Assessing the diagnostic power of variables measured with a detection limit

Abstract

The phenomenon of the limit of detection (LoD) often happens in many practical situations because of technique and instrument limitations. In the literature, some reports show that, in general, to apply conventional methods to evaluate the diagnostic power of variables while ignoring LoD could be seriously biased. Although the area under the receiver operating characteristic (ROC) curve can be estimated consistently if the distribution of variables are known. In practical situation, such information is usually not available. On the other hand, it has been proved that the area under ROC curve of a variable with a LoD and no distribution assumptions is usually biased no matter what kinds of replacement strategies are used. However, there is a lack of similar studies on the partial area under ROC curve (pAUC), and because this measure is usually preferred in practice, it is of interest to examine whether the estimate of pAUC of a variable measured with a LoD behaves the same. In this study, we found that for some LoD scenarios, and even without distribution assumption, consistent estimate of pAUC can be constructed. When the consistent estimate of pAUC cannot be obtained, the bias can be ineffectual in practical situations, and the proposed estimator can be a good approximation of pAUC. Numerical studies using simulated data sets and real data examples are reported.

Appendix

Proof of Theorem 2.1

From (2) and (5), we know that

$$\begin{aligned} pAUC(u)= & {} P\left( Y\ge X, X\ge S_{\bar{D}}^{-1}(u)\right) =P\left( Y\ge X,X\ge c\right) \\ pAUC^*(u)= & {} P\left( \tilde{Y}\ge \tilde{X}, \tilde{X} \ge S_{\bar{D*}}^{-1}(u)\right) =P\left( \tilde{Y}\ge \tilde{X}, \tilde{X}\ge c^*\right) , \end{aligned}$$

where $c=S_{\bar{D}}^{-1}(u)$ and $c^*=S_{\bar{D*}}^{-1}(u)$ are $1-u$ quantiles of X and $\tilde{X}$, respectively. By total probability formula, we have

$$\begin{aligned} pAUC(u)= & {} P\left( Y\ge X,X\ge c|Y\ge d,X\ge d\right) P\left( Y\ge d,X\ge d\right) \nonumber \\&+\,P\left( Y\ge X,X\ge c|Y\ge d,X\le d\right) P\left( Y\ge d,X\le d\right) \nonumber \\&+\,P\left( Y\ge X,X\ge c|Y\le d,X\ge d\right) P\left( Y\le d,X\ge d\right) \nonumber \\&+\,P\left( Y\ge X,X\ge c|Y\le d,X\le d\right) P\left( Y\le d,X\le d\right) \nonumber \\:= & {} {I}+{J}+{K}+{L}, \end{aligned}$$

(8)

$$\begin{aligned} {pAUC^*(u)}= & {} P\left( Y\ge X,X\ge c^*|Y\ge d,X\ge d\right) P\left( Y\ge d,X\ge d\right) \nonumber \\&+\,P\left( Y\ge \widetilde{X},\widetilde{X}\ge c^*|Y\ge d,X\le d\right) P\left( Y\ge d,X\le d\right) \nonumber \\&+\,P\left( \widetilde{Y}\ge X,X\ge c^*|Y\le d,X\ge d\right) P\left( Y\le d,X\ge d\right) \nonumber \\&+\,P\left( \widetilde{Y}\ge \widetilde{X},\widetilde{X}\ge c^*|Y\le d,X\le d\right) P\left( Y\le d,X\le d\right) \nonumber \\:= & {} I^*+J^*+K^*+L^*. \end{aligned}$$

(9)

By the definitions of $\widetilde{X}$ and $\widetilde{Y}$, we know that $\widetilde{X}=r$ and $\widetilde{Y}=r$ when $X<d$ and $Y<d$. Then components of (9) can be rewritten as

$$\begin{aligned}&J^*=P\left( Y\ge r,r\ge c^*|Y\ge d,X\le d\right) P\left( Y\ge d,X\le d\right) ,\nonumber \\&K^*=P\left( X\le r,X\ge c^*|Y\le d,X\ge d\right) P\left( Y\le d,X\ge d\right) ,\nonumber \\&L^*=P\left( \widetilde{Y}\ge \widetilde{X},r\ge c^*|Y\le d,X\le d\right) P\left( Y\le d,X\le d\right) . \end{aligned}$$

(10)

(1) When $S_{\bar{D}}(d)>u$ and $S_{\bar{D*}}(d) > u$, both of $c=S_{\bar{D}}^{-1}(u)$ and $c^*=S_{\bar{D*}}^{-1}(u)$ are not smaller than the lower bound d. Therefore, it follows from definition of quantile and continuous property of X that $c=c^*\ge d$. Thus, from continuous properties of X and Y and $r\le d$, we have $I={I^*}$, $J={J^*}=0$, $K={K^*}=0$, $L={L^*}=0$, which suggests $pAUC^*(u)={pAUC(u)}$.

(2) When $S_{\bar{D}}(d)>u$ and $S_{\bar{D*}}(d)<u$, the inequality $c>d>c^*$ holds. Thus, $I^*>I$, $J=0$, ${J^*}\ge 0$, $K^*=K=0$, $L=0$, ${L^*}\ge 0$. Hence, we get $pAUC^*(u)-{pAUC(u)}>0$. Similarly, If $S_{\bar{D}}(d)<u$ and $S_{\bar{D*}}(d)>u$, then $c<d<c^*$ and $pAUC^*(u)-{pAUC(u)}<0$. In conclusion, under situation (2), $|pAUC^*(u)-{pAUC(u)}|>0$.

