A new bounded log-linear regression model

Abstract

In this paper we introduce a new regression model in which the response variable is bounded by two unknown parameters. A special case is a bounded alternative to the four parameter logistic model. The four parameter model which has unbounded responses is widely used, for instance, in bioassays, nutrition, genetics, calibration and agriculture. In reality, the responses are often bounded although the bounds may be unknown, and in that situation, our model reflects the data-generating mechanism better. Complications arise for the new model, however, because the likelihood function is unbounded, and the global maximizers are not consistent estimators of unknown parameters. Although the two sample extremes, the smallest and the largest observations, are consistent estimators for the two unknown boundaries, they have a slow convergence rate and are asymptotically biased. Improved estimators are developed by correcting for the asymptotic biases of the two sample extremes in the one sample case; but even these consistent estimators do not obtain the optimal convergence rate. To obtain efficient estimation, we suggest using the local maximizers of the likelihood function, i.e., the solution to the likelihood equations. We prove that, with probability approaching one as the sample size goes to infinity, there exists a solution to the likelihood equation that is consistent at the rate of the square root of the sample size and it is asymptotically normally distributed.

Appendix: Technical details

1.1 Proof of Proposition 1

Proof

First, following the idea in Sect. 2.3 of Galambos (1978), for any $t$,

$$\begin{aligned}&\lim _{n\rightarrow \infty } \text {Pr}\left[ \left\{ \frac{\log (B_0-Y)-\log (Y-A_0)-\mu _0}{\sigma _0}\right\} _{(n)}<r_n+s_nt\right] =e^{-e^{-t}}\\&\quad =\lim _{n\rightarrow \infty } \text {Pr}\left\{ \left( \frac{B_0-Y}{Y-A_0}\right) _{(n)}<e^{\mu _0+\sigma _0r_n+\sigma _0s_nt}\right\} \\&\quad =\lim _{n\rightarrow \infty } \text {Pr}\left\{ \frac{B_0-Y_{(1)}}{Y_{(1)}-A_0}<e^{\mu _0+\sigma _0r_n}+\sigma _0s_ne^{\mu _0+\sigma _0r_n}(1+v\sigma _0s_nt)t\right\} \!, \end{aligned}$$

where $|v|\le 1$. Since $v\sigma _0s_nt\rightarrow 0$ as $n\rightarrow \infty$, it follows that

$$\begin{aligned} \lim _{n\rightarrow \infty } \text {Pr}\left( \frac{B_0-Y_{(1)}}{Y_{(1)}-A_0}<c_n+d_nt\right) =e^{-e^{-t}}, \end{aligned}$$

where $c_n=e^{\mu _0+\sigma _0r_n}$ and $d_n=\sigma _0s_ne^{\mu _0+\sigma _0r_n}$.

Then, for any $t\ne 0$,

$$\begin{aligned} \text {Pr}\left( \frac{B_0-Y_{(1)}}{Y_{(1)}-A_0}<c_n+d_nt\right)&=\text {Pr}\left( \frac{Y_{(1)}-A_0}{B_0-A_0}>\frac{1}{1+c_n+d_nt}\right) \\&=\text {Pr}\left[ \frac{Y_{(1)}-A_0}{B_0-A_0}>\frac{c_n-1}{c_n^2}-\left\{ \frac{d_n}{c_n^2}-\frac{(1+d_nt)}{c_n^2(1+c_n+d_nt)t}\right\} t\right] \!. \end{aligned}$$

It can be shown that $[{(1+d_nt)}/\{c_n^2(1+c_n+d_nt)t\}]/({d_n}/{c_n^2})\rightarrow 0$ and $({1}/{c_n^2})/({d_n}/{c_n^2})$ $\rightarrow 0$ as $n\rightarrow \infty$. So from Lemma 2.2.2 in Galambos (1978),

$$\begin{aligned} \lim _{n\rightarrow \infty } \text {Pr}\left\{ \frac{B_0-Y_{(1)}}{Y_{(1)}-A_0}<c_n+d_nt\right\} =\lim _{n\rightarrow \infty } \text {Pr}\left\{ \frac{Y_{(1)}-A_0}{B_0-A_0}>\frac{1}{c_n}-\frac{d_n}{c_n^2}t\right\} \!. \end{aligned}$$

When $t=0$, the result can be verified by using the properties of the extreme order statistics of normal distribution directly. The second equation can be proved similarly. $\square$

1.2 Derivation of the Fisher information

The lemma below is useful in deriving the Fisher information.

