Mathematics for the least-squares slope

Mathematics for the least-squares slope#

This page follows on from the page on mean squared deviations. Like that page, is assumes a lot more maths than the standard flow of the course. In particular, it assumes that you know the basics of finding the derivative of a function.

Please make sure you have read and understand the mean squared deviation page, because this page builds on that argument.

You should also check the page on means and slopes for the problem we are trying to solve.

In our problem, we have $n$ $x$ values $x_{1}, x_{2}, . . ., x_{n}$ , that we want to use to predict $n$ corresponding $y$ values $y_{1}, y_{2}, . . ., y_{n}$ . For example, in terms of the mean and slopes page, we have 158 hemoglobin concentration values, so $n = 158$ and we can write our hemoglobin values as $x_{1}, x_{2}, . . ., x_{158}$ . We have 158 packed cell volume values, and we can write these as $y_{1}, y_{2}, . . ., y_{158}$ .

We decide we will use a straight line going through the origin to predict our $y$ points from our $x$ points. We define this line with its slope $s$ . This is the number of units that $y$ increases for every unit increase in $x$ . Our predicted values will therefore be $s x_{1}, s x_{2}, . . ., s x_{n}$ .

We want to choose $s$ such that it minimizes the sum of squared prediction errors. We define the prediction error for the first value as the actual value $y_{1}$ minus the prediction for that value $s x_{1}$ . We have $n$ prediction errors $y_{1} - s x_{1}, y_{2} - s x_{2}, . . ., y_{n} - s x_{n}$ . The thing we want to minimize is the sum of squared prediction error for a particular slope $s$ , defined as:

(3)#

S S E_{s} ≜ \sum (y_{i} - s x_{i})^{2}

The $≜$ symbol means is defined as.

This is the general formula for the specific plot we saw at the end of mean and slope page, where the value for $s$ is on the horizontal axis, and the value for $S S E_{s}$ is on the vertical axis.

We want to find the value of $s$ that gives the smallest value for $S S E_{s}$ .

The plot turned out to be U-shaped; we want to find the horizontal axis location ( $s$ value) corresponding to the bottom of the U (minimum of the corresponding $S S E_{s}$ values).

We follow the same scheme as for the mean squared deviations page; we transform the formula in (3) above into a formula for the gradient of the line that (3) represents, by taking the derivative. When the derivative of equation (3) is equal to zero, it means the gradient of (3) is 0, and this is true when we are at a peak or a trough of (3). We want the trough.

First we expand (3), and use the laws of sums to simplify the result:

\begin{array}{r} S S E_{s} ≜ \sum (y_{i} - s x_{i})^{2} ⟹ \\ = \sum (y_{i}^{2} - 2 y_{i} s x_{i} + s^{2} x_{i}^{2}) ⟹ \\ = \sum y_{i}^{2} - 2 s \sum y_{i} x_{i} + s^{2} \sum x_{i}^{2} \end{array}

Now differentiate with respect to $s$ :

(4)#

\frac{\partial S S E_{s}}{\partial s} = - 2 \sum y_{i} x_{i} + 2 s \sum x_{i}^{2}

Find the zero(s) for equation (4):

\begin{array}{r} 2 \sum y_{i} x_{i} = 2 s \sum x_{i}^{2} ⟹ \\ \frac{\sum y_{i} x_{i}}{\sum x_{i}^{2}} = s \end{array}

Equation (4) only has one zero.

We take the second derivative of (4) to see if the solution to $s$ is at a trough (with a positive second derivative) or a peak (with a negative second derivative).

\frac{\partial^{2} S S E_{s}}{\partial s^{2}} = 2 \sum x_{i}^{2}

$\sum x_{i}^{2}$ is always positive; this means that the second derivative is always positive, and therefore, it is also positive at our zero point $s = \frac{\sum y_{i} x_{i}}{\sum x_{i}^{2}}$ . So, equation (3) only has a one trough, at $s = \frac{\sum y_{i} x_{i}}{\sum x_{i}^{2}}$ , and no peaks.

This is the value $s$ for the slope that minimizes the sum of squared errors, also called the sum of squared deviations, also called the sum of squared prediction errors.