On Limiting Likelihood Ratio Processes
of some Change-Point Type Statistical Models

Different change-point type models encountered in statistical inference for stochastic processes give rise to different limiting likelihood ratio processes. In this paper we consider two of these processes. The first one is the random process $Z_{\rho}$ on $\mathbb{R}$ defined by

$$\ln Z_{\rho}(x)=\begin{cases}\vphantom{{\mathchoice{\mbox{$\displaystyle\left)\vrule height=11.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\textstyle\left)\vrule height=11.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptstyle\left)\vrule height=5.63496pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptscriptstyle\left)\vrule height=2.875pt,depth=0.0pt,width=0.0pt\right.\n@space$}}}}\rho\,\Pi_{+}(x)-x,&\text{if }x\geqslant 0,\\ \vphantom{{\mathchoice{\mbox{$\displaystyle\left)\vrule height=11.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\textstyle\left)\vrule height=11.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptstyle\left)\vrule height=5.63496pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptscriptstyle\left)\vrule height=2.875pt,depth=0.0pt,width=0.0pt\right.\n@space$}}}}-\rho\,\Pi_{-}(-x)-x,&\text{if }x\leqslant 0,\\ \end{cases}$$

(1)

where $\rho>0$, and $\Pi_{+}$ and $\Pi_{-}$ are two independent Poisson processes on $\mathbb{R}_{+}$ with intensities $1/(e^{\rho}-1)$ and $1/(1-e^{-\rho})$ respectively. We also consider the random variables

$$\zeta_{\rho}=\frac{\int_{\mathbb{R}}x\,Z_{\rho}(x)\;dx}{\int_{\mathbb{R}}\,Z_{\rho}(x)\;dx}\quad\text{and}\quad\xi_{\rho}=\mathop{\rm argsup}\limits_{x\in\mathbb{R}}Z_{\rho}(x)$$

(2)

related to this process, as well as to their second moments $B_{\rho}=\mathbf{E}\zeta_{\rho}^{2}$ and $M_{\rho}=\mathbf{E}\xi_{\rho}^{2}$.

The process $Z_{\rho}$ (up to a linear time change) arises in some non-regular, namely change-point type, statistical models as the limiting likelihood ratio process, and the variables $\zeta_{\rho}$ and $\xi_{\rho}$ (up to a multiplicative constant) as the limiting distributions of the Bayesian estimators and of the maximum likelihood estimator respectively. In particular, $B_{\rho}$ and $M_{\rho}$ (up to the square of the above multiplicative constant) are the limiting variances of these estimators, and the Bayesian estimators being asymptotically efficient, the ratio $E_{\rho}=B_{\rho}/M_{\rho}$ is the asymptotic efficiency of the maximum likelihood estimator in these models.

The main such model is the below detailed model of i.i.d. observations in the situation when their density has a jump (is discontinuous). Probably the first general result about this model goes back to Chernoff and Rubin [1]. Later, it was exhaustively studied by Ibragimov and Khasminskii in [10, Chapter 5] $\bigl{(}$see also their previous works [7] and [8]$\bigr{)}$.

Denote $\mathbf{P}_{\theta}^{n}$ the distribution (corresponding to the parameter $\theta$) of the observation $X^{n}$. As $n\to\infty$, the normalized likelihood ratio process of this model defined by

$$Z_{n}(u)=\frac{d\mathbf{P}_{\theta+\frac{u}{n}}^{n}}{d\mathbf{P}_{\theta}^{n}}(X^{n})=\prod_{i=1}^{n}\frac{f\left(X_{i}-\theta-\frac{u}{n}\right)}{f(X_{i}-\theta)}$$

converges weakly in the space $\mathcal{D}_{0}(-\infty,+\infty)$ (the Skorohod space of functions on $\mathbb{R}$ without discontinuities of the second kind and vanishing at infinity) to the process $Z_{a,b}$ on $\mathbb{R}$ defined by

$$\ln Z_{a,b}(u)=\begin{cases}\vphantom{{\mathchoice{\mbox{$\displaystyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\textstyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptstyle\left)\vrule height=7.10497pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptscriptstyle\left)\vrule height=3.625pt,depth=0.0pt,width=0.0pt\right.\n@space$}}}}\ln(\frac{a}{b})\,\Pi_{b}(u)-(a-b)\,u,&\text{if }u\geqslant 0,\\ \vphantom{{\mathchoice{\mbox{$\displaystyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\textstyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptstyle\left)\vrule height=7.10497pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptscriptstyle\left)\vrule height=3.625pt,depth=0.0pt,width=0.0pt\right.\n@space$}}}}-\ln(\frac{a}{b})\,\Pi_{a}(-u)-(a-b)\,u,&\text{if }u\leqslant 0,\\ \end{cases}$$

