Appendix
1.1 Properties of the scheme
Proof of Proposition 1
By integrating (8) by parts two times we get the following:
$$\begin{aligned} \mathbb {E}\left[ {\tilde{Z}}_{i+1}|{\tilde{Z}}_{i} \right]= & {} {\tilde{Z}}_i + J^{-1}({\tilde{Z}}_i; \theta )\left( e^{J({\tilde{Z}}_i; \theta )\varDelta _n}- I \right) A({\tilde{Z}}_i; \theta ) \nonumber \\&+\frac{1}{2}J^{-2}({\tilde{Z}}_i; \theta )\left( e^{J({\tilde{Z}}_i; \theta )\varDelta _n}- I-J({\tilde{Z}}_i; \theta )\varDelta _n \right) b^2({\tilde{Z}}_i;\sigma )\partial ^2_{yy}A({\tilde{Z}}_i; \theta )\nonumber \\ \end{aligned}$$
(25)
Recall that the matrix exponent for some square matrix M is given by \(e^M = \sum _{l=0}^\infty \frac{M^l}{l!}\). Then (25) can be simplified as:
$$\begin{aligned} \mathbb {E}\left[ {\tilde{Z}}_{i+1}|{\tilde{Z}}_{i} \right]= & {} {\tilde{Z}}_i + J^{-1}({\tilde{Z}}_i; \theta )\left( I + \varDelta _nJ({\tilde{Z}}_i; \theta ) \right. \\&\left. + \frac{\varDelta _n^2}{2}J^2({\tilde{Z}}_i; \theta ) - I+ O(\varDelta ^3_n) \right) A({\tilde{Z}}_i; \theta ) \\&+\frac{1}{2}J^{-2}({\tilde{Z}}_i; \theta )\left( I + \varDelta _nJ({\tilde{Z}}_i; \theta ) + \frac{\varDelta _n^2}{2}J^2({\tilde{Z}}_i; \theta ) \right. \\&\left. - I-\varDelta _n J({\tilde{Z}}_i; \theta )+ O(\varDelta ^3_n) \right) b^2({\tilde{Z}}_i;\sigma )\partial ^2_{yy}A({\tilde{Z}}_i; \theta ) \\= & {} {\tilde{Z}}_i+ \varDelta _n A({\tilde{Z}}_i; \theta ) + \frac{\varDelta _n^2}{2}J({\tilde{Z}}_i; \theta ) A({\tilde{Z}}_i; \theta ) \\&+ \frac{\varDelta _n^2}{4}b^2({\tilde{Z}}_i;\sigma )\partial ^2_{yy}A({\tilde{Z}}_i; \theta ) + O(\varDelta ^3_n) \end{aligned}$$
Writing the above expression component-wise gives the proposition. \(\square \)
Proof of Proposition 2
Let us consider each integral of (10) separately. Denote:
$$\begin{aligned} {\mathcal {W}}_{(i+1)\varDelta _n} = \int _{i\varDelta _n}^{(i+1)\varDelta _n} e^{J({\tilde{Z}}_i; \theta )((i+1)\varDelta _n - s) }B({\tilde{Z}}_i; \sigma ) d W_s. \end{aligned}$$
Recall that the Jacobian of system (3) is given by (6) and the definition of the matrix exponent, we have:
$$\begin{aligned} {\mathcal {W}}_{(i+1)\varDelta _n}= & {} \int _{i\varDelta _n}^{(i+1)\varDelta _n}(I + J({\tilde{Z}}_i; \theta ) ((i+1)\varDelta _n - s) + {O}(\varDelta _n^2) ) B({\tilde{Z}}_i; \sigma ) d W_s \\= & {} \int _{i\varDelta _n}^{(i+1)\varDelta _n} \left[ \left( \begin{matrix} 1 + \partial _xa_1({\tilde{Z}}_i; \theta ^{(1)})((i+1)\varDelta _n - s) &{} \partial _ya_1({\tilde{Z}}_i; \theta ^{(1)})((i+1)\varDelta _n - s) \\ \partial _xa_2({\tilde{Z}}_i; \theta ^{(2)})((i+1)\varDelta _n - s) &{} 1 + \partial _ya_2({\tilde{Z}}_i; \theta ^{(2)})((i+1)\varDelta _n - s) \end{matrix}\right) \right. \\&\left. + {O}(\varDelta _n^2)\right] \left( \begin{matrix} 0&{} 0 \\ 0&{} 1 \end{matrix} \right) b({\tilde{Z}}_i; \sigma ) dW_s \\= & {} b({\tilde{Z}}_i; \sigma )\left[ \begin{matrix} 0 &{} \partial _y a_1({\tilde{Z}}_i; \theta ^{(1)}) \int _{i\varDelta _n}^{(i+1)\varDelta _n}((i+1)\varDelta _n - s)dW_s + {O}(\varDelta _n^2) \\ 0 &{} \int _{i\varDelta _n}^{(i+1)\varDelta _n}dW_s +\partial _y a_2({\tilde{Z}}_i; \theta ^{(2)}) \int _{i\varDelta _n}^{(i+1)\varDelta _n}((i+1)\varDelta _n - s)dW_s + {O}(\varDelta _n^2) \end{matrix}\right] \end{aligned}$$
Then we can calculate \(\mathbb {E}\left[ {\mathcal {W}}_{(i+1)\varDelta _n} {\mathcal {W}}^\prime _{(i+1)\varDelta _n}\right] \):
