# Regularization for regular people

The idea of “regularization" is often spoken about in the world of Machine Learning as a way of tuning models or preventing overfitting, but the precise meaning of this is usually not spoken about, or at least not in terms that any regular person would be happy to hear.

In this blog post, we'll take a look at the meaning of regularization and work out in full detail how the two main kinds of regularization allow us to answer the question:

$\text{What is the “best" line going through a given point in the plane?}$

## The Spirit of Regularization

We live in a modern and complicated world. Things are happening around us everyday, and the more complicated the world gets the more choices we have to make. Choices are usually advertised as a good thing, but at least from time-to-time they can be overwhelming. Like when you walk into a supermarket thinking to yourself “I just want to get some bread and something to drink", but then find yourself confronted with dozens of kinds to choose from. The problem changes from finding something to drink to finding the best thing to drink, and so you spend at least 5 minutes looking at all of the choices to evaluate which of these might be the best. You might check to see that the drink isn't too expensive, doesn't have too much sugar, has enough real fruit juice to be called juice, or some not-so-bad tradeoff between these factors.

Once you get home and finally have time to relax with your drink, you might think to yourself that “Hey, I didn't really need all of that choice. What I really wanted was something real, not too expensive, and not too sweet." If someone just handed that to me then I'd be happy.

That's kind of what regularization tries to do. It hands you something “pretty good" by cutting down on the amount of freedom that you have.

## Flavors of Regularization

In the world of mathematical modelling, we usually have some data and we're trying to decide among possible ways of “best understanding" it from a certain perspective. The perspective we use to understand the data is our choice of mathematical model, and the way we have of best understanding it is from this perspective what machine learning calls the “best fit" model. It's the best choice of how to undersand the data with this model.

For purposes of this post, we'll focus on the perspective of understanding data given by points in the $$(x,y)$$-plane by trying to draw a line through them that comes as close to predicting the $$y$$-values of all of them as closely as possible. This perspective formally goes under the name of Linear Regression.

In general, once we have three points it's impossible to find a line that goes through all three of them, and the “best" line is given by minimizing the “sum of the squares error" between the $$y$$-values of the line and the points for each of their $$x$$-values. There are several well-known ways of finding this best line, but we'll settle here for a beautiful picture.

In addition to this “standard" minimization problem of finding the best line, there are two flavors of regularization that people often impose to give preference to lines whose coefficients are small. These are somewhat unintuitively called $$\text{“}L^1\text{"}$$ and $$\text{“}L^2\text{"}$$-regularization (pronounced as “L one" and “L two", or equally unintuitively called “Lasso" and “Ridge" regression). They involve imposing a certain kind of penalty for choosing lines whose coefficients are large. Mathematically this involves adding a penalty that is either the $$\text{“}L^1\text{"}$$-norm or $$\text{“}L^2\text{"}$$-norm of the coefficients, and has the general cost function formulas

$J_2(m,b) = \underbrace{\sum_{i=1}^n ((mx^{(i)} + b) - y^{(i)})^2}_{\text{Sum of squared errors} } + \underbrace{\lambda(m^2 + b^2)}_{\text{L^2 term} }$

$J_1(m,b) = \underbrace{\sum_{i=1}^n ((mx^{(i)} + b) - y^{(i)})^2}_{\text{Sum of squared errors} } + \underbrace{\lambda(|m| + |b|)}_{\text{L^1 term} }$

If this is not so enlightning, don't worry, most people would agree with you! In the next section we'll explain what this means and what it does in the simplest possible example -- the underdetermined line through one point. For now I'll just remind you of the often repeated general wisdom about these regularizations: $$\text{“}L^1\text{"}$$ is for feature selection and $$\text{“}L^2\text{"}$$ is for small coefficients.

## The “best" line through one point

Imagine that you have one point in the plane. One lonely isolated data point. Let's say it's the point $$P = (2,1)$$ for now just to be concrete (i.e. $$x=2$$ and $$y=1$$). If this is the only data point we have, then it's not very hard to find a line (of the form $$y = mx + b$$ for some numbers $$m$$ and $$b$$ going through this point. In fact there are lots of lines through the point $$P -$$ infinitely many of them $$-$$ and we can see this from playing with the following picture:

In the language of mathematical modelling, we would say that the problem of finding a line through the point $$P$$ is overfit (since it passes exactly through all data points) and underdetermined (since there is not a unique line that does the job). Because there are many lines that pass through $$P$$, and they all seem to do the job, it's not possible to choose which among them is the “best" without imposing additional conditions to help say what we mean by “best". Here is where the various flavors of regularization come in to help us to make that choice.

It's actually very helpful to be able to somehow see all lines in the plane at once, and then determine which of these lines in the plane passes through the given point $$P$$. One great way to do this is to think about lines in the plane as all being of the form $$y=mx+b$$ for some numbers $$m$$ and $$b$$ (i.e. the slope and intercept of the line) and imagine a separate $$(m,b)$$-plane. This is the “parameter space" of lines in the plane, and here any point $$(m,b)$$ in the $$(m,b)$$-plane corresponds to one line (i.e. $$y=mx+b$$) in the usual $$(x,y)$$-plane. It's kind of fun to be able to talk about all lines in the plane at once, but this “parameter space" language also allows us to write down explicitly which lines in the plane pass through the given point $$P$$. It's the points in the $$(m,b)$$-plane where the associated line $$y = mx + b$$ contains the given point $$P = (x,y) = (2,1)$$, or equivalently, the points in the $$(m,b)$$-plane satisfying the equation $$1 = m \cdot 2 + b$$.

