Nonparametric statistical inference for P(X < Y < Z)

Abstract

Let X, Y and Z be three independent random variables from three different populations. The stress-strength model P(X < Y < Z), the volume under the three-class ROC surface, has extensive applications in various areas since it provides a global measure of differences between or among populations. In this paper, we suggest to make statistical inference for P(X < Y < Z) via two methods, the nonparametric normal approximation and the jackknife empirical likelihood, since the usual empirical likelihood method for U-statistics is too complicated to apply. The results of the simulation studies indicate that these two methods work promisingly compared to other existing methods. Some classical and real data sets were analyzed using these two proposed methods. Practically, for simplicity, the nonparametric normal approximation method should be preferred; for better statistical results, one is suggested to use the JEL method although it is more complex than the normal approximation one.

Appendix: Proof of Theorem 2.2

In this part, we provide the technique details to prove Theorem 2.2. In fact, we prove it for the general three-sample U-statistic U. Before proceeding to the proof of Theorem 2.2, we list some results that will be used. Referring to the proofs below, without loss of generality, we may assume that n ₁ ≤ n ₂ ≤ n ₃ thereafter.

As a direct consequence of Theorem 2.1, we have

$$U-\theta_0=O_p(n^{-1/2}_1). \label{append5} $$

(A.1)

The following Lemma guarantees the existence and uniqueness of the solution to equation (2.8).

Lemma 5.1

Suppose that $\sigma_{1,0,0}^2>0$ , $\sigma_{0,1,0}^2> 0$ , $\sigma_{0,0,1}^2> 0, \liminf_{n\to \infty}(n_1/$ n ₂) > 0, and $\liminf_{n\to \infty}(n_2/n_3)>0$ . Then as n ₁→ ∞, we have

$$ P\left\{\underset{1\leq i\leq n}{\emph{min}}(\widehat V_i-E\widehat V_i)<0<\underset{1\leq i\leq n}{\emph{max}}(\widehat V_i-E\widehat V_i)\right\}\longrightarrow1. $$

Proof

The proof is similar to the two-sample case. The reader is referred to Jing et al. (2009) for the detailed proof.

Let $S_n=n^{-1}\sum_{i=1}^{n}(\widehat V_i-E\widehat V_i)^2$ and $Q_n=\underset{1\leq i\leq n}{\text{max}}|\widehat V_i-E\widehat V_i|$. Under the conditions of Lemma 5.1, some calculations reveal that, as n ₁→ ∞, with probability 1

$$S_n=nS^2_{n_1,n_2,n_3}+o(1),~~Q_n=o(n^{1/2}) \label{append8} $$

(A.2)

and

$$ n^{-1}\sum_{i=1}^{n}|\widehat V_i-E\widehat V_i|^3\leq S_n\times Q_n\leq o(n^{1/2}). \label{append9} $$

Proof of Theorem 2.2. By Lemma 5.1, the solution to equation (2.8) exists and is unique. We next show that this solution γ satisfies $|\gamma|=O_p(n^{-1/2})$. Noting that, (2.8) together with the fundamental inequality |x±y| ≥ |x| − |y| leads to

$$\begin{array}{lll} 0=|f(\gamma)|&=\frac{1}{n}\left|\sum_{i=1}^{n}(\widehat V_i-E\widehat V_i)-\gamma\sum_{i=1}^{n}\frac{(\widehat V_i-E\widehat V_i)^2}{1+\gamma(\widehat V_i-E\widehat V_i)}\right|\\ &\geq\frac{|\gamma|S_n}{1+|\gamma|Q_n}-\frac{1}{n}\left|\sum_{i=1}^{n}\widehat V_i-\theta_0\right|. \end{array}$$

By (10), the second term is $O_p(n_1^{-1/2})$. By (A.1), $S_n=nS^2_{n_1,n_2,n_3}+o(1)~a.s.$, it follows that $|\gamma|(1+|\gamma|Q_n)^{-1}=O_p(n^{-1/2})$, hence by (A.2) again, $|\gamma|=O_p(n^{-1/2})$. Further, if let $\beta_i=\gamma(\widehat V_i-E\widehat V_i)$, then

$$\begin{array}{ll} \underset{1\leq i\leq n}{\text{max}}|\beta_i|&=|\gamma|\underset{1\leq i\leq n}{\text{max}}|\widehat V_i-E\widehat V_i| \\ &=O_p(n^{-1/2})o(n^{1/2})=o_p(1). \label{append10} \end{array} $$

(A.3)

On the one hand, expanding (2.8), we get

$$\begin{array}{lll} 0&=f(\gamma)\\ &=\frac{1}{n}\sum_{i=1}^n\widehat V_i-\theta_0-\gamma S_n+\frac{1}{n}\sum_{i=1}^n\frac{(\widehat V_i-E\widehat V_i)\beta_i^2}{1+\beta_i}, \end{array}$$

where the last term is bounded by

$$ \frac{1}{n}\sum_{i=1}^n\frac{|\widehat V_i-E\widehat V_i|^3}{|1+\beta_i|}\gamma^2=o(n^{1/2})O_p(n^{-1})O_p(1)=o_p(n^{-1/2}). $$

Therefore, we may write

$$\gamma=\left(\frac{1}{n}\sum_{i=1}^n\widehat V_i-\theta_0\right)S_n^{-1}+\tau=(U-\theta_0)S_n^{-1}+\tau \label{append11} $$

(A.4)

where $|\tau|=o_p(n^{-1/2})$.

On the other hand, in virtue of (A.3) and by a Taylor expansion, we have $\text{log}(1+\beta_i)=\beta_i-\beta_i^2/2+\alpha_i$, where for some finite A > 0, $P\{|\alpha_i|\leq A|\beta_i|^3,1\leq i\leq n\}\rightarrow1$ as n→ ∞. Then plugging (A.4) and (2.7) into (2.6), we get

$$\begin{array}{lll} -2\text{log}R(\theta_0)&=-2\sum_{i=1}^n\text{log}(np_i)=2\sum_{i=1}^n\text{log}(1+\beta_i)\\ &=2n\gamma(U-\theta_0)-nS_n\gamma^2+2\sum_{i=1}^n\alpha_i\\ &=\frac{n(U-\theta_0)^2}{S_n}-nS_n\tau^2+2\sum_{i=1}^n\alpha_i, \end{array}$$

where

$$\begin{array}{rll}|-nS_n\tau^2|&=n(nS_{n_1,n_2,n_3}^2+o(1))o_p(n^{-1})=o_p(1),\\ |\sum_{i=1}^n\alpha_i|&\leq A|\gamma|^3\sum_{i=1}^n|\widehat V_i-\theta_0|^3=O_p(n^{-3/2})o(n^{3/2})=o_p(1) \end{array}$$

and by Theorem 2.1 and (A.2), as n→ ∞,

$$ \frac{n(U-\theta_0)^2}{S_n}\overset{d}{\longrightarrow}\chi_1^2. $$

Hence, from Slutsky’s theorem, we have $-2\text{log}R(\theta_0)\rightarrow_d\chi_1^2$, which completes the proof.