# Elimination of features based on high covariance without affecting performance?

I ran into a question where the answer ran me into a big doubt.

Suppose we have a dataset $$A=$${$$x1,x2,y$$} in which $$x1$$ and $$x2$$ are our features and $$y$$ is the label.

Also, suppose that the covariance matrix between these three random variables are as follows:

$$\begin{array}{|c|c|c|} \hline &x1&x2&y\\ \hline x1&a&d&e\\ \hline x2&d&b&f\\\hline y&e&f&c\\\hline \end{array}$$

When $$|d| \gg 0$$, someone may ignore either of the two features, without losing any performance.

The claim was marked as a true question. Is there any idea why the mentioned proposition is True and performance was not affected?

## 2 Answers

$$|d| \gg 0$$ means there is a very strong correlation between $$x_1$$ and $$x_2$$. This means one can be expressed (almost completely) in terms of the other, thus one of two is almost redundant.

A simple example:

Consider that $$x_2$$ is simply a copy of $$x_1$$, ie $$x_2=x_1$$. Does $$x_2$$ offer any new information about the the label $$y$$ apart from the information $$x_1$$ provides? No, it is clear.

Now consider that $$x_2$$ is simply a shifted version of $$x_1$$, ie $$x_2=c_0+x_1$$. Does it now offer any new information apart from what $$x_1$$ provides? Again no, by simply changing the reference point $$c_0$$, which is equivalent to changing the baseline of the measuring system, does not by itself offer new information.

Thirdly consider that $$x_2$$ is a re-scaled version of $$x_1$$, ie $$x_2 = s \cdot x_1$$. Does it offer now different information than what $$x_1$$ provides? Again, no since scaling is equivalent to changing the measuring unit of $$x_1$$, but this is simply a re-naming of units, no new information is added.

One can see that any transformation of the form $$x_2 = c_0 + s \cdot x_1$$ does not offer any new information from what $$x_1$$ already provides, by the arguments above.

If one wants to be more realistic one can add some noise, ie $$x_2 = c_0 + s \cdot x_1 + \epsilon$$, where $$\epsilon$$ is some random variable with zero mean and very small variance.

A correlation value $$|d|$$ which is quite high actualy means that $$x_2$$ can be expressed as such a transformation of $$x_1$$, ie $$x_2 = c_0 + s \cdot x_1 + \epsilon$$ (or vice-versa) (there are some non-trivial exceptions to this). So $$x_2$$ offers no new information apart from what $$x_1$$ offers and is redundant.

So if such a high correlation exists between variables/features, one of them can be always eliminated without diminishing performance as being redundant (in the sense that, as far as second-order statistics are concerned, the two variables are identical).

This is one of the main rationales of feature selection and dimensionality reduction in machine learning.

• I agree that the question poser almost surely had that explanation in mind as the answer. But I disagree with the answer, at least the "always" qualification. Your "nontrivial exceptions" link suggests that at least sometimes the two variables contribute useful information when both included. – Ben Reiniger Jan 8 at 15:58
• @BenReiniger, "always" is qualified in the sense of use of 2nd-order statistics only. In this sense and for algorithms relying only on 2nd-order statistics, the 2 variables are indeed equivalent. – Nikos M. Jan 8 at 16:11

I primarily agree with @NikosM., but take issue with the question's assertion (my emphasis)

without losing any performance

Having very high correlation means they are very nearly in a linear relationship, but it is possible that the deviation from linearity is not random, and in fact is predictive. Here's a simple example:

import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
from sklearn.model_selection import cross_val_score, KFold

x1 = np.linspace(0, 10, num=1000)
x2 = x1 + np.sqrt(x1)
X = np.vstack([x1, x2]).T
y = np.sqrt(x1)  # so y = x2 - x1

plt.plot(x1, x2);

print(np.corrcoef(x1, x2))  # = 0.9991

lr = LinearRegression()
cv = KFold(shuffle=True, random_state=314)

print(cross_val_score(lr, X, y, cv=cv).mean())
print(cross_val_score(lr, x1.reshape(-1,1), y, cv=cv).mean())
print(cross_val_score(lr, x2.reshape(-1,1), y, cv=cv).mean())


outputs

1.0
0.9592797987922872
0.9741089428341769


and the plot of $$x_1$$ vs $$x_2$$: • In the notation in Nikos's answer, in my example we have $x_2=c_0 + s\cdot x_1+\epsilon$, but $\epsilon\approx\sqrt{x_1}$ is predictive; in this concocted example, that actually is the target. – Ben Reiniger Jan 14 at 16:38
• If they are "linearly dependent," then their correlation coefficient is $\pm1$, and you can indeed ignore one. When their corr-coef is close to but not equal to $\pm1$, as in this example, removing one degrades performance (not a lot perhaps, but clearly some). If their corr-coef is not close to $\pm1$, then we're out of scope of the question. – Ben Reiniger Jan 14 at 17:43