$$\begin{aligned} \mathbb {E}\left[ {\mathcal {W}}_{(i+1)\varDelta _n} {\mathcal {W}}^\prime _{(i+1)\varDelta _n}\right] = b^2({\tilde{Z}}_i; \sigma ) \mathbb {E} \left( \begin{matrix} \varSigma ^{(1)}_{\varDelta _n} &{}\quad \varSigma ^{(12)}_{\varDelta _n} \\ \varSigma ^{(12)}_{\varDelta _n} &{} \quad \varSigma ^{(2)}_{\varDelta _n} \end{matrix}\right) + {O}(\varDelta _n^4), \end{aligned}$$
where entries are given by:
$$\begin{aligned} \varSigma ^{(1)}_{\varDelta _n}&= \left( \partial _y a_1({\tilde{Z}}_i; \theta ^{(1)}) \right) ^2 \left[ \int _{i\varDelta _n}^{(i+1)\varDelta _n}((i+1)\varDelta _n - s)dW_s\right] ^2 \\ \varSigma ^{(12)}_{\varDelta _n}&= \left( \partial _y a_1({\tilde{Z}}_i; \theta ^{(1)}) \int _{i\varDelta _n}^{(i+1)\varDelta _n}((i+1)\varDelta _n - s)dW_s \right) \\&\quad \times \left( \int _{i\varDelta _n}^{(i+1)\varDelta _n}dW_s +\partial _y a_2 ({\tilde{Z}}_i; \theta ^{(2)}) \int _{i\varDelta _n}^{(i+1)\varDelta _n}((i+1)\varDelta _n - s)dW_s \right) \\ \varSigma ^{(2)}_{\varDelta _n}&= \left( \int _{i\varDelta _n}^{(i+1)\varDelta _n}dW_s +\partial _y a_2({\tilde{Z}}_i; \theta ^{(2)}) \int _{i\varDelta _n}^{(i+1)\varDelta _n}((i+1)\varDelta _n - s)dW_s \right) ^2 \end{aligned}$$
The first entry can be easily calculated by the Itô isometry:
$$\begin{aligned} \mathbb {E}[ \varSigma ^{(1)}_{\varDelta _n} ]= & {} \left( \partial _y a_1({\tilde{Z}}_i; \theta ^{(1)}) \right) ^2 \mathbb {E} \left[ \int _{i\varDelta _n}^{(i+1)\varDelta _n}((i+1)\varDelta _n - s)dW_s\right] ^2 \\= & {} \left( \partial _y a_1({\tilde{Z}}_i; \theta ^{(1)}) \right) ^2 \int _{i\varDelta _n}^{(i+1)\varDelta _n}((i+1)\varDelta _n - s)^2ds = \left( \partial _y a_1({\tilde{Z}}_i; \theta ^{(1)}) \right) ^2 \frac{\varDelta _n^3}{3} \end{aligned}$$
Now consider the product of two stochastic integrals in the terms \(\varSigma ^{(12)}_{\varDelta _n}\) and \(\varSigma ^{(2)}_{\varDelta _n}\). Assume for simplicity that \(t=0\). From the properties of the stochastic integrals (Karatzas and Shreve 1987), it is straightforward to see that:
$$\begin{aligned}&\mathbb {E}\left[ \lim _{n\rightarrow \infty }\sum _{t_i, t_{i-1}\in [0, \varDelta _n]} (\varDelta _n - s) (W_{t_i} - W_{t_{i-1}}) \sum _{t_i, t_{i-1}\in [0, \varDelta _n]} (W_{t_i} - W_{t_{i-1}}) \right] \\&\quad = \lim _{n\rightarrow \infty }\sum _{t_i, t_{i-1}\in [0, \varDelta _n]} (\varDelta _n - s) \mathbb {E}\left[ (W_{t_i} - W_{t_{i-1}})^2 \right] = \int _0^{\varDelta _n} (\varDelta _n - s)ds = \frac{\varDelta _n^2}{2} \end{aligned}$$
That gives the proposition. \(\square \)
1.2 Auxiliary results
We start with an important Lemma which links the sampling and the probabilistic law of the continuous process:
Lemma 4
(Kessler 1997) Let \(\varDelta _n\rightarrow 0\) and \(n\varDelta _n \rightarrow \infty \), let \(f\in \mathbb {R} \times \varTheta \rightarrow \mathbb {R}\) be such that f is differentiable with respect to z and \(\theta \), with derivatives of polynomial growth in z uniformly in \(\theta \). Then:
$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n f(Z_i; \theta ) \overset{\mathbb {P}_0}{\longrightarrow } \int f(z; \theta ) \nu _0(dz) \text { as } n\rightarrow \infty \text { uniformly in } \theta . \end{aligned}$$
Lemma is proven in Kessler (1997) for the one-dimensional case. Its proof is based only on ergodicity of the process and the assumptions analogous to ours, and not on the discretization scheme or dimensionality. So it can be generalized to a multi-dimensional case.
