# Perfect multicollinearity with a cubic term in the model?

I'm trying to figure out why adding a cubic term in the model doesn't guarantee a perfect multicollinearity. If $$X$$ is known, then $$X^3$$ is known in both magnitude and sign and vice versa. It may not be the case between $$X$$ and $$X^2$$ in terms of sign.

• Multicollinearity usually refers to a linear relation between two variables. – nope Oct 11 '19 at 10:44
• A standard tool in mathematics is based on such considerations: if a nontrivial linear relation $c_0+c_1X+c_2X^2+c_3X^3=0$ holds, that means every component of $X$ is a root of the polynomial $c_0+c_1x+c_2x^2+c_3x^3,$ whence (by the Fundamental Theorem of Algebra) there are at most three distinct possible values for the components of $X.$ Since that's not generally true--many datasets have many more distinct values of their variables than that--it cannot be generally true that $1,X,X^2,X^3$ are collinear. This idea appears in my analysis at stats.stackexchange.com/a/408855/919, e.g. – whuber Oct 11 '19 at 14:50

Multicollinearity refers to the situation in which the regressor matrix $$Z$$ does not have full column rank $$k$$.

This is the case if it is possible to linearly combine the columns $$z_1,\ldots,z_k$$ into the zero vector with a vector $$a=(a_1,\ldots,a_k)'$$ other than the trivial zero vector $$0$$, i.e., $$a_1z_1+\ldots+a_kz_k=0$$ for $$a\neq0$$. If, say, $$z_1\equiv X=(-1,0,1,2)'$$, then $$z_2\equiv X^3=(-1,0,1,8)'$$. You will not find values $$a_1,a_2$$ other than zeros that produce $$a_1\begin{pmatrix}-1\\0\\1\\2\end{pmatrix}+a_2\begin{pmatrix}-1\\0\\1\\8\end{pmatrix}=0.$$ If $$z_2$$ were some multiple or fraction of $$z_1$$, it would be possible, so that we would have multicollinearity.

As an aside, if your regressor $$X$$ is a dummy variable, we do have multicollinearity with powers of $$X$$, as powers of $$0$$ and $$1$$ are of course also $$0$$ and $$1$$.

Try, e.g.,

X <- -1:2
lm(rnorm(4)~X+I(X^3)-1)

X <- sample(c(0,1),10, replace = T)
lm(rnorm(10)~X+I(X^3)-1)


You often will get multicollinnearity issue with cubes but not the perfect kind. In your case a perfect multicollinnearity can be defined as: $$\alpha x+x^3=c$$. This is not true by definition. However, when $$x<<1$$ you get $$x-x^3\approx 0$$ because $$x^3\approx x$$. Therefore, sometimes you may get perfect multicollinearity warning or design matrix condition number too big warning due to rounding, but this will not happen with every data set.