Lemma 1

From Lemma 2 of Stein (1981), we obtain that, if $E|h'(Z)|<\infty$ for a normal random variable $Z\sim N(\mu ,\sigma ^2)$ and some differentiable function $h$. Then

$$\begin{aligned} E\{(Z-\mu )h(Z)\}=\sigma ^2E\{ h'(Z)\}. \end{aligned}$$

The log-likelihood function of model (1) based on one observation $(\mathbf{x},Y)$ is

$$\begin{aligned} \ell (\varvec{\theta },\mathbf{x},Y)&= -\frac{\log {(2\pi )}}{2}-\log \sigma +\log (B-A)-\log (Y-A)\\&\,\quad -\log (B-Y)-\frac{\left\{ \log (B-Y)-\log (Y-A)-\mathbf{x}^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right\} ^2}{2\sigma ^2} \end{aligned}$$

for $Y\in (A, B)$ and 0 otherwise. Let $Z=\log (B-Y)-\log (Y-A)$. By direct calculation,

$$\begin{aligned} \frac{\partial \ell (\varvec{\theta },\mathbf{x},Y)}{\partial A}&=\frac{1}{Y-A+\frac{(Y-A)^2}{B-Y}}-\frac{\left\{ \log (B-Y)-\log (Y-A)-\mathbf{x}^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right\} }{\sigma ^2(Y-A)};\\ \frac{\partial ^2\ell (\varvec{\theta },\mathbf{x},Y)}{\partial A^2}&=\frac{1+2\frac{Y-A}{B-Y}}{\left\{ Y-A+\frac{(Y-A)^2}{B-Y}\right\} ^2}-\frac{\left\{ \log (B-Y)-\log (Y-A)-\mathbf{x}^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right\} }{\sigma ^2(Y-A)^2}\\&-\frac{1}{\sigma ^2(Y-A)^2}=\frac{e^{2Z}+2e^{Z}}{(B-A)^2}-\frac{(Z-\mathbf{x}^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta })(1+e^{Z})^2}{\sigma ^2(B-A)^2}-\frac{(1+e^{Z})^2}{\sigma ^2(B-A)^2}. \end{aligned}$$

Then from Lemma 1,

$$\begin{aligned} \begin{aligned} E\left\{ \frac{\partial ^2\ell (\varvec{\theta },\mathbf{x},Y)}{\partial A^2}\right\}&=-E\left\{ \frac{e^{2Z}}{(B-A)^2}+\frac{1+2e^{Z}+e^{2Z}}{\sigma ^2(B-A)^2}\right\} \\&=-\frac{e^{2\mathbf{x}^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }+2\sigma ^2}}{(B-A)^2}-\frac{1+2e^{\mathbf{x}^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }+\frac{\sigma ^2}{2}}+e^{2\mathbf{x}^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }+2\sigma ^2}}{\sigma ^2(B-A)^2}. \end{aligned} \end{aligned}$$

Other elements of the Fisher information can be derived similarly.

1.3 Proof of Theorems for the regression model

The proof of Theorem 1 begins with some lemmas.

Lemma 2

For constant sequences $v_n\downarrow v$ and $w_n\uparrow w$ as $n\rightarrow \infty$, let $\xi _{v_n}\in (v_{n+1},v_n)$ and $\xi _{w_n}\in (w_n,w_{n+1})$. If a continuous function sequence $f_n(\cdot )>0$, which is decreasing in $n$, satisfies $n^{1+\alpha }f_n(\xi _{v_n})\rightarrow 0$ and $n^{1+\alpha }f_n(\xi _{w_n})\rightarrow 0$ for $\alpha >0$ as $n\rightarrow \infty$, then

$$\begin{aligned} \limsup _n\int \limits _{v_n}^{w_n}f_n(x)\mathrm{d}x<\infty . \end{aligned}$$

Proof

Let $S_n=\int _{v_n}^{w_n}f_n(x)\mathrm{d}x$. Then

$$\begin{aligned} S_{n}-S_{n-1}&=\int \limits _{v_n}^{w_n}f_n(x)\mathrm{d}x-\int \limits _{v_{n-1}}^{w_{n-1}}f_{n-1}(x)\mathrm{d}x\\&\le (v_{n-1}-v_n)f_{n-1}(\xi _{v_{n-1}})+(w_n-w_{n-1})f_{n-1}(\xi _{w_{n-1}})\\&=\frac{(v_{n-1}-v_n)n^{1+\alpha }f_{n-1}(\xi _{v_{n-1}})+(w_n-w_{n-1})n^{1+\alpha }f_{n-1}(\xi _{w_{n-1}})}{n^{1+\alpha }}\\&=o\left( \frac{1}{n^{1+\alpha }}\right) \!. \end{aligned}$$

So $\limsup _nS_n=\limsup _n\sum _{i=1}^n(S_n-S_{n-1})$ is finite.

Lemma 3

For any $\alpha >0$, let $\delta _n=n^{-\alpha }$. Then for any $k_1\ge 0$ and $k_2\ge 0$, there exists a constant $M$ such that

$$\begin{aligned} \begin{aligned}&\lim _{n\rightarrow \infty }\text {Pr}\left\{ \frac{1}{n}\sum _{i=1}^{n}\frac{|\log (B-Y_i)|^{k_1}}{(Y_i-A)^{k_2}}<M\right\} =1,\\&\lim _{n\rightarrow \infty }\text {Pr}\left\{ \frac{1}{n}\sum _{i=1}^{n}\frac{|\log (Y_i-A)|^{k_1}}{(B-Y_i)^{k_2}}<M\right\} =1 \end{aligned} \end{aligned}$$

(10)

uniformly in $A$ and $B$ so long as $|A-A_0|<\delta _n$ and $|B-B_0|<\delta _n$.