Proposition 4 in combination with the continuous ergodic theorem and Lemma 4 allow us to establish the following important result:
Lemma 5
Let \(f: \mathbb {R}^2 \times \varTheta \rightarrow \mathbb {R} \) be a function with the derivatives of polynomial growth in x, uniformly in \(\theta \). Assume \(\varDelta _n \rightarrow 0 \text { and } n\varDelta _n \rightarrow \infty \). Then:
-
(i)
\( \frac{1}{n\varDelta ^3_n} \sum _{i = 0}^{n-1}\frac{ f(Z_i; \theta )}{(\partial _{y} a_1(Z_i; \theta ^{(1)}_0))^2}\left( X_{i+1} - {\bar{A}}_1(Z_i;\theta ^{(1)}_0, \theta ^{(2)},\sigma )\right) ^2 \overset{\mathbb {P}_0}{\longrightarrow } \frac{1}{3} \int f(z; \theta ) b^2(z; \sigma _0) \nu _0(dz) \)
-
(ii)
\( \frac{1}{n\varDelta _n} \sum _{i = 0}^{n-1}f(Z_i; \theta )\left( Y_{i+1} - Y_i\right) ^2 \overset{\mathbb {P}_0}{\longrightarrow } \int f(z; \theta ) b^2(z; \sigma _0) \nu _0(dz)\)
-
(iii)
\( \frac{1}{n\varDelta ^2_n} \sum _{i = 0}^{n-1}\frac{ f(Z_i; \theta )}{\partial _{y} a_1(Z_i; \theta ^{(1)}_0)}\left( X_{i+1} - {\bar{A}}_1(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( Y_{i+1} - Y_i\right) \overset{\mathbb {P}_0}{\longrightarrow } \frac{1}{2} \int f(z; \theta ) b^2(z; \sigma _0) \nu _0(dz)\)
Proof
We consider only the cross-term (iii), since the results for the first and the second term are analogous to Ditlevsen and Samson (2017) (upon replacing the bounds from Proposition 3 by 4). Thanks to Proposition 4 we know that:
$$\begin{aligned}&\mathbb {E} \left[ \frac{1}{n\varDelta ^2} \frac{f(Z_i; \theta )}{\partial _{y} a_1(Z_i; \theta ^{(1)}_0)}\left( X_{i+1} - {\bar{A}}_1(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( Y_{i+1} - Y_i\right) |{\mathcal {F}}_i\right] \\&\quad =\frac{1}{2n} f(Z_{i}; \theta )b^2(Z_i; \sigma _0) + {O}(\varDelta _n). \end{aligned}$$
Then from Lemma 4 it follows that for \(n\rightarrow \infty \) uniformly in \(\theta \):
$$\begin{aligned}&\sum _{i=0}^{n-1}\mathbb {E} \left[ \frac{1}{n\varDelta ^2} \frac{f(Z_i; \theta )}{\partial _{y} a_1(Z_i; \theta ^{(1)}_0)}\left( X_{i+1} - {\bar{A}}_1(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( Y_{i+1} - Y_i\right) |{\mathcal {F}}_i\right] \overset{\mathbb {P}_0}{\longrightarrow } \\&\quad \frac{1}{2}\int f(z; \theta )b^2(z; \sigma _0)\nu _0(dz) \end{aligned}$$
\(\square \)
Let us introduce an auxiliary Lemma which establishes the convergence in probability for the first moments:
Lemma 6
Let \(f: \mathbb {R}^{2} \times \varTheta \rightarrow \mathbb {R}\) be a function with derivatives of polynomial growth in x, uniformly in \(\theta \). Assume \(\varDelta _n \rightarrow 0\) and \(n\varDelta _n \rightarrow \infty \). Then the following convergence results hold:
-
(i)
\( \frac{1}{n\varDelta _n} \sum _{i=0}^{n-1} f(Z_i; \theta ) (X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )) \overset{\mathbb {P}_0}{\longrightarrow } 0 \)
-
(ii)
\( \frac{1}{n\varDelta _n} \sum _{i=0}^{n-1} f(Z_i; \theta ) (Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}_0, \sigma )) \overset{\mathbb {P}_0}{\longrightarrow } 0 \)
uniformly in \(\theta \).
Proof
Consider (ii). Expectation of the sum tends to zero for \(\varDelta _n \rightarrow 0\) and \(n\varDelta _n \rightarrow \infty \) due to Proposition 4. Convergence for \(\theta ^{(1)}\) is due to Lemma 9 in Genon-Catalot and Jacod (1993) and uniformity in \(\theta ^{(1)}\) follows the proof of Lemma 10 in Kessler (1997). The second assertion is proven in the same way. For (i) see Lemma 3 in Ditlevsen and Samson (2017). \(\square \)
We also need the following Lemma for proving the asymptotic normality of the estimators.
