ABESS Algorithm: Details¶

Introduction¶

With the abess library, users can use the ABESS algorithm to efficiently solve many best subset selection problems. The aim of this page is providing a complete and coherent documentation for ABESS algorithm under linear model such that users can easily understand the ABESS algorithm, thereby facilitating the usage of abess software.

linear regression¶

Sacrifices¶

Consider the $$\ell_{0}$$ constraint minimization problem,

$\min _{\boldsymbol{\beta}} \mathcal{L}_{n}(\beta), \quad \text { s.t }\|\boldsymbol{\beta}\|_{0} \leq \mathrm{s},$

where $$\mathcal{L}_{n}(\boldsymbol \beta)=\frac{1}{2 n}\|y-X \beta\|_{2}^{2} .$$ Without loss of generality, we consider $$\|\boldsymbol{\beta}\|_{0}=\mathrm{s}$$. Given any initial set $$\mathcal{A} \subset \mathcal{S}=\{1,2, \ldots, p\}$$ with cardinality $$|\mathcal{A}|=s$$, denote $$\mathcal{I}=\mathcal{A}^{\mathrm{c}}$$ and compute:

$\hat{\boldsymbol{\beta}}=\arg \min _{\boldsymbol{\beta}_{\mathcal{I}}=0} \mathcal{L}_{n}(\boldsymbol{\beta}).$

We call $$\mathcal{A}$$ and $$\mathcal{I}$$ as the active set and the inactive set, respectively.

Given the active set $$\mathcal{A}$$ and $$\hat{\boldsymbol{\beta}}$$, we can define the following two types of sacrifices:

1. Backward sacrifice: For any $$j \in \mathcal{A}$$, the magnitude of discarding variable $$j$$ is,

$\xi_{j}=\mathcal{L}_{n}\left(\hat{\boldsymbol{\beta}}^{\mathcal{A} \backslash\{j\}}\right)-\mathcal{L}_{n}\left(\hat{\boldsymbol{\beta}}^{\mathcal{A}}\right)=\frac{X_{j}^{\top} X_{j}}{2 n}\left(\hat{\boldsymbol\beta}_{j}\right)^{2},$

2. Forward sacrifice: For any $$j \in \mathcal{I}$$, the magnitude of adding variable $$j$$ is,

$\zeta_{j}=\mathcal{L}_{n}\left(\hat{\boldsymbol{\beta}^{\mathcal{A}}}\right)-\mathcal{L}_{n}\left(\hat{\boldsymbol{\beta}}^{\mathcal{A}}+\hat{t}^{\{j\}}\right)=\frac{X_{j}^{\top} X_{j}}{2 n}\left(\frac{\hat{\boldsymbol d}_{j}}{X_{j}^{\top} X_{j} / n}\right)^{2}.$
where $$\hat{t}=\arg \min _{t} \mathcal{L}_{n}\left(\hat{\boldsymbol{\beta}}^{\mathcal{A}}+t^{\{j\}}\right), \hat{\boldsymbol d}_{j}=X_{j}^{\top}(y-X \hat{\boldsymbol{\beta}}) / n$$. Intuitively, for $$j \in \mathcal{A}$$ (or $$j \in \mathcal{I}$$ ), a large $$\xi_{j}$$ (or $$\zeta_{j}$$) implies the $$j$$ th variable is potentially important.

Algorithm¶

Best-Subset Selection with a Given Support Size¶

Unfortunately, it is noteworthy that these two sacrifices are incomparable because they have different sizes of support set. However, if we exchange some "irrelevant" variables in $$\mathcal{A}$$ and some "important" variables in $$\mathcal{I}$$, it may result in a higher-quality solution. This intuition motivates our splicing method. Specifically, given any splicing size $$k \leq s$$, define

$\mathcal{A}_{k}=\left\{j \in \mathcal{A}: \sum_{i \in \mathcal{A}} \mathrm{I}\left(\xi_{j} \geq \xi_{i}\right) \leq k\right\},$

to represent $$k$$ least relevant variables in $$\mathcal{A}$$ and,

$\mathcal{I}_{k}=\left\{j \in \mathcal{I}: \sum_{i \in \mathcal{I}} \mid\left(\zeta_{j} \leq \zeta_{i}\right) \leq k\right\},$

