# square things in statistics- generalized rationale [duplicate]

Why do you square things in stats? I have run across this a lot, in both data mining and statistics classes, but no one has ever been able to give me an answer. One specific example is when summing the deviation scores in statistics you have to square them (otherwise the sum is 0). Why do you square them rather then using something else, like absolute value.

Difference between prior question: If you have an answer for the problem above, does your answer apply to most statistics stuff that does this? If not, why not.

• see also this for some additional insight – Glen_b Dec 11 '14 at 2:18
• (In general, I don't think this thread should be reopened. However, I'm halfway inclined to try to have it reopened just so I could award a bounty to @BenKuhn's answer below, & then re-close it.) – gung - Reinstate Monica Nov 21 '16 at 21:16
• Yes please- that was an incredibly detailed answer! I would love to reward him for that one. – Rilcon42 Nov 22 '16 at 0:37

## 4 Answers

$$\newcommand{\predicted}{{\rm predicted}}\newcommand{\actual}{{\rm actual}}\newcommand{\Var}{{\rm Var}}$$ You're right that one could instead choose to use the absolute error--in fact, the absolute error is often closer to what you "care about" when making predictions from your model. For instance, if you buy a stock expecting its future price to be $$P_{\predicted}$$ and its future price is $$P_{\actual}$$ instead, you lose money proportional to $$(P_{\predicted} - P_{\actual})$$, not its square! The same is true in many other contexts.

# So why squared error?

The squared error has many nice mathematical properties. Echoing the other answerers here, I would say that many of them are merely "convenient"--we might choose to use the absolute error instead if it didn't pose technical issues when solving problems. For instance:

• If $$X$$ is a random variable, then the estimator of $$X$$ that minimizes the squared error is the mean, $$E(X)$$. On the other hand, the estimator that minimizes the absolute error is the median, $$m(X)$$. The mean has much nicer properties than the median; for instance, $$E(X + Y) = E(X) + E(Y)$$, but there is no general expression for $$m(X + Y)$$.

• If you have a vector $$\vec X = (X_1, X_2)$$ estimated by $$\vec x = x_1, x_2$$, then for the squared error it doesn't matter whether you consider the components separately or together: $$||\vec X - \vec x||^2 = (X_1 - x_1)^2 + (X_2 - x_2)^2$$, so the squared error of the components just adds. You can't do that with absolute error. This means that the squared error is independent of re-parameterizations: for instance, if you define $$\vec Y_1 = (X_1 + X_2, X_1 - X_2)$$, then the minimum-squared-deviance estimators for $$Y$$ and $$X$$ are the same, but the minimum-absolute-deviance estimators are not.

• For independent random variables, variances (expected squared errors) add: $$\Var(X + Y) = \Var(X) + \Var(Y)$$. The same is not true for expected absolute error.

• For a sample from a multivariate Gaussian distribution (where probability density is exponential in the squared distance from the mean), all of its coordinates are Gaussian, no matter what coordinate system you use. For a multivariate Laplace distribution (like a Gaussian but with absolute, not squared, distance), this isn't true.

• The squared error of a probabilistic classifier is a proper scoring rule. If you had an oracle telling you the actual probability of each class for each item, and you were being scored based on your Brier score, your best bet would be to predict what the oracle told you for each class. This is not true for absolute error. (For instance, if the oracle tells you that $$P(Y=1) = 0.9$$, then predicting that $$P(Y=1) = 0.9$$ yields an expected score of $$0.9\cdot 0.1 + 0.1 \cdot 0.9 = 0.18$$; you should instead predict that $$P(Y=1) = 1$$, for an expected score of $$0.9\cdot 0 + 0.1 \cdot 1 = 0.1$$.)

Some mathematical coincidences or conveniences involving the squared error are more important, though. They don't pose technical problem-solving issues; rather, they give us intrinsic reasons why minimizing the square error might be a good idea:

• When fitting a Gaussian distribution to a set of data, the maximum-likelihood fit is that which minimizes the squared error, not the absolute error.

• When doing dimensionality reduction, finding the basis that minimizes the squared reconstruction error yields principal component analysis, which is nice to compute, coordinate-independent, and has a natural interpretation for multivariate Gaussian distributions (finding the axes of the ellipse that the distribution makes). There's a variant called "robust PCA" that is sometimes applied to minimizing absolute reconstruction error, but it seems to be less well-studied and harder to understand and compute.

