all redundant features goes to one. We assume there are $q \;(1\le q < d)$ non-redundant features denoted as $X[j_1], X[j_2], \cdots X[j_q]$. The redundant features are denoted as $X[j_{q+1]}, X[j_{q+2}], \cdots X[j_d]$.

Before stating these two conditions, we need to discuss how a partition can decrease the tree impurity. Suppose we are partitioning on node $A$ at feature $X[j]$, resulting in two new leaf nodes: $A_1 = \{X\in A: X[j] \le x_j\}$ and $A_2 = \{X\in A: X[j] > x_j\}$. Then the weighted conditional expectation of $Y$ in node $A_1, A_2$ is $$\label{p1alpha fs}
p_{h, \alpha} = \frac{\alpha \mathbb{E}(Y|X\in A_h)}{1-\mathbb{E}(Y|X\in A_h) + \alpha\mathbb{E}(Y|X\in A_h)},\quad h=1,2.$$ Thus the impurity of $A_1, A_2$ is $I_\alpha(A_1, \mathbb{P}) = 1 - p_{1, \alpha}^2 - (1-p_{1, \alpha})^2 = 2p_{1, \alpha}(1-p_{1,\alpha}),$ and $I_\alpha(A_2, \mathbb{P}) = 2p_{2,\alpha}(1-p_{2,\alpha}).$ Similarly, denoting the weighted conditional expectation of $Y$ in node $A$ as $p_\alpha$, the impurity of $A$ is $I_\alpha(A,\mathbb{P}) = 2p_\alpha(1-p_\alpha)$. The impurity decrease on node $A$ is $$\label{delta Ialpha}
\Delta I_\alpha(A, \mathbb{P}) = 2p_\alpha(1-p_\alpha) - 2p_{1,\alpha}(1-p_{1,\alpha})\mathbb{P}_\alpha(A_1|A) - 2p_{2,\alpha}(1-p_{2,\alpha})\mathbb{P}_\alpha(A_2|A),$$ Noting $A_1, A_2$ is determined by $x_j$, $\Delta I_\alpha(A, \mathbb{P})$ can be considered as a function of $x_j$. Denoting $\Delta I_\alpha(A,\mathbb{P}) = h_{A,j}(x_j)$, then the maximal impurity decrease at feature 