We noted three issues about our article “A new confidence interval based on the theory of U-statistics for the area under the curve”, Modern Stochastics: Theory and Applications 13 (2026), 57–82:

1.

The proof of Lemma 3 is not correct. We have not thought about the finiteness of the integral ∫ { v ∈ R p | β T v = 0 } ⟨ v , u ⟩ ‖ β ‖ ∫ R p q ( w ) ⋅ π ( β 0 T w ) ⋅ q ( v + w ) ⋅ ( 1 − π ( β 0 T ( v + w ) ) ) dw dv appearing on page 76, line 2. However, if both the positive part and the negative part of that integral are infinite, the proof of Lemma 3 breaks down. We are not sure whether Lemma 3 holds under the regularity assumptions made in the article. Below we present a proof using different regularity assumptions.

The definition of Model 2 has to be modified as follows:

Model 2.

Let Y ˆ 2 , i , i = 1 , … , n, be the fitted values of a logistic regression model (see Section 2) with true parameter β 0 ∈ R p , applied to i.i.d. data points X 2 , i , i = 1 , … , n, and let I 2 , i , 1 , … , n, be the known true classification. We assume β 0 ≠ 0. We shall assume that the design points have finite second moment, E ‖ X 2 , i ‖ 2 < ∞ , i = 1 , … , n , and that P ( X 2 , i ∈ H ) = 0 for every affine hyperplane H ⊆ R p perpendicular to β 0 and for every linear hyperplane H ⊆ R p (no matter whether it is perpendicular to β 0 or not). Moreover, the cardinality m of the training set and the cardinality n of the test set should fulfill

(1) lim inf n → ∞ m n > 0 , lim sup n → ∞ m n < ∞ .

Lemma 3 and its proof have to be modified as follows:

Lemma 3.

Put z ( β ) : = P ( β T X 1 < β T X 2 , I 1 = 0 , I 2 = 1 ) + 1 2 P ( β T X 1 = β T X 2 , I 1 = 0 , I 2 = 1 ) . Then, under Model 2, z is differentiable in β 0 with d dβ z ( β ) | β = β 0 .

Notice, first, that the second summand in the definition of z has to be mentioned explicitely now, since while P ( β 0 T X 1 = β 0 T X 2 , I 1 = 0 , I 2 = 1 ) = 0 still holds, it is not sure that P ( β T X 1 = β T X 2 , I 1 = 0 , I 2 = 1 ) = 0 holds for β ≠ β 0 any more. Notice, moreover, that we do not get differentiability (not even continuity) for β ≠ β 0 any more.

Proof.

Observe z ( β ) : = 1 2 P ( β T X 1 ≤ β T X 2 , I 1 = 0 , I 2 = 1 ) + 1 2 P ( β T X 1 > β T X 2 , I 1 = 1 , I 2 = 0 ) , since ( X 1 , I 1 , X 2 , I 2 ) = ( X 2 , I 2 , X 1 , I 1 ) in distribution.

Let ( β n ) n ∈ N be a sequence in R p ∖ { β 0 } converging to β 0 .

Recall π ( t ) = 1 1 + exp { t } so that P ( I 1 = 0 | X 1 ) = π ( β 0 T X 1 ). Then z ( β n ) − z ( β 0 ) = E [ 1 2 P ( β n T X 1 ≤ β n T X 2 , I 1 = 0 , I 2 = 1 | X 1 , X 2 ) + 1 2 P ( β n T X 1 > β n T X 2 , I 1 = 1 , I 2 = 0 | X 1 , X 2 ) − 1 2 P ( β 0 T X 1 ≤ β 0 T X 2 , I 1 = 0 , I 2 = 1 | X 1 , X 2 ) − 1 2 P ( β 0 T X 1 > β 0 T X 2 , I 1 = 1 , I 2 = 0 | X 1 , X 2 ) ] = 1 2 ∫ R p ∫ R p 1 { β n T x 1 ≤ β n T x 2 } ⋅ π ( β 0 T x 1 ) ⋅ ( 1 − π ( β 0 T x 2 ) ) + 1 { β n T x 1 > β n T x 2 } ⋅ ( 1 − π ( β 0 T x 1 ) ) ⋅ π ( β 0 T x 2 ) − 1 { β 0 T x 1 ≤ β 0 T x 2 } ⋅ π ( β 0 T x 1 ) ⋅ ( 1 − π ( β 0 T x 2 ) ) + 1 { β 0 T x 1 > β 0 T x 2 } ⋅ ( 1 − π ( β 0 T x 1 ) ) ⋅ π ( β 0 T x 2 ) dμ ( x 1 ) dμ ( x 2 ) . Obviously, the integrand is zero outside the symmetric difference Δ n of { β n T x 1 ≤ β n T x 2 } and { β 0 T x 1 ≤ β 0 T x 2 }. It is also zero on the set Δ = : = { β 0 T x 1 = β 0 T x 2 }. On the whole of R p × R p , the absolute value of the integrand can be bounded from above by | π ( β 0 T x 1 ) ⋅ ( 1 − π ( β 0 T x 2 ) ) − π ( β 0 T x 2 ) ⋅ ( 1 − π ( β 0 T x 1 ) ) | ≤ | π ( β 0 T x 1 ) − π ( β 0 T x 2 ) | + | ( 1 − π ( β 0 T x 2 ) ) − ( 1 − π ( β 0 T x 1 ) ) | ≤ 2 ⋅ 1 / 4 ⋅ | β 0 T x 1 − β 0 T x 2 | , since π is Lipschitz continuous with Lipschitz constant 1 4 .

