The Cucconi statistic for Type-I censored data

Abstract

Various nonparametric test statistics have been proposed for censored data. Two-sample nonparametric testing plays an important role in biometry. While most of two-sample nonparametric tests intend to detect a shift in location or in scale, the two-sample Cucconi test statistic is suitable for the joint comparison of both parameters. The Cucconi test statistic is extended to the left- and right-censored data based on the theory of ties. We derive the limiting distribution of the Cucconi test statistic for censored data. We conduct simulation studies to investigate the convergence of the Cucconi test statistic to the limiting distribution and the power of the proposed statistic with various population distributions. The method is illustrated with an analysis using real data.

Appendices

Proof of Theorem 1

First, we introduce Lemma 1 to show the Theorem 1.

Lemma 1

(Cucconi 1968) Let $Y_1,Y_2$ be the random variables from the standard normal distribution and $\rho _*=\mathrm{Corr}(Y_1,Y_2)$. Then,

$$\begin{aligned} T=\frac{Y_1^2+Y_2^2-2\rho _* Y_1Y_2}{2(1-\rho _*^2)} \end{aligned}$$

is distributed as the standard exponential distribution.

If we obtain that $U_L$ ($U_R$) and $V_L$ ($V_R$) have the asymptotic normality for the left- or right-censored data, it is easily to prove the theorem by using Lemma 1. Let $R_{1j}^*$ be the rank of the uncensored $X_1$’s in $N-r_1-r_2$ uncorrelated observations. In general, the sum of $R{^{*}_{1j}}^{\beta }$ ($\beta >0$) is asymptotically distributed according to normal introduced by Halperin (1960).

1.1 Proof for the left-censored data

$Q^{L}_{1}$ and $Q^{L}_{2}$ are decomposed as follows:

$$\begin{aligned} Q^{L}_{1}= & {} \frac{r_1(r+1)(2r+1)}{6} + \sum _{j=r_1+1}^{n_1} R^2_{1j} = \frac{r_1(r+1)(2r+1)}{6} + \sum _{j=1}^{n'_1} (r+R^{*}_{1j})^2\\= & {} \frac{r_1(r+1)(2r+1)}{6} + \sum _{j=1}^{n'_1} \left( R^{*}_{1j} -\frac{N'+1}{2} \right) ^2 + (N+r+1)\sum _{j=1}^{n'_1} R^{*}_{1j} \\&-\frac{1}{4} n'_1(N+r+1)(N-3r+1) \\= & {} A_1+A_2+A_3+A_4\quad (\text {say}).\\ Q^{L}_{2}= & {} \frac{r_1}{r} \sum _{j=1}^{r} (N+1-j)^2+ \sum _{j=r_1+1}^{n_1} (N+1-R_{1j})^2 \\= & {} Q^{L}_{1} +n_1(N+1)^2 -r_1(N+1)(r+1) -2r (N+1)n'_1 \\&-2(N+1) \sum _{j=1}^{n'_1} R^*_{1j}\\= & {} Q^{L}_{1}+B_1+B_2+B_3+B_4\quad (\mathrm{say}), \end{aligned}$$

where $N'=N-r$, $n'_1=n_1-r_1$. Define that

$$\begin{aligned} M_1 =\sum _{j=1}^{n'_1} \left( R^*_{1j} - \frac{N'+1}{2} \right) ^2 ,\quad W =\sum _{j=1}^{n'_1} R^*_{1j}. \end{aligned}$$

Furthermore, put

$$\begin{aligned} t_1=\frac{M_1-\text {E}[M_1]}{\sqrt{\text {var}[M_1]}},\quad w=\frac{W-\text {E}[W]}{\sqrt{\text {var}[W]}},\quad \tau =\frac{r_1-\mathrm{E}[r_1]}{\sqrt{\mathrm{var}[r_1]}}. \end{aligned}$$

The equations $A_1, \ldots , A_4$, $B_1, \ldots , B_4$ are rewritten by using the constants a, b, p, q and the random variables t, w, $\tau$. Then we obtain

Therefore, we re-formulated the $Q^{L}_{1}$ and $Q^{L}_{2}$ as follows:

$$\begin{aligned} Q^{L}_{1}= & {} \frac{1}{3}N^3 a + \frac{1}{6\sqrt{5}}N^{\frac{5}{2}} a^{\frac{1}{2}} b^{\frac{1}{2}} q^{\frac{5}{2}} t_1 + \frac{1}{2\sqrt{3}} N^{\frac{5}{2}} (1+p) a^{\frac{1}{2}} b^{\frac{1}{2}} q^{\frac{3}{2}} w \\&- \frac{1}{3} N^{\frac{5}{2}} (1+p) a^{\frac{1}{2}} b^{\frac{1}{2}} p^{\frac{1}{2}} q^{\frac{1}{2} }\tau +O(N^2). \\ Q^{L}_{2}= & {} Q^{L}_{1} -\frac{1}{\sqrt{3}} N^{\frac{5}{2}} a^{\frac{1}{2}} b^{\frac{1}{2}} q^{\frac{3}{2}} w - N^{\frac{5}{2}} (1-2p) a^{\frac{1}{2}} b^{\frac{1}{2}} p^{\frac{1}{2}} q^{\frac{1}{2}} \tau +O(N^2). \end{aligned}$$