Lemma 7
Assume (A1)–(A4) and \(n\varDelta _n\rightarrow \infty \) and \(n\varDelta _n^2\rightarrow 0\). Then for any bounded function \(f(z;\theta )\in \mathbb {R}^2\times \varTheta \rightarrow \mathbb {R}\) the following holds:
-
(i)
\(\frac{1}{\sqrt{n\varDelta _n^3}}\sum _{i=0}^{n-1}f(Z_i;\theta )(X_{i+1} - {\bar{A}}_1(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma ))\)\(\overset{{\mathcal {D}}}{\longrightarrow } {\mathcal {N}} (0, \frac{1}{3}\nu _0(b^2(z;\sigma _0)(\partial _y a_1(z; \theta ^{(1)}_0) )^2 f^2(z;\theta )))\)
-
(ii)
\(\frac{1}{\sqrt{n} \varDelta _n^3}\sum _{i=0}^{n-1}f(Z_i;\theta )(X_{i+1} - {\bar{A}}_1(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma ))^2 \)\(- \frac{1}{\sqrt{n}}\sum _{i=0}^{n-1} f(Z_i; \theta ) \frac{1}{3}b^2(z;\sigma _0)(\partial _y a_1(z; \theta ^{(1)}_0) )^2 \overset{{\mathcal {D}}}{\longrightarrow } {\mathcal {N}}\left( 0, \frac{2}{9}\nu _0\left( b^4(z;\sigma _0)(\partial _y a_1(z; \theta ^{(1)}_0) )^4 f^2(z;\theta )\right) \right) \)
-
(iii)
\(\frac{1}{\sqrt{n\varDelta _n}}\sum _{i=0}^{n-1}f(Z_i;\theta )(Y_{i+1} - Y_i)\overset{{\mathcal {D}}}{\longrightarrow } {\mathcal {N}}\left( 0, \nu _0\left( b^2(z;\sigma _0) f^2(z;\theta )\right) \right) \)
-
(iv)
\(\frac{1}{\sqrt{n}\varDelta _n}\sum _{i=0}^{n-1}f(Z_i;\theta )(Y_{i+1} - Y_i)^2 - \frac{1}{\sqrt{n}}\sum _{i=0}^{n-1} f(Z_i; \theta ) b^2(Z_i; \sigma _0)\)\(\overset{{\mathcal {D}}}{\longrightarrow } {\mathcal {N}}\left( 0, 2\nu _0\left( b^4(z;\sigma _0) f^2(z;\theta )\right) \right) \)
-
(v)
\(\frac{1}{\sqrt{n}\varDelta _n^2}\sum _{i=0}^{n-1}f(Z_i;\theta )(X_{i+1} - {\bar{A}}_1(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(Y_{i+1} - Y_i) \)\(- \frac{1}{\sqrt{n}}\sum _{i=0}^{n-1}f(Z_i;\theta )\frac{1}{2}b^2(Z_i; \sigma _0)\partial _ya_1(Z_i; \theta ^{(1)}_0) \overset{{\mathcal {D}}}{\longrightarrow } {\mathcal {N}}\left( 0, \frac{4}{3}\nu _0\left( f(z;\theta )b^4(z; \sigma _0)(\partial _ya_1(z; \theta ^{(1)}_0))^2 \right) \right) \)
Proof
We focus on the proof of (v), since (i)–(iv) closely follow Lemmas 4–5 in Ditlevsen and Samson (2017). To simplify the proof for the cross-term, we recall that the representation (15) can be transformed so that the two noise terms are independent. For example, we can use an analogue of such a decomposition proposed in Pokern et al. (2007):
$$\begin{aligned} X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )&= b(Z_i; \sigma _0)\partial _ya_1(Z_i; \theta ^{(1)}_0)\left( \frac{\varDelta _n^{3\over 2}}{\sqrt{12}}\eta ^1_i +\frac{\varDelta _n^{3\over 2}}{2}\eta ^2_i \right) + \delta ^1_i \\ Y_{i+1} - Y_i&= \varDelta _n a_2(Z_i; \theta ^{(2)})+ b(Z_i; \sigma _0)\varDelta _n^{1\over 2}\eta ^2_i+ \delta ^2_i, \end{aligned}$$
where \(\delta ^1_i\) and \(\delta ^2_i\) are error terms such that \(\mathbb {E}[\delta ^k_i|{\mathcal {F}}_i]=O(\varDelta _n^2)\) and \(\mathbb {E}[(\delta ^k_i)^2|{\mathcal {F}}_i]=O(\varDelta _n^4)\) (see Proposition 4), and \(\eta ^1_i\) and \(\eta ^2_i\) are standard independent normal variables.