Proof

We gives the details of proof for the first quantity in (10). The proof for the other one is similar.

$$\begin{aligned} \frac{\left| \log (B-Y_i)\right| ^{k_1}}{(Y_i-A)^{k_2}}&= \frac{|\log (B-B_0+B_0-Y_i)|^{k_1}}{(Y_i-A_0+A_0-A)^{k_2}}I{(Y_i-A_0>2\delta _n,B_0-Y_i>2\delta _n)}+o_p(1)\\&= \frac{|\log (B-B_0+B_0-Y_i)|^{k_1}}{(Y_i-A_0+A_0-A)^{k_2}}I{(Y_i-A_0>2\delta _n,1-2\delta _n>B_0\!-\!Y_i>2\delta _n)}\\&+\,\frac{|\log (B-B_0+B_0-Y_i)|^{k_1}}{(Y_i-A_0+A_0-A)^{k_2}}I{(Y_i-A_0>2\delta _n,B_0-Y_i>1)}+o_p(1)\\&< \frac{I{(Y_i-A_0>2\delta _n,1-2\delta _n>B_0-Y_i>2\delta _n)}}{(B_0-Y_i-\delta _n)^{k_1}(Y_i-A_0-\delta _n)^{k_2}}\\&+\,\frac{(B_0-Y_i+\delta _n)^{k_1}}{(Y_i-A_0-\delta _n)^{k_2}}I{(Y_i-A_0>2\delta _n,B_0-Y_i>1)}+o_p(1)\\&< \frac{I{(B_0-2\delta _n>Y_i>A_0+2\delta _n)}}{(B_0-Y_i-\delta _n)^{k_1}(Y_i-A_0-\delta _n)^{k_2}}\\&+\,\frac{(B_0-A_0+1)^{k_1}}{(Y_i-A_0-\delta _n)^{k_2}}I{\left( B_0-1>Y_i>A_0+2\delta _n\right) }+o_p(1)\\&= C_{in1}+C_{in2}+o_p(1). \end{aligned}$$

Note that

$$\begin{aligned}&(2\pi )^{\frac{1}{2}}E(C_{in1})=\frac{1}{\sigma _0}\int \limits _{A_0+2\delta _n}^{B_0-2\delta _n}\frac{1}{(B_0-y-\delta _n)^{k_1}(y-A_0-\delta _n)^{k_2}}\\&\qquad \times \frac{B_0-A_0}{(B_0-y)(y-A_0)}\exp \left[ -\frac{\left\{ \log \left( \frac{B_0-y}{y-A_0}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_0\right\} ^2}{2\sigma _0^2}\right] \mathrm{d}y\\&\quad \le \frac{1}{\sigma _0}\int \limits _{A_0+2\delta _n}^{B_0-2\delta _n}\frac{1}{(B_0-y-\delta _n)^{k_1+1}(y-A_0-\delta _n)^{k_2+1}}\\&\qquad \times \exp \left[ -\frac{\frac{1}{4}\left\{ \log \left( \frac{B_0-y}{y-A_0}\right) \right\} ^2-(\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_0)^2}{2\sigma _0^2}\right] \mathrm{d}y\\&\quad =\frac{\exp \left\{ \frac{(\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_0)^2}{2\sigma _0^2}\right\} }{\sigma _0}\int \limits _{A_0+2\delta _n}^{B_0-2\delta _n}\frac{1}{(B_0-y-\delta _n)^{k_1+1}(y-A_0-\delta _n)^{k_2+1}}\\&\qquad \times \exp \left[ -\frac{\left\{ \log \left( \frac{B_0-y}{y-A_0}\right) \right\} ^2}{8\sigma _0^2}\right] \mathrm{d}y. \end{aligned}$$

From Lemma 2, $\limsup _nE(C_{in1})$ is bounded by a finite constant, say $C_1$. Similarly, it can also be shown that $\limsup _nE(C_{in2})$ is bounded by a finite constant, say $C_2$. So using the formula $X_n=E(X_n)+O_P\{{\text {var}(X_n)^{1/2}}\}$, we have

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^{n}\frac{|\log (B-Y_i)|^{k_1}}{(Y_i-A)^{k_2}}=\frac{1}{n}\sum _{i=1}^nE(C_{in1})+\frac{1}{n}\sum _{i=1}^nE(C_{in2}) + O_P\left( n^{-\frac{1}{2}}\right) + o_P(1). \end{aligned}$$

Thus any $M$ that is greater than $C_1+C_2$ satisfys the requirement.

Lemma 4

If assumptions 1- 3 hold, then $-n^{-1}\partial {^2\ell _n(\varvec{\theta })}/(\partial {\varvec{\theta }\partial \varvec{\theta }^{{ {\mathrm{\scriptscriptstyle T} }}}})\rightarrow \mathcal I (\varvec{\theta }_0)$ in probability uniformly over $\Vert \varvec{\theta }-\varvec{\theta }_0\Vert <\delta _n$.

