Steps for calculus solution to least square estimate
In order to find minimal values of a least squares function, we may follow the following steps:
Construct the sum-of-squares function as:
\[S(\beta) = \sum_{i=1}^n(y_i- \beta x_i)^2\]
Differentiate with respect \(\beta\);
\[\frac{dS}{d\beta}.\]
Set equal to \(0\) and solve for \(\beta\)
\[\frac{dS}{d\beta}=0.\]
Check that you have found a minimum by computing second derivatives.
Common notation and sum of square functions
In order to derive the least squares estimate, we need some notation on formula.
Let \(\bar{x}\) and \(\bar{y}\) denote the sample means of the \(x_i\)’s and \(y_i\)’s respectively for \(i=1,\ldots,n\).
Let \[S_{xx} = \sum_{i=1}^n(x_i-\bar{x})^2\] which can be written as \(\sum_{i=1}^nx_i^2-(\sum_{i=1}^nx_i)^2/n\). This is often called the corrected sum of squares of \(x\).
Similarly define \[S_{yy} = \sum_{i=1}^n(y_i-\bar{y})^2\] the corrected sum of squares for \(y\).
Let \[S_{xy} =\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y}) = \sum_{i=1}^nx_iy_i-\bigg(\sum_{i=1}^nx_i\sum_{i=1}^ny_i\bigg)/n\] which is called the corrected sum of products of \(x\) and \(y\).
Least squares estimates for a simple linear regression
The least squares estimates for a simple linear regression is given by \[\begin{aligned} \hat{\beta} &= \frac{S_{xy}}{S_{xx}},\nonumber\\ \hat{\alpha} &= \bar{y}-\frac{S_{xy}}{S_{xx}}\bar{x}.\nonumber\end{aligned}\]
Now let us show this.
\(\newline\)
Model: \(y_i = \alpha+\beta x_i+\epsilon_i\)
\(S(\alpha,\beta) = \sum_{i=1}^n(y_i-\alpha-\beta x_i)^2\)
\(\newline\)
Here there we have two parameters to estimate \(\alpha\) and \(\beta\).
\[\begin{aligned} S(\alpha,\beta) &= \sum_iy_i^2-2\sum (\alpha+\beta x_i)y_i+\sum_i(\alpha+\beta x_i)^2\nonumber\\ &=\sum_iy_i^2-2\alpha\sum y_i-2\beta\sum x_iy_i+n\alpha^2+2\alpha\beta\sum_ix_i+\beta^2\sum_i x_i^2\nonumber\end{aligned}\]
Thus
\[\begin{aligned} \frac{\partial S}{\partial \alpha}&=2n\alpha+2\beta\sum_ix_i-2\sum_iy_i\nonumber\\ &=2\left(n\alpha+\beta\sum_ix_i-\sum_iy_i\right)\nonumber\\ \frac{\partial S}{\partial \beta} &= -2\sum_ix_iy_i+2\alpha\sum_ix_i+2\beta\sum_ix_i^2\nonumber\\ &=2\left(\alpha\sum_ix_i+\beta\sum_ix_i^2-\sum_ix_iy_i\right)\nonumber\end{aligned}\]
i.e. we must solve
\[\begin{aligned} n\alpha+\beta\sum_ix_i &= \sum_iy_i\\ \alpha\sum_ix_i+\beta\sum_ix_i^2 &= \sum_ix_iy_i\end{aligned}\]
Divide (1) by n:
\[\alpha = \bar{y}-\bar{x}\beta\]
Substitute in (2):
\[\begin{aligned} \bar{y}\sum_ix_i-\bar{x}\beta\sum_ix_i+\beta\sum_ix_i^2 &=\sum_ix_iy_i\nonumber\\ n\bar{x}\bar{y}-\beta n\bar{x}^2+\beta\sum_ix_i^2 &=\sum_ix_iy_i\nonumber\\ \beta &= \frac{\sum_ix_iy_i-n\bar{x}\bar{y}}{\sum_ix_i^2-n\bar{x}^2}\nonumber\\ &=\frac{\sum_i(x_i-\bar{x})(y_i-\bar{y})}{\sum_i(x_i-\bar{x})^2}\nonumber\end{aligned}\]
Once again we denote estimates with a hat (\(\hat{.}\)) symbol. Thus,
\[\hat{\beta} = \frac{\sum_i(x_i-\bar{x})(y_i-\bar{y})}{\sum_i(x_i-\bar{x})^2}\]
and
\[\hat{\alpha} = \bar{y}-\hat{\beta}\bar{x}\]
Since \(S_{xx} = \sum_i(x_i-\bar{x})^2\), \(S_{xy} = \sum_i(x_i-\bar{x})(y_i-\bar{y})\) then
\[\begin{aligned} \hat{\beta} &= \frac{S_{xy}}{S_{xx}}\nonumber\\ \hat{\alpha} &= \bar{y}-\frac{S_{xy}}{S_{xx}}\bar{x}.\nonumber\end{aligned}\]
In the above approach, the second derivatives of the sum-of-squares functions w.r.t the parameters should be computed to verify that a minimum has been found.