We must have | β 0 T ( x 1 − x 2 ) | ≤ | ( β n − β 0 ) T ( x 1 − x 2 ) | on Δ n , since | β 0 T ( x 1 − x 2 ) | > | ( β n − β 0 ) T ( x 1 − x 2 ) | implies that β 0 T ( x 1 − x 2 ) and β n T ( x 1 − x 2 ) have the same sign and thus ( x 1 , x 2 ) ∉ Δ n . Using that the Cauchy-Schwarz inequality gives | ( β n − β 0 ) T ( x 1 − x 2 ) | ≤ ‖ β n − β 0 ‖ ⋅ ‖ x 1 − x 2 ‖, we get | z ( β n ) − z ( β 0 ) | ≤ 1 4 ‖ β n − β 0 ‖ ⋅ ∫ R p ∫ R p 1 { ( x 1 , x 2 ) ∈ Δ n ∖ Δ = } ‖ x 1 − x 2 ‖ dμ ( x 1 ) dμ ( x 2 ) . Now lim n → ∞ 1 { ( x 1 , x 2 ) ∈ Δ n ∖ Δ = } = 0 for any x 1 , x 2 ∈ R p implies lim n → ∞ | z ( β n ) − z ( β 0 ) | ‖ β n − β 0 ‖ ≤ 1 4 lim n → ∞ ∫ R p ∫ R p 1 { ( x 1 , x 2 ) ∈ Δ n ∖ Δ = } ‖ x 1 − x 2 ‖ dμ ( x 1 ) dμ ( x 2 ) = 0 , since we are allowed to apply the dominated convergence theorem due to ∫ R p ∫ R p 1 { ( x 1 , x 2 ) ∈ Δ n ∖ Δ = } ‖ x 1 − x 2 ‖ dμ ( x 1 ) dμ ( x 2 ) ≤ ∫ R p ‖ x 1 ‖ dμ ( x 1 ) + ∫ R p ‖ x 2 ‖ dμ ( x 2 ) < ∞ . This implies that z is differentiable in β 0 with derivative 0. □

Equality (4) on page 79 has now to be derived in a different way. Put D ¯ ( β ) = { x ∈ S p − 1 | β T x ≤ 0 , β 0 T x ≥ 0 } ∪ { x ∈ S p − 1 | β T x ≥ 0 , β 0 T x ≤ 0 } and D = = { x ∈ S p − 1 | β 0 T x = 0 } . Then the strong consistency of β ˆ (Corollary 3 of [1]) implies ⋂ n ∈ N ⋃ N ≥ n D ¯ ( β ˆ N ) ⊆ D = a.s. Hence P ( n ⋅ | U 1 ( β ˆ ) − E [ U 1 ( β ˆ ) | β ˆ ] − U 1 ( β 0 ) + E U 1 ( β 0 ) | > ϵ ) = E [ P ( n ⋅ | U 1 ( β ˆ ) − E [ U 1 ( β ˆ ) | β ˆ ] − U 1 ( β 0 ) + E U 1 ( β 0 ) | > ϵ | β ˆ ) ] ≤ E [ 8 ϵ 2 ⋅ P ( X 2 − X 1 ‖ X 2 − X 1 ‖ ∈ D ¯ ( β ˆ ) | β ˆ ) + 8 n ϵ 2 ] ≤ E [ 8 ϵ 2 ⋅ P ( X 2 − X 1 ‖ X 2 − X 1 ‖ ∈ ⋃ N ≥ n D ¯ ( β ˆ N ) ∪ D = | β ˆ ) + 8 n ϵ 2 ] ⟶ E [ 8 ϵ 2 ⋅ P ( X 2 − X 1 ‖ X 2 − X 1 ‖ ∈ D = | β ˆ ) ] = 8 ϵ 2 ⋅ P ( β 0 T X 1 = β 0 T X 2 ) = 0 , where the first inequality is derived by arguments of page 78 of the article.

On page 79, line 14/15, the central limit theorem of [2] cannot be used. Instead one has to use Corollary 3 of [1].