Proof

The first element of ${\partial ^2\ell _n(\varvec{\theta })}/({\partial \varvec{\theta }\partial \varvec{\theta }^{{ {\mathrm{\scriptscriptstyle T} }}}})$ is

$$\begin{aligned} \frac{\partial ^2\ell _n(\varvec{\theta })}{\partial A^2} =\sum _{i=1}^n\left[ \frac{1-\frac{1}{\sigma ^2}}{(Y_i-A)^2}-\frac{1}{(B-A)^2}-\frac{\left\{ \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right\} }{\sigma ^2(Y_i-A)^2}\right] \!. \end{aligned}$$

So it is straightforward to get

$$\begin{aligned}&\frac{1}{n}\left| \frac{\partial ^2\ell _n(\varvec{\theta })}{\partial A^2}-\frac{\partial ^2\ell _n(\varvec{\theta }_0)}{\partial A^2}\right| \\&\quad \le \frac{1}{n}\sum _{i=1}^n\left| \frac{1}{(B-A)^2}-\frac{1}{(B_0-A_0)^2}\right| + \frac{1}{n}\sum _{i=1}^n\left| \frac{1+\frac{1}{\sigma ^2}}{(Y_i-A)^2}-\frac{1+\frac{1}{\sigma _0^2}}{(Y_i-A_0)^2}\right| \\&\quad \quad +\frac{1}{n}\sum _{i=1}^n\left| \frac{\left\{ \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right\} }{\sigma ^2(Y_i-A)^2}-\frac{\left\{ \log \left( \frac{B_0-Y_i}{Y_i-A_0}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_0\right\} }{\sigma _0^2(Y_i-A_0)^2}\right| \\&\quad ={\varDelta }_1+{\varDelta }_2+{\varDelta }_3. \end{aligned}$$

${\varDelta }_1$ goes to 0 as $\delta _n$ goes to 0. By straightforward but tedious calculation, we obtain

$$\begin{aligned} {\varDelta }_3&\le \frac{1}{n}\sum _{i=1}^n\left| \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right| \left| \frac{1}{\sigma ^2(Y_i-A)^2}-\frac{1}{\sigma _0^2(Y_i-A_0)^2}\right| \\&\quad \quad +\frac{1}{n}\sum _{i=1}^n\left| \frac{\log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }}{\sigma _0^2(Y_i-A_0)^2}-\frac{\log \left( \frac{B_0-Y_i}{Y_i-A_0}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_0}{\sigma _0^2(Y_i-A_0)^2}\right| \\&\le \left| \frac{1}{\sigma ^2}-\frac{1}{\sigma _0^2}\right| \times \frac{1}{n}\sum _{i=1}^n\frac{\left| \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right| }{(Y_i-A)^2}\\&\quad \quad +\frac{1}{n}\sum _{i=1}^n\frac{\left| \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right| }{\sigma _0^2}\times \left| \frac{1}{(Y_i-A)^2}-\frac{1}{(Y_i-A_0)^2}\right| \\&\quad \quad +\frac{1}{n}\sum _{i=1}^n\frac{\left| \log \left( \frac{B-Y_i}{B_0-Y_i}\right) \right| }{\sigma _0^2(Y_i-A_0)^2}+ \frac{1}{n}\sum _{i=1}^n\frac{\left| \log \left( \frac{Y_i-A}{Y_i-A_0}\right) \right| }{\sigma _0^2(Y_i-A_0)^2}+ \frac{1}{n}\sum _{i=1}^n\frac{\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}(\varvec{\beta }-\varvec{\beta }_0)}{\sigma _0^2(Y_i-A_0)^2}\\&\le \frac{1}{n}\sum _{i=1}^n\frac{\left| \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right| }{(Y_i-A)^2}\times \left| \frac{1}{\sigma ^2}-\frac{1}{\sigma _0^2}\right| + \frac{4B|A-A_0|}{\sigma _0^2}\\&\qquad \times \frac{1}{n}\sum _{i=1}^n\frac{\left| \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right| }{(Y_i-A)^2(Y_i-A_0)^2}+\frac{1}{n}\sum _{i=1}^n\frac{|B-B_0|}{\sigma _0^2(B^*-Y_i)(Y_i-A_0)^2}\\&\quad \quad + \frac{1}{n}\sum _{i=1}^n\frac{|A-A_0|}{\sigma _0^2(Y_i-A^*)(Y_i-A_0)^2}+ \frac{1}{n}\sum _{i=1}^n\frac{\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}(\varvec{\beta }-\varvec{\beta }_0)}{\sigma _0^2(Y_i-A_0)^2}\\&={\varDelta }_{3.1}+{\varDelta }_{3.2}+{\varDelta }_{3.3}+{\varDelta }_{3.4}+{\varDelta }_{3.5}, \end{aligned}$$

where $A^*$ is between $A$ and $A_0$ and $B^*$ is between $B$ and $B_0$. Now we look into each term in the last equation above.

$$\begin{aligned} {\varDelta }_{3.2}\le \frac{2B|A-A_0|}{\sigma _0^2} \times \frac{1}{n}\sum _{i=1}^n\left| \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right| \left\{ \frac{1}{(Y_i-A)^4}+\frac{1}{(Y_i-A_0)^2}\right\} \!.\nonumber \\ \end{aligned}$$