Then Proposition 4 gives that \(\mathbb {E}\left[ \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( Y_{i+1} - Y_i\right) |{\mathcal {F}}_i \right] = \frac{\varDelta _n^2}{2} b(Z_i; \sigma _0)\partial _ya_1(Z_i; \theta ^{(1)}_0) + O(\varDelta _n^3)\), and then \( \mathbb {E}\left[ f(Z_i; \theta ) \left( \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( Y_{i+1} - Y_i\right) - \frac{\varDelta _n^2}{2}b^2(Z_i; \sigma _0)\partial _ya_1(Z_i; \theta ^{(1)}_0) \right) |{\mathcal {F}}_i \right] = 0\). With slightly more tedious computations (which are omitted) we get also that
$$\begin{aligned}&\mathbb {E}\left[ \left( \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( Y_{i+1} - Y_i\right) -\frac{\varDelta _n^2}{2}b^2(Z_i; \sigma _0)(\partial _ya_1(Z_i; \theta ^{(1)}_0)) \right) ^2 |{\mathcal {F}}_i \right] \\&\quad =\frac{4\varDelta ^4_n}{3}b^4(Z_i; \sigma _0)(\partial _ya_1(Z_i; \theta ^{(1)}_0))^2 + O(\varDelta _n^5) \end{aligned}$$
Then we obtain:
$$\begin{aligned}&\frac{1}{\sqrt{n}\varDelta _n^2}\sum _{i=0}^{n-1}f(Z_i;\theta )\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( Y_{i+1} - Y_i\right) \\&\quad \quad - \frac{1}{\sqrt{n}}\sum _{i=0}^{n-1}f(Z_i;\theta )\frac{1}{2}b^2(Z_i; \sigma _0)\partial _ya_1(Z_i; \theta ^{(1)}_0) \\&\quad = \frac{1}{\sqrt{n} \varDelta _n^2}\sum _{i=0}^{n-1}f(Z_i;\theta ) \left( b(Z_i; \sigma _0)\partial _ya_1(Z_i; \theta ^{(1)}_0)\left( \frac{\varDelta _n^{3\over 2}}{\sqrt{ 12}}\eta ^1_i +\frac{\varDelta _n^{3\over 2}}{2}\eta ^2_i \right) + \delta ^1_i\right) \\&\quad \quad \times \left( \varDelta _n a_2(Z_i; \theta ^{(2)})+ b(Z_i; \sigma _0)\varDelta _n^{1\over 2}\eta ^2_i + \delta ^2_i\right) \\&\quad - \frac{1}{\sqrt{n}}\sum _{i=0}^{n-1}f(Z_i;\theta )\frac{1}{2}b^2(Z_i; \sigma _0)\partial _ya_1(Z_i; \theta ^{(1)}_0) \end{aligned}$$
Since \(\frac{\varDelta _n}{n}\rightarrow 0\) by design we see that
$$\begin{aligned}&\frac{1}{n\varDelta ^4_n} \mathbb {E}\left[ \sum _{i=0}^{n-1} f^2(Z_i; \theta )\left( \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( Y_{i+1} - Y_i\right) \right. \right. \\&\quad \left. \left. -\frac{\varDelta ^2_n}{2}b(Z_i; \sigma _0)(\partial _ya_1(Z_i; \theta ^{(1)}_0)) \right) ^2\right] \rightarrow \frac{4}{3}\nu _0\left( f^2(z;\theta )b^4(z; \sigma _0)(\partial _ya_1(z; \theta ^{(1)}_0)^2\right) \end{aligned}$$
Further, since
\(\mathbb {E}\left[ f^4(Z_i; \theta )\left( \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( Y_{i+1} - Y_i\right) -\frac{\varDelta _n^2}{2}b^2(Z_i; \sigma _0)\right. \right. \left. \left. (\partial _ya_1(Z_i; \theta ^{(1)}_0)) \right) ^4 |{\mathcal {F}}_i \right] \) is bounded by (A2), we have
$$\begin{aligned}&\frac{1}{n^2\varDelta ^8_n}\mathbb {E}\left[ \sum _{i=0}^{n-1}f^4(Z_i; \theta )\left( \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( Y_{i+1} - Y_i\right) \right. \right. \\&\quad \left. \left. -\frac{\varDelta _n^2}{2}b^2(Z_i; \sigma _0)(\partial _ya_1(Z_i; \theta ^{(1)}_0)) \right) ^4 |{\mathcal {F}}_i \right] \rightarrow 0. \end{aligned}$$
Therefore, we can apply again the Theorem 3.2 from Hall and Heyde (1980) and obtain the statement (v). \(\square \)
Remark 1
Note that the results for the convergence in distribution for the increments of the second coordinate hold without any assumption on the parameters of the function \(a_2(z; \theta ^{(2)})\). It is due to the fact that the order of the noise dominates the order of the drift term (which is not the case in first coordinate, where the noise is propagated with the higher order). As a consequence, the convergence of a functional \(\sum _{i=0}^{n-1}f(Z_i;\theta )(Y_{i+1}-{{\bar{A}}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ))\) holds, with a proper scaling, for any value of \(\theta \).