to represent $$k$$ most relevant variables in $$\mathcal{I} .$$

Then, we splice $$\mathcal{A}$$ and $$\mathcal{I}$$ by exchanging $$\mathcal{A}_{k}$$ and $$\mathcal{I}_{k}$$ and obtain a new active set:$$\tilde{\mathcal{A}}=\left(\mathcal{A} \backslash \mathcal{A}_{k}\right) \cup \mathcal{I}_{k}.$$ Let $$\tilde{\mathcal{I}}=\tilde{\mathcal{A}}^{c}, \tilde{\boldsymbol{\beta}}=\arg \min _{\boldsymbol{\beta}_{\overline{\mathcal{I}}=0}} \mathcal{L}_{n}(\boldsymbol{\beta})$$, and $$\tau_{s}>0$$ be a threshold. If $$\tau_{s}<\mathcal{L}_{n}(\hat{\boldsymbol\beta})-\mathcal{L}_{n}(\tilde{\boldsymbol\beta})$$, then $$\tilde{A}$$ is preferable to $$\mathcal{A} .$$
The active set can be updated iteratively until the loss function cannot be improved by splicing. Once the algorithm recovers the true active set, we may splice some irrelevant variables, and then the loss function may decrease slightly. The threshold $$\tau_{s}$$ can reduce this unnecessary calculation. Typically, $$\tau_{s}$$ is relatively small, e.g. $$\tau_{s}=0.01 s \log (p) \log (\log n) / n.$$
Algorithm 1: BESS.Fix(s): Best-Subset Selection with a given support size $$s$$.¶
1. Input: $$X, y$$, a positive integer $$k_{\max }$$, and a threshold $$\tau_{s}$$.

2. Initialize:

$\mathcal{A}^{0}=\left\{j: \sum_{i=1}^{p} \mathrm{I}\left(\left|\frac{X_{j}^{\top} y}{\sqrt{X_{j}^{\top} X_{j}}}\right| \leq \left| \frac{X_{i}^{\top} y}{\sqrt{X_{i}^{\top} X_{i}}}\right| \leq \mathrm{s}\right\}, \mathcal{I}^{0}=\left(\mathcal{A}^{0}\right)^{c}\right.$

and $$\left(\boldsymbol\beta^{0}, d^{0}\right):$$

$\begin{split}&\boldsymbol{\beta}_{\mathcal{I}^{0}}^{0}=0,\\ &d_{\mathcal{A}^{0}}^{0}=0,\\ &\boldsymbol{\beta}_{\mathcal{A}^{0}}^{0}=\left(\boldsymbol{X}_{\mathcal{A}^{0}}^{\top} \boldsymbol{X}_{\mathcal{A}^{0}}\right)^{-1} \boldsymbol{X}_{\mathcal{A}^{0}}^{\top} \boldsymbol{y},\\ &d_{\mathcal{I}^{0}}^{0}=X_{\mathcal{I}^{0}}^{\top}\left(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta}^{0}\right).\end{split}$
1. For $$m=0,1, \ldots$$, do

$\left(\boldsymbol{\beta}^{m+1}, \boldsymbol{d}^{m+1}, \mathcal{A}^{m+1}, \mathcal{I}^{m+1}\right)= \text{Splicing} \left(\boldsymbol{\beta}^{m}, \boldsymbol{d}^{m}, \mathcal{A}^{m}, \mathcal{I}^{m}, k_{\max }, \tau_{s}\right).$

If $$\left(\mathcal{A}^{m+1}, \mathcal{I}^{m+1}\right)=\left(\mathcal{A}^{m},\mathcal{I}^{m}\right)$$, then stop.

End For

2. Output $$(\hat{\boldsymbol{\beta}}, \hat{\boldsymbol{d}}, \hat{\mathcal{A}}, \hat{\mathcal{I}})=\left(\boldsymbol{\beta}^{m+1}, \boldsymbol{d}^{m+1} \mathcal{A}^{m+1}, \mathcal{I}^{m+1}\right).$$

Algorithm 2: Splicing $$\left(\boldsymbol\beta, d, \mathcal{A}, \mathcal{I}, k_{\max }, \tau_{s}\right)$$¶
1. Input: $$\boldsymbol{\beta}, \boldsymbol{d}, \mathcal{A}, \mathcal{I}, k_{\max }$$, and $$\tau_{\mathrm{s}} .$$

2. Initialize: $$L_{0}=L=\frac{1}{2 n}\|y-X \beta\|_{2}^{2}$$, and set

$\xi_{j}=\frac{X_{j}^{\top} X_{j}}{2 n}\left(\beta_{j}\right)^{2}, \zeta_{j}=\frac{X_{j}^{\top} X_{j}}{2 n}\left(\frac{d_{j}}{X_{j}^{\top} X_{j} / n}\right)^{2}, j=1, \ldots, p.$
3. For $$k=1,2, \ldots, k_{\max }$$, do

