Boundary Ties and Sign-at-Zero Conventions

Why Boundary Conventions Matter

A real-valued classification score determines a label by its sign, but the score $0$ has no positive or negative sign. A prediction rule must therefore specify what happens at zero. This convention matters for the classification error of a particular score. Calibration, however, is usually defined with an infimum over a constrained set of scores that includes the boundary. The distinction lets the calibration gap describe the cost of approaching a wrong-sign decision without making its value depend on how a tie is broken.

Write $\eta=P(Y=+1\mid X=x)$ for the conditional probability of the positive label at a fixed $x$, and let $a\in\mathbb{R}$ be the score. For a margin loss $\phi$, the conditional surrogate risk is

C_\eta(a)=\eta\phi(a)+(1-\eta)\phi(-a).

The first term is the loss when $Y=+1$, and the second is the loss when $Y=-1$. In this tutorial, assume $\phi:\mathbb{R}\to[0,\infty)$ is finite-valued and convex. In particular, $C_\eta(a)$ is finite and continuous in $a$. Nonnegativity will also ensure that the infima used below are finite.

Definition: Fix the tie rule $g_+(a)=+1$ when $a\geq 0$ and $g_+(a)=-1$ when $a<0$. The conditional zero-one risk of a prediction $g\in\{-1,+1\}$ is $R_{01}(\eta,g)=\eta\mathbf{1}\{g=-1\}+(1-\eta)\mathbf{1}\{g=+1\}$. A Bayes prediction minimizes this risk. The conditional classification regret is the prediction's risk minus the minimum risk.

When $\eta>1/2$, $+1$ is the unique Bayes label; when $\eta<1/2$, $-1$ is the unique Bayes label. When $\eta=1/2$, both labels minimize the conditional risk. Thus a tie rule affects the predicted label at $a=0$, but cannot create classification regret at $\eta=1/2$.

Definition: The unrestricted conditional surrogate infimum is $H(\eta)=\inf_{a\in\mathbb{R}}C_\eta(a)$. The constrained infimum and calibration gap are $$ H^-(\eta)=\inf_{\{a:\,a(\eta-\tfrac12)\leq 0\}}C_\eta(a), \qquad \Delta_\phi(\eta)=H^-(\eta)-H(\eta). $$

The constraint includes scores on the boundary $a=0$. If $\eta>1/2$, it selects $a\leq0$, the nonpositive side of the score axis; if $\eta<1/2$, it selects $a\geq0$. At $\eta=1/2$, every real score satisfies the constraint. These are closed sets, so the definition does not depend on whether a particular prediction rule assigns zero to the positive or negative class.

Boundary Scores and Classification Regret

Under the positive-at-zero convention, a score is classified incorrectly when $\eta>1/2$ and $a<0$, or when $\eta<1/2$ and $a\geq0$. The first set is open at zero, whereas the constrained sets in $H^-$ are closed. Since $C_\eta$ is continuous, adding the boundary point to an open half-line does not change its infimum: values at zero are limits of values from that half-line. Thus $H^-$ captures the infimum of the surrogate risk over wrong predictions, including the limiting boundary score. The other side already includes zero under this tie convention.

The conditional zero-one regret has a particularly direct form. If $\eta\ne1/2$, a prediction incurs regret only when it differs from the unique Bayes label, and the regret in that case is $|2\eta-1|$. At the tie $\eta=1/2$, either prediction is Bayes, so the regret is zero for every score.

Result: Under the positive-at-zero tie rule, the conditional classification regret at score $a$ is $$ R_{01}(\eta,g_+(a))-\min_{g\in\{-1,+1\}}R_{01}(\eta,g) = |2\eta-1|\mathbf{1}\{g_+(a)\text{ is not a Bayes label}\}. $$ At $\eta=1/2$, the right-hand side is zero regardless of the prediction.

To verify the formula, the risk of predicting $+1$ is $1-\eta$, and the risk of predicting $-1$ is $\eta$. The smaller risk is $\min(\eta,1-\eta)$. If $\eta>1/2$, predicting $-1$ has excess risk $\eta-(1-\eta)=2\eta-1$, while predicting $+1$ has zero excess. If $\eta<1/2$, predicting $+1$ has excess risk $(1-\eta)-\eta=1-2\eta$, while predicting $-1$ has zero excess. At equality the two risks are equal. This proves the result.