(3) When $S_{\bar{D}}(d)<u$ and $S_{\bar{D*}}(d)<u$, we have $c<d$ and $c^*<d$. This situation can be splitted into two parts, (i) $c^*\le c<d$ and (ii) $c<c^*<d$.

If (i) holds, then $I=I^*,~J<J^*,~K=K^*=0,~L<L^*$, which indicates that $pAUC^*(u)-{pAUC(u)}>0$. And if (ii) holds, then $I={I^*},~J\le J^*,~K={K^*}=0$ and $L<{L^*}$. So $(pAUC^*(u)-{pAUC(u)})>0$. Hence, the conclusion holds. $\square$

Proof of Theorem 2.2

Similar to proof of Theorem 2.1, we have

$$\begin{aligned} pAUC(u)= & {} P\left( Y\ge X, X\ge S_{\bar{D}}^{-1}(u)\right) =P\left( Y\ge X,X\ge c\right) \\ pAUC^*(u)= & {} P\left( \tilde{Y}\ge \tilde{X}, \tilde{X} \ge S_{\bar{D*}}^{-1}(u)\right) =P\left( \tilde{Y}\ge \tilde{X}, \tilde{X}\ge c^*\right) , \end{aligned}$$

where $c=S_{\bar{D}}^{-1}(u)$ and $c^*=S_{\bar{D*}}^{-1}(u)$ are $1-u$ quantiles of X and $\tilde{X}$, respectively. By total probability formula, we have

$$\begin{aligned} pAUC(u)= & {} P\left( Y\ge X,X\ge c|Y\ge d,X\ge d\right) P\left( Y\ge d,X\ge d\right) \nonumber \\&+\,P\left( Y\ge X,X\ge c|Y\ge d,X\le d\right) P\left( Y\ge d,X\le d\right) \nonumber \\&+\,P\left( Y\ge X,X\ge c|Y\le d,X\ge d\right) P\left( Y\le d,X\ge d\right) \nonumber \\&+\,P\left( Y\ge X,X\ge c|Y\le d,X\le d\right) P\left( Y\le d,X\le d\right) \nonumber \\:= & {} {I}+{J}+{K}+{L}, \end{aligned}$$

(11)

$$\begin{aligned} {pAUC^*(u)}= & {} P\left( \widetilde{Y}\ge \widetilde{X},\widetilde{X}\ge c^*|Y\ge d,X\ge d\right) P\left( Y\ge d,X\ge d\right) \nonumber \\&+\,P\left( \widetilde{Y}\ge X,X\ge c^*|Y\ge d,X\le d\right) P\left( Y\ge d,X\le d\right) \nonumber \\&+\,P\left( Y\ge \widetilde{X},\widetilde{X}\ge c^*|Y\le d,X\ge d\right) P\left( Y\le d,X\ge d\right) \nonumber \\&+\,P\left( Y\ge X, X\ge c^*|Y\le d,X\le d\right) P\left( Y\le d,X\le d\right) \nonumber \\:= & {} I^*+J^*+K^*+L^*. \end{aligned}$$

(12)

By the definitions of $\widetilde{X}$ and $\widetilde{Y}$, we know that $\widetilde{X}=r$ and $\widetilde{Y}=r$ when $X>d$ and $Y>d$. Then components of (12) can be rewritten as

$$\begin{aligned}&I^*=P\left( r\ge r,r\ge c^*|Y\ge d,X\ge d\right) P\left( Y\ge d,X\ge d\right) ,\\&J^*=P\left( r\ge X,X\ge c^*|Y\ge d,X\le d\right) P\left( Y\ge d,X\le d\right) ,\\&K^*=P\left( Y\ge r,r\ge c^*|Y\le d,X\ge d\right) P\left( Y\le d,X\ge d\right) . \end{aligned}$$

(1) When $S_{\bar{D}}(d)<u$, $c=S_{\bar{D}}^{-1}(u)$ is smaller than d and then $c^*=c$. Therefore, ${I^*}=P(Y\ge d,X\ge d)$, ${J^*}=P(X\ge c|Y\ge d,X\le d)P(Y\ge d,X\le d)$, ${K^*}=0$ and ${L^*}=P(Y\ge X, X\ge c|Y\le d,X\le d)P(Y\le d,X\le d)$, which suggests that ${pAUC^*(u)}$ is free of choice of the replacement value r and is a constant. Moreover, by Eq. (11), ${pAUC^*(u)}-pAUC(u)= P(Y\ge d,X\ge d)-P(Y\ge X,X\ge c|Y\ge d,X\ge d)P(Y\ge d,X\ge d) =P(Y<X,Y\ge d,X\ge d)$.

(2) when $S_{\bar{D}}(d)>u$, $c^*=r$ and ${I^*}=P(Y\ge d,X\ge d)$, ${J^*}=0$, ${K^*}=0$ and ${L^*}=P(Y\ge X, X\ge r|Y\le d,X\le d)P(Y\le d,X\le d)=0$. So the ${pAUC^*(u)}=P(Y\ge d,X\ge d)$ becomes a constant not varying with the replacement value r. Hence, the conclusion holds. $\square$