where $\Pi_{b}$ and $\Pi_{a}$ are two independent Poisson processes on $\mathbb{R}_{+}$ with intensities $b$ and $a$ respectively. The limiting distributions of the Bayesian estimators and of the maximum likelihood estimator are given by

$$\zeta_{a,b}=\frac{\int_{\mathbb{R}}u\,Z_{a,b}(u)\;du}{\int_{\mathbb{R}}\,Z_{a,b}(u)\;du}\quad\text{and}\quad\xi_{a,b}=\mathop{\rm argsup}\limits_{u\in\mathbb{R}}Z_{a,b}(u)$$

respectively. The convergence of moments also holds, and the Bayesian estimators are asymptotically efficient. So, $\mathbf{E}\zeta_{a,b}^{2}$ and $\mathbf{E}\xi_{a,b}^{2}$ are the limiting variances of these estimators, and $\mathbf{E}\zeta_{a,b}^{2}/\mathbf{E}\xi_{a,b}^{2}$ is the asymptotic efficiency of the maximum likelihood estimator.

Now let us note, that up to a linear time change, the process $Z_{a,b}$ is nothing but the process $Z_{\rho}$ with $\rho=\left|\ln(\frac{a}{b})\right|$. Indeed, by putting $u=\frac{x}{a-b}$ we get

	$\displaystyle\ln Z_{a,b}(u)$	$\displaystyle=\begin{cases}\vphantom{{\mathchoice{\mbox{$\displaystyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\textstyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptstyle\left)\vrule height=7.10497pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptscriptstyle\left)\vrule height=3.625pt,depth=0.0pt,width=0.0pt\right.\n@space$}}}}\ln(\frac{a}{b})\,\Pi_{b}(\frac{x}{a-b})-x,&\text{if }\frac{x}{a-b}\geqslant 0,\\ \vphantom{{\mathchoice{\mbox{$\displaystyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\textstyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptstyle\left)\vrule height=7.10497pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptscriptstyle\left)\vrule height=3.625pt,depth=0.0pt,width=0.0pt\right.\n@space$}}}}-\ln(\frac{a}{b})\,\Pi_{a}(-\frac{x}{a-b})-x,&\text{if }\frac{x}{a-b}\leqslant 0,\\ \end{cases}$
		$\displaystyle=\ln Z_{\rho}(x)=\ln Z_{\rho}\bigl{(}(a-b)\,u\bigr{)}.$

So, we have

$$\zeta_{a,b}=\frac{\zeta_{\rho}}{a-b}\quad\text{and}\quad\xi_{a,b}=\frac{\xi_{\rho}}{a-b}\,,$$

and hence

$$\mathbf{E}\zeta_{a,b}^{2}=\frac{B_{\rho}}{(a-b)^{2}}\;,\quad\mathbf{E}\xi_{a,b}^{2}=\frac{M_{\rho}}{(a-b)^{2}}\quad\text{and}\quad\frac{\mathbf{E}\zeta_{a,b}^{2}}{\mathbf{E}\xi_{a,b}^{2}}=E_{\rho}\,.$$

Some other models where the process $Z_{\rho}$ arises occur in the statistical inference for inhomogeneous Poisson processes, in the situation when their intensity function has a jump (is discontinuous). In Kutoyants [14, Chapter 5] $\bigl{(}$see also his previous work [12]$\bigr{)}$ one can find several examples, one of which is detailed below.

Denote $\mathbf{P}_{\theta}^{T}$ the distribution (corresponding to the parameter $\theta$) of the observation $X^{T}$. As $T\to\infty$, the normalized likelihood ratio process of this model defined by

$$Z_{T}(u)=\frac{d\mathbf{P}_{\theta+\frac{u}{T}}^{T}}{d\mathbf{P}_{\theta}^{T}}(X^{T})=\exp\biggl{\{}\int_{0}^{T}\!\!\ln\frac{S_{\theta+\frac{u}{T}}(t)}{S_{\theta}(t)}\,dX(t)-\int_{0}^{T}\!\!\left[S_{\theta+\frac{u}{T}}(t)-S_{\theta}(t)\right]\!dt\biggr{\}}$$

converges weakly in the space $\mathcal{D}_{0}(-\infty,+\infty)$ to the process $Z_{\tau,S_{-},S_{+}}$ on $\mathbb{R}$ defined by

