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GAN objectives and the ideal discriminator — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Separated supports saturate the ideal discriminator

Example

Let $X = {a, b}$ with counting measure $μ (E) = ∣ E ∣$ . Let $P_{d a t a}$ put mass $1$ at $a$ and $P_{G}$ put mass $1$ at $b$ , so their densities are $p (a) = 1$ , $p (b) = 0$ , $q (a) = 0$ , and $q (b) = 1$ . The ideal discriminator is $D^{*} (a) = 1$ and $D^{*} (b) = 0$ , the density-defined Jensen–Shannon divergence is $\ln 2$ , and the optimized minimax value is $C (P_{d a t a}, P_{G}) = 0$ . Thus disjoint supports give a saturated ideal discriminator; this calculation makes no claim about finite-network gradient behavior.

Example references

Optimal discriminator and Jensen–Shannon identity

Density-defined Jensen–Shannon divergence

Goodfellow et al., §4.1, Proposition 1 and Theorem 1, equations (2)–(6), complete 9-page paper

Li, §1.2, equations (4), (10)–(11), complete 6-page lecture note

Facts & Assumptions

Given: $X = {a, b}$ , counting measure $μ (E) = ∣ E ∣$ , $P_{d a t a} ({a}) = 1$ , $P_{G} ({b}) = 1$ , and zero mass at the other point for each law.

[F1]

Optimal discriminator and Jensen–Shannon divergence in the unrestricted GAN game: $D^{*} = p / (p + q)$ when $p + q > 0$ , $D^{*} = 1 / 2$ when $p + q = 0$ , and $C = - \ln 4 + 2 {JS}_{μ} (P, Q)$ .

[F2]

Density-defined Jensen–Shannon divergence: for $m = (p + q) / 2$ and $r = p / m$ , $s = q / m$ on $m > 0$ , ${JS}_{μ} (P, Q) = \frac{1}{2} \int m ϕ (r) d μ + \frac{1}{2} \int m ϕ (s) d μ$ , with $ϕ (0) = 0$ and $ϕ (t) = t \ln t$ .

Verification

Given: The two-point laws and counting measure in the Example.

1.1givenF2construct

Their densities are $p (a) = 1, p (b) = 0, q (a) = 0, q (b) = 1$ . Thus $m (a) = m (b) = 1 / 2$ , $(r (a), s (a)) = (2, 0)$ and $(r (b), s (b)) = (0, 2)$ .

2.1givenF1step 1.1algebra

By [F1], $D^{*} (a) = 1 / (1 + 0) = 1$ and $D^{*} (b) = 0 / (0 + 1) = 0$ .

2.2F2step 1.1algebra

Since $ϕ (0) = 0$ and $ϕ (2) = 2 \ln 2$ , the two integrals in [F2] each equal $\ln 2$ before their factor $1 / 2$ ; hence ${JS}_{μ} (P_{d a t a}, P_{G}) = \ln 2$ .

3.1F1step 2.1step 2.2algebra∎

By [F1] and step 2.2, $C (P_{d a t a}, P_{G}) = - \ln 4 + 2 \ln 2 = 0$ . The discriminator takes only the saturated endpoint values on the two supports, as claimed.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A finite adversarial gradient game can cycle

Example

Let $θ, φ \in R$ be a generator and discriminator parameter, and let the generator minimize and the discriminator maximize the shared payoff $L (θ, φ) = θ φ$ . Starting from any $(θ_{0}, φ_{0}) \neq (0, 0)$ , use one exact unit-step generator gradient descent update followed by one exact unit-step discriminator gradient ascent update, with the discriminator update evaluated at the new generator parameter:

θ_{n + 1} = θ_{n} - \partial_{θ} L (θ_{n}, φ_{n}), φ_{n + 1} = φ_{n} + \partial_{φ} L (θ_{n + 1}, φ_{n}) .

The resulting parameter sequence repeats after six steps and does not converge. This scalar bilinear game is a finite-parameter adversarial-game illustration, not the GAN cross-entropy objective or a neural GAN; it shows why a distribution-space optimum alone is not a finite-parameter training-convergence theorem.

Example references

Unrestricted distribution-space optimum and its scope

Goodfellow et al., §4.2, final paragraph on finite MLP parameterization, complete 9-page paper

Facts & Assumptions

Given: $θ_{n}, φ_{n} \in R$ , payoff $L (θ, φ) = θ φ$ , generator descent first, discriminator ascent second at the updated $θ$ , unit step size, and a nonzero initial pair.

[F1]

Optimal discriminator and Jensen–Shannon divergence in the unrestricted GAN game: the proved GAN optimum is for the unrestricted distribution-space objective, and it asserts no convergence for finite neural alternating optimization.

Verification

Given: The scalar payoff and exact alternating update rule in the Example, with $(θ_{0}, φ_{0}) \neq (0, 0)$ .

1.1givenalgebra

For $h \neq 0$ , $(L (θ + h, φ) - L (θ, φ)) / h = ((θ + h) φ - θ φ) / h = φ$ , so $\partial_{θ} L = φ$ . Likewise $(L (θ, φ + h) - L (θ, φ)) / h = θ$ , so $\partial_{φ} L = θ$ .

2.1givenstep 1.1algebra

Substitution into the update gives $θ_{n + 1} = θ_{n} - φ_{n}$ and $φ_{n + 1} = φ_{n} + θ_{n + 1} = θ_{n}$ . Thus $x_{n + 1} = M x_{n}$ for $x_{n} = (θ_{n}, φ_{n})^{T}$ and $M = (\begin{array}{cc} 1 & - 1 \\ 1 & 0 \end{array})$ .

3.1step 2.1algebra

Matrix multiplication gives $M^{2} = (\begin{array}{cc} 0 & - 1 \\ 1 & - 1 \end{array})$ and $M^{3} = - I$ . Hence $x_{n + 3} = - x_{n}$ and $x_{n + 6} = x_{n}$ for every $n$ .

4.1givenF1step 3.1algebra∎

Since $x_{0} \neq 0$ , the subsequence $(x_{6 k})_{k \geq 0}$ is constantly $x_{0}$ and $(x_{6 k + 3})_{k \geq 0}$ is constantly $- x_{0} \neq x_{0}$ . These subsequences have different limits, so $(x_{n})$ does not converge. The computed cycle is for this explicitly defined toy game; [F1] shows it supplies no conclusion about convergence of GAN neural training.

GAN objectives and the ideal discriminator — Examples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

Separated supports saturate the ideal discriminator

Example

Example references

Facts & Assumptions

Verification

A finite adversarial gradient game can cycle

Example

Example references

Facts & Assumptions

Verification

Sources