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One cyclic coordinate-descent cycle for correlated-design Lasso

Example

Let $n = 2$ , $X_{1} = (1, 0)^{T}$ , $X_{2} = (1, 1)^{T}$ , $y = (2, 1)^{T}$ , and $τ = 1 / 4$ in Lasso estimator for fixed design. Starting at $θ^{0} = (0, 0)$ , one exact cyclic sweep in the order $1, 2$ gives $θ^{1} = (3 / 2, 0)$ and $θ^{2} = (3 / 2, 1 / 2)$ .

Facts & Assumptions

Given: The design and response in the Example, positive Lasso penalty $τ = 1 / 4$ , cyclic coordinate order $1, 2$ , and sweep budget $K = 1$ .

[F1]

Cyclic exact block minimization: a cyclic sweep replaces each selected block by a global minimizer of the objective slice with the other coordinates fixed.

[F2]

Lasso estimator for fixed design: $Q_{y} (θ) = \frac{1}{2} ∥ y - X θ ∥_{2}^{2} + \frac{1}{2} ∥ θ ∥_{1}$ for these $n, τ$ .

[F3]

Exact Lasso coordinate update, including a zero column: a nonzero column update is $S_{n τ} (X_{j}^{T} r_{j}) / ∥ X_{j} ∥_{2}^{2}$ .

Proof

1.1F1F2F3givenalgebra

Initially $r_{1} = y$ , so $X_{1}^{T} r_{1} = 2$ , $∥ X_{1} ∥_{2}^{2} = 1$ , and $n τ = 1 / 2$ . The update lemma gives $θ_{1}^{1} = S_{1 / 2} (2) = 3 / 2$ ; holding $θ_{2} = 0$ gives residual $(1 / 2, 1)$ and objective $Q_{y} (3 / 2, 0) = \frac{1}{2} (1 / 4 + 1) + \frac{1}{2} (3 / 2) = 11 / 8$ .

2.1F1F2F3step 1.1algebra∎

For the second block, $r_{2} = y - X_{1} (3 / 2) = (1 / 2, 1)$ , so $X_{2}^{T} r_{2} = 3 / 2$ and $∥ X_{2} ∥_{2}^{2} = 2$ . Thus $θ_{2}^{2} = S_{1 / 2} (3 / 2) / 2 = 1 / 2$ . The final residual is $(0, 1 / 2)$ and $Q_{y} (3 / 2, 1 / 2) = \frac{1}{2} (1 / 4) + \frac{1}{2} (2) = 9 / 8$ . The values decrease from $Q_{y} (0, 0) = 5 / 2$ to $11 / 8$ to $9 / 8$ , as the exact updates require.

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Sources

Stephen J. Wright, Coordinate Descent Algorithms, §2 (standard reference, not scraped)
P. Tseng, Convergence of a Block Coordinate Descent Method for Nondifferentiable Minimization, §§2 and 4 (standard reference, not scraped)