1.3 Consistency and asymptotic normality of the LL contrast estimator
Proof
Consider
$$\begin{aligned} \frac{{\varDelta _n}}{n} \left[ {\mathcal {L}}_{n, \varDelta _n}(\theta ^{(1)}, \theta ^{(2)}, \sigma ^2; Z_{0:n}) - {\mathcal {L}}_{n, {\varDelta _n}}(\theta ^{(1)}_0, \theta ^{(2)}, \sigma ^2; Z_{0:n}) \right] = T_1 + T_2 + T_3 + T_4, \end{aligned}$$
where the terms are given as follows:
$$\begin{aligned} T_1&= \frac{6{\varDelta _n}}{n{\varDelta _n^3}}\sum _{i=0}^{n-1}\left[ \frac{ \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ) \right) ^2}{b^2(Z_i; \sigma )\left( \partial _ya_1(Z_i; \theta ^{(1)})\right) ^2} - \frac{ \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) ^2}{b^2(Z_i; \sigma )\left( \partial _ya_1(Z_i; \theta ^{(1)}_0)\right) ^2} \right] \\ T_2&= -\frac{6{\varDelta _n}}{n{\varDelta _n^2}}\sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma ) }\\&\qquad \left[ \frac{\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ) \right) \left( Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma )\right) }{\partial _ya_1(Z_i; \theta ^{(1)})} \right. \\&\qquad \left. -\,\frac{\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) \left( Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) }{\partial _ya_1(Z_i; \theta ^{(1)}_0)} \right] \\ T_3&= \frac{2\varDelta _n}{n\varDelta _n} \sum _{i=0}^{n-1} \left[ \frac{ \left( Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ) \right) ^2}{b^2(Z_i; \sigma )} - \frac{ \left( Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) ^2}{b^2(Z_i; \sigma )} \right] \\ T_4&= \frac{{\varDelta _n}}{n}\sum _{i=0}^{n-1} \log \left( \frac{\partial _ya_1(Z_i; \theta ^{(1)})}{\partial _ya_1(Z_i; \theta ^{(1)}_0)}\right) \end{aligned}$$
Consider term \(T_1\):
$$\begin{aligned} T_1= & {} \frac{6{\varDelta _n}}{n{\varDelta _n^3}}\sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )} \left[ \frac{ \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) + {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) - {\bar{A}}_1(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ) \right) ^2}{\left( \partial _ya_1(Z_i; \theta ^{(1)})\right) ^2} \right. \\&-\,\left. \frac{ \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) ^2}{\left( \partial _ya_1(Z_i; \theta ^{(1)}_0)\right) ^2} \right] \\= & {} \frac{6{\varDelta _n}}{n{\varDelta _n^3}}\sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )}\left[ \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) ^2\left[ \frac{1}{\left( \partial _ya_1(Z_i; \theta ^{(1)})\right) ^2} \right. \right. \\&-\,\left. \left. \frac{1}{\left( \partial _ya_1(Z_i; \theta ^{(1)}_0)\right) ^2} \right] + \frac{2\varDelta _n}{\left( \partial _ya_1(Z_i; \theta ^{(1)})\right) ^2} \right. \\&\times \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) (a_1(Z_i; \theta ^{(1)}_0) - a_1(Z_i; \theta ^{(1)})) \\&+\left. \frac{\varDelta _n^2}{(\partial _ya_1(Z_i; \theta ^{(1)}))^2}\left( a_1(Z_i; \theta ^{(1)}_0) - a_1(Z_i; \theta ^{(1)})\right) ^2\right] . \end{aligned}$$
Recalling Lemmas 4–6 we have that:
$$\begin{aligned}&\frac{6}{n{\varDelta _n^2}}\sum _{i=0}^{n-1} \frac{\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) ^2}{b^2(Z_i; \sigma )} \\&\quad \left[ \frac{1}{\left( \partial _ya_1(Z_i; \theta ^{(1)})\right) ^2} - \frac{1}{\left( \partial _ya_1(Z_i; \theta ^{(1)}_0)\right) ^2} \right] \overset{\mathbb {P}_0}{\longrightarrow } 0 \\&\frac{6}{n{\varDelta _n} }\sum _{i=0}^{n-1}\frac{1}{b^2(Z_i; \sigma )(\partial _ya_1(Z_i; \theta ^{(1)}))^2} \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) \\&\quad (a_1(Z_i; \theta ^{(1)}_0) - a_1(Z_i; \theta ^{(1)})) \overset{\mathbb {P}_0}{\longrightarrow } 0 \\&\frac{6}{n}\sum _{i=0}^{n-1}\frac{ (a_1(Z_i; \theta ^{(1)}_0) - a_1(Z_i; \theta ^{(1)}))^2}{b^2(Z_i; \sigma )(\partial _ya_1(Z_i; \theta ^{(1)}))^2} \overset{\mathbb {P}_0}{\longrightarrow } 6 \int \frac{ (a_1(z; \theta ^{(1)}_0) - a_1(z; \theta ^{(1)}))^2}{b^2(z; \sigma )(\partial _ya_1(z; \theta ^{(1)}))^2} \nu _0(dz). \end{aligned}$$