$$\ln Z_{\tau,S_{-},S_{+}}=\begin{cases}\vphantom{{\mathchoice{\mbox{$\displaystyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\textstyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptstyle\left)\vrule height=7.10497pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptscriptstyle\left)\vrule height=3.625pt,depth=0.0pt,width=0.0pt\right.\n@space$}}}}\ln\bigl{(}\frac{S_{+}}{S_{-}}\bigr{)}\,\Pi_{S_{-}}\bigl{(}\frac{u}{\tau}\bigr{)}-(S_{+}-S_{-})\,\frac{u}{\tau}\,,&\text{if }u\geqslant 0,\\ \vphantom{{\mathchoice{\mbox{$\displaystyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\textstyle\left)\vrule height=14.5pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptstyle\left)\vrule height=7.10497pt,depth=0.0pt,width=0.0pt\right.\n@space$}}{\mbox{$\scriptscriptstyle\left)\vrule height=3.625pt,depth=0.0pt,width=0.0pt\right.\n@space$}}}}-\ln\bigl{(}\frac{S_{+}}{S_{-}}\bigr{)}\,\Pi_{S_{+}}\bigl{(}-\frac{u}{\tau}\bigr{)}-(S_{+}-S_{-})\,\frac{u}{\tau}\,,&\text{if }u\leqslant 0,\\ \end{cases}$$

where $\Pi_{S_{-}}$ and $\Pi_{S_{+}}$ are two independent Poisson processes on $\mathbb{R}_{+}$ with intensities $S_{-}$ and $S_{+}$ respectively. The limiting distributions of the Bayesian estimators and of the maximum likelihood estimator are given by

$$\zeta_{\tau,S_{-},S_{+}}=\frac{\int_{\mathbb{R}}u\,Z_{\tau,S_{-},S_{+}}(u)\;du}{\int_{\mathbb{R}}\,Z_{\tau,S_{-},S_{+}}(u)\;du}\quad\text{and}\quad\xi_{\tau,S_{-},S_{+}}=\mathop{\rm argsup}\limits_{u\in\mathbb{R}}Z_{\tau,S_{-},S_{+}}(u)$$

respectively. The convergence of moments also holds, and the Bayesian estimators are asymptotically efficient. So, $\mathbf{E}\zeta_{\tau,S_{-},S_{+}}^{2}$ and $\mathbf{E}\xi_{\tau,S_{-},S_{+}}^{2}$ are the limiting variances of these estimators, and $\mathbf{E}\zeta_{\tau,S_{-},S_{+}}^{2}/\mathbf{E}\xi_{\tau,S_{-},S_{+}}^{2}$ is the asymptotic efficiency of the maximum likelihood estimator.

Now let us note, that up to a linear time change, the process $Z_{\tau,S_{-},S_{+}}$ is nothing but the process $Z_{\rho}$ with $\rho=\bigl{|}\ln\bigl{(}\frac{S_{+}}{S_{-}}\bigr{)}\bigr{|}$. Indeed, by putting $u=\frac{\tau x}{S_{+}-S_{-}}$ we get

$$Z_{\tau,S_{-},S_{+}}(u)=Z_{\rho}(x)=Z_{\rho}\left(\frac{S_{+}-S_{-}}{\tau}\,u\right).$$

So, we have

$$\zeta_{\tau,S_{-},S_{+}}=\frac{\tau\,\zeta_{\rho}}{S_{+}-S_{-}}\quad\text{and}\quad\zeta_{\tau,S_{-},S_{+}}=\frac{\tau\,\xi_{\rho}}{S_{+}-S_{-}}\,,$$

and hence

$$\mathbf{E}\zeta_{\tau,S_{-},S_{+}}^{2}=\frac{\tau^{2}\,B_{\rho}}{(S_{+}-S_{-})^{2}}\;,\quad\mathbf{E}\xi_{\tau,S_{-},S_{+}}^{2}=\frac{\tau^{2}\,M_{\rho}}{(S_{+}-S_{-})^{2}}\quad\text{and}\quad\frac{\mathbf{E}\zeta_{\tau,S_{-},S_{+}}^{2}}{\mathbf{E}\xi_{\tau,S_{-},S_{+}}^{2}}=E_{\rho}\,.$$

The second limiting likelihood ratio process considered in this paper is the random process

$$Z_{0}(x)=\exp\left\{W(x)-\frac{1}{2}\left|x\right|\right\},\quad x\in\mathbb{R},$$

(3)

where $W$ is a standard two-sided Brownian motion. In this case, the limiting distributions of the Bayesian estimators and of the maximum likelihood estimator (up to a multiplicative constant) are given by

$$\zeta_{0}=\frac{\int_{\mathbb{R}}x\,Z_{0}(x)\;dx}{\int_{\mathbb{R}}\,Z_{0}(x)\;dx}\quad\text{and}\quad\xi_{0}=\mathop{\rm argsup}\limits_{x\in\mathbb{R}}Z_{0}(x)$$

(4)

respectively, and the limiting variances of these estimators (up to the square of the above multiplicative constant) are $B_{0}=\mathbf{E}\zeta_{0}^{2}$ and $M_{0}=\mathbf{E}\xi_{0}^{2}$.

