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Exercises 19 and 20 involve a design matrix \(X\) with two or more columns and a least-squares solution \(\hat \beta \) of \({\bf{y}} = X\beta \). Consider the following numbers.

(i) \({\left\| {X\hat \beta } \right\|^2}\)鈥攖he sum of the squares of the 鈥渞egression term.鈥 Denote this number by .

(ii) \({\left\| {{\bf{y}} - X\hat \beta } \right\|^2}\)鈥攖he sum of the squares for error term. Denote this number by \(SS\left( E \right)\).

(iii) \({\left\| {\bf{y}} \right\|^2}\)鈥攖he 鈥渢otal鈥 sum of the squares of the \(y\)-values. Denote this number by \(SS\left( T \right)\).

Every statistics text that discusses regression and the linear model \(y = X\beta + \in \) introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume that the mean of the -values is zero. In this case, \(SS\left( T \right)\) is proportional to what is called the variance of the set of -values.

19. Justify the equation \(SS\left( T \right) = SS\left( R \right) + SS\left( E \right)\). (Hint: Use a theorem, and explain why the hypotheses of the theorem are satisfied.) This equation is extremely important in statistics, both in regression theory and in the analysis of variance.

Short Answer

Expert verified

The equation \(SS\left( T \right) = SS\left( R \right) + SS\left( E \right)\) is justified.

Step by step solution

01

Find \(SS\left( T \right)\)

The given residual vector is \( \in = {\bf{y}} - X\hat \beta \) which is orthogonal to \(\text{Col}X\), while \({\bf{\hat y}} = X\hat \beta \) is in \({\rm{Col}}X\).

As, \( \in = {\bf{y}} - X\hat \beta \) and \({\bf{\hat y}} = X\hat \beta \) are orthogonal, apply the orthogonal theorem and find \(\).

\(\begin{aligned}SS\left( T \right) &= {\left\| {\bf{y}} \right\|^2}\\ &= {\left\| {{\bf{\hat y}} + \in } \right\|^2}\\ &= {\left\| {{\bf{\hat y}}} \right\|^2} + {\left\| \in \right\|^2}\end{aligned}\)

Use \({\bf{\hat y}} = X\hat \beta \) and \( \in = {\bf{y}} - X\hat \beta \) into the obtained expression.

\(SS\left( T \right) = {\left\| {X\hat \beta } \right\|^2} + {\left\| {{\bf{y}} - X\hat \beta } \right\|^2}{\rm{ }}\left( 1 \right)\)

02

Find \(SS\left( R \right) + SS\left( E \right)\)

Find\(SS\left( R \right) + SS\left( E \right)\).

\(SS\left( R \right) + SS\left( E \right) = {\left\| {X\hat \beta } \right\|^2} + {\left\| {{\bf{y}} - X\hat \beta } \right\|^2}{\rm{ }}\left( 2 \right)\)

From equations (1) and (2),

\(SS\left( T \right) = SS\left( R \right) + SS\left( E \right)\)

Hence, the equation \(SS\left( T \right) = SS\left( R \right) + SS\left( E \right)\) is justified.

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Most popular questions from this chapter

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

8. \(\left\| {\mathop{\rm x}\nolimits} \right\|\)

Find an orthonormal basis of the subspace spanned by the vectors in Exercise 3.

Given data for a least-squares problem, \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\), the following abbreviations are helpful:

\(\begin{aligned}{l}\sum x = \sum\nolimits_{i = 1}^n {{x_i}} ,{\rm{ }}\sum {{x^2}} = \sum\nolimits_{i = 1}^n {x_i^2} ,\\\sum y = \sum\nolimits_{i = 1}^n {{y_i}} ,{\rm{ }}\sum {xy} = \sum\nolimits_{i = 1}^n {{x_i}{y_i}} \end{aligned}\)

The normal equations for a least-squares line \(y = {\hat \beta _0} + {\hat \beta _1}x\) may be written in the form

\(\begin{aligned}{c}{{\hat \beta }_0} + {{\hat \beta }_1}\sum x = \sum y \\{{\hat \beta }_0}\sum x + {{\hat \beta }_1}\sum {{x^2}} = \sum {xy} {\rm{ (7)}}\end{aligned}\)

Derive the normal equations (7) from the matrix form given in this section.

Find an orthogonal basis for the column space of each matrix in Exercises 9-12.

9. \(\left[ {\begin{aligned}{{}{}}3&{ - 5}&1\\1&1&1\\{ - 1}&5&{ - 2}\\3&{ - 7}&8\end{aligned}} \right]\)

A healthy child鈥檚 systolic blood pressure (in millimetres of mercury) and weight (in pounds) are approximately related by the equation

\({\beta _0} + {\beta _1}\ln w = p\)

Use the following experimental data to estimate the systolic blood pressure of healthy child weighing 100 pounds.

\(\begin{array} w&\\ & {44}&{61}&{81}&{113}&{131} \\ \hline {\ln w}&\\vline & {3.78}&{4.11}&{4.39}&{4.73}&{4.88} \\ \hline p&\\vline & {91}&{98}&{103}&{110}&{112} \end{array}\)

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