Maximum likelihood estimation for a special exponential family under random double-truncation

Abstract

Doubly-truncated data often appear in lifetime data analysis, where samples are collected under certain time constraints. Nonparametric methods for doubly-truncated data have been studied well in the literature. Alternatively, this paper considers parametric inference when samples are subject to double-truncation. Efron and Petrosian (J Am Stat Assoc 94:824–834, 1999) proposed to fit a parametric family, called the special exponential family, with doubly-truncated data. However, non-trivial technical aspects, such as parameter space, support of the density, and computational algorithms, have not been discussed in the literature. This paper fills this gap by providing the technical aspects, including adequate choices of parameter space as well as support, and reliable computational algorithms. Simulations are conducted to verify the suggested techniques, and real data are used for illustration.

Appendix: The data generations for simulations

1.1 Data generations for one-parameter SEF

First, consider the case $\eta >0$. We let

$$\begin{aligned} U\sim f_U (u)&= \eta _u \exp \{\eta _u (u-\tau _2 )\}, -\infty <u<\tau _2, \eta _u >0, \\ V\sim f_V (v)&= \eta _v \exp \{\eta _v (v-\tau _2 )\}, -\infty <v<\tau _2, \eta _v >0, \\ Y\sim f_Y (y)&= \eta \exp \{\eta (y-\tau _2 )\}, -\infty <y<\tau _2, \eta >0. \end{aligned}$$

In this case, data are generated by using inverse transformations

$$\begin{aligned} U=\tau _2 +\frac{1}{\eta _u }\log (W_1), \quad V=\tau _2 +\frac{1}{\eta _v }\log (W_2), \quad Y=\tau _2 +\frac{1}{\eta }\log (W_3), \end{aligned}$$

where $W_1, W_2, W_3 \sim U(0, 1)$. Then, the inclusion probability is

$$\begin{aligned} P(U\le Y\le V)\!=\!\int \limits _{-\infty }^{\tau _2 } {\int \limits _{-\infty }^v {\int \limits _{-\infty }^y {f_Y (y)f_V (v)} } } f_U (u) dudydv =\frac{\eta \cdot \eta _v }{(\eta +\eta _u )(\eta +\eta _u +\eta _v )}. \end{aligned}$$

$\eta _u =0$ yields $P(-\infty <Y\le V)=\eta _v / (\eta _v +\eta )$ and $\eta _v =\infty$ yields $P(U\le Y<\tau _2 )=\eta / (\eta +\eta _u )$. One can get $P(U\le Y\le V)\approx 0.5$ by making $P(U\le Y)=0.75$ and $P(Y\le V)=0.75$.

For instance, with fixed $\eta =3$, we find $\eta _u$ and $\eta _v$ as follows:

• 1. Set $P(U\le Y)=\eta / (\eta +\eta _u )=0.75$, and then obtain $\eta _u =1$.

• 2. Set $P(Y\le V)=\eta _v / (\eta _v +\eta )=0.75$, and then obtain $\eta _v =9$.

Accordingly, the inclusion probability becomes $P(U\le Y\le V)=0.5031447$.

Another case is $\eta <0$, where the range of $Y$ is $y\in [\tau _1, \infty )$. We consider

$$\begin{aligned} U\sim f_U (u)&= -\eta _u \exp \{\eta _u (u-\tau _1 )\}, \tau _1 <u<\infty , \eta _u <0, \\ V\sim f_V (v)&= -\eta _v \exp \{\eta _v (v-\tau _1 )\}, \tau _1 <v<\infty , \eta _v <0, \\ Y\sim f_Y (y)&= -\eta \exp \{\eta (y-\tau _1 )\}, \tau _1 <y<\infty , \eta <0. \end{aligned}$$

In this case, data are generated by using inverse transformations

$$\begin{aligned} U=\tau _1 +\frac{1}{\eta _u }\log (1-W_1), \quad V=\tau _1 +\frac{1}{\eta _v }\log (1-W_2), \quad Y=\tau _1 +\frac{1}{\eta }\log (1-W_3), \end{aligned}$$

where $W_1, W_2, W_3 \sim U(0, 1)$. Then, the inclusion probability is

$$\begin{aligned} P(U\le Y\le V)=\int \limits _\tau ^\infty {\int \limits _u^\infty {\int \limits _y^\infty {f_Y (y)f_V (v)} } } f_U (u) dvdydu =\frac{\eta \cdot \eta _u }{(\eta +\eta _v )(\eta +\eta _u +\eta _v )}. \end{aligned}$$

$\eta _u =\infty$ yields $P(0\le Y\le V)=\eta /(\eta +\eta _v )$ and $\eta _v =0$ yields $P(U\le Y<\infty )=\eta _u /(\eta +\eta _u )$. One can get $P(U\le Y\le V)\approx 0.5$ by making $P(U\le Y)=0.75$ and $P(Y\le V)=0.75$.