The models where the process $Z_{0}$ arises occur in various fields of statistical inference for stochastic processes. A well-known example is the below detailed model of a discontinuous signal in a white Gaussian noise exhaustively studied by Ibragimov and Khasminskii in [10, Chapter 7.2] $\bigl{(}$see also their previous work [9]$\bigr{)}$, but one can also cite change-point type models of dynamical systems with small noise $\bigl{(}$see Kutoyants [12] and [13, Chapter 5]$\bigr{)}$, those of ergodic diffusion processes $\bigl{(}$see Kutoyants [15, Chapter 3]$\bigr{)}$, a change-point type model of delay equations $\bigl{(}$see Küchler and Kutoyants [11]$\bigr{)}$, an i.i.d. change-point type model $\bigl{(}$see Deshayes and Picard [3]$\bigr{)}$, a model of a discontinuous periodic signal in a time inhomogeneous diffusion $\bigl{(}$see Höpfner and Kutoyants [6]$\bigr{)}$, and so on.

Denote $\mathbf{P}_{\theta}^{\varepsilon}$ the distribution (corresponding to the parameter $\theta$) of the observation $X^{\varepsilon}$. As $\varepsilon\to 0$, the normalized likelihood ratio process of this model defined by

	$\displaystyle Z_{\varepsilon}(u)$	$\displaystyle=\frac{d\mathbf{P}_{\theta+\varepsilon^{2}u}^{\varepsilon}}{d\mathbf{P}_{\theta}^{\varepsilon}}(X^{\varepsilon})$
		$\displaystyle=\exp\biggl{\{}\frac{1}{\varepsilon}\int_{0}^{1}\bigl{[}S(t-\theta-\varepsilon^{2}u)-S(t-\theta)\bigr{]}\,dW(t)$
		$\displaystyle\phantom{=\exp\biggl{\{}}-\frac{1}{2\,\varepsilon^{2}}\int_{0}^{1}\bigl{[}S(t-\theta-\varepsilon^{2}u)-S(t-\theta)\bigr{]}^{2}\,dt\biggr{\}}$

converges weakly in the space $\mathcal{C}_{0}(-\infty,+\infty)$ (the space of continuous functions vanishing at infinity equipped with the supremum norm) to the process $Z_{0}(r^{2}u)$, $u\in\mathbb{R}$. The limiting distributions of the Bayesian estimators and of the maximum likelihood estimator are $r^{-2}\zeta_{0}$ and $r^{-2}\xi_{0}$ respectively. The convergence of moments also holds, and the Bayesian estimators are asymptotically efficient. So, $r^{-4}B_{0}$ and $r^{-4}M_{0}$ are the limiting variances of these estimators, and $E_{0}$ is the asymptotic efficiency of the maximum likelihood estimator.

Let us also note that Terent’yev in [20] determined explicitly the distribution of $\xi_{0}$ and calculated the constant $M_{0}=26$. These results were taken up by Ibragimov and Khasminskii in [10, Chapter 7.3], where by means of numerical simulation they equally showed that $B_{0}=19.5\pm 0.5$, and so $E_{0}=0.73\pm 0.03$. Later in [5], Golubev expressed $B_{0}$ in terms of the second derivative (with respect to a parameter) of an improper integral of a composite function of modified Hankel and Bessel functions. Finally in [18], Rubin and Song obtained the exact values $B_{0}=16\,\zeta(3)$ and $E_{0}=8\,\zeta(3)/13$, where $\zeta$ is Riemann’s zeta function defined by

$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}}\;.$$

The random variables $\zeta_{\rho}$ and $\xi_{\rho}$ and the quantities $B_{\rho}$, $M_{\rho}$ and $E_{\rho}$, $\rho>0$, are much less studied. One can cite Pflug [16] for some results about the distribution of the random variables

$$\mathop{\rm argsup}\limits_{x\in\mathbb{R}_{+}}Z_{\rho}(x)\quad\text{and}\quad\mathop{\rm argsup}\limits_{x\in\mathbb{R}_{-}}Z_{\rho}(x)$$

related to $\xi_{\rho}$.

In this paper we establish that the limiting likelihood ratio processes $Z_{\rho}$ and $Z_{0}$ are related. More precisely, we show that as $\rho\to 0$, the process $Z_{\rho}(y/\rho)$, $y\in\mathbb{R}$, converges weakly in the space $\mathcal{D}_{0}(-\infty,+\infty)$ to the process $Z_{0}$. So, the random variables $\rho\,\zeta_{\rho}$ and $\rho\,\xi_{\rho}$ converge weakly to the random variables $\zeta_{0}$ and $\xi_{0}$ respectively. We show equally that the convergence of moments of these random variables holds, that is, $\rho^{2}B_{\rho}\to 16\,\zeta(3)$, $\rho^{2}M_{\rho}\to 26$ and $E_{\rho}\to 8\,\zeta(3)/13$.

These are the main results of the present paper, and they are presented in Section 2, where we also briefly discuss the second possible asymptotics $\rho\to+\infty$. The necessary lemmas are proved in Section 3. Finally, some numerical simulations of the quantities $B_{\rho}$, $M_{\rho}$ and $E_{\rho}$ for $\rho\in\left]0,\infty\right[$ are presented in Section 4.

