 1.0 \\ 8 & 0.37800 & 7.6 & 5.1 & 3.4 \\ 9 & 0.49400 & 10.0 & 0.0 & 0.3 \\ 10 & 0.45600 & 8.4 & 3.7 & 4.1 \\ 11 & 0.45200 & 9.3 & 3.6 & 2.0 \\ 12 & 0.11200 & 7.7 & 2.8 & 7.1 \\ 13 & 0.43200 & 9.8 & 4.2 & 2.0 \\ 14 & 0.10100 & 7.3 & 2.5 & 6.8 \\ 15 & 0.23200 & 8.5 & 2.0 & 6.6 \\ 16 & 0.30600 & 9.5 & 2.5 & 5.0 \\ 17 & 0.09230 & 7.4 & 2.8 & 7.8 \\ 18 & 0.11600 & 7.8 & 2.8 & 7.7 \\ 19 & 0.07640 & 7.7 & 3.0 & 8.0 \\ 20 & 0.43900 & 10.3 & 1.7 & 4.2 \\ 21 & 0.09440 & 7.8 & 3.3 & 8.5 \\ 22 & 0.11700 & 7.1 & 3.9 & 6.6 \\ 23 & 0.07260 & 7.7 & 4.3 & 9.5 \\ 24 & 0.04120 & 7.4 & 6.0 & 10.9 \\ 25 & 0.25100 & 7.3 & 2.0 & 5.2 \\ 26 & 0.00002 & 7.6 & 7.8 & 20.7 \\ \hline \end{array} $$

A chemical engineer is investigating how the amount of conversion of a product from a raw material \((y)\) depends on reaction temperature \(\left(x_{1}\right)\) and the reaction time \(\left(x_{2}\right) .\) He has developed the following regression models: 1\. \(\hat{y}=100+2 x_{1}+4 x_{2}\) 2\. \(\hat{y}=95+1.5 x_{1}+3 x_{2}+2 x_{1} x_{2}\) Both models have been built over the range \(0.5 \leq x_{2} \leq 10 .\) (a) What is the predicted value of conversion when \(x_{2}=2 ?\) Repeat this calculation for \(x_{2}=8 .\) Draw a graph of the predicted values for both conversion models. Comment on the effect of the interaction term in model 2 . (b) Find the expected change in the mean conversion for a unit change in temperature \(x_{1}\) for model 1 when \(x_{2}=5 .\) Does this quantity depend on the specific value of reaction time selected? Why? (c) Find the expected change in the mean conversion for a unit change in temperature \(x_{1}\) for model 2 when \(x_{2}=5 .\) Repeat this calculation for \(x_{2}=2\) and \(x_{2}=8\). Does the result depend on the value selected for \(x_{2}\) ? Why?

