Jackknife empirical likelihood for the difference of two volumes under ROC surfaces

Abstract

The volume under a surface (VUS) is an effective measure for evaluating the discriminating power of a diagnostic test with three ordinal diagnostic groups. In this paper, we investigate the difference of two correlated VUS’s to compare two treatments for discrimination of three-class classification data. A jackknife empirical likelihood (JEL) procedure is employed to avoid the variance estimation in the existing methods. We prove that the limiting distribution of the empirical log-likelihood ratio statistic follows a $\chi ^2$ distribution. Extensive numerical studies show that the JEL confidence intervals outperform those based on the normal approximation method. The proposed method is also applied to the Alzheimer’s disease data.

Appendix: Proof of Theorem 1

The variance $Var(U_n)$ can be estimated by a consistent estimator ${\hat{\sigma }^2}$ as in Sen (1960) and Arvesen (1969),

$$\begin{aligned} {\hat{\sigma }^2}= & {} \dfrac{1}{n_1(n_1-1)}\displaystyle \sum _{i=1}^{n_1}(V_{i,0,0}-\bar{V}_{\cdot ,0,0})^2 +\dfrac{1}{n_2(n_2-1)}\displaystyle \sum _{j=1}^{n_2}(V_{0,j,0}-\bar{V}_{0,\cdot ,0})^2\\&+\dfrac{1}{n_3(n_3-1)}\displaystyle \sum _{k=1}^{n_3}(V_{0,0,k}-\bar{V}_{0,0,\cdot })^2. \end{aligned}$$

Lemma 1

We have the following conclusions.

• (a) The U-statistic $U_n\buildrel a.s. \over \rightarrow \theta _0$ as $min(n_1, n_2, n_3)\rightarrow \infty$;

• (b) Suppose that $\sigma ^2_{1,0,0}>0$, $\sigma ^2_{0,1,0}>0$, $\sigma ^2_{0,0,1}>0$, and denote $S^2_{n_1, n_2, n_3}=\sigma ^2_{1,0,0}/n_1+\sigma ^2_{0,1,0}/n_2+\sigma ^2_{0,0,1}/n_3$. As $\quad min(n_1, n_2, n_3)\rightarrow \infty ,$

  $$\begin{aligned} \dfrac{U_n-\theta _0}{S_{n_1, n_2, n_3}}\buildrel d \over \rightarrow N(0,1),\quad \end{aligned}$$

  (3)

  and

  $$\begin{aligned} \hat{\sigma }^2-S_{n_1, n_2, n_3}^2=o_p((min(n_1, n_2, n_3))^{-1}). \end{aligned}$$

  (4)

The proof of part (a) and Eqs. (3) and (4) can be found in Arvesen (1969) and Kowalski and Tu (2007).

Lemma 2

Let $S_n=n^{-1}\sum \nolimits _{l=1}^n(\hat{V}_l-E\hat{V}_l)^2$. We assume the same conditions as (a) and (b) in Theorem 1. Then as $n_1\rightarrow \infty$,

$$\begin{aligned} S_n=nS^2_{n_1,n_2,n_3}+o_p(1). \end{aligned}$$

Proof of Lemma 2

For $1\le l\le n_1$, it is clear that

$$\begin{aligned} {\hat{V_l}}-E{\hat{V_l}} =\dfrac{n(n-1)}{(n-3)n_1}(V_{l,0,0}-U_n)+\dfrac{n(n-2n_1-1)}{(n-3)n_1}(U_n-\theta _0), \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\dfrac{1}{n_1}\displaystyle \sum _{l=1}^{n_1}(V_{l,0,0}-U_n)(U_n-\theta _0)\\ =&\,(U_n{-}\theta _0)\left\{ \dfrac{1}{n_1n_2n_3}\displaystyle \sum _{i=1}^{n1}\sum _{j=1}^{n2}\sum _{k=1}^{n3} [I(X_{1i}{<}Y_{1j}{<}Z_{1k})-I(X_{2i}{<}Y_{2j}<Z_{2k})]-U_n\right\} \\ =&\,0 .\end{aligned} \end{aligned}$$

As Pan et al. (2013) and Wang (2010) did, we have that

$$\begin{aligned} \displaystyle \sum _{l=1}^{n_1}(\hat{V}_l{-}E\hat{V}_l)^2 =\left[ \dfrac{n(n{-}1)}{(n{-}3)n_1}\right] ^2\displaystyle \sum _{l=1}^{n_1}(V_{l,0,0}-U_n)^2 {+}\left[ \dfrac{n(n{-}2n_1{-}1)}{(n{-}3)n_1}\right] ^2n_1(U_n-\theta _0)^2. \end{aligned}$$

