Chapter 11, Problem 96SE

a.

To determine

Show the expression ${\displaystyle \sum _{i=1}^n\left(y_i-{\widehat{y}}_i\right)}=0$  is correct.

Explanation of Solution

Given info:

It is given that Y is the response variable and x is the predictor variable.

Calculation:

Let, Y be the response variable and x be the predictor variable, then the simple linear regression model can be expressed as $Y={\beta }_0+{\beta }_{1}x+\epsilon$, where, ${\beta }_0$ and ${\beta }_1$ are the intercept and slope coefficient respectively. $\epsilon$ is the random error with mean zero and variance ${\sigma }^2$.

From the linear regression it is known that,

$\begin{array}{l}{y}_i={\beta }_0+{\beta }_1{x}_i+{\epsilon }_i\\ {\widehat{y}}_i={\widehat{\beta }}_0+{\widehat{\beta }}_1{x}_i\end{array}$

From the normal equation,${\displaystyle \sum _{i=1}^n{y}_i}=n{\widehat{\beta }}_0+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i\right)}$

$\begin{array}{c}{\displaystyle \sum _{i=1}^n{\widehat{y}}_i}={\displaystyle \sum _{i=1}^n\left({\widehat{\beta }}_0+{\widehat{\beta }}_1{x}_i\right)}\\ ={\displaystyle \sum _{i=1}^n{\widehat{\beta }}_0}+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i\right)}\\ =n{\widehat{\beta }}_0+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i\right)}\end{array}$

Substituting the values,

$\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(y_i-{\widehat{y}}_i\right)}={\displaystyle \sum _{i=1}^n{y}_i}-{\displaystyle \sum _{i=1}^n{\widehat{y}}_i}\\ =n{\widehat{\beta }}_0+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i\right)}-n{\widehat{\beta }}_0+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i\right)}\\ =0\end{array}$

Hence, the expression ${\displaystyle \sum _{i=1}^n\left(y_i-{\widehat{y}}_i\right)}=0$ is correct.

b.

To determine

Show the expression ${\displaystyle \sum _{i=1}^n\left(y_i-{\widehat{y}}_i\right)x_i}=0$  is correct.

Explanation of Solution

Calculation:

From the linear regression it is known that,

$\begin{array}{l}{y}_i={\beta }_0+{\beta }_1{x}_i+{\epsilon }_i\\ {\widehat{y}}_i={\widehat{\beta }}_0+{\widehat{\beta }}_1{x}_i\end{array}$

$\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(y_i-{\widehat{y}}_i\right)x_i}\\ ={\displaystyle \sum _{i=1}^n{y}_i{x}_i}-{\displaystyle \sum _{i=1}^n{\widehat{y}}_i{x}_i}\end{array}$

From the normal equation,${\displaystyle \sum _{i=1}^n{x}_i{y}_i}={\widehat{\beta }}_0{\displaystyle \sum _{i=1}^n\left(x_i\right)}+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i{}^2\right)}$

$\begin{array}{c}{\displaystyle \sum _{i=1}^n{\widehat{y}}_i{x}_i}={\displaystyle \sum _{i=1}^n\left({\widehat{\beta }}_0+{\widehat{\beta }}_1{x}_i\right)x_i}\\ ={\displaystyle \sum _{i=1}^n{\widehat{\beta }}_0}{x}_i+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x^2{}_i\right)}\\ ={\widehat{\beta }}_0{\displaystyle \sum _{i=1}^n{x}_i}+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x^2{}_i\right)}\end{array}$

Substituting the values,

$\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(y_i-{\widehat{y}}_i\right)x_i}={\displaystyle \sum _{i=1}^n{y}_i{x}_i}-{\displaystyle \sum _{i=1}^n{\widehat{y}}_i{x}_i}\\ ={\widehat{\beta }}_0{\displaystyle \sum _{i=1}^n\left(x_i\right)}+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i{}^2\right)}-{\widehat{\beta }}_0{\displaystyle \sum _{i=1}^n{x}_i}-{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x^2{}_i\right)}\\ =0\end{array}$

Hence, the expression ${\displaystyle \sum _{i=1}^n\left(y_i-{\widehat{y}}_i\right)x_i}=0$ is correct.

c.

To determine

Show the expression $\frac{1}{n}{\displaystyle \sum _{i=1}^n{\widehat{y}}_i}=\overline{y}$  is correct.

Explanation of Solution

Calculation:

From the linear regression it is known that,

$\begin{array}{l}{\widehat{y}}_i={\widehat{\beta }}_0+{\widehat{\beta }}_1{x}_i\\ {\widehat{\beta }}_0=\overline{y}-{\widehat{\beta }}_1\overline{x}\end{array}$

$\begin{array}{c}{\displaystyle \sum _{i=1}^n{\widehat{y}}_i}={\displaystyle \sum _{i=1}^n\left({\widehat{\beta }}_0+{\widehat{\beta }}_1{x}_i\right)}\\ ={\displaystyle \sum _{i=1}^n{\widehat{\beta }}_0}+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i\right)}\\ =n{\widehat{\beta }}_0+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i\right)}\end{array}$

Substituting the values,

$\begin{array}{c}\frac{1}{n}{\displaystyle \sum _{i=1}^n{\widehat{y}}_i}=\frac{n{\widehat{\beta }}_0+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i\right)}}{n}\\ =\frac{n\left(\overline{y}-{\widehat{\beta }}_1\overline{x}\right)+{\widehat{\beta }}_1\times {\displaystyle \sum _{i=1}^n\left(x_i\right)}}{n}\\ =\frac{n\overline{y}-n{\widehat{\beta }}_1\overline{x}+{\widehat{\beta }}_{1}n\overline{x}}{n}\\ =\frac{n\overline{y}}{n}\\ =\overline{y}\end{array}$

Hence, the expression $\frac{1}{n}{\displaystyle \sum _{i=1}^n{\widehat{y}}_i}=\overline{y}$ is correct.