Note that the moment of $Q^{L}_{1}$ and $Q^{L}_{2}$ are expressed as following formula:

$$\begin{aligned} \mathrm{E}[Q^{L}_{1}]= & {} \mathrm{E}[Q^{L}_{2}]=\frac{n_1(N+1)(2N+1)}{6} = \frac{N^3}{3}a + O(N^2), \end{aligned}$$

(3)

$$\begin{aligned} \text {var}[Q^{L}_{1}]= & {} \frac{n_1n_2(N+1)(2N+1)(8N+11)}{180} \nonumber \\&-\frac{n_1n_2}{180N(N-1)}r(r^2-1)(2r+1)(8r+11)\nonumber \\= & {} \frac{1}{180}ab N^5\left( 1+\frac{1}{N} \right) \left( 2+\frac{1}{N} \right) \left( 8+\frac{11}{N} \right) \nonumber \\&-\frac{ab N^5}{180(1-\frac{1}{N})}p \left( p^2-\frac{1}{N^2} \right) \left( 2p+\frac{1}{N} \right) \left( 8p+\frac{1}{N} \right) \nonumber \\= & {} \frac{4N^5}{45}ab Q^2 +O(N^4), \end{aligned}$$

(4)

$$\begin{aligned} \text {var}[Q^{L}_{1}]= & {} \frac{n_1n_2(N+1)(2N+1)(8N+11)}{180} \nonumber \\&-\frac{n_1n_2}{180N(N-1)}r(r^2-1)(2r+1)(8r+11) \nonumber \\&- \frac{n_1n_2}{3N(N-1)}r(r^2-1)(N+1)(N-r)\nonumber \\= & {} \frac{1}{180}ab N^5\left( 1+\frac{1}{N} \right) \left( 2+\frac{1}{N} \right) \left( 8+\frac{11}{N} \right) \nonumber \\&-\frac{ab N^5}{180\left( 1-\frac{1}{N}\right) }p \left( p^2-\frac{1}{N^2} \right) \left( 2p+\frac{1}{N} \right) \left( 8p+\frac{1}{N} \right) \nonumber \\&-\frac{ab N^5}{3\left( 1-\frac{1}{N}\right) }pq \left( p^2-\frac{1}{N^2} \right) \nonumber \\= & {} \frac{4N^5}{45}ab Q{'}^2 +O(N^4), \end{aligned}$$

(5)

where $Q=\sqrt{1-p^5},$$Q'=\sqrt{1-\frac{15}{4}p^3q -p^5}$. Hence, the statistics $U_L$ and $V_L$ are given by

$$\begin{aligned} U_L= & {} \frac{Q^{L}_{1}-\text {E}[Q^{L}_{1}] }{\sqrt{\text {var}[Q^{L}_{1}]}}= \frac{q^{\frac{5}{2}}}{4Q} t + \frac{\sqrt{15}(1+p) q^{\frac{3}{2}}}{4Q}w -\frac{\sqrt{5}(1+p) p^{\frac{1}{2}} q^{\frac{1}{2}}}{2Q} \tau +O(N^{-\frac{1}{2}}), \\ V_L= & {} \frac{Q^{L}_{2}-\mathrm{E}[Q^{L}_{2}] }{\sqrt{\text {var}[Q^{L}_{2}]}} = \frac{q^{\frac{5}{2}}}{Q'} t-\frac{\sqrt{15}q^{\frac{5}{2}}}{2Q'}w -\frac{\sqrt{5}p^{\frac{1}{2}} q^{\frac{1}{2}}(4-2p-p^2)}{2Q'}\tau +O(N^{-\frac{1}{2}}). \end{aligned}$$

Note that t, w and $\tau$ are the asymptotically normal and mutually independent. By taking into account $\text {E}[U_L]=\text {E}[V_L]=0$, $\text {var}[U_L]=\text {var}[V_L]=1$, the proof is completed to show for the case of the left-censored data by using Lemma 1.