Consider the process $X_{\rho}(y)=Z_{\rho}(y/\rho)$, $y\in\mathbb{R}$, where $\rho>0$ and $Z_{\rho}$ is defined by (1). Note that

$$\frac{\int_{\mathbb{R}}y\,X_{\rho}(y)\;dy}{\int_{\mathbb{R}}\,X_{\rho}(y)\;dy}=\rho\,\zeta_{\rho}\quad\text{and}\quad\mathop{\rm argsup}\limits_{y\in\mathbb{R}}X_{\rho}(y)=\rho\,\xi_{\rho}\,,$$

where the random variables $\zeta_{\rho}$ and $\xi_{\rho}$ are defined by (2). Remind also the process $Z_{0}$ on $\mathbb{R}$ defined by (3) and the random variables $\zeta_{0}$ and $\xi_{0}$ defined by (4). Recall finally the quantities $B_{\rho}=\mathbf{E}\zeta_{\rho}^{2}$, $M_{\rho}=\mathbf{E}\xi_{\rho}^{2}$, $E_{\rho}=B_{\rho}/M_{\rho}$, $B_{0}=\mathbf{E}\zeta_{0}^{2}=16\,\zeta(3)$, $M_{0}=\mathbf{E}\xi_{0}^{2}=26$ and $E_{0}=B_{0}/M_{0}=8\,\zeta(3)/13$. Now we can state the main result of the present paper.

The results concerning the random variable $\zeta_{\rho}$ are direct consequence of Ibragimov and Khasminskii [10, Theorem 1.10.2] and the following three lemmas.

Note that these lemmas are not sufficient to establish the weak convergence of the process $X_{\rho}$ in the space $\mathcal{D}_{0}(-\infty,+\infty)$ and the results concerning the random variable $\xi_{\rho}$. However, the increments of the process $\ln X_{\rho}$ being independent, the convergence of its restrictions (and hence of those of $X_{\rho}$) on finite intervals $[A,B]\subset\mathbb{R}$ $\bigl{(}$that is, convergence in the Skorohod space $\mathcal{D}[A,B]$ of functions on $[A,B]$ without discontinuities of the second kind$\bigr{)}$ follows from Gihman and Skorohod [4, Theorem 6.5.5], Lemma 2 and the following lemma.

Now, Theorem 1 follows from the following estimate on the tails of the process $X_{\rho}$ by standard argument.

All the above lemmas will be proved in the next section, but before let us discuss the second possible asymptotics $\rho\to+\infty$. One can show that in this case, the process $Z_{\rho}$ converges weakly in the space $\mathcal{D}_{0}(-\infty,+\infty)$ to the process $Z_{\infty}(u)=e^{-u}\,\mathbb{1}_{\{u>\eta\}}$, $u\in\mathbb{R}$, where $\eta$ is a negative exponential random variable with $\mathbf{P}\{\eta<t\}=e^{t}$, $t\leqslant 0$. So, the random variables $\zeta_{\rho}$ and $\xi_{\rho}$ converge weakly to the random variables

$$\zeta_{\infty}=\frac{\int_{\mathbb{R}}u\,Z_{\infty}(u)\;du}{\int_{\mathbb{R}}\,Z_{\infty}(u)\;du}=\eta+1\quad\text{and}\quad\xi_{\infty}=\mathop{\rm argsup}\limits_{u\in\mathbb{R}}Z_{\infty}(u)=\eta$$

respectively. One can equally show that, moreover, for any $k>0$ we have

$$\mathbf{E}\zeta_{\rho}^{k}\to\mathbf{E}\zeta_{\infty}^{k}\quad\text{and}\quad\mathbf{E}\xi_{\rho}^{k}\to\mathbf{E}\xi_{\infty}^{k},$$

and in particular, denoting $B_{\infty}\!=\mathbf{E}\zeta_{\infty}^{2}$, $M_{\infty}\!=\mathbf{E}\xi_{\infty}^{2}$ and $E_{\infty}\!=B_{\infty}/M_{\infty}$, we finally have $B_{\rho}\to B_{\infty}\!=\mathbf{E}(\eta+1)^{2}=1$, $M_{\rho}\to M_{\infty}\!=\mathbf{E}\eta^{2}=2$ and $E_{\rho}\to E_{\infty}\!=1/2$.

Let us note that these convergences are natural, since the process $Z_{\infty}$ can be considered as a particular case of the process $Z_{\rho}$ with $\rho=+\infty$ if one admits the convention $+\infty\cdot 0=0$.

