Correlation and variable importance in random forests

Abstract

This paper is about variable selection with the random forests algorithm in presence of correlated predictors. In high-dimensional regression or classification frameworks, variable selection is a difficult task, that becomes even more challenging in the presence of highly correlated predictors. Firstly we provide a theoretical study of the permutation importance measure for an additive regression model. This allows us to describe how the correlation between predictors impacts the permutation importance. Our results motivate the use of the recursive feature elimination (RFE) algorithm for variable selection in this context. This algorithm recursively eliminates the variables using permutation importance measure as a ranking criterion. Next various simulation experiments illustrate the efficiency of the RFE algorithm for selecting a small number of variables together with a good prediction error. Finally, this selection algorithm is tested on the Landsat Satellite data from the UCI Machine Learning Repository.

Appendix: Proofs

1.1 Proof of Proposition 1

The random variable $X_j^{\prime }$ and the vector ${\mathbf {X}}_{(j)}$ are defined as in Section 2:

$$\begin{aligned} I(X_j)&{=} \mathbb {E}\left[ (Y {-} f({\mathbf {X}}) {+} f({\mathbf {X}}) {-} f({\mathbf {X}}_{(j)}))^2\right] {-} \mathbb {E}\left[ (Y{-}f({\mathbf {X}}))^2\right] \\&= \mathbb {E}[(f({\mathbf {X}}) - f({\mathbf {X}}_{(j)}))^2] + 2 \mathbb {E}\left[ \varepsilon (f({\mathbf {X}}) {-} f({\mathbf {X}}_{(j)})) \right] \\&= \mathbb {E}[(f({\mathbf {X}}) - f({\mathbf {X}}_{(j)}))^2], \end{aligned}$$

since $\mathbb {E}[ \varepsilon f({\mathbf {X}}) ] = \mathbb {E}[ f({\mathbf {X}}) \mathbb {E}[ \varepsilon | {\mathbf {X}} ] ] = 0$ and $\mathbb {E}[ \varepsilon f({\mathbf {X}}_{(j}) ] = \mathbb {E}( \varepsilon ) \mathbb {E}[ f({\mathbf {X}}_{(j}) ] = 0$. Since the model is additive, we have:

$$\begin{aligned} I(X_j)&= \mathbb {E}[(f_j(X_j) - f_j(X_j^{\prime }))^2]\\&= 2\mathbb {V}[f_j(X_j)], \end{aligned}$$

as $X_j$ and $X_j^{\prime }$ are independent and identically distributed. For the second statement of the proposition, using the fact that $f_j(X_j)$ is centered we have:

$$\begin{aligned} \mathbb {C}[Y, f_j(X_j)]&= \mathbb {E}\left[ f_j(X_j) \mathbb {E}[Y| {\mathbf {X}}] \right] \\&= \mathbb {E}[f_j(X_j) \sum _{k=1}^p f_k(X_k)]\\&= \mathbb {V}[f_j(X_j)] + \sum _{k\ne j} \mathbb {E}\left[ f_j(X_j) f_k(X_k) \right] \\&= \frac{I (X_j)}{2} + \sum _{k\ne j} \mathbb {C}\left[ f_j(X_j), f_k(X_k) \right] . \end{aligned}$$

1.2 Proof of Proposition 2

This proposition is an application of Proposition 1 for a particular distribution. We only show that $\alpha = C^{-1} \varvec{\tau }$ in that case.

Since $({\mathbf {X}}, Y)$ is a normal multivariate vector, the conditional distribution of Y over ${\mathbf {X}}$ is also normal and the conditional mean $f({\mathbf {x}}) = \mathbb {E}[Y|{\mathbf {X}}={\mathbf {x}}]$ is a linear function: $f({\mathbf {x}}) = \sum _{j=1}^p \alpha _j x_j$ (see for instance Rao 1973, p. 522). Then, for any $j \in \{1, \dots , p \}$,

$$\begin{aligned} \tau _j&= \mathbb {E}[X_jY] \\&= \mathbb {E}[\; X_j \mathbb {E}[Y | {\mathbf {X}}] \;] \\&= \alpha _1 \mathbb {E}[X_1X_j] + \cdots + \alpha _j \mathbb {E}[X_j^2] + \cdots + \alpha _p \mathbb {E}[X_p X_j] \\&= \alpha _1 c_{1j} + \cdots + \alpha _j c_{jj}+ \cdots + \alpha _p c_{pj}. \end{aligned}$$

The vector $\alpha$ is thus solution of the equation ${\varvec{\tau }} = C \alpha$ and the expected result is proven since the covariance matrix C is invertible.

1.3 Proof of Proposition 3

The correlation matrix C is assumed to have the form $C = (1-c) I_p + c \mathbbm {1}\mathbbm {1}^t$. We show that the invert of C can be decomposed in the same way. Let $M = a I_p + b \mathbbm {1}\mathbbm {1}^t$ where a and b are real numbers to be chosen later. Then

$$\begin{aligned} C M= & {} \big ( (1-c)I_p + c \mathbbm {1}\mathbbm {1}^t \big ) \big ( a I_p + b \mathbbm {1}\mathbbm {1}^t \big ) \\= & {} a (1-c) I_p + b (1-c) \mathbbm {1}\mathbbm {1}^t + a c \mathbbm {1}\mathbbm {1}^t + b c \mathbbm {1}\mathbbm {1}^t \mathbbm {1}\mathbbm {1}^t \\= & {} a (1-c) I_p + (b (1-c) + ac + pbc )\mathbbm {1}\mathbbm {1}^t, \end{aligned}$$

since $\mathbbm {1}^t \mathbbm {1}= p$. Thus, $C M = I_d$ if and only if

$$\begin{aligned} \left\{ \begin{array}{l} a (1-c) = 1 \\ b (1-c) + ac + pbc = 0, \end{array} \right. \end{aligned}$$

which is equivalent to

$$\begin{aligned} \left\{ \begin{array}{l} a = \dfrac{1}{(1-c)} \\ b = \dfrac{- c}{(1-c)(1-c+pc)}. \end{array} \right. \end{aligned}$$

Consequently, $M^{-1}_{jk} = C^{-1}_{jk} = b$ if $j \ne k$ and $M^{-1}_{jk} = C^{-1}_{jj} = a+b$. Finally we find that for any $j \in \{1\dots p\}$:

$$\begin{aligned}{}[C^{-1} {\varvec{\tau }}]_j= & {} \tau _0 (a+b) + \tau _0 b (p-1) \\= & {} \tau _0 ( a + pb ) \\= & {} \tau _0 \bigg ( \dfrac{1}{(1-c)} - \dfrac{pc}{(1-c)(1-c+pc)} \bigg ) \\= & {} \dfrac{\tau _0}{1 - c + pc}. \end{aligned}$$

The second point derives from Proposition 2.