(11)

The right hand side term in (11) goes to 0 in probability uniformly since the second factor is bound with probability tending to 1 by Lemma 3 and the boundedness of $x_i$. Similarly, ${\varDelta }_{3.1}, {\varDelta }_{3.3}, {\varDelta }_{3.4}$ and ${\varDelta }_{3.5}$ can be shown to go to 0 in probability uniformly which implies ${\varDelta }_3$ goes to 0 in probability uniformly. Similarly but more easily, ${\varDelta }_1$ and ${\varDelta }_2$ can be showen to converge to 0 in probability uniformly, which implies $n^{-1}\left| \partial {^2\ell _n(\varvec{\theta })}/\partial {A^2}-\partial {^2\ell _n(\varvec{\theta }_0)}/\partial {A^2}\right| \rightarrow 0$ in probability uniformly. By similiar arguments, other components of ${\partial ^2\ell _n(\varvec{\theta })}/({\partial \varvec{\theta }\partial \varvec{\theta }^{{ {\mathrm{\scriptscriptstyle T} }}}})$ can be shown to have the same property. This implys that $-n^{-1}\partial {^2\ell _n(\varvec{\theta })}/(\partial \varvec{\theta }\partial \varvec{\theta }^{{ {\mathrm{\scriptscriptstyle T} }}})\rightarrow \mathcal I (\varvec{\theta }_0)$ in probability uniformly over $\Vert \varvec{\theta }-\varvec{\theta }_0\Vert <\delta _n$. $\square$

The following lemma is the Lemma 5 of Smith (1985). We state it for integrity and skip the proof.

Lemma 5

Let $h$ be a continuously differentiable real-valued function of $p+1$ real variables and let $H$ denote the gradient vector of $h$. Suppose that the scalar product of $\mathbf{u}$ and $H(\mathbf{u})$ is negative whenever $\Vert \mathbf{u}\Vert =1$. Then $h$ has a local maximum, at which $H=0$, for some $\mathbf{u}$ with $\Vert \mathbf{u}\Vert <1$.

Proof

(of Theorem 1) It suffices to show for any $\epsilon$, there exists a constant $c$ such that

$$\begin{aligned} \text {Pr}\left\{ \mathbf{u}^{{ {\mathrm{\scriptscriptstyle T} }}}\frac{\partial \ell _n\left( \varvec{\theta }_0+{n^{-{1}/{2}}c\mathbf{u}}\right) }{\partial \varvec{\theta }}<0\right\} >1-\epsilon \end{aligned}$$

(12)

for any vector $\mathbf{u}$ such that $\Vert \mathbf{u}\Vert =1$. Using Taylor’s expansion,

$$\begin{aligned} \frac{\partial \ell _n\left( \varvec{\theta }_0+{n^{-{1}/{2}}c\mathbf{u}}\right) }{\partial \varvec{\theta }}&=\frac{\partial \ell _n\left( \varvec{\theta }_0\right) }{\partial \varvec{\theta }}+cn^{-1/2}\frac{\partial ^2\ell _n\left( \varvec{\theta }_0+{n^{-1/2}c\mathbf{u}^*}\right) }{\partial \varvec{\theta }\partial \varvec{\theta }^{{ {\mathrm{\scriptscriptstyle T} }}}}\mathbf{u}\\&=\frac{\partial \ell _n\left( \varvec{\theta }_0\right) }{\partial \varvec{\theta }}-cn^{1/2}\mathcal I (\varvec{\theta }_0)\mathbf{u}+ n^{1/2}\epsilon _{n,\mathbf{u}}, \end{aligned}$$

where $\mathbf{u}^*$ is a vector satisfying $\Vert \mathbf{u}^*\Vert \le 1$ and, by Lemma 3, $\epsilon _{n,\mathbf{u}}\rightarrow 0$ in probability uniformly over $\Vert \mathbf{u}\Vert \le 1$ as $n\rightarrow \infty$. It follows that

$$\begin{aligned} n^{-1/2}\mathbf{u}^{{ {\mathrm{\scriptscriptstyle T} }}}\frac{\partial \ell _n\left( \varvec{\theta }_0+n^{-1/2}\mathbf{u}\right) }{\partial \varvec{\theta }}=n^{-1/2}\mathbf{u}^{{ {\mathrm{\scriptscriptstyle T} }}}\frac{\partial \ell _n\left( \varvec{\theta }_0\right) }{\partial \varvec{\theta }}-c\mathbf{u}^{{ {\mathrm{\scriptscriptstyle T} }}}\mathcal I (\varvec{\theta }_0)\mathbf{u}+ \mathbf{u}^{{ {\mathrm{\scriptscriptstyle T} }}}\epsilon _{n,\mathbf{u}}. \end{aligned}$$

(13)

Note that $n^{-1/2}\mathbf{u}^{{ {\mathrm{\scriptscriptstyle T} }}}\partial {\ell _n\left( \varvec{\theta }_0\right) }/\partial {\varvec{\theta }}$ is $O_P(1)$. So the second term dominates the first term in (13) for large enough $c$. This proves Eq. (12) and the result follows from Lemma 5.