Now consider \(T_2\), which can be rewritten as:
$$\begin{aligned}&-\frac{6}{n{\varDelta _n}}\sum _{i=0}^{n-1} \frac{\left( Y_{i+1} - Y_i + O(\varDelta _n)\right) }{ b^2(Z_i; \sigma )} \\&\quad \quad \times \left[ \frac{\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) + {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) - {\bar{A}}_1(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ) \right) }{\partial _ya_1(Z_i; \theta ^{(1)})} \right. \\&\quad \quad \left. -\,\frac{\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) }{\partial _ya_1(Z_i; \theta ^{(1)}_0)} \right] \\&\quad =-\frac{6}{n {\varDelta _n}} \sum _{i=0}^{n-1} \frac{\left( Y_{i+1} - Y_i + O(\varDelta _n)\right) }{b^2(Z_i; \sigma )} \left[ \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \right. \\&\quad \quad \times \left[ \frac{1}{\partial _ya_1(Z_i; \theta ^{(1)})} - \frac{1}{\partial _ya_1(Z_i; \theta ^{(1)}_0)}\right] \\&\quad \quad \left. + \frac{{\varDelta _n}}{(\partial _ya_1(Z_i; \theta ^{(1)}))} (a_1(Z_i; \theta ^{(1)}_0) - a_1(Z_i; \theta ^{(1)}))\right] . \end{aligned}$$
Then we use the fact that the expectation of \(\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \) is of order \(\varDelta _n^2\) and of increments \(Y_{i+1} - Y_i\) is of \(\varDelta _n\), and by Lemma 5 we obtain:
$$\begin{aligned}&-\frac{6}{n{\varDelta _n} } \sum _{i=0}^{n-1} \frac{\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) \left( Y_{i+1} - Y_i + O(\varDelta _n)\right) }{b^2(Z_i; \sigma )(\partial _ya_1(Z_i; \theta ^{(1)}_0))} \left[ \frac{(\partial _ya_1(Z_i; \theta ^{(1)}_0))}{(\partial _ya_1(Z_i; \theta ^{(1)}))} - 1 \right] \\&\quad \overset{\mathbb {P}_0}{\longrightarrow } 0. \end{aligned}$$
The same holds for \(T_4\). Consider then \(T_3\):
$$\begin{aligned}&\frac{2\varDelta _n}{n\varDelta _n} \sum _{i=0}^{n-1} \left[ 2\frac{ \left( Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ) \right) ^2}{b^2(Z_i; \sigma )} \right. \\&\quad \quad \left. - \frac{ \left( Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) ^2}{b^2(Z_i; \sigma )} \right] \\&\quad =\frac{2}{n} \sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )} \left[ \left( Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) \right. \\&\quad \quad \times \left( {\bar{A}}_2(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ) \right) \\&\quad \left. - \left( {\bar{A}}_2(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ) \right) ^2 \right] \end{aligned}$$
This term is of order \(O(\varDelta ^3_n)\) (since \(\theta ^{(1)}\) is contained only in terms of order \(\varDelta ^2_n\)), thus it converges to zero as \(\varDelta _n\rightarrow 0\). Thus, we indeed have
$$\begin{aligned}&\lim _{n\rightarrow \infty , {\varDelta _n} \rightarrow 0} \frac{{\varDelta _n}}{n} \left[ {\mathcal {L}}_{n, \varDelta _n}(\theta ^{(1)}, \theta ^{(2)}, \sigma ^2; Z_{0:n}) - {\mathcal {L}}_{n, {\varDelta _n}}(\theta ^{(1)}_0, \theta ^{(2)}, \sigma ^2; Z_{0:n}) \right] \overset{\mathbb {P}_0}{\longrightarrow } \nonumber \\&\quad 6 \int \frac{ (a_1(z; \theta ^{(1)}_0) - a_1(z; \theta ^{(1)}))^2}{b^2(z; \sigma )(\partial _ya_1(z; \theta ^{(1)}))^2} \nu _0(dz). \end{aligned}$$
(26)
\(\square \)
Proof
[Theorem 1 (consistency and asymptotic normality of \(\theta ^{(1)}\))] Throughout the proof we assume that \(\theta ^{(1)}\in \mathbb {R}\) in order to simplify the notations.
Consistency It follows essentially from Lemma 1. Indeed, the result of the Lemma (and the fact that the parameter space is compact) implies that we can find a subsequence \({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} \) which converges to some value \(\theta ^{(1)}_\infty \). However, the minimum of the expression in Lemma 1 is attained for \(\theta ^{(1)}_0\). Then by identifiability of the drift function we have the consistency, that is \({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} \rightarrow \theta ^{(1)}_0\).