Worked Example: A Zero Score Under Two Tie Rules

Let $\eta=0.8$ and $a=0$. Under the positive-at-zero rule, the prediction is $+1$. Its conditional risk is $1-\eta=0.2$, which is the Bayes risk, so its regret is $0$. Under the alternative rule $g_-(a)=+1$ for $a>0$ and $g_-(a)=-1$ for $a\leq0$, the prediction is $-1$. Its risk is $\eta=0.8$, and its regret is $0.8-0.2=0.6=|2\eta-1|$.

The two predictions differ only because the tie convention differs. The constrained set for the surrogate infimum is $a\leq0$ in either case, and contains the boundary score $a=0$. The classification regret of that particular score is therefore convention-dependent, while the constrained calibration gap is defined independently of the tie rule.

Why the Gap Is Well-Defined at the Probability Tie

At $\eta=1/2$, the constraint in $H^-$ becomes $a\cdot0\leq0$, which holds for every real $a$. Thus $H^-(1/2)$ and $H(1/2)$ are infima over the same set. To conclude that their difference is zero, the infima must be finite. Under the stated assumption that $\phi$ is finite-valued and nonnegative, $C_\eta(a)\geq0$, while $C_\eta(0)=\phi(0)<\infty$. The unrestricted infimum is therefore finite. The constrained set always contains zero, so its infimum is finite as well.

Result: If $\phi$ is finite-valued and nonnegative, then $H(\eta)$ and $H^-(\eta)$ are finite for every $\eta\in[0,1]$, $H^-(\eta)\geq H(\eta)$, and $$ H^-(\tfrac12)=H(\tfrac12), \qquad \Delta_\phi(\tfrac12)=0. $$

Proof. Since $C_\eta(a)\geq0$ and $C_\eta(0)=\phi(0)<\infty$, the infimum over all scores satisfies $0\leq H(\eta)\leq\phi(0)$. The constrained set contains $a=0$, so $0\leq H^-(\eta)\leq\phi(0)$ too. Because the constrained set is a subset of $\mathbb{R}$, its infimum cannot be smaller than the infimum over all scores: $H^-(\eta)\geq H(\eta)$. At $\eta=1/2$, the constraint holds for every $a$, so the two infima are equal and their finite difference is zero. $\square$

The finiteness condition is substantive. If one allowed, for example, $\phi(t)=-t^2$, then $C_{1/2}(a)=-a^2$, and both infima would be $-\infty$. Equality of those extended-real infima would not define a difference $\Delta_\phi(1/2)$. Nonnegative losses rule out this problem.

Worked Example: The Hinge Loss at a Non-Tie Boundary

Take $\phi(t)=\max(1-t,0)$ and $\eta=0.7$. Then $C_{0.7}(a)=0.7\max(1-a,0)+0.3\max(1+a,0)$. For $-1\leq a\leq1$, this is $0.7(1-a)+0.3(1+a)=1-0.4a$. For $a\geq1$, it is $0.3(1+a)$; for $a\leq-1$, it is $0.7(1-a)$.

On $[-1,1]$, $1-0.4a$ decreases as $a$ increases, so its minimum there is $0.6$ at $a=1$. On $[1,\infty)$, $0.3(1+a)$ increases, so its minimum is also $0.6$ at $a=1$. On $(-\infty,-1]$, $0.7(1-a)$ decreases as $a$ increases, so its minimum there is $1.4$ at $a=-1$. Hence $H(0.7)=0.6$.

Here $a(\eta-\tfrac12)\leq0$ means $a\leq0$. On $[-1,0]$, $1-0.4a$ is minimized at $a=0$, with value $1$; on $(-\infty,-1]$, the minimum is $1.4$. Therefore $H^-(0.7)=1$ and $\Delta_\phi(0.7)=1-0.6=0.4$. The boundary score $a=0$ is included in the constrained set. Under the positive-at-zero rule it predicts $+1$, so it is not a wrong classification at this $\eta$; nevertheless, wrong scores $a<0$ approach zero, and their surrogate-risk infimum is $1$.