Proof

(of Theorem 2, part 1) For any $\varvec{\theta }_1\in S$, $E\ell _n(\varvec{\theta }_1)\!<\!\infty$, so $E[\ell _n(\varvec{\theta }_1)-\ell _n(\varvec{\theta }_0)]\!<\!0$ by Jensen’s inequality. This implies the existence of $\xi _{\varvec{\theta }_1}$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\text {Pr}\left\{ \ell _n(\varvec{\theta }_1)-\ell _n(\varvec{\theta }_0)<-\xi _{\varvec{\theta }_1}\right\} =1. \end{aligned}$$

For $\varvec{\theta }$ and $\eta$ such that $|\varvec{\theta }-\varvec{\theta }_1|<\eta <|\varvec{\theta }_1-\varvec{\theta }_0|-\delta$,

$$\begin{aligned}&|\ell _n(\varvec{\theta })-\ell _n(\varvec{\theta }_1)|\\&\quad \le |\log \sigma -\log \sigma _1| +\frac{1}{n}\sum _{i=1}^n\left| \log \left( \frac{1}{B-Y_i}+\frac{1}{Y_i-A}\right) -\log \left( \frac{1}{B_1-Y_i}+\frac{1}{Y_i-A_1}\right) \right| \\&\quad \quad +\frac{1}{n}\sum _{i=1}^n\left| \frac{\left\{ \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right\} ^2}{\sigma ^2}-\frac{\left\{ \log \left( \frac{B_1-Y_i}{Y_i-A_1}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1\right\} ^2}{\sigma _1^2}\right| \\&\quad ={\varDelta }_4+{\varDelta }_5+{\varDelta }_6. \end{aligned}$$

${\varDelta }_4$ can be made smaller than ${\xi _{\varvec{\theta }_1}}/{4}$ by choosing $\eta$ small enough. By the mean value theorem,

$$\begin{aligned} {\varDelta }_5=&\frac{1}{n}\sum _{i=1}^n\left| \frac{1}{B^*-Y_i}\frac{Y_i-A^*}{B^*-A^*}(B-B_1)+\frac{1}{Y_i-A^*}\frac{B^*-Y_i}{B^*-A^*}(A-A_1)\right| \\ \le&\frac{1}{n}\sum _{i=1}^n\left\{ \frac{B_0-A_1+\eta }{B_0-A_0}\frac{|B-B_1|}{B_0-Y_i}+\frac{B_1-A_0+\eta }{B_0-A_0}\frac{|A-A_1|}{Y_i-A_0}\right\} \!, \end{aligned}$$

for some $A^*$ between $A_0$ and $A_1$ and $B^*$ between $B_0$ and $B_1$. So $E({\varDelta }_5)$ can be make arbitrarily small by choosing small enough $\eta$, which implies

$$\begin{aligned} \lim _{n\rightarrow \infty }\text {Pr}\left( {\varDelta }_5<\frac{\xi _{\varvec{\theta }_1}}{4}\right) =1 \end{aligned}$$

for small enough $\eta$.

$$\begin{aligned} {\varDelta }_6&\le \frac{1}{n}\sum _{i=1}^n\left| \frac{\left\{ \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right\} ^2}{\sigma ^2}-\frac{\left\{ \log \left( \frac{B_1-Y_i}{Y_i-A_1}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1\right\} ^2}{\sigma ^2}\right| \\&\quad +\frac{1}{n}\sum _{i=1}^n\left\{ \log \left( \frac{B_1-Y_i}{Y_i-A_1}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1\right\} ^2\left| \frac{1}{\sigma ^2}-\frac{1}{\sigma _1^2}\right| \\&\le \frac{1}{n}\sum _{i=1}^n\left| \frac{\log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }+\log \left( \frac{B_1-Y_i}{Y_i-A_1}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1}{\sigma ^2}\right| \\&\quad \times \left\{ \frac{|A-A_0|}{A_0-Y_i}+\frac{|B-B_0|}{B_0-Y_i}+|\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }-\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1|\right\} \\&\quad +\frac{1}{n}\sum _{i=1}^n\left\{ \log \left( \frac{B_1-Y_i}{Y_i-A_1}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1\right\} ^2\left| \frac{1}{\sigma ^2}-\frac{1}{\sigma _1^2}\right| . \end{aligned}$$

So, for small enough $\eta$, we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }\text {Pr}\left( {\varDelta }_6<\frac{\xi _{\varvec{\theta }_1}}{4}\right) =1. \end{aligned}$$

Combining results for ${\varDelta }_4$, ${\varDelta }_5$ and ${\varDelta }_6$,

$$\begin{aligned} \lim _{n\rightarrow \infty }\text {Pr}\left\{ \sup _{|\varvec{\theta }-\varvec{\theta }_1|<\eta }\ell _n(\varvec{\theta })-\ell _n(\varvec{\theta }_0)<-\frac{\xi _{\varvec{\theta }_1}}{4}\right\} =1. \end{aligned}$$

