Chapter 7, Problem 20SE

Use Equation (7.34) (page 545) to show that ${\mu }_{{\hat{\beta }}_1}={\beta }_1$.

To determine

Show that ${\mu }_{{\widehat{\beta }}_1}={\beta }_1$ by using Equation 7.34.

Explanation of Solution

Calculation:

The equation 7.34 is,

${\widehat{\beta }}_1={\displaystyle \sum _{i=1}^n\left[\frac{\left(x_i-\overline{x}\right)}{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}\right]}{y}_i$

The denominator of the equation is,

$\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}={\displaystyle \sum _{i=1}^n{x}_i}-n\overline{x}\\ =\cancel{n\overline{x}}-\cancel{n\overline{x}}\\ =0\end{array}$

Since $\overline{x}$ is constant,

$\begin{array}{c}{\displaystyle \sum _{i=1}^n\overline{x}\left(x_i-\overline{x}\right)}=\overline{x}{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}\\ =\overline{x}\left(0\right)\\ =0\end{array}$

Express that ${\displaystyle \sum _{i=1}^n{x}_i}\left(x_i-\overline{x}\right)={\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}$.

$\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i}\left(x_i-\overline{x}\right)={\displaystyle \sum _{i=1}^n{x}_i}\left(x_i-\overline{x}\right)-{\displaystyle \sum _{i=1}^n\overline{x}\left(x_i-\overline{x}\right)}\\ ={\displaystyle \sum _{i=1}^n\left[x_i\left(x_i-\overline{x}\right)-\overline{x}\left(x_i-\overline{x}\right)\right]}\\ ={\displaystyle \sum _{i=1}^n\left[x_i{}^2-x_i\overline{x}-\overline{x}{x}_i+{\overline{x}}^2\right]}\\ ={\displaystyle \sum _{i=1}^n\left[x_i{}^2-2x_i\overline{x}+{\overline{x}}^2\right]}\end{array}$

$={\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}$

The mean of $y_i$ is,

${\mu }_{y_i}={\beta }_0+{\beta }_1{x}_i$

Therefore,

$\begin{array}{c}{\mu }_{{\widehat{\beta }}_1}={\displaystyle \sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)}{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}}{\mu }_{y_i}\\ ={\displaystyle \sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)}{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}}\left({\beta }_0+{\beta }_1{x}_i\right)\\ =\frac{1}{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}\left({\beta }_0+{\beta }_1{x}_i\right)\\ =\frac{1}{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}\left[{\beta }_0{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}+{\beta }_1{\displaystyle \sum _{i=1}^n\left(x_i\right)\left(x_i-\overline{x}\right)}\right]\end{array}$

On simplification,

$\begin{array}{c}{\mu }_{{\widehat{\beta }}_1}=\frac{1}{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}\left[{\beta }_0\left(0\right)+{\beta }_1{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}\right]\left[\begin{array}{l}\text{Since,}{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}=0\text{and}\\ {\displaystyle \sum _{i=1}^n{x}_i}\left(x_i-\overline{x}\right)={\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}\end{array}\right]\\ =\frac{1}{\cancel{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}}\left[{\beta }_1\cancel{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}\right]\\ ={\beta }_1\end{array}$

Hence, ${\mu }_{{\widehat{\beta }}_1}={\beta }_1$.