Least Squares — the Gory Details
How Do We Find the Line of Best Fit?
Copyright © 2002–2025 by Stan Brown, BrownMath.com
Copyright © 2002–2025 by Stan Brown, BrownMath.com
You have a set of observed points (x,y). You’ve plotted them and they seem to be pretty much linear. How do you find the line that best fits those points?
“Don’t be silly,” you say. “Put them into a TI-83 or Excel and look at the answer.”
Okay, you got me. I’ll ask a deeper question: How does the calculator find the answer? What are the underlying equations?
See also: The least-squares method involves summations. If you’re shaky on your ∑ (sigma) notation, see “∑ Means Add ’em Up”.
To answer that question, first we have to agree on what we mean by the “best fit” of a line to a set of points. Why do we say that the line on the left fits the points better than the line on the right? And can we say that some other line might fit them better still?
Intuitively, we think of a close fit as a good fit. We look for a line with little space between the line and the points it’s supposed to fit. We would say that the best fitting line is the one that has the least space between itself and the data points, which represent actual measurements.
Okay, what do we mean by “least space”? There are three ways to measure the space between a point and a line: vertically in the y direction, horizontally in the x direction, and on a perpendicular to the line. We choose to measure the space vertically. Why? Because our whole purpose in making a regression line is to use it to predict the y value for a given x, and the vertical distances are how far off the predictions would be for the points we actually measured.
That vertical deviation, or prediction error, is called the residual, y−ŷ. Since the line predicts a y value (symbol ŷ) for every x value, and there’s an actual measured y value for every x value, there is a residual for every x value in the data set.
For example, suppose the line is y=3x+2 and we have a measured data point (2,9). The actual y is 9 and the line predicts ŷ=3·2+2=8. Subtracting, we can say that the residual for x=2, or the residual for the point (2,9), is 9−8 = 1. This is a positive number because the actual value is greater than the predicted value, and the line passes below the data point (2,9). In general, between any given point (x,y) and the line y=mx+b the residual (vertical gap) is y−(mx+b).
But each residual could be negative or positive, depending on whether the line passes above or below that point. We can’t simply add up residuals, because then a line would be considered good if it fell way below some points as long as it fell way above others. To prevent that, we’ll square each residual, and add up the squares. (This also has the desirable effect that a few small deviations are more tolerable than one or two big ones.)
And at long last we can say exactly what we mean by the line of best fit. If we compute the residual for every point, square each one, and add up the squares, we say the line of best fit is the line for which that sum is the least. Since it’s a sum of squares, the method is called the method of least squares.
It’s always a giant step in finding something to get clear on what it is you’re looking for, and we’ve done that. The best-fit line, as we have decided, is the line that minimizes the sum of squares of residuals. For any given line y=mx+b, we can write that sum as
E(m,b) = ∑(y − ŷ)²
where ŷ is the predicted value for a given x, namely mx+b, and y is the actual value measured for that given x.
E is a function of m and b because the sum of squared residuals is different for different lines y=mx+b. E is not a function of x and y because the data points are what they are, and don’t change within any given problem.
Do we just try a bunch of lines, compute their E values, and pick the line with the lowest E value? No, it would be a lot of work without proving anything — a lose-lose — because we could never be sure that there wasn’t some other line with still a lower E.
Instead, we use a powerful and common trick in mathematics: We assume we know the line, and use its properties to help us find its identity. Here’s how that works.
What is the line of best fit? It’s y=mx+b, because any line (except a vertical one) is y=mx+b. We happen not to know m and b just yet, but we can use the properties of the line to find them.
What is the chief property of the line? It is that E is less for this line than for any other line that might pass through the same set of points. In other words, E(m,b) is minimized by varying m and b. Let’s look at how we can write an expression for E in terms of m and b, and of course using the measured data points (x,y).
The squared residual for any one point follows from the definition I gave earlier:
residual² = (y − ŷ)² = (y − (mx+b))²
Since (A−B)² = (B−A)², let’s reverse the subtraction to get rid of a layer of parentheses:
residual² = (y − ŷ)² = (mx + b − y)²
Now multiply:
residual² = (y − ŷ)² = m²x² + 2bmx + b² − 2mxy − 2by + y²
The sum of squared residuals for a line y=mx+b is found by summing over all points:
E(m,b) = ∑residual² = ∑(y − ŷ)²
E(m,b) = ∑(m²x² + 2bmx + b² − 2mxy − 2by + y²)
E(m,b) = m²∑x² + 2bm∑x + nb² − 2m∑xy − 2b∑y + ∑y²
Once we find the m and b that minimize E(m,b), we’ll know the exact equation of the line of best fit.
