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A scalar regularized LM step controls noise at fixed damping

Example

Assume Countable Choice. Let $ε, α, δ > 0$ , let $σ \in (0, 1)$ , and consider the scalar forward map $F_{ε} : R \to R$ , $F_{ε} (t) = ε t$ at a current point $x$ with exact datum $y = F_{ε} (x)$ and noisy datum $y^{δ}$ satisfying $0 < r : = y^{δ} - F_{ε} (x) \leq δ$ . The regularized linearized step with fixed damping $α$ is the unique minimizer of $h ⟼ \frac{1}{2} ∣ ε h - r ∣^{2} + \frac{α}{2} ∣ h ∣^{2},$ namely $h_{α} = ε r / (ε^{2} + α)$ ; hence $∣ h_{α} ∣ \leq δ / (2 \sqrt{α})$ , uniformly in $ε$ . The unregularized Gauss–Newton increment, defined by $ε h_{0} = r$ , is $h_{0} = r / ε$ and satisfies only $∣ h_{0} ∣ \leq δ / ε$ . If instead $α$ is selected by the residual-forcing equality $∣ ε h_{α} - r ∣ = σ r$ , then $α = σ ε^{2} / (1 - σ)$ and $h_{α} = (1 - σ) r / ε$ , so the forcing-selected increment is not uniformly bounded over $ε ↓ 0$ at a fixed noise bound. For the finite-dimensional safeguarded LM trial on this same scalar linear residual, with exact arithmetic, $D = 1$ , current damping $λ = α$ , damping floor $0 < λ_{\min} \leq α$ , factors $γ = 2$ , $η_{1} = 1 / 4$ , $η_{2} = 3 / 4$ , trust radius $Δ > ∣ h_{α} ∣$ , one-attempt budget, and a supplied exact reduced QR factorization of the augmented matrix, the trial is accepted with ratio $1$ and unchanged damping.

Example references

Residual-forcing regularized LM step

Finite-budget safeguarded LM trial

Christian Clason, Regularization of Inverse Problems, §11.2

Barbara Kaltenbacher, Inverse Problems – Applications and Solution Strategies, PDF p. 63

Facts & Assumptions

Given: Countable Choice; $ε, α, δ > 0$ ; $0 < σ < 1$ ; $F_{ε} (t) = ε t$ ; exact datum $y = F_{ε} (x)$ ; and $0 < r = y^{δ} - F_{ε} (x) \leq δ$ .

[A1]

Countable Choice is the axiom required by the general Hilbert-space regularized LM definition (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The regularized LM method selects $α_{n}$ by the displayed residual-forcing equality (Regularized nonlinear Levenberg–Marquardt with discrepancy stopping).

[F2]

In exact mode the safeguarded LM trial minimizes the damped least-squares objective with the supplied scaling matrix (Finite-budget Levenberg-Marquardt with a trust-ratio safeguard).

[F3]

An exact safeguarded trial is accepted only when the actual-to-predicted ratio is at least $η_{1}$ ; damping is reduced only when the ratio is at least $η_{2}$ and the scaled step reaches the trust boundary (Finite-budget Levenberg-Marquardt with a trust-ratio safeguard).

[F4]

Each positive real has a positive square root, and $2 ε \sqrt{α} \leq ε^{2} + α$ follows from $(ε - \sqrt{α})^{2} \geq 0$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Verification

1.1givenalgebra

Completing the square gives $\frac{1}{2} ∣ ε h - r ∣^{2} + \frac{α}{2} ∣ h ∣^{2} = \frac{ε^{2} + α}{2} {(h - \frac{ε r}{ε^{2} + α})}^{2} + \frac{r^{2}}{2} - \frac{ε^{2} r^{2}}{2 (ε^{2} + α)} .$ Since $ε^{2} + α > 0$ , the unique minimizer is $h_{α} = ε r / (ε^{2} + α)$ .

2.1F4givenstep 1.1algebra

By [F4], $2 ε \sqrt{α} \leq ε^{2} + α$ , so $∣ h_{α} ∣ = ε r / (ε^{2} + α) \leq r / (2 \sqrt{α}) \leq δ / (2 \sqrt{α})$ , with a bound independent of $ε$ . The unregularized linearized equation $ε h_{0} = r$ instead gives $∣ h_{0} ∣ = r / ε \leq δ / ε$ , which is unbounded as $ε ↓ 0$ for fixed $r > 0$ .

2.2F1step 1.1algebra

The linearized residual after the regularized step is $ε h_{α} - r = - α r / (ε^{2} + α)$ . Since $r > 0$ , the residual-forcing equality is equivalent to $α / (ε^{2} + α) = σ$ , hence $α = σ ε^{2} / (1 - σ)$ and $h_{α} = (1 - σ) r / ε$ . This forcing-selected step grows like $1 / ε$ when $r$ is held fixed; the fixed- $α$ bound above does not give a uniform bound for this varying-parameter family.

2.3F2F4step 1.1givenconstructalgebra

For the finite safeguarded procedure, the augmented matrix is the column $A = (ε, \sqrt{α})^{T}$ . Put $s : = \sqrt{ε^{2} + α}$ , $Q : = A / s$ , and $R : = (s)$ . Then $s > 0$ , $Q^{T} Q = 1$ , and $Q R = A$ , so $A = Q R$ is an exact reduced QR factorization. Its augmented residual vector is $b = (- r, 0)^{T}$ , and the exact damped least-squares minimizer is the same $p = h_{α}$ from step 1.1.

3.1A1F3step 1.1step 2.2step 2.3givenalgebra∎

Take residual map $g (t) = ε t - y^{δ}$ , Jacobian $J = ε$ , $D = 1$ , current damping $λ = α$ , a floor $0 < λ_{\min} \leq α$ , factors $γ = 2$ , $η_{1} = 1 / 4$ , $η_{2} = 3 / 4$ , trust radius $Δ > ∣ h_{α} ∣$ , and one allowed attempt. The exact gradient is nonzero since $J^{T} g (x) = - ε r \neq 0$ . The model residual equals the actual residual because $g$ is affine. The predicted reduction is positive because $∣ ε h_{α} - r ∣ = α r / (ε^{2} + α) < r$ , so the actual-to-predicted ratio is $1$ . The trial lies inside the trust radius, hence it is accepted and does not meet the boundary condition for lowering damping; it returns the same damping $λ = α$ .

Remarks

The fixed-damping bound and the residual-forcing parameter rule are different comparisons. The finite safeguarded procedure is used here for its exact finite-dimensional damped-step mechanics; its trust-ratio update does not replace the residual-forcing rule in the regularizing Hilbert-space iteration.

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Sources

Christian Clason, Regularization of Inverse Problems, §11.2 (standard reference, not scraped)
Barbara Kaltenbacher, Inverse Problems – Applications and Solution Strategies, PDF p. 63 (standard reference, not scraped)