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Duplicate predictors give a least-squares line, a ridge point, and a lasso segment

Example

Take $n = 1$ , $p = 2$ , $X = (1 1)$ , and $y = 1$ . The OLS minimizers are exactly ${(β_{1}, β_{2}) : β_{1} + β_{2} = 1}$ . Ridge with parameter $1$ has the unique minimizer $(1 / 3, 1 / 3)$ . Lasso with the empirical-loss convention and penalty $λ = 1 / 4$ has the full minimizer set ${(β_{1}, β_{2}) : β_{1}, β_{2} \geq 0, β_{1} + β_{2} = 3 / 4}$ .

Example references

The Elements of Statistical Learning, §§3.2 and 3.4

Rigollet and Hütter, §2.2

Facts & Assumptions

Given: The fixed design $X = (1 1)$ , response $y = 1$ , the OLS and ridge objectives, and Lasso penalty $λ = 1 / 4$ .

[F1]

OLS minimizers solve the normal equations and ridge has its unique closed-form minimizer (The full ordinary least-squares solution set and the unique ridge solution).

[F2]

The Lasso objective is $∥ y - X β ∥_{2}^{2} / (2 n) + λ ∥ β ∥_{1}$ (Fixed-design least-squares, ridge, lasso, group, and fused objectives).

Verification

1.1F1givenalgebra

Here $X^{T} X = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ and $X^{T} y = (1, 1)^{T}$ . The OLS normal equations reduce to $β_{1} + β_{2} = 1$ , giving the stated full line.

2.1F1step 1.1algebra

For ridge parameter $1$ , $X^{T} X + I_{2} = (\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array})$ , whose inverse sends $(1, 1)^{T}$ to $(1 / 3, 1 / 3)^{T}$ .

3.1F2givenalgebra∎

Write $t = β_{1} + β_{2}$ . Since $∣ β_{1} ∣ + ∣ β_{2} ∣ \geq ∣ t ∣$ , the Lasso objective is at least $\frac{1}{2} (1 - t)^{2} + \frac{1}{4} ∣ t ∣$ , with equality exactly when the coordinates have the same weak sign. The scalar function decreases up to $t = 3 / 4$ and increases after it: on $t > 0$ its derivative is $t - 3 / 4$ , and on $t < 0$ its derivative is $t - 5 / 4 < 0$ . Thus its unique minimizing sum is $t = 3 / 4 > 0$ , where equality in the triangle inequality requires both coordinates nonnegative. This gives exactly the stated segment of minimizers.

Diagram

The axes are $β_{1}$ and $β_{2}$ . The dashed segment is the first-quadrant portion of the OLS line; the solid segment is the full Lasso minimizer set; the point is the ridge minimizer. The algebraic verification above proves the exact sets and point; the diagram is illustrative.

diagram failed to render:
\begin{tikzpicture}[scale=3]
  \draw[->] (-0.08,0) -- (1.15,0) node[right] {$\beta_1$};
  \draw[->] (0,-0.08) -- (0,1.12) node[above] {$\beta_2$};
  \draw[dashed,gray,thick] (0,1) -- (1,0) node[pos=.22,above right] {OLS};
  \draw[blue,very thick] (0,0.75) -- (0.75,0) node[pos=.25,above right] {Lasso};
  \fill[red] (0.3333,0.3333) circle (0.018) node[above right] {ridge};
  \foreach \x in {0.25,0.5,0.75,1.0} {\draw (\x,0.012) -- (\x,-0.012) node[below=2pt] {\x};}
  \foreach \y in {0.25,0.5,0.75,1.0} {\draw (0.012,\y) -- (-0.012,\y) node[left=2pt] {\y};}
\end{tikzpicture}

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Sources

Hastie, Tibshirani, and Friedman, The Elements of Statistical Learning, 2nd ed., corrected 12th printing, §§3.2 and 3.4 (standard reference, not scraped)
Philippe Rigollet and Jan-Christian Hütter, High-Dimensional Statistics: Lecture Notes, §2.2 (standard reference, not scraped)