For instance, with fixed $\eta =-1$, we find $\eta _u$ and $\eta _v$ as follows:

• 1. Set $P(U\le Y)=\eta _u /(\eta +\eta _u )=0.75$, and then obtain $\eta _u =-3$.

• 2. Set $P(Y\le V)=\eta /(\eta +\eta _v )=0.75$, and then obtain $\eta _v =-1/3$.

Accordingly, the inclusion probability becomes $P(U\le Y\le V)=0.5235602$.

1.2 Data generation for two-parameter SEF

We consider $U\sim N(\mu _u, 1)$, $V\sim N(\mu _v, 1)$ and $Y\sim N(\mu , 1)$. One can obtain $P(U\le Y\le V)\approx 0.5$ by making left-truncated percentage is equal to right-truncated percentage. Since the normal distribution is symmetric, we set $\mu _u =\mu -\Delta$ and $\mu _v =\mu +\Delta$. Then,

$$\begin{aligned} P(U\le Y\le V)=\int \limits _{-\infty }^\infty {\varphi (y-\mu )\cdot \Phi (y-\mu +\Delta )\cdot \{1-\Phi (y-\mu -\Delta )\}} \, dy. \end{aligned}$$

The desired value $\Delta >0$ is chosen numerically. For instance, with fixed $\mu =5$, the desired value is $\Delta$= 0.91, which makes $P(U\le Y\le V)\approx 0.5$ (Fig. 8). Then, we choose $\mu _u =4.09$ and $\mu _v =5.91$. With this setting, we have $P(U\le Y\le V)=0.5076142$.

Fig. 8

An example for how to choose the value $\Delta$ under the two-parameter SEF

1.3 Data generations for cubic SEF

For the cubic SEF with $\eta _3 >0$, we consider $U\sim N(\mu _u , 1)$, $V\sim N(\mu _v, 1)$ and

$$\begin{aligned} Y\sim f_{\varvec{\eta }} (y)=\exp [\eta _1 y+\eta _2 y^{2}+\eta _3 y^{3}-\phi (\varvec{\eta })],\quad y\in {\mathcal {Y}}=(-\infty , \tau _2 ], \end{aligned}$$

where $\phi (\varvec{\eta })=\log \{\int _{\mathcal {Y}} {\exp (\eta _1 y+\eta _2 y^{2}+\eta _3 y^{3})\, dy} \}$. Data are generated by using an inverse transformation, which numerically solves $1-S_\eta (y)=W$, where $W\sim U(0, 1)$. One can get $P(U\le Y\le V)\approx 0.5$ by making left-truncated and right-truncated percentages equal by setting $\mu _u =\eta _1 -\Delta$, $\mu _v =\eta _1 +\Delta$. Then,

$$\begin{aligned} P(U\le Y\le V)=\int \limits _{-\infty }^{\tau _2 } {\{1-\Phi (y-\eta _1 -\Delta )\}\Phi (y-\eta _1 +\Delta )} f_\eta (y)\, dy. \end{aligned}$$

The desired value $\Delta >0$ is chosen numerically. For instance, for fixed $\eta _1 =5$, $\eta _2 =-0.5$, $\eta _3 =0.005$ and $\tau _2 =8$, the value of $\Delta$ is 1.01 (Fig. 9). Hence, we choose $\mu _u =3.99$ and $\mu _v =6.01$. Accordingly, the inclusion probability becomes $P(U\le Y\le V)=0.5035228$.

Fig. 9

An example for how to choose the value $\Delta$ under the cubic SEF

The other case $\eta _3 <0$ is similar. Under $\eta _1 =5$, $\eta _2 =-0.5$, $\eta _3 =-0.005$, and $\tau _1 =2$, the desired value is $\Delta =0.91$ (see Fig. 9). Hence, we choose $\mu _u =4.09$ and $\mu _v =5.91$. Accordingly, the inclusion probability becomes $P(U\le Y\le V)=0.5027334$.