Theorem 1

Let Y0 and Y1 denote the biomarker in the negative population and in the positive population, respectively. Assume that Yi is normally distributed with mean μi and standard deviation σi, for i ∈ {0,1}. Assume that two independent samples of m of i.i.d. observations from Y0 and n i.i.d. observations from Y1 are available. The sample sizes satisfy

A1.-n/m→nλ>0.

Then, we have the following weak convergences,

1. n⋅{JˆgP−Jg}⟶LnσJP⋅Z and,

2. n⋅{tˆYP−tY}⟶LnσtYP⋅Z,

where Z is a standard normal random variable and σJP and σtYP are given in eqs. (3) and (4), respectively.

Proof

Results in [14] allow to derive that the binormal estimation of the gROC curve are based on intervals of the form (−∞,a/(1−b2)−x]∪[a/(1−b2)+x,∞), where x is a non-negative real number, a=(μ1−μ0)/σ0 and b=σ1/σ0. Then, direct calculations lead to deduce that the interval leading to the the value of 1−specificity for Jg is achieved is of the form

These points are the ones where the normal densities associated to the positive and negative populations cross each other. Therefore, the Youden index is

Let ξˆi and Sˆi be the sample mean and the sample standard deviation, which estimate μi and σi, respectively (i ∈ {0,1}). Then, we directly have that JˆgP=Jg(ξˆ1,Sˆ1,ξˆ0,Sˆ0). In addition, we have the following convergences in distribution:

Besides, the independence among the four random variables is also well-known [25]. Hence, the multivariate delta method ensures the weak convergence stated in (i).

In order to prove (ii), we just consider the value of 1−specificity for which the point associated to the Youden index is achieved,

Arguing as above, and taking into account that tˆYP=tY(ξˆ1,Sˆ1,ξˆ0,Sˆ0), the result can be obtained again by applying the multivariate delta method.   □

Theorem 2

Let Y0 and Y1 be the random variables modelling the biomarker behavior in the positive and negative and positive populations, respectively. Assume that the resulting gROC curve, Rg, satisfies

A2.-Rg has two continuous derivatives on some subinterval (a0,b0)⊂[0,1], such as tY∈(a0,b0), where tY=argmaxt∈[0,1]{Rg(t)−t}.

A3.-|rg′(tY)|=a>0, where rg′(t)=∂rg(t)/∂t and rg(t)=∂Rg(t)/∂t.

Assume that two independent samples of m of i.i.d. observations from Y0 and n i.i.d. observations from Y1 are available.

Then, if the sample sizes satisfy A1, the following weak convergences hold

1. n⋅{Jˆg−Jg}⟶LnσJ⋅Z    and

2. r′(tY)24⋅(1+λ)1/3⋅n1/3⋅{tˆY−tY}⟶Lnargmaxz∈R{Z(n)(z)−z2},

where Z is a standard normal random variable, {Z(n)(z)}{−∞<z<∞} is a two-sided Brownian motion and σJ2=Rg(tY)[1−Rg(tY)]+λ⋅tY⋅[1−tY].

Proof

Let C[Rg] the subset containing all the pairs (uL,uU) such that there exists t ∈ [0,1] satisfying that Rg(t)=F(uL)+1−F(uU). Then,

Let (xL,xU) be the pair of points that leads to Jg, that is, Jg=F(xL)−F(xU)+G(xU)−G(xL). Therefore,

where

On the one hand, assumption A2 allows to apply the Hungarian embedding [26] to derive that the random variables Zˆn(xL,xU) and

where {B1(n)(t)}{0≤t≤1} and {B2(m)(t)}{0≤t≤1} are two independent Brownian bridges, have the same asymptotic distribution. Taking into account basic properties of the Brownian bridge and A1, it can easily be shown that the random variable eq. (6) and

where Z is a standard normal, also coincide in distribution.

On the other hand, from A2 and A3, and since rg(tY)=1, for t close to tY, Rg(t)−Rg(tY) can be approximated by t−tY. The Brownian bridge and the two-sided Brownian motion properties [27] guarantee that, for t close to tY and under A3, the random variable {Hˆn(uL,uU,xL,xU)−H(uL,uU,xL,xU)} has the same asymptotic distribution of

where {Z1(n)(z)}{−∞<z<∞} and {Z2(m)(z)}{−∞<z<∞} are independent two-sided Brownian motions and {Z(n)(z)}{−∞<z<∞} is the weighted sum of those independent two-sided Brownian motions and, therefore, a two-sided Brownian motion as well.

Also, by assumptions A2 and A3, we have the approximation

Therefore, from eqs. (7) and (8), we have that the random variable

weakly converges to the distribution of the random variable

where z=(t−tY)/γ with γ=(4⋅(1+λ)/rg′(tY)2)1/3⋅n−1/3, κ=(2(1+λ)2/rg′(tY))1/3 and {Z(n)(z)}{−∞<z<∞} is a two-sided Brownian motion. We obtain i) directly from the equality eq. (5) and the convergences in eqs. (6) and (9).

On the other hand, if tˆY is the point which maximizes (Rˆg(t)−t), then from eqs. (5) and (9), (tˆY−tY)/γ has the same asymptotic distribution that argmaxz∈R{Z(n)(z)−z2}. Result in ii) is derived from the equality