For $n_1< l\le n_1+n_2$, one has that

$$\begin{aligned} \displaystyle \sum _{l=n_1+1}^{n_1+n_2}(\hat{V}_l-E\hat{V}_l)^2= & {} \left[ \dfrac{n(n-1)}{(n-3)n_2}\right] ^2\displaystyle \sum _{l=n_1}^{n_1+n_2}(V_{0,l,0}-U_n)^2\\&+\left[ \dfrac{n(n-2n_2-1)}{(n-3)n_2}\right] ^2n_2(U_n-\theta _0)^2. \end{aligned}$$

For $n_1+n_2< l\le n$, we have that (see Pan et al. 2013)

$$\begin{aligned} \displaystyle \sum _{l=n_1+n_2+1}^{n}(\hat{V}_l-E\hat{V}_l)^2= & {} \left[ \dfrac{n(n-1)}{(n-3)n_3}\right] ^2\displaystyle \sum _{l=n_1+n_2}^{n}(V_{0,0,l}-U_n)^2\\&+\left[ \dfrac{n(n-2n_3-1)}{(n-3)n_3}\right] ^2n_3(U_n-\theta _0)^2. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} S_n =&\,\dfrac{1}{n}\left[ \dfrac{n(n-1)}{(n-3)}\right] ^2 \left[ \dfrac{1}{n_1^2}\displaystyle \sum _{l=1}^{n_1}(V_{l,0,0}-\bar{V}_{\cdot ,0,0})^2 +\dfrac{1}{n_2^2}\displaystyle \sum _{l=n_1+1}^{n_1+n_2}(V_{0,l,0}-\bar{V}_{0,\cdot ,0})^2\right. \\&\left. +\dfrac{1}{n_{3}^2}\displaystyle \sum _{l=n_1+n_2+1}^{n}(V_{0,0,l}-\bar{V}_{0,0,\cdot })^2\right] \\&+\dfrac{1}{n}\left[ \dfrac{n}{(n-3)}\right] ^2 \left[ \dfrac{(n-2n_1-1)^2}{n_1}+\dfrac{(n-2n_2-1)^2}{n_2}+\dfrac{(n-2n_3-1)^2}{n_3}\right] (U_n- \theta _0)^2 .\end{aligned} \end{aligned}$$

(5)

From the LLN of U-statistic, we have the conclusion $U_n-\theta _0=O_p(n_1^{-1/2})$. The second term in Eq. (5) is

$$\begin{aligned} \dfrac{n}{(n-3)^2}&\left[ \dfrac{(n-2n_1-1)^2}{n_1}+\dfrac{(n-2n_2-1)^2}{n_2}+\dfrac{(n-2n_3-1)^2}{n_3}\right] (U_n- \theta _0)^2\\&\quad =O_p(n^{-1}). \end{aligned}$$

Moreover, the 1st term of Eq. (5) is (cf. Wang 2010)

$$\begin{aligned} \begin{aligned}&n\left( \dfrac{n-1}{n-3}\right) ^2 \left[ \dfrac{1}{n_1^2}\displaystyle \sum _{l=1}^{n_1}\left( V_{l,0,0}-\bar{V}_{\cdot ,0,0}\right) ^2 +\dfrac{1}{n_2^2}\displaystyle \sum _{l=n_1+1}^{n_1+n_2}\left( V_{0,l,0}-\bar{V}_{0,\cdot ,0}\right) ^2\right. \\&\left. +\dfrac{1}{n_3^2}\displaystyle \sum _{l=n_1+n_2+1}^{n}(V_{0,0,l}-\bar{V}_{0,0,\cdot })^2\right] \\ =&\,n\hat{\sigma }^2+o_p(1) .\end{aligned} \end{aligned}$$

Using Eq. (4), we prove Lemma 2. $\square$

Lemma 3

Let $Q_n=\displaystyle \max _{1\le l\le n}|\hat{V}_l-\theta _0|$. Under the same conditions as in Lemma 2, we have $Q_n=o_p(n^{1/2})$ and $n^{-1}\sum \nolimits _{l=1}^n|\hat{V}_l-\theta _0|^3=o_p(n^{1/2})$.