For any compact set $K, S_\delta \cap K$ can be covered by a finite number of neighborhoods of points in $S_\delta$, so it follows that

$$\begin{aligned} \lim _{n\rightarrow \infty }\text {Pr}\left\{ \sup _{S_\delta \cap K}\ell _n(\varvec{\theta })-\ell _n(\varvec{\theta }_0)<-\xi _{m}\right\} =1. \end{aligned}$$

Proof

(of Theorem 2, part 2) First, if $A_0$ and $B_0$ are known, model (1) can be transformed to a linear model with normal random error with unknown mean and variance. It follows that

$$\begin{aligned} \lim _{n\rightarrow \infty }\text {Pr}\left\{ \sup _{\Vert \varvec{\beta }-\varvec{\beta }_0\Vert >\delta \, |\sigma -\sigma _0|>\delta }\ell _n(\varvec{\beta },\sigma ,A_0,B_0)-\ell _n(\varvec{\theta }_0)<-\xi \right\} =1. \end{aligned}$$

(14)

For $\varvec{\beta }_1, \sigma _1, \eta$ and $(\varvec{\beta },\sigma ,A,B)\in T$ such that $(\varvec{\beta }_1,\sigma _1, A,B)\in T$, $\Vert \varvec{\beta }-\varvec{\beta }_1\Vert <\eta$, $|\sigma -\sigma _1|<\eta$ and $\delta <\eta$,

$$\begin{aligned}&|\ell _n(\varvec{\beta },\sigma ,A,B)-\ell _n(\varvec{\beta }_1,\sigma _1,A_0,B_0)|\nonumber \\&\quad \le |\log \sigma -\log \sigma _1|+|\log (B-A)-\log (B_0-A_0)|\nonumber \\&\qquad +\frac{1}{n}\sum _{i=1}^n|\log (B-Y_i)-\log (B_0-Y_i)| +\frac{1}{n}\sum _{i=1}^n\left| \log (Y_i-A)-\log (Y_i-A_0)\right| \nonumber \\&\qquad +\frac{1}{2n}\sum _{i=1}^n\left| \frac{\left\{ \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right\} ^2}{\sigma ^2}-\frac{\left\{ \log \left( \frac{B_0-Y_i}{Y_i-A_0}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1\right\} ^2}{\sigma _1^2}\right| \nonumber \\&={\varDelta }_7+{\varDelta }_8+{\varDelta }_9+{\varDelta }_{10}+{\varDelta }_{11}. \end{aligned}$$

(15)

The terms ${\varDelta }_7$ and ${\varDelta }_8$ can be made smaller than ${\xi }/{8}$ by choosing $\eta$ small enough. By the mean value theorem,

$$\begin{aligned} {\varDelta }_9=\frac{1}{n}\sum _{i=1}^n\left| \frac{B-B_0}{B^*-Y_i}\right| \le \frac{|B-B_0|}{n}\sum _{i=1}^n\frac{1}{\min (B,B_0)-Y_i} \end{aligned}$$

with probability tending to 1. If $B\ge B_0$,

$$\begin{aligned} n^{-1}\sum _{i=1}^n\frac{1}{\left| \min (B,B_0)-Y_i\right| }\le n^{-1}\sum _{i=1}^n\frac{1}{(B_0-Y_i)}, \end{aligned}$$

and the right hand side of the upper inequality goes to the limit of

$$\begin{aligned} \frac{1+n^{-1}\sum _{i=1}^ne^{-\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }+\sigma ^2/2}}{(B_0-A_0)} \end{aligned}$$

in probability. If $B_0-\delta _n<B<B_0$, Lemma 3 provides that there exists some constant $M^*$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\text {Pr}\left( \frac{1}{n}\sum _{i=1}^n\frac{1}{\left| B-Y_i\right| }<M^*\right) =1 \end{aligned}$$

for small enough $\eta$. This implies that for small enough $\eta$,

$$\begin{aligned} \lim _{n\rightarrow \infty }\text {Pr}\left( {\varDelta }_9<\frac{\xi }{8}\right) =1. \end{aligned}$$

(16)

The same result can be found for ${\varDelta }_{10}$ using similar arguments.

$$\begin{aligned} {\varDelta }_{11}\le&\frac{1}{2n}\sum _{i=1}^n\left| \frac{\left\{ \log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }\right\} ^2}{\sigma ^2}-\frac{\left\{ \log \left( \frac{B_0-Y_i}{Y_i-A_0}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1\right\} ^2}{\sigma ^2}\right| \\&+\frac{1}{2n}\sum _{i=1}^n\left\{ \log \left( \frac{B_0-Y_i}{Y_i-A_0}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1\right\} ^2\left| \frac{1}{\sigma ^2}-\frac{1}{\sigma _1^2}\right| \\ \le&\frac{1}{2n}\sum _{i=1}^n\left| \frac{\log \left( \frac{B-Y_i}{Y_i-A}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }+\log \left( \frac{B_0-Y_i}{Y_i-A_0}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1}{\sigma ^2}\right| \times |\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }-\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1|\\&+\frac{1}{2n}\sum _{i=1}^n\left\{ \log \left( \frac{B_0-Y_i}{Y_i-A_0}\right) -\mathbf{x}_i^{{ {\mathrm{\scriptscriptstyle T} }}}\varvec{\beta }_1\right\} ^2\left| \frac{1}{\sigma ^2}-\frac{1}{\sigma _1^2}\right| . \end{aligned}$$

