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Logistic and exponential losses satisfy the tight-tail condition

Statement

For $ℓ_{\exp} (u) = e^{- u}$ and $ℓ_{\log} (u) = \log (1 + e^{- u})$ , the negative derivatives $- ℓ_{\exp}^{}$ and $- ℓ_{\log}^{}$ have tight exponential tails with $c = a = μ_{+} = μ_{-} = 1$ and $u_{+} = u_{-} = 1$ in the definition of a tight exponential tail.

Statement references

Tight exponential-tail definition

Daniel Soudry et al., The Implicit Bias of Gradient Descent on Separable Data, Definition 2 and §2

Matus Telgarsky, Deep Learning Theory lecture notes, §10.2

Facts & Assumptions

Given: The loss functions $ℓ_{\exp} (u) = e^{- u}$ and $ℓ_{\log} (u) = \log (1 + e^{- u})$ .

[F1]

Tight exponential tail for a positive loss derivative: it suffices to bound $f (u)$ above by $(1 + e^{- u}) e^{- u}$ and below by $(1 - e^{- u}) e^{- u}$ for every $u > 1$ .

Proof

technique · differentiate the two losses and compare their negative derivatives with the defining bounds

1.1givenF1algebra

The exponential loss has $- ℓ_{\exp}^{} (u) = e^{- u}$ . For $u > 1$ , $0 < 1 - e^{- u} < 1$ , so $(1 - e^{- u}) e^{- u} \leq e^{- u} \leq (1 + e^{- u}) e^{- u}$ , which is exactly the pair of bounds in [F1].

2.1givenF1algebra∎

The logistic loss has $- ℓ_{\log}^{} (u) = e^{- u} / (1 + e^{- u})$ . For $r = e^{- u} \in (0, 1)$ , $1 - r \leq 1 / (1 + r) \leq 1$ because $1 - (1 - r) (1 + r) = r^{2} \geq 0$ and $1 \leq 1 + r$ . Multiplying by $r > 0$ gives $(1 - e^{- u}) e^{- u} \leq - ℓ_{\log}^{} (u) \leq e^{- u} \leq (1 + e^{- u}) e^{- u}$ for $u > 1$ , as required by [F1].

Depends on

Tight exponential tail for a positive loss derivative

Used by

Separable logistic and exponential gradient descent select the hard-margin direction Theorem

Dependency tree · one level

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Sources

Daniel Soudry et al., The Implicit Bias of Gradient Descent on Separable Data, Definition 2 and §2 (standard reference, not scraped)
Matus Telgarsky, Deep Learning Theory lecture notes, §10.2 (standard reference, not scraped)