This means that the lines passing through the point $$P = (2,1)$$ are described in the $$(m,b)$$ “parameter space" by the points satisfying the equation $$2m + b = 1$$. Since there are infinitely many points on the line $$2m+b=1$$, we again see that there are infinitely many lines in the $$(x,y)$$-plane passing through the point $$P = (2,1)$$.

With this picture in hand, we're ready to see the effect of various flavors of regualrization on determining their own version of a “best" line.

## The $$L^2$$-best line

From the perspective of $$L^2$$-regularization, the best line is the one that minimizes the $$L^2$$ cost function

$J_2(m,b) = \underbrace{\sum_{i=1}^n ((mx^{(i)} + b) - y^{(i)})^2}_{\text{Sum of squared errors} } + \underbrace{\lambda(m^2 + b^2)}_{\text{L^2 term} }$

This formula might feel a little unwieldy, but in our case there is just one data point, and since any line we consider passes through the data poing the $$y$$-value of the line and the point agree (at the $$x$$-value of the point). This means that the sum of squared errors is zero, so we're really minimizing the function $$J_2(m,b) = \lambda(m^2 + b^2)$$. Here no matter what the positive constant $$\lambda$$ happens to be, the function will be minimized at the same point. Therefore we can set $$\lambda = 1$$ and we're reduced to trying to find the line in the $$(x,y)$$-plane through the point $$P=(2,1)$$ where the function $$J_2(m,b) = m^2 + b^2$$ is the smallest.

Now here the cool part is that we know what the level sets of this function $$J_2$$ looks like in the $$(m,b)$$-plane. They are given by circles centered at the origin, and the value of $$J_2$$ grows as the radius of the circle grows. Therefore to find the “$$L^2$$-best" line through $$P=(2,1)$$ we need to find the smallest radius circle centered at the origin that touches the line $$2m + b = 1$$ in the $$(m,b)$$-plane.

By drawing this, we can see that the circle tangent to the line $$2m + b = 1$$ at this point, and so this radius of the circle is perpendicular to the line. Therefore this intersection point lies on the line with inverse-reciprocal slope $$(\frac{-1}{\text{slope}} = \frac{-1}{-2} = \frac{1}{2})$$ and this line goes through the origin, so the intersection point lies on the line $$b = \frac{m}{2}$$. Solving both equations gives that the intersection point is $$(m,b) = (\frac{2}{5}, \frac{1}{5})$$, so the $$L^2$$-best line through the point $$P = (2,1)$$ is the line $$y = \frac{2}{5}x + \frac{1}{5}$$.

Notice that this is the line through $$P$$ whose sum of squared of coefficients is smallest, or equivalently the line $$y = mx+b$$ whose coefficient vector $$(m,b)$$ is the shortest among all possible lines through $$P$$.

## The $$L^1$$-best line

From an $$L^1$$-regularization perspective, the best line is the one that minimizes the L^1 cost function

$J_1(m,b) = \underbrace{\sum_{i=1}^n ((mx^{(i)} + b) - y^{(i)})^2}_{\text{Sum of squared errors} } + \underbrace{\lambda(|m| + |b|)}_{\text{L^1 term} }$

For the same reasons as in the $$L^2$$ case (i.e. the data is overfit, so we can take $$\lambda=1$$) the sum of squared errors term is zero and we are left with minimizing the function $$J_1(m,b) = |m| + |b|$$. The level sets of this function are square diamonds centered at the origin, and (as in the $$L^2$$ case) the size of $$J_1(m,b)$$ grows with the width of the diamond. Therefore the minimum value of $$J_1$$ for lines passing through the point $$P = (2,1)$$ occurs where the smallest diamond touches the line $$2m + b = 1$$ in the $$(m,b)$$-plane.

From the picture we see that this happens when $$(m,b) = (\frac{1}{2}, 0)$$, so the $$L^1$$-best line through the point $$P=(2,1)$$ is the line $$y = \frac{1}{2}x$$.

We notice that this has selected the feature of the slope $$m$$, and has selected against the $$y$$-intercept $$b$$ (i.e. it preferred to have $$m \neq 0$$ and $$b=0$$). This gives an example of why it is said that $$L^1$$-regularization performs feature selection.

## Conclusion

Finding the “best" line through a point is just one example of where regularization helps to choose a model, but this example should help to justify the general intuition about $$L^2$$- and $$L^1$$-regularization. In these cases we see that $$L^2$$-regularization tries to make the coefficient vector small, while $$L^1$$-regularization helps to make the coefficients small but also has a strong preference for some coefficients to be zero.

While in general it's not as easy to visually see what's going on in a regularization, one can imagine that it's something very similar to what we can see in this example. Hopefully this kind of intuition is helpful to the reader when deciding what kinds of regularization(s) to use when approaching a model, and particularly a model that is overfit to some given data. Thanks for reading!

## Exercises

For the reader looking to try their hand at working out some examples on their own, I'd recommend the following exercises:

• Give the $$L^2$$ and $$L^1$$ regularized lines through the point $$(1,3)$$.

• Give the $$L^2$$ and $$L^1$$ regularized lines through the point $$(1,1)$$ -- noting that there may not be a unique choice!

• Give the $$L^2$$ and $$L^1$$ regularized lines through the point $$(0,0)$$.

• For which points $$P$$ in the $$(x,y)$$-plane does $$L^1$$ regularization select for the slope $$m$$? for the slope $$b$$? for neither/both?

• For which points $$P$$ is the best $$L^1$$ regularized line through $$P$$ not unique? What conditions does $$L^1$$-regularization impose at these points?

Written on October 25, 2015