Heat treating is often used to carburize metal parts such as gears. The thickness of the carburized layer is considered a crucial feature of the gear and contributes to the overall reliability of the part. Because of the critical nature of this feature, two different lab tests are performed on each furnace load. One test is run on a sample pin that accompanies each load. The other test is a destructive test that cross-sections an actual part. This test involves running a carbon analysis on the surface of both the gear pitch (top of the gear tooth) and the gear root (between the gear teeth). Table \(\mathrm{E} 12-6\) shows the results of the pitch carbon analysis test for 32 parts. $$ \begin{array}{cccccc} \hline \text { TEMP } & \text { SOAKTIME } & \text { SOAKPCT } & \text { DIFFTIME } & \text { DIFFPCT } & \text { PITCH } \\ \hline 1650 & 0.58 & 1.10 & 0.25 & 0.90 & 0.013 \\ 1650 & 0.66 & 1.10 & 0.33 & 0.90 & 0.016 \\ 1650 & 0.66 & 1.10 & 0.33 & 0.90 & 0.015 \\ 1650 & 0.66 & 1.10 & 0.33 & 0.95 & 0.016 \\ 1600 & 0.66 & 1.15 & 0.33 & 1.00 & 0.015 \\ 1600 & 0.66 & 1.15 & 0.33 & 1.00 & 0.016 \\ 1650 & 1.00 & 1.10 & 0.50 & 0.80 & 0.014 \\ 1650 & 1.17 & 1.10 & 0.58 & 0.80 & 0.021 \\ 1650 & 1.17 & 1.10 & 0.58 & 0.80 & 0.018 \\ 1650 & 1.17 & 1.10 & 0.58 & 0.80 & 0.019 \\ 1650 & 1.17 & 1.10 & 0.58 & 0.90 & 0.021 \\ 1650 & 1.17 & 1.10 & 0.58 & 0.90 & 0.019 \end{array} $$ $$ \begin{array}{cccccc} \hline \text { TEMP } & \text { SOAKTIME } & \text { SOAKPCT } & \text { DIFFTIME } & \text { DIFFPCT } & \text { PITCH } \\ \hline 1650 & 1.17 & 1.15 & 0.58 & 0.90 & 0.021 \\ 1650 & 1.20 & 1.15 & 1.10 & 0.80 & 0.025 \\ 1650 & 2.00 & 1.15 & 1.00 & 0.80 & 0.025 \\ 1650 & 2.00 & 1.10 & 1.10 & 0.80 & 0.026 \\ 1650 & 2.20 & 1.10 & 1.10 & 0.80 & 0.024 \\ 1650 & 2.20 & 1.10 & 1.10 & 0.80 & 0.025 \\ 1650 & 2.20 & 1.15 & 1.10 & 0.80 & 0.024 \\ 1650 & 2.20 & 1.10 & 1.10 & 0.90 & 0.025 \\ 1650 & 2.20 & 1.10 & 1.10 & 0.90 & 0.027 \\ 1650 & 2.20 & 1.10 & 1.50 & 0.90 & 0.026 \\ 1650 & 3.00 & 1.15 & 1.50 & 0.80 & 0.029 \\ 1650 & 3.00 & 1.10 & 1.50 & 0.70 & 0.030 \\ 1650 & 3.00 & 1.10 & 1.50 & 0.75 & 0.028 \\ 1650 & 3.00 & 1.15 & 1.66 & 0.85 & 0.032 \\ 1650 & 3.33 & 1.10 & 1.50 & 0.80 & 0.033 \\ 1700 & 4.00 & 1.10 & 1.50 & 0.70 & 0.039 \\ 1650 & 4.00 & 1.10 & 1.50 & 0.70 & 0.040 \\ 1650 & 4.00 & 1.15 & 1.50 & 0.85 & 0.035 \\ 1700 & 12.50 & 1.00 & 1.50 & 0.70 & 0.056 \\ 1700 & 18.50 & 1.00 & 1.50 & 0.70 & 0.068 \end{array} $$ The regressors are furnace temperature (TEMP), carbon concentration and duration of the carburizing cycle (SOAKPCT, SOAKTIME), and carbon concentration and duration of the diffuse cycle (DIFFPCT, DIFFTIME). (a) Fit a linear regression model relating the results of the pitch carbon analysis test (PITCH) to the five regressor variables. (b) Estimate \(\sigma^{2}\). (c) Find the standard errors \(\operatorname{se}\left(\hat{\boldsymbol{\beta}}_{j}\right)\) (d) Use the model in part (a) to predict PITCH when TEMP = \(1650,\) SOAKTIME \(=1.00,\) SOAKPCT \(=1.10,\) DIFFTIME \(=\) \(1.00,\) and \(\mathrm{DIFFPCT}=0.80\)

The electric power consumed each month by a chemical plant is thought to be related to the average ambient temperature \(\left(x_{1}\right)\), the number of days in the month \(\left(x_{2}\right)\), the average product purity \(\left(x_{3}\right),\) and the tons of product produced \(\left(x_{4}\right)\). The past year's historical data are available and are presented in Table \(\mathrm{E} 12-2\) (a) Fit a multiple linear regression model to these data. (b) Estimate \(\sigma^{2}\). (c) Compute the standard errors of the regression coefficients. Are all of the model parameters estimated with the same precision? Why or why not? (d) Predict power consumption for a month in which \(x_{1}=75^{\circ} \mathrm{F}\), \(x_{2}=24\) days, \(x_{3}=90 \%,\) and \(x_{4}=98\) tons. $$ \begin{array}{ccccc} \hline y & x_{1} & x_{2} & x_{3} & x_{4} \\ \hline 240 & 25 & 24 & 91 & 100 \\ 236 & 31 & 21 & 90 & 95 \\ 270 & 45 & 24 & 88 & 110 \\ 274 & 60 & 25 & 87 & 88 \\ 301 & 65 & 25 & 91 & 94 \\ 316 & 72 & 26 & 94 & 99 \\ 300 & 80 & 25 & 87 & 97 \\ 296 & 84 & 25 & 86 & 96 \\ 267 & 75 & 24 & 88 & 110 \\ 276 & 60 & 25 & 91 & 105 \\ 288 & 50 & 25 & 90 & 100 \\ 261 & 38 & 23 & 89 & 98 \\ \hline \end{array} $$

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