# Looking deeper

One might well ask whether there is some deep mathematical truth underlying the many different conveniences of the squared error. As far as I know, there are a few (which are related in some sense, but not, I would say, the same):

## Differentiability

The squared error is everywhere differentiable, while the absolute error is not (its derivative is undefined at 0). This makes the squared error more amenable to the techniques of mathematical optimization. To optimize the squared error, you can just set its derivative equal to 0 and solve; to optimize the absolute error often requires more complex techniques.

## Inner products

The squared error is induced by an inner product on the underlying space. An inner product is basically a way of "projecting vector $$x$$ along vector $$y$$," or figuring out "how much does $$x$$ point in the same direction as $$y$$." In finite dimensions this is the standard (Euclidean) inner product $$\langle a, b\rangle = \sum_i a_ib_i$$. Inner products are what allow us to think geometrically about a space, because they give a notion of:

• a right angle ($$x$$ and $$y$$ are right angles if $$\langle x, y\rangle = 0$$);
• and a length (the length of $$x$$ is $$||x|| = \sqrt{\langle x, x\rangle}$$).

By "the squared error is induced by the Euclidean inner product" I mean that the squared error between $$x$$ and $$y$$ is $$||x-y||$$, the Euclidean distance between them. In fact the Euclidean inner product is in some sense the "only possible" axis-independent inner product in a finite-dimensional vector space, which means that the squared error has uniquely nice geometric properties.

For random variables, in fact, you can define is a similar inner product: $$\langle X, Y\rangle = E(XY)$$. This means that we can think of a "geometry" of random variables, in which two variables make a "right angle" if $$E(XY) = 0$$. Not coincidentally, the "length" of $$X$$ is $$E(X^2)$$, which is related to its variance. In fact, in this framework, "independent variances add" is just a consequence of the Pythagorean Theorem:

\begin{align} \Var(X + Y) &= ||(X - \mu_X)\, + (Y - \mu_Y)||^2 \\ &= ||X - \mu_X||^2 + ||Y - \mu_Y||^2 \\ &= \Var(X)\quad\ \ \, + \Var(Y). \end{align}

# Beyond squared error

Given these nice mathematical properties, would we ever not want to use squared error? Well, as I mentioned at the very beginning, sometimes absolute error is closer to what we "care about" in practice. For instance, if your data has tails that are fatter than Gaussian, then minimizing the squared error can place too much weight on outlying points.

The absolute error is less sensitive to such outliers. (For instance, if you observe an outlier in your sample, it changes the squared-error-minimizing mean proportionally to the magnitude of the outlier, but hardly changes the absolute-error-minimizing median at all!) And although the absolute error doesn't enjoy the same nice mathematical properties as the squared error, that just means absolute-error problems are harder to solve, not that they're objectively worse in some sense. The upshot is that as computational methods have advanced, we've become able to solve absolute-error problems numerically, leading to the rise of the subfield of robust statistical methods.

In fact, there's a fairly nice correspondence between some squared-error and absolute-error methods:

Squared error           | Absolute error
========================|============================
Mean                    | Median
Variance                | Expected absolute deviation
Gaussian distribution   | Laplace distribution
Linear regression       | Quantile regression
PCA                     | Robust PCA
Ridge regression        | LASSO


As we get better at modern numerical methods, no doubt we'll find other useful absolute-error-based techniques, and the gap between squared-error and absolute-error methods will narrow. But because of the connection between the squared error and the Gaussian distribution, I don't think it will ever go away entirely.

It's because of the close connection between many statistical methods and geometric concepts such as projections, distances, and the Pythagorean Theorem. For example, suppose that you view the data values $(x_1,x_2,\ldots,x_n)$ as a point in $n$-dimensional space. Then the sample SD is $1/\sqrt {n-1}$ times the distance between this point and the point of means $(\bar x,\bar x,\ldots,\bar x)$. And the sums of squares in one-way anova really do satisfy the Pythagorean Theorem, framed in a similar way.

Because it makes the math easier. One can use other techniques for example for linear regression. Thes other methods tend to be more complicated in implementation details and have less elegant closed form solutions. Thus they are often ignored until a project demands they be used.

Honestly, it's because it makes the math easier than if absolute value were used. Laplace in fact tried to use absolute value instead of squared differences. It makes things quite annoying. Here's a link to a description of the Laplace distrubtion http://en.wikipedia.org/wiki/Laplace_distribution. Before computers using absolute value instead of squared differences made life hard for the statistician.