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An exact primal state can have a wrong finite-unroll derivative

Example

Fix $q \in (0, 1)$ and define $G (z, θ) = q z + (1 - q) θ$ for $z, θ \in R$ , with initialization $z_{0} (θ) = 0$ , recurrence $z_{k + 1} (θ) = G (z_{k} (θ), θ)$ , and outer objective $g (z, θ) = z$ . The exact fixed-point branch is $z^{*} (θ) = θ$ , so the exact hypergradient of $J (θ) = g (z^{*} (θ), θ)$ is $1$ . For every finite $K \geq 1$ , the $K$ -step iterate is $z_{K} (θ) = (1 - q^{K}) θ$ . At $θ_{0} = 0$ it equals the exact state and its fixed-point residual $r_{K} : = z_{K} (0) - G (z_{K} (0), 0)$ is zero, but its derivative is $1 - q^{K} \neq 1$ . Thus an exact primal state and zero primal residual do not force the finite-unroll derivative to equal the implicit hypergradient.

Example references

Stationary branch and hypergradient

Finite-iteration and exact-branch distinction

Finite-iteration sensitivity result

Facts & Assumptions

Given: $q \in (0, 1)$ , $G (z, θ) = q z + (1 - q) θ$ , $z_{0} (θ) = 0$ , and $g (z, θ) = z$ . For comparison with the stationary-branch formula, set $f (θ, z) : = \frac{1}{2} (z - θ)^{2}$ on $Θ = W = R$ ; both $f$ and $g$ are polynomials, so they meet the $C^{2}$ and $C^{1}$ assumptions on these open domains. The stationarity equation is $z - θ = 0$ , which is equivalent to $z = G (z, θ)$ because $1 - q > 0$ . Compare the exact fixed-point derivative with the derivative of a finite iterate at $θ_{0} = 0$ .

[F1]

Smooth bilevel stationary problems and their hypergradients: the exact stationary branch defines the outer objective whose derivative is the implicit hypergradient.

[F2]

Finite fixed-point iteration sensitivity error bounds and the limits of residual control: a finite contraction iterate differentiates its finite composition, and its derivative can differ from the fixed-point sensitivity even with zero primal residual.

[F3]

Hypergradient of a smooth local stationary branch: for a $C^{2}$ inner objective and $C^{1}$ outer objective with invertible stationary Hessian, the local branch hypergradient is $\partial_{θ} g - B^{T} H^{- T} \nabla_{z} g$ .

Verification

1.1givenF1algebra

Solving $z = G (z, θ)$ gives $(1 - q) z = (1 - q) θ$ . Since $1 - q > 0$ , the unique fixed point is $z^{*} (θ) = θ$ , and therefore $J (θ) = z^{*} (θ)$ has derivative $1$ . Also $G_{z} = q < 1$ .

1.2givenalgebraconstruct

For $K = 0$ , $z_{0} (θ) = (1 - q^{0}) θ = 0$ . If $z_{K} (θ) = (1 - q^{K}) θ$ , then $z_{K + 1} = q (1 - q^{K}) θ + (1 - q) θ = (1 - q^{K + 1}) θ$ . Induction gives the displayed finite-iterate formula for every $K \geq 1$ .

2.1F1F3givenstep 1.1algebra

The inner objective has $\partial_{z} f = z - θ$ , so its stationary branch is the same $z^{*} (θ) = θ$ as the fixed-point branch. Here $H = 1$ , $B = - 1$ , $\partial_{θ} g = 0$ , and $\nabla_{z} g = 1$ . Formula [F3] therefore gives $0 - (- 1) (1)^{- 1} (1) = 1$ , verifying the exact hypergradient under the bilevel prerequisites.

2.2givenstep 1.2algebra

At $θ_{0} = 0$ , step 1.2 gives $z_{K} (0) = 0 = z^{*} (0)$ and the fixed-point residual is $z_{K} (0) - G (z_{K} (0), 0) = 0 - G (0, 0) = 0$ . The stopped outer objective is $J_{K} (θ) = g (z_{K} (θ), θ) = (1 - q^{K}) θ$ , so $J_{K}^{'} (0) = 1 - q^{K}$ .

3.1F2step 2.1step 2.2algebra∎

Because $0 < q < 1$ and $K \geq 1$ , $q^{K} > 0$ , hence $1 - q^{K} \neq 1$ . Thus the zero residual at the exact state does not make the finite-unroll derivative equal to the exact hypergradient.

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Sources

MIT OpenCourseWare 18.S096, Matrix Calculus for Machine Learning and Beyond, Lecture 4 Part 2, §§6.3.1–6.3.2 (standard reference, not scraped)