\[\begin{aligned} \frac{\partial^2 S}{\partial \alpha^2}&=2n\\ \\ \frac{\partial^2 S}{\partial \beta^2} &= 2\sum_ix_i^2\nonumber \end{aligned}\]
since \(n>0\) and \(\sum_ix_i^2 >0\) then in both cases we have found minimal values.
Protein in pregnancy
Data were collected through interest in whether (and, if so, in what way) the level of protein changes in expectant mothers throughout pregnancy. Observations have been taken on 19 healthy women. Each woman was at a different stage of pregnancy (gestation). Fit a simple linear regression that describes the relationship between the mothers’ protein levels and the gestation length, and then estimate the parameters in this model
| Protein level (mgml\(^{-1})\) | Gestation (weeks) |
|---|---|
| 0.38 | 11 |
| 0.58 | 12 |
| 0.51 | 13 |
| 0.38 | 15 |
| 0.58 | 17 |
| 0.67 | 18 |
| 0.84 | 19 |
| 0.56 | 21 |
| 0.78 | 22 |
| 0.86 | 25 |
| 0.65 | 27 |
| 0.74 | 28 |
| 0.83 | 29 |
| 0.99 | 30 |
| 0.84 | 31 |
| 1.04 | 33 |
| 0.92 | 34 |
| 1.18 | 35 |
| 0.92 | 36 |
Data:\((y_i,x_i), \quad i=1,\dots,n\)
\(y_i\), protein level of mother \(i\)
\(x_i\), gestation of baby \(i\) (in weeks)
Model:\(\mathrm{E}(Y_i)=\beta_0 + \beta_1 x_i\)
\(\newline\)
Find the least square estimates using the following summary statistics:
\(n=19,\) \(\sum x_i =456,\) \(\sum x_i^2 =12164,\) \(\sum y_i = 14.25\), \(\sum x_i y_i =369.87\) and \(\sum y_i^2 =11.55\)
\[\begin{aligned} S_{xx} &= \sum_{i=1}^nx_i^2-(\sum_{i=1}^nx_i)^2/n \\ &= 12164 - 456^2/19\\ &= 1220 \end{aligned} \quad \quad \quad \quad \quad \quad \begin{aligned} S_{xy} &= \sum_{i=1}^nx_iy_i-(\sum_{i=1}^nx_i\sum_{i=1}^ny_i)/n \\ &= 369.87 - (456 \times 14.25)/19\\ &= 27.87 \end{aligned} \]
\[\begin{aligned} \hat{\beta} &= \frac{S_{xy}}{S_{xx}}\\ &= \frac{27.87}{1220}\\ &= 0.02284 \end{aligned} \quad \quad \quad \quad \quad \quad \quad \quad \quad \begin{aligned} \alpha &= \bar{y}-\frac{S_{xy}}{S_{xx}}\bar{x}\\ &= 14.25/19 - 27.87/1220 \times 456/19\\ &= 0.20174 .\nonumber \end{aligned}\]