Note also that the process $Z_{\infty}$ (up to a linear time change) is the limiting likelihood ratio process of Model 1 (Model 2) in the situation when $a\cdot b=0$ ($S_{-}\cdot S_{+}$=0). In this case, the variables $\zeta_{\infty}=\eta+1$ and $\xi_{\infty}=\eta$ (up to a multiplicative constant) are the limiting distributions of the Bayesian estimators and of the maximum likelihood estimator respectively. In particular, $B_{\infty}=1$ and $M_{\infty}=2$ (up to the square of the above multiplicative constant) are the limiting variances of these estimators, and the Bayesian estimators being asymptotically efficient, $E_{\infty}=1/2$ is the asymptotic efficiency of the maximum likelihood estimator.

First we prove Lemma 2. Note that the restrictions of the process $\ln X_{\rho}$ (as well as those of the process $\ln Z_{0}$) on $\mathbb{R}_{+}$ and on $\mathbb{R}_{-}$ are mutually independent processes with stationary and independent increments. So, to obtain the convergence of all the finite-dimensional distributions, it is sufficient to show the convergence of one-dimensional distributions only, that is,

$$\ln X_{\rho}(y)\Rightarrow\ln Z_{0}(y)=W(y)-\frac{\left|y\right|}{2}=\mathcal{N}\Bigl{(}-\frac{\left|y\right|}{2}\,,\left|y\right|\Bigr{)}$$

for all $y\in\mathbb{R}$. Here and in the sequel “$\Rightarrow$” denotes the weak convergence of the random variables, and $\mathcal{N}(m,V)$ denotes a “generic” random variable distributed according to the normal law with mean $m$ and variance $V$.

Let $y>0$. Then, noting that $\displaystyle\Pi_{+}\Bigl{(}\frac{y}{\rho}\Bigr{)}$ is a Poisson random variable of parameter $\displaystyle\lambda=\frac{y}{\rho\,(e^{\rho}-1)}\to\infty$, we have

	$\displaystyle\ln X_{\rho}(y)$	$\displaystyle=\rho\,\Pi_{+}\Bigl{(}\frac{y}{\rho}\Bigr{)}-\frac{y}{\rho}=\rho\,\sqrt{\frac{y}{\rho\,(e^{\rho}-1)}}\ \frac{\Pi_{+}\bigl{(}\frac{y}{\rho}\bigr{)}-\lambda}{\sqrt{\lambda}}+\frac{y}{e^{\rho}-1}-\frac{y}{\rho}$
		$\displaystyle=\sqrt{y}\,\sqrt{\frac{\rho}{e^{\rho}-1}}\ \frac{\Pi_{+}\bigl{(}\frac{y}{\rho}\bigr{)}-\lambda}{\sqrt{\lambda}}-y\,\frac{e^{\rho}-1-\rho}{\rho\,(e^{\rho}-1)}\Rightarrow\mathcal{N}\left(-\frac{y}{2}\,,y\right),$

since

$$\frac{\rho}{e^{\rho}-1}=\frac{\rho}{\rho+o(\rho)}\to 1,\qquad\frac{e^{\rho}-1-\rho}{\rho\,(e^{\rho}-1)}=\frac{\rho^{2}/2+o(\rho^{2})}{\rho\,\bigr{(}\rho+o(\rho)\bigr{)}}\to\frac{1}{2}$$

and

$$\frac{\Pi_{+}\bigl{(}\frac{y}{\rho}\bigr{)}-\lambda}{\sqrt{\lambda}}\Rightarrow\mathcal{N}(0,1).$$

Similarly, for $y<0$ we have

	$\displaystyle\ln X_{\rho}(y)$	$\displaystyle=-\rho\,\Pi_{-}\Bigl{(}\frac{-y}{\rho}\Bigr{)}-\frac{y}{\rho}=\rho\,\sqrt{\frac{-y}{\rho\,(1-e^{-\rho})}}\ \frac{\lambda^{\prime}-\Pi_{-}\bigl{(}\frac{-y}{\rho}\bigr{)}}{\sqrt{\lambda^{\prime}}}-\frac{-y}{1-e^{-\rho}}-\frac{y}{\rho}$
		$\displaystyle=\sqrt{-y}\,\sqrt{\frac{\rho}{1-e^{-\rho}}}\ \frac{\lambda^{\prime}-\Pi_{-}\bigl{(}\frac{-y}{\rho}\bigr{)}}{\sqrt{\lambda^{\prime}}}+y\,\frac{e^{-\rho}-1+\rho}{\rho\,(1-e^{-\rho})}\Rightarrow\mathcal{N}\left(\frac{y}{2}\,,-y\right),$

and so, Lemma 2 is proved.