Asymptotic normality The proof follows the standard pattern (see Kessler 1997; Genon-Catalot et al. 1999; Ditlevsen and Samson 2017). First, we write the Taylor expansion of the function (17). Then we have:
$$\begin{aligned}&\int \frac{\varDelta _n}{n}\frac{\partial ^2}{\partial \theta ^{(1)}\partial \theta ^{(1)}}{\mathcal {L}}_{n, \varDelta _n}\left( \theta ^{(1)}_0 + u({{\hat{\theta }}}^{(1)}_{n,\varDelta _n} - \theta _0), \theta ^{(2)}, \sigma ; z \right) du \cdot \sqrt{\frac{n}{\varDelta _n}}({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} - \theta ^{(1)}_0) \\&\quad =- \sqrt{\frac{\varDelta _n}{n}}\frac{\partial }{\partial \theta ^{(1)}}{\mathcal {L}}_{n, \varDelta _n}(\theta ^{(1)}_0, \theta ^{(2)}, \sigma ; z) \end{aligned}$$
Note that the values of \(\theta ^{(2)}\) and \(\sigma \) may be taken arbitrary. Now we have to compute the first and the second order derivatives of (17). We omit the dependency on parameters in the expression for partial derivatives to make it readable and study the convergence of the first order derivative:
$$\begin{aligned}&\frac{\partial }{\partial \theta ^{(1)}}{\mathcal {L}}_{n, \varDelta _n}(\theta ^{(1)}_0,\theta ^{(2)}, \sigma ; z) = \sum _{i=1}^{n-1}\left[ \frac{2\partial ^2_{y,\theta ^{(1)}}a_1}{\partial _ya_1} - \frac{6}{b^2(Z_i; \sigma ) \partial _ya_1}\right. \nonumber \\&\quad \left. \left[ \frac{2(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))^2}{\varDelta _n^3 (\partial _ya_1)^2}\partial ^2_{y,\theta ^{(1)}}a_1 + \frac{2(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(\partial _{\theta ^{(1)}}a_1) }{\varDelta _n^2(\partial _ya_1)} \right. \right. \nonumber \\&\quad -\frac{(Y_{i+1}-\bar{A}_2(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(\partial _{\theta ^{(1)}}a_1) }{\varDelta _n} \nonumber \\&\quad \left. \left. -\frac{(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(Y_{i+1}-\bar{A}_2(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(\partial ^2_{y,\theta ^{(1)}}a_1)}{\varDelta _n^2 (\partial _ya_1)} \right] \right] \end{aligned}$$
(27)
Under assumption (A5) the only non-zero terms are the following:
$$\begin{aligned}&\frac{\partial }{\partial \theta ^{(1)}}{\mathcal {L}}_{n, \varDelta _n}(\theta ^{(1)}_0,\theta ^{(2)}, \sigma ; z)\\&\quad = \sum _{i=1}^{n-1} - \frac{6}{b^2(Z_i; \sigma ) \partial _ya_1}\left[ \frac{2(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(\partial _{\theta ^{(1)}}a_1) }{\varDelta _n^2(\partial _ya_1)} \right. \\&\quad \left. -\frac{(Y_{i+1}-{{\bar{A}}}_2(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(\partial _{\theta ^{(1)}}a_1) }{\varDelta _n} \right] \end{aligned}$$
Applying Lemma 7, we get:
$$\begin{aligned}&\frac{1}{\sqrt{n\varDelta _n^3}} \sum _{i=1}^{n-1} \left[ \frac{12 (\partial _{\theta ^{(1)}}a_1)}{b^2(Z_i; \sigma ) (\partial _ya_1)}(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )) \right] \\&\quad \overset{{\mathcal {D}}}{\longrightarrow } {\mathcal {N}}\left( 0, 36\nu _0\left( \frac{b^2(z;\sigma _0)}{b^4(z;\sigma )}(\partial _{\theta ^{(1)}}a_1)^2\right) \right) \\&\quad \frac{1}{\sqrt{n\varDelta _n}} \sum _{i=1}^{n-1} \left[ \frac{6(\partial _{\theta ^{(1)}}a_1)}{b^2(Z_i; \sigma ) }\frac{(Y_{i+1}-{{\bar{A}}}_2(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma )) }{(\partial _ya_1)} \right] \\&\quad \overset{{\mathcal {D}}}{\longrightarrow } {\mathcal {N}}\left( 0, 36\nu _0\left( \frac{b^2(z;\sigma _0)}{b^4(z;\sigma )}\frac{(\partial _{\theta ^{(1)}}a_1)^2}{(\partial _ya_1)^2}\right) \right) \end{aligned}$$
Thus, we have the following convergence in law:
$$\begin{aligned} \sqrt{\frac{\varDelta _n}{n}}\frac{\partial }{\partial \theta ^{(1)}}{\mathcal {L}}_{n, \varDelta _n}(\theta ^{(1)}_0,\theta ^{(2)}, \sigma ; z)\overset{{\mathcal {D}}}{\longrightarrow } {\mathcal