Incidentally, why is there no ∑ in the third term of the final expression for E(m,b)? Because b² in the previous line is a property of the line that we’re looking for and doesn’t vary from point to point. Adding up b² once for each of the n points gives nb².
As soon as you hear “minimize”, you think “calculus”. (Well, you do if you’ve taken calculus!) And indeed calculus can find m and b. Surprisingly, we can also find m and b using plain algebra.
It’s not entirely clear who invented the method of least squares. Most authors attach it to the name of Karl Friedrich Gauss (1777–1855), who first published on the subject in 1809.
But the Frenchman Adrien Marie Legendre (1752–1833) “published a clear explanation of the method, with a worked example, in 1805” according to Stephen Stigler in Statistics on the Table (Cambridge, Massachusetts; Harvard University Press, 1999; see Chapter 17). In setting up the new metric system of measurement, the meter was to be fixed at a ten-millionth of the distance from the North Pole through Paris to the Equator. Surveyors had measured portions of that arc, and Legendre invented the method of least squares to get the best measurement for the whole arc.
Using calculus, a function has its minimum where the derivative is 0. Since we need to adjust both m and b, we take the partial derivative of E with respect to m, and separately with respect to b, and set both to 0:
Em = 2m∑x² + 2b∑x − 2∑xy = 0 ⇒ m∑x² + b∑x = ∑xy
and
Eb = 2m∑x + 2nb −2∑y = 0 ⇒ m∑x + bn = ∑y
Each equation then gets divided by the common factor 2, and the terms not involving m or b are moved to the other side. With a little thought you can recognize the result as two simultaneous equations in m and b, namely:
(∑x²)m + (∑x)b = ∑xy and (∑x)m + nb = ∑y
The summation expressions are all just numbers, the results of summing x and y in various combinations.
These simultaneous equations can be solved like any others: by substitution or by linear combination. Let’s try substitution. The second equation looks easy to solve for b:
b = (∑y − m∑x) / n
Substitute that in the other equation and you eventually come up with
m = ( n∑xy − (∑x)(∑y) ) / ( n∑x² − (∑x)² )
b = ( (∑x²)(∑y) − (∑x)(∑xy) ) / ( n∑x² − (∑x)² )
And that is very probably what your calculator (or Excel) does: Add up all the x’s, all the x², all the xy, and so on, and compute the coefficients. It’s tedious, but not hard. (Usually these equations are presented in the shortcut form shown later.)
There is a second derivative test for two variables, but it’s more complicated than the second derivative test for one variable. You have a minimum E for particular values of m and b if three conditions are all met:
(a) The first partial derivatives Em and Eb must both be 0. That’s how we came up with m and b in the first place, so this condition is met.
(b) The determinant of the Hessian matrix must be positive. For independent variables m and b, that determinant is defined in terms of second partial derivatives as
D = Emm Ebb − (Emb)²
which evaluates to
D = (2∑x²)(2n) − (2∑x)²
D = 4n∑x² − 4(∑x)² = 4[n∑x² − (∑x)²]
The average of the x’s is x̅ = ∑x/n, so ∑x = nx̅ and
D = 4(n∑x² − n²x̅²) = 4n(∑x² − nx̅²)
Remember, we need to show that this is positive in order to be sure that our m and b minimize the sum of squared residuals E(m,b). To show that, consider the sum of the squares of deviations between each x value and the average of all x’s:
∑(x−x̅)² = ∑(x² − 2xx̅ + x̅²)
∑(x−x̅)² = ∑x² − 2x̅∑x + nx̅²
But ∑x/n = x̅, so ∑x = nx̅ and
∑(x−x̅)² = ∑x² − 2nx̅² + nx̅²
∑(x−x̅)² = ∑x² − nx̅²
Look back at D = 4n(∑x²−nx̅²). 4n is positive, since the number of points n is positive. The quantity in parentheses must be positive because it equals ∑(x−x̅)², which is a sum of squares. D is the product of two positive numbers, so D itself is positive, and this condition is met.
(c) The second partial derivative with respect to either variable must be positive. We have Ebb = 2n, which is positive, and therefore this condition is met. (It doesn’t matter which variable you use; if condition (b) is met, then either both second derivatives are positive or both are negative.)