Now we turn to the proof of Lemma 4 (we will prove Lemma 3 just after). For $y>0$ we can write

$$\mathbf{E}X_{\rho}^{1/2}(y)=\mathbf{E}\exp\left(\frac{\rho}{2}\,\Pi_{+}\Bigl{(}\frac{y}{\rho}\Bigr{)}-\frac{y}{2\rho}\right)=\exp\left(-\frac{y}{2\rho}\right)\,\mathbf{E}\exp\left(\frac{\rho}{2}\,\Pi_{+}\Bigl{(}\frac{y}{\rho}\Bigr{)}\right).$$

Note that $\displaystyle\Pi_{+}\Bigl{(}\frac{y}{\rho}\Bigr{)}$ is a Poisson random variable of parameter $\displaystyle\lambda=\frac{y}{\rho\,(e^{\rho}-1)}$ with moment generating function $M(t)=\exp\bigl{(}\lambda\,(e^{t}-1)\bigr{)}$. So, we get

	$\displaystyle\mathbf{E}X_{\rho}^{1/2}(y)$	$\displaystyle=\exp\left(-\frac{y}{2\rho}\right)\,\exp\left(\frac{y}{\rho\,(e^{\rho}-1)}\,(e^{\rho/2}-1)\right)$
		$\displaystyle=\exp\left(-\frac{y}{2\rho}+\frac{y}{\rho\,(e^{\rho/2}+1)}\right)=\exp\left(-y\,\frac{e^{\rho/2}-1}{2\rho\,(e^{\rho/2}+1)}\right)$
		$\displaystyle=\exp\left(-y\,\frac{e^{\rho/4}-e^{-\rho/4}}{2\rho\,(e^{\rho/4}+e^{-\rho/4})}\right)=\exp\left(-y\,\frac{\tanh(\rho/4)}{2\rho}\right).$

For $y<0$ we obtain similarly

	$\displaystyle\mathbf{E}X_{\rho}^{1/2}(y)$	$\displaystyle=\mathbf{E}\exp\left(-\frac{\rho}{2}\,\Pi_{-}\Bigl{(}\frac{-y}{\rho}\Bigr{)}-\frac{y}{2\rho}\right)$
		$\displaystyle=\exp\left(-\frac{y}{2\rho}\right)\,\exp\left(\frac{-y}{\rho\,(1-e^{-\rho})}\,(e^{-\rho/2}-1)\right)$
		$\displaystyle=\exp\left(-\frac{y}{2\rho}+\frac{y}{\rho\,(1+e^{-\rho/2})}\right)=\exp\left(y\,\frac{1-e^{-\rho/2}}{2\rho\,(1+e^{-\rho/2})}\right)$
		$\displaystyle=\exp\left(y\,\frac{\tanh(\rho/4)}{2\rho}\right).$

Thus, for all $y\in\mathbb{R}$ we have

$$\mathbf{E}X_{\rho}^{1/2}(y)=\exp\left(-\left|y\right|\frac{\tanh(\rho/4)}{2\rho}\right),$$

(5)

and since

$$\frac{\tanh(\rho/4)}{2\rho}=\frac{\rho/4+o(\rho)}{2\rho}\to\frac{1}{8}$$

as $\rho\to 0$, for any $c\in\left]\,0\,{,}\;1/8\,\right[$ we have $\mathbf{E}X_{\rho}^{1/2}(y)\leqslant\exp\bigl{(}-c\left|y\right|\bigr{)}$ for all sufficiently small $\rho$ and all $y\in\mathbb{R}$. Lemma 4 is proved.

Further we verify Lemma 3. We first consider the case $y_{1},y_{2}\in\mathbb{R}_{+}$ (say $y_{1}\geqslant y_{2}$). Using (5) and taking into account the stationarity and the independence of the increments of the process $\ln X_{\rho}$ on $\mathbb{R}_{+}$, we can write

	$\displaystyle\mathbf{E}\left|X_{\rho}^{1/2}(y_{1})-X_{\rho}^{1/2}(y_{2})\right|^{2}$	$\displaystyle=\mathbf{E}X_{\rho}(y_{1})+\mathbf{E}X_{\rho}(y_{2})-2\,\mathbf{E}X_{\rho}^{1/2}(y_{1})X_{\rho}^{1/2}(y_{2})$
		$\displaystyle=2-2\,\mathbf{E}X_{\rho}(y_{2})\,\mathbf{E}\frac{X_{\rho}^{1/2}(y_{1})}{X_{\rho}^{1/2}(y_{2})}$
		$\displaystyle=2-2\,\mathbf{E}X_{\rho}^{1/2}(y_{1}-y_{2})$
		$\displaystyle=2-2\exp\left(-\left|y_{1}-y_{2}\right|\frac{\tanh(\rho/4)}{2\rho}\right)$
		$\displaystyle\leqslant\left|y_{1}-y_{2}\right|\frac{\tanh(\rho/4)}{\rho}\leqslant\frac{1}{4}\left|y_{1}-y_{2}\right|.$

The case $y_{1},y_{2}\in\mathbb{R}_{-}$ can be treated similarly.