So we obtain, for small enough $\eta$,

$$\begin{aligned} \lim _{n\rightarrow \infty }\text {Pr}\left( {\varDelta }_{11}<\frac{\xi }{8}\right) =1. \end{aligned}$$

(17)

Combining (14), (15), (16) and (17), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\text {Pr}\left\{ \sup \ell _n(a,b,\sigma ,A,B)-\ell _n(\varvec{\theta }_0)<-\frac{3\xi }{8}\right\} =1, \end{aligned}$$

where the supermum is taken over all $\varvec{\theta }$ satisfying $(\varvec{\beta }_1,\sigma _1,A,B)\in T, \Vert \varvec{\beta }-\varvec{\beta }_1\Vert <\eta$ and $|\sigma -\sigma _1|<\eta$ for fixed $\varvec{\beta }_1$ and $\sigma _1$. This result can be extended directly to any finite set of values of $\varvec{\beta }_1$ and $\sigma _1$, and then to any compact sets of values of $\varvec{\beta }_1$ and $\sigma _1$.

Proof

(of Theorem 3) By Taylor expansion,

$$\begin{aligned} 0=\frac{\partial \ell _n({\widehat{\varvec{\theta }}}_n)}{\partial \varvec{\theta }}=\frac{\partial \ell _n\left( \varvec{\theta }_0\right) }{\partial \varvec{\theta }}+\frac{\partial ^2\ell _n({\widehat{\varvec{\theta }}}^*)}{\partial \varvec{\theta }\partial \varvec{\theta }^{{ {\mathrm{\scriptscriptstyle T} }}}}({\widehat{\varvec{\theta }}}_n-\varvec{\theta }_0), \end{aligned}$$

where ${\widehat{\varvec{\theta }}}^*$ is between $\varvec{\theta }_0$ and ${\widehat{\varvec{\theta }}}_n$. From Lemma 4, $n^{-1}\partial {^2\ell _n({\widehat{\varvec{\theta }}}^*)}/(\partial {\varvec{\theta }\partial \varvec{\theta }^{{ {\mathrm{\scriptscriptstyle T} }}}})\rightarrow -\mathcal I (\varvec{\theta }_0)$ in probability. So

$$\begin{aligned} n^{1/2}({\widehat{\varvec{\theta }}}_n-\varvec{\theta }_0)=\left\{ \mathcal I (\varvec{\theta }_0)\right\} ^{-1}n^{-1/2}\frac{\partial \ell _n\left( \varvec{\theta }_0\right) }{\partial \varvec{\theta }}+o_P(1). \end{aligned}$$

(18)

Note $n^{-1/2}\partial {\ell _n\left( \varvec{\theta }_0\right) }/\partial {\varvec{\theta }}=n^{-1/2}\sum _{i=1}^n\partial \ell (\varvec{\theta }_0,\mathbf{x}_i,Y_i)/\partial \varvec{\theta }$ is summation of independent random vectors and its variance converges to $\mathcal I (\varvec{\theta }_0)$. Also we have for $t>0$,

$$\begin{aligned}&\frac{1}{n}\sum _{i=1}^nE\left[ \left\| \frac{\partial \ell (\varvec{\theta }_0,\mathbf{x}_i,Y_i)}{\partial \varvec{\theta }}\right\| ^2I{\left\{ \left\| \frac{\partial \ell (\varvec{\theta }_0,\mathbf{x}_i,Y_i)}{\partial \varvec{\theta }}\right\| >n^{1/2}\epsilon \right\} }\right] \\&\quad \le \frac{1}{n}\frac{1}{(n^{1/2}\epsilon )^t}\sum _{i=1}^nE\left[ \left\| \frac{\partial \ell (\varvec{\theta }_0,\mathbf{x}_i,Y_i)}{\partial \varvec{\theta }}\right\| ^{2+t}I{\left\{ \left\| \frac{\partial \ell (\varvec{\theta }_0,\mathbf{x}_i,Y_i)}{\partial \varvec{\theta }}\right\| >n^{1/2}\epsilon \right\} }\right] \nonumber \\&\quad \le \frac{1}{n}\frac{1}{(n^{1/2}\epsilon )^t}\sum _{i=1}^nE\left[ \left\| \frac{\partial \ell (\varvec{\theta }_0,\mathbf{x}_i,Y_i)}{\partial \varvec{\theta }}\right\| ^{2+t}\right] \rightarrow 0 \quad \text { as }\quad n\rightarrow \infty . \end{aligned}$$

By the multivariate central limit theorem (cf. Rao 1973; Serfling 1980),

$$\begin{aligned} n^{-1/2}\frac{\partial \ell _n\left( \varvec{\theta }_0\right) }{\partial \varvec{\theta }}\rightarrow N\{0,\mathcal I (\varvec{\theta }_0)\} \end{aligned}$$

(19)

in distribution. Combining (18), (19) and applying Slutsky’s theorem, the result follows.