Thus all three conditions are met, apart from pathological cases like all points having the same x value, and the m and b you get from solving the equations do minimize the total of the squared residuals, E(m,b).
But you don’t need calculus to solve every minimum or maximum problem. Look back again at the equation for E, which is the quantity we want to minimize:
E(m,b) = m²∑x² + 2bm∑x + nb² − 2m∑xy − 2b∑y + ∑y²
Now that may look intimidating, but remember that all the sigmas are just constants, formed by adding up various combinations of the (x,y) of the original points. In fact, collecting like terms reveals that E is really just a parabola with respect to m or b:
E(m) = (∑x²)m² + (2b∑x − 2∑xy)m + (nb² − 2b∑y + ∑y²)
E(b) = nb² + (2m∑x − 2∑y)b + (m²∑x² − 2m∑xy + ∑y²)
Both these parabolas are open upward. (Why? because the coefficients of the m² and b² terms are positive. The sum of x² must be positive unless all x’s are 0; and of course n, the number of points, is positive.) Since the parabolas are open upward, each one has a minimum at its vertex.
Where is the vertex for each of these parabolas? Well, recall that a parabola y=px²+qx+r has its vertex at -q/2p. These are parabolas in m and b, not in x, but you can find the vertex of each one the same way:
The vertex of E(m) is at m = ( −2b∑x + 2∑xy ) / 2∑x² = ( ∑xy − b∑x ) / ∑x²
The vertex of E(b) is at b = ( −2m∑x + 2∑y ) / 2n = ( ∑y − m∑x ) / n
Now there are two equations in m and b. Substitute one into the other one, perhaps the second into the first, and the solution is
m = ( n∑xy − (∑x)(∑y) ) / ( n∑x² − (∑x)² )
b = ( (∑x²)(∑y) − (∑x)(∑xy) ) / ( n∑x² − (∑x)² )
These are exactly the equations obtained by the calculus method.
The formula for m is bad enough, and the formula for b is a monstrosity. But if you compute m first, then it’s easier to compute b using m:
m = ( n∑xy − (∑x)(∑y) ) / ( n∑x² − (∑x)² )
b = ( ∑y − m∑x ) / n
Just to make things more concrete, here’s an example. Suppose that x is dial settings in your freezer, and y is the resulting temperature in °F. Here’s the full calculation:
| n=5 | x | y | x² | xy |
|---|---|---|---|---|
| 0 | 6 | 0 | 0 | |
| 2 | −1 | 4 | −2 | |
| 3 | −3 | 9 | −9 | |
| 5 | −10 | 25 | −50 | |
| 6 | −16 | 36 | −96 | |
| ∑ | 16 | −24 | 74 | −157 |
Now compute m and b:
5(−157) − (16)(−24)
m = -------------------
5(74) − 16²
−401
m = ----
114
m = −3.5175
−24 − (−3.5175)(16)
b = --------------------
5
32.28
b = ----- = 6.456
5“These values agree precisely with the regression equation calculated by a TI-83 for the same data,” he said smugly.
Some authors give a different form of the solutions for m and b, such as:
m = ∑(x−x̅)(y−y̅) / ∑(x−x̅)² and b = y̅ − mx̅
where x̅ and y̅ are the average of all x’s and average of all y’s.
These formulas are equivalent to the ones we derived earlier. (Can you prove that? Remember that nx̅ is ∑ x, and similarly for y.) While the m formula looks simpler, it requires you to compute mean x and mean y first. If you do that, here’s how the numbers work out:
| n=5 | x | y | x−x̅ | y−y̅ | (x−x̅)² | (x−x̅)(y−y̅) |
|---|---|---|---|---|---|---|
| 0 | 6 | −3.2 | 10.8 | 10.24 | −34.56 | |
| 2 | −1 | −1.2 | 3.8 | 1.44 | −4.56 | |
| 3 | −3 | −0.2 | 1.8 | −0.04 | −0.36 | |
| 5 | −10 | 1.8 | −5.2 | 3.24 | −9.36 | |
| 6 | −16 | 2.8 | −11.2 | 7.84 | −31.36 | |
| ∑ | 16 | −24 | 0 | 0 | 22.8 | −80.2 |
| mean | 3.2 | −4.8 |
Whew! Once you’ve got through that, m and b are only a little more work:
m = −80.2 / 22.8 = −3.5175
b = −4.8 − (−3.5175)(3.2) = 6.456
The simplicity of the alternative formulas is definitely deceptive.
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