1.2 Proof for the right-censored data

The equation $R_{1j}=R^*_{1j}$ holds for the right-censored data. Define that

$$\begin{aligned} M_2 =\sum _{j=1}^{n'_1} R^{*2}_{1j} \quad \text {and} \quad t_2=\frac{M_2-\text {E}[M_2]}{\sqrt{\text {var}[M_2]}}. \end{aligned}$$

$Q^{R}_{1}$ and $Q^{R}_{2}$ are decomposed as follows:

$$\begin{aligned} Q^{R}_{1}= & {} \sum _{j=1}^{n'_1} R_{1j}^{2} +\frac{r_1(6N^2-6Nr+6N+2r^2-3r+1)}{6} \\= & {} \sum _{j=1}^{n'_1} R_{1j}^{*2} +\frac{r_1(6N^2-6Nr+6N+2r^2-3r+1)}{6} \\= & {} C_1+C_2\quad (\mathrm{say}), \\ Q^{R}_{2}= & {} \sum _{i=1}^{n_1} (N+1-R_{1j})^2 + \frac{r_1(r+1)(2r+1)}{6} \\= & {} \sum _{j=1}^{n_1} (N+1-R^*_{1j})^2 + \frac{r_1(r+1)(2r+1)}{6} \\= & {} \frac{r_1(r+1)(2r+1)}{6} + \sum _{j=1}^{n_1} \left( R^*_{1j} -\frac{N'+1}{2} \right) ^2 -(N+r+1) \sum _{j=1}^{n'_1} R^*_{1j} \\&+\frac{1}{4}n'_1(N+r+1)(3N-r+3) \\= & {} A_1+A_2-A_3+D_1\quad (\mathrm{say}). \end{aligned}$$

Since it is rewritten by utilizing the constants a, b, p, q and the random variables t, w, $\tau$, we obtain

Therefore, we re-formulated $Q^{R}_{1}$ and $Q^{R}_{2}$ as follows:

$$\begin{aligned} Q^{R}_{1}= & {} \frac{1}{3}N^3 a + \frac{2}{3\sqrt{5}}N^{\frac{5}{2}} a^{\frac{1}{2}} b^{\frac{1}{2}} q^{\frac{5}{2}} t_2 + \frac{1}{3}N^{\frac{5}{2}} (2-p) a^{\frac{1}{2}} b^{\frac{1}{2}} p^{\frac{1}{2}} q^{\frac{1}{2}} \tau +O(N^2), \\ Q^{R}_{2}= & {} \frac{1}{3}N^3 a + \frac{1}{6\sqrt{5}}N^{\frac{5}{2}} a^{\frac{1}{2}} b^{\frac{1}{2}} q^{\frac{5}{2}} t_1 - \frac{1}{2\sqrt{3}}N^{\frac{5}{2}} (1+p)a^{\frac{1}{2}} b^{\frac{1}{2}} q^{\frac{3}{2}} w \\&- \frac{1}{3} N^{\frac{5}{2}} (1+p)a^{\frac{1}{2}} b^{\frac{1}{2}} p^{\frac{1}{2}} q^{\frac{1}{2}}\tau +O(N^2). \end{aligned}$$

By taking into account for $\text {var}[Q_{\xi }^{R}]=\text {var}[Q_{3-\xi }^{L}]$, $\xi =1,2$ and the Eqs. (3), (4), (5), we have

$$\begin{aligned} U_R= & {} \frac{Q^{R}_{1}-\text {E}[Q^{R}_{1}] }{\sqrt{\text {var}[Q^{R}_{1}]}} = \frac{q^{\frac{5}{2}}}{Q'} t_2 + \frac{\sqrt{5}(2-p)p^{\frac{1}{2}}q^{\frac{1}{2}}}{2Q'}\tau +O(N^{-\frac{1}{2}}), \\ V_R= & {} \frac{Q^{R}_{2}-\text {E}[Q^{R}_{2}] }{\sqrt{\text {var}[Q^{R}_{2}]}} = \frac{q^{\frac{5}{2}}}{4Q}t_1-\frac{\sqrt{15}(1+p)q^{\frac{3}{2}}}{4Q}w-\frac{\sqrt{5}(1+p)p^{\frac{1}{2}}q^{\frac{1}{2}}}{2Q}\tau +O(N^{-\frac{1}{2}}), \end{aligned}$$

where $Q=\sqrt{1-p^5}$, $Q'=\sqrt{1-\small {\tfrac{4}{15}}p^3q-p^5}$. It is noteworthy that $t_2$ is asymptotically distributed according to N(0, 1) and $t_2$ and $\tau$ are mutually independent and $\text {E}[U_R]=\text {E}[V_R]=0$, $\text {var}[U_R]=\text {var}[V_R]=1$. Conclusively, the proof is completed to show for the case of the right-censored data by applying Lemma 1.

Proof of Theorem 2

The correlation coefficient between $U_L$ ($U_R$) and $V_L$ ($V_R$) is approximated to

where $Q=\sqrt{1-p^5}$, $Q'=\sqrt{1-\small {\tfrac{4}{15}}p^3q-p^5}$. Note that $p=1-q$. Then the behavior of $\rho$ on $p\in (0,1)$ is as follows:

$$\begin{aligned} f(0)=-\frac{7}{8}, \quad \lim _{p\rightarrow 1-0} f(p) = -1, \quad f'(p)<0, \quad \left\{ p\in [0,1]; \ |f(p)|=0 \right\} = \phi . \end{aligned}$$

Therefore, f(p) is strictly monotonic decreasing and $|f(p)| \ne 1$ on (0, 1).