Finally, if $y_{1}y_{2}\leqslant 0$ (say $y_{2}\leqslant 0\leqslant y_{1}$), we have

	$\displaystyle\mathbf{E}\left|X_{\rho}^{1/2}(y_{1})-X_{\rho}^{1/2}(y_{2})\right|^{2}$	$\displaystyle=2-2\,\mathbf{E}X_{\rho}^{1/2}(y_{1})\,\mathbf{E}X_{\rho}^{1/2}(y_{2})$
		$\displaystyle=2-2\exp\left(-\left|y_{1}\right|\frac{\tanh(\rho/4)}{2\rho}-\left|y_{2}\right|\frac{\tanh(\rho/4)}{2\rho}\right)$
		$\displaystyle=2-2\exp\left(-\left|y_{1}-y_{2}\right|\frac{\tanh(\rho/4)}{2\rho}\right)$
		$\displaystyle\leqslant\frac{1}{4}\left|y_{1}-y_{2}\right|,$

and so, Lemma 3 is proved.

Now let us check Lemma 5. First let $y_{1},y_{2}\in\mathbb{R}_{+}$ (say $y_{1}\geqslant y_{2}$) such that $\Delta=\left|y_{1}-y_{2}\right|<h$. Then

	$\displaystyle\mathbf{P}\Bigl{\{}\bigl{|}\ln X_{\rho}(y_{1})-\ln X_{\rho}(y_{2})\bigr{|}>\varepsilon\Bigr{\}}$	$\displaystyle\leqslant\frac{1}{\varepsilon^{2}}\,\mathbf{E}\bigl{|}\ln X_{\rho}(y_{1})-\ln X_{\rho}(y_{2})\bigr{|}^{2}$
		$\displaystyle=\frac{1}{\varepsilon^{2}}\,\mathbf{E}\bigl{|}\ln X_{\rho}(\Delta)\bigr{|}^{2}$
		$\displaystyle=\frac{1}{\varepsilon^{2}}\,\mathbf{E}\left|\rho\,\Pi_{+}\Bigl{(}\frac{\Delta}{\rho}\Bigr{)}-\frac{\Delta}{\rho}\right|^{2}$
		$\displaystyle=\frac{1}{\varepsilon^{2}}\left(\rho^{2}(\lambda+\lambda^{2})+\frac{\Delta^{2}}{\rho^{2}}-2\lambda\Delta\right)$
		$\displaystyle=\frac{1}{\varepsilon^{2}}\left(\beta(\rho)\,\Delta+\gamma(\rho)\,\Delta^{2}\right)$
		$\displaystyle<\frac{1}{\varepsilon^{2}}\left(\beta(\rho)\,h+\gamma(\rho)\,h^{2}\right),$

where $\lambda=\frac{\Delta}{\rho\,(e^{\rho}-1)}$ is the parameter of the Poisson random variable $\Pi_{+}\bigl{(}\!\frac{\Delta}{\rho}\!\bigr{)}$,

	$\displaystyle\beta(\rho)$	$\displaystyle=\frac{\rho}{(e^{\rho}-1)}=\frac{\rho}{\rho+o(\rho)}\to 1$
and
	$\displaystyle\gamma(\rho)$	$\displaystyle=\frac{1}{(e^{\rho}-1)^{2}}+\frac{1}{\rho^{2}}-\frac{2}{\rho\,(e^{\rho}-1)}=\left(\frac{1}{\rho}-\frac{1}{e^{\rho}-1}\right)^{2}$
		$\displaystyle=\biggl{(}\frac{e^{\rho}-1-\rho}{\rho\,(e^{\rho}-1)}\biggr{)}^{2}=\biggl{(}\frac{\rho^{2}/2+o(\rho^{2})}{\rho\,\bigl{(}\rho+o(\rho)\bigr{)}}\biggr{)}^{2}\to\frac{1}{4}$

as $\rho\to 0$. So, we have

	$\displaystyle\lim_{\rho\to 0}\ \sup_{\left|y_{1}-y_{2}\right|<h}\mathbf{P}\Bigl{\{}\bigl{|}\ln X_{\rho}(y_{1})-\ln X_{\rho}(y_{2})\bigr{|}>\varepsilon\Bigr{\}}$	$\displaystyle\leqslant\lim_{\rho\to 0}\frac{1}{\varepsilon^{2}}\left(\beta(\rho)\,h+\gamma(\rho)\,h^{2}\right)$
		$\displaystyle=\frac{1}{\varepsilon^{2}}\left(h+\frac{h^{2}}{4}\right),$

and hence

$$\lim_{h\to 0}\ \lim_{\rho\to 0}\ \sup_{\left|y_{1}-y_{2}\right|<h}\mathbf{P}\Bigl{\{}\bigl{|}\ln X_{\rho}(y_{1})-\ln X_{\rho}(y_{2})\bigr{|}>\varepsilon\Bigr{\}}=0,$$

where the supremum is taken only over $y_{1},y_{2}\in\mathbb{R}_{+}$.