Worked Example: Logistic Loss at the Probability Tie

Take $\phi(t)=\log(1+e^{-t})$ and $\eta=1/2$. The conditional risk is $$ C_{1/2}(a)=\tfrac12\log(1+e^{-a})+\tfrac12\log(1+e^a). $$

Using $(1+e^{-a})(1+e^a)=2+e^a+e^{-a}\geq4$, we obtain $C_{1/2}(a)=\tfrac12\log\{(1+e^{-a})(1+e^a)\}\geq\log 2$. Equality holds at $a=0$, since the product there is $4$. Thus $H(1/2)=\log2$. Every score satisfies the constrained condition at $\eta=1/2$, so $H^-(1/2)=\log2$ and $\Delta_\phi(1/2)=0$. Regardless of the zero-score tie rule, the classification regret at this probability tie is also zero.

Worked Example: Logistic Loss Just Below the Tie

Let $\eta=0.3$, again with $\phi(t)=\log(1+e^{-t})$. Differentiating the conditional risk gives $$ C'_{0.3}(a)=\frac{0.7e^a-0.3}{1+e^a}. $$

The derivative is zero exactly when $e^a=3/7$, so the unrestricted minimizer is $a_*=\log(3/7)<0$. At this minimizer, direct substitution gives $C_{0.3}(a_*)=-0.3\log(0.3)-0.7\log(0.7)$, and hence this is $H(0.3)$. The constrained set is $a\geq0$. The derivative is positive for $a\geq0$, because $0.7e^a-0.3\geq0.4>0$, so the constrained minimum occurs at $a=0$ and equals $\log2$. Therefore $$ \Delta_\phi(0.3)=\log2+0.3\log(0.3)+0.7\log(0.7)\approx0.0823. $$

At $a=0$, the positive-at-zero rule predicts $+1$, although the Bayes label is $-1$. Its classification regret is $1-2(0.3)=0.4$. The calibration gap is not this classification regret: it is the difference between two surrogate-risk infima.

Common Pitfalls and Practical Interpretation

Treating zero as if it had an automatic sign. A classifier must state its tie rule. At $\eta\ne1/2$, the rule can change the regret of a zero score; at $\eta=1/2$, neither label has regret.
Using a strict wrong-sign constraint without checking the boundary. The constrained-infimum definition uses a closed set. For continuous conditional risk, its infimum agrees with the infimum over the corresponding open wrong-sign side when zero is only a limiting point. This makes the boundary explicit.
Subtracting infinite infima. Equality of two infima is not enough to define their difference if both are infinite. The finite, nonnegative loss assumption guarantees finite conditional infima here.
Interpreting a zero calibration gap at $\eta=1/2$ as a defect. At that probability, every label is Bayes-optimal, so there is no classification regret to penalize. The zero gap expresses that fact; it does not assert that all scores minimize the surrogate risk.
Confusing calibration gap with the regret of one score. A regret is evaluated at a chosen prediction; $\Delta_\phi(\eta)$ compares a constrained surrogate infimum with the unrestricted one.

The boundary convention matters most when interpreting individual predictions and when translating between strict mistakes and constrained optimization. The closed constraint gives a stable, tie-rule-independent definition: it includes zero, remains meaningful at $\eta=1/2$ under finite nonnegative losses, and records the least surrogate risk attainable on the non-Bayes side or its boundary.

Check Your Understanding

Work through these questions using the positive-at-zero convention unless a question specifies otherwise.

For $\eta=0.65$ and $a=0$, what label is predicted, and what is the conditional classification regret?
Why does the constraint $a(\eta-\tfrac12)\leq0$ include every score when $\eta=1/2$?
What assumption in this tutorial ensures that $H(\eta)$ and $H^-(\eta)$ are finite?
For $\eta<1/2$, which scores belong to the constrained set, and which scores are classification errors under the positive-at-zero rule?
Why is the logistic calibration gap at $\eta=0.3$ not the same quantity as the regret of the zero score?