$\begin{split}\mathcal{A}_{k}=\left\{j \in \mathcal{A}: \sum_{i \in \mathcal{A}} \mathrm{I}\left(\xi_{j} \geq \xi_{i}\right) \leq k\right\},\\ \mathcal{I}_{k}=\left\{j \in \mathcal{I}: \sum_{i \in \mathcal{I}} \mathrm{I}\left(\zeta_{j} \leq \zeta_{i}\right) \leq k\right\}.\end{split}$

Let $$\tilde{\mathcal{A}}_{k}=\left(\mathcal{A} \backslash \mathcal{A}_{k}\right) \cup \mathcal{I}_{k}, \tilde{\mathcal{I}}_{k}=\left(\mathcal{I} \backslash \mathcal{I}_{k}\right) \cup \mathcal{A}_{k}$$ and solve:

$\begin{split}\tilde{\boldsymbol{\beta}}_{{\mathcal{A}}_{k}}=\left(\boldsymbol{X}_{\mathcal{A}_{k}}^{\top} \boldsymbol{X}_{{\mathcal{A}}_{k}}\right)^{-1} \boldsymbol{X}_{{\mathcal{A}_{k}}}^{\top} y, \quad \tilde{\boldsymbol{\beta}}_{{\mathcal{I}}_{k}}=0\\ \tilde{\boldsymbol d}_{\mathcal{I}^k}=X_{\mathcal{I}^k}^{\top}(y-X \tilde{\beta}) / n,\quad \tilde{\boldsymbol d}_{\mathcal{A}^k} = 0.\end{split}$

Compute: $$\mathcal{L}_{n}(\tilde{\boldsymbol\beta})=\frac{1}{2 n}\|y-X \tilde{\boldsymbol\beta}\|_{2}^{2}.$$ If $$L>\mathcal{L}_{n}(\tilde{\boldsymbol\beta})$$, then

$\begin{split}(\hat{\boldsymbol{\beta}}, \hat{\boldsymbol{d}}, \hat{\mathcal{A}}, \hat{\mathcal{I}})=\left(\tilde{\boldsymbol{\beta}}, \tilde{\boldsymbol{d}}, \tilde{\mathcal{A}}_{k}, \tilde{\mathcal{I}}_{k}\right)\\ L=\mathcal{L}_{n}(\tilde{\boldsymbol\beta}).\end{split}$

End for

4. If $$L_{0}-L<\tau_{s}$$, then $$(\hat{\boldsymbol\beta}, \hat{d}, \hat{A}, \hat{I})=(\boldsymbol\beta, d, \mathcal{A}, \mathcal{I}).$$

5. Output $$(\hat{\boldsymbol{\beta}}, \hat{\boldsymbol{d}}, \hat{\mathcal{A}}, \hat{\mathcal{I}})$$.

Determining the Best Support Size with SIC¶

In practice, the support size is usually unknown. We use a datadriven procedure to determine s. For any active set $$\mathcal{A}$$, define an $$\mathrm{SIC}$$ as follows:

$\operatorname{SIC}(\mathcal{A})=n \log \mathcal{L}_{\mathcal{A}}+|\mathcal{A}| \log (p) \log \log n,$

where $$\mathcal{L}_{\mathcal{A}}=\min _{\beta_{\mathcal{I}}=0} \mathcal{L}_{n}(\beta), \mathcal{I}=(\mathcal{A})^{c}$$. To identify the true model, the model complexity penalty is $$\log p$$ and the slow diverging rate $$\log \log n$$ is set to prevent underfitting. Theorem 4 states that the following ABESS algorithm selects the true support size via SIC.

Let $$s_{\max }$$ be the maximum support size. We suggest $$s_{\max }=o\left(\frac{n}{\log p}\right)$$ as the maximum possible recovery size. Typically, we set $$s_{\max }=\left[\frac{n}{\log p \log \log n}\right]$$ where $$[x]$$ denotes the integer part of $$x$$.

Algorithm 3: ABESS.¶
1. Input: $$X, y$$, and the maximum support size $$s_{\max } .$$

2. For $$s=1,2, \ldots, s_{\max }$$, do

$\left(\hat{\boldsymbol{\beta}}_{s}, \hat{\boldsymbol{d}}_{s}, \hat{\mathcal{A}}_{s}, \hat{\mathcal{I}}_{s}\right)= \text{BESS.Fixed}(s).$

End for

3. Compute the minimum of SIC:

$s_{\min }=\arg \min _{s} \operatorname{SIC}\left(\hat{\mathcal{A}}_{s}\right).$
4. Output $$\left(\hat{\boldsymbol{\beta}}_{s_{\min}}, \hat{\boldsymbol{d}}_{s_{\min }}, \hat{A}_{s_{\min }}, \hat{\mathcal{I}}_{s_{\min }}\right) .$$

Now, enjoy the data analysis with abess library:

import abess

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