Proof of Lemma 3

For $1\le l\le n_1$, we have (see Wang 2010)

$$\begin{aligned} |\hat{V}_l-E\hat{V}_l| \le \Big |\dfrac{n}{n_1}\dfrac{n-1}{n-3}V_{l,0,0}\Big | +\Big |\dfrac{n}{n_1}\dfrac{n-1}{n-3}U_n\Big |+\Big |\dfrac{n(n-2n_1-1)}{(n-3)n_1}(U_n-\theta _0)\Big |. \end{aligned}$$

Note that $|V_{l,0,0}|\le \tilde{H}_n$, and $|U_n|\le \tilde{H}_n$, where

$$\begin{aligned} \tilde{H}_n=\displaystyle \max _{1\le i\le n_1<l\le n_1+n_2<k\le n}|h(X_i,Y_j,Z_k)|. \end{aligned}$$

Therefore, $|\hat{V}_l-E\hat{V}_l| \le c^*\tilde{H}_n+c^*\tilde{H}_n+ c^{*}|U_n-\theta |$, where $c^*$ is a constant. Similar to Pan et al. (2013) and Wang (2010), we have $\tilde{H}_n=o_p(n^{1/2})$ and $U_n-\theta _0 =O_p(n^{-1/2})$. Therefore, $|\hat{V}_l-E\hat{V}_l|=o_p(n^{1/2})$ for $1\le l \le n_1$. For $n_1<l\le n_1+n_2$, $|\hat{V}_l-E\hat{V}_l| \le 2c^*\tilde{H}_n+c^*|U_n-\theta _0|.$ Thus, $|\hat{V}_l-E\hat{V}_l|=o_p(n^{1/2})$ for $n_1<l\le n_1+n_2$. For $n_1+n_2<l\le n$, we have $|\hat{V}_l-E\hat{V}_l|=o_p(n^{1/2})$ similarly as before. Thus $Q_n=o_p(n^{1/2}).$ Hence,

$$\begin{aligned} n^{-1}\displaystyle \sum _{l=1}^n|\hat{V}_l-E \hat{V}_l|^3 = o_p(n^{1/2})(nS^2_{n_1,n_2,n_3}+o_p(1)) = o_p(n^{1/2}). \end{aligned}$$

$\square$

Proof of Theorem 1

The proof follows the same lines of Pan et al. (2013). Recall $U_n={1}/{n}\displaystyle \sum _{l=1}^n\hat{V}_l$ and $\theta _0={1}/{n}\displaystyle \sum _{l=1}^n E \hat{V}_l$. Then

$$\begin{aligned} |U_n-\theta _0| \ge |\gamma |\dfrac{1}{1+|\gamma |\max |\hat{V}_l-E \hat{V}_l|}\dfrac{1}{n}\displaystyle \sum _{l=1}^n(\hat{V}_l-E \hat{V}_l)^2 \ge |\gamma |\dfrac{S_n}{1+|\gamma |Q_n} . \end{aligned}$$

We have that $|\gamma |=O_p(n^{-1/2})$. Using Taylor’s expansion, one has that

$$\begin{aligned} \begin{aligned} -2\log R(\theta _0) =&\,2\displaystyle \sum _{l=1}^n\left\{ \gamma (\hat{V}_l-E \hat{V}_l)-\dfrac{1}{2}\left[ \gamma (\hat{V}_l-E \hat{V}_l)\right] ^2\right\} +o_p(1) .\end{aligned}\end{aligned}$$

(6)

Let $F_0= {1}/{n}\displaystyle \sum _{l=1}^n\dfrac{\gamma ^2(\hat{V}_l-E \hat{V}_l)^3}{1+\gamma (\hat{V}_l-E \hat{V}_l)}$. By Eq. (2), we have that

$$\begin{aligned} \begin{aligned} 2n\gamma (U_n-\theta _0)-nS_n\gamma ^2 = n\dfrac{(U_n-\theta _0)^2}{S_n} - \dfrac{n F_0^2}{S_n} .\end{aligned} \end{aligned}$$

One can obtain $F_0 =o_p(n^{-1/2})$. Thus Eq. (6) can be re-expressed as follows,

$$\begin{aligned} -2\log R(\theta _0)= n\dfrac{(U_n-\theta _0)^2}{S_n}+o_p(1). \end{aligned}$$

Combining Lemmas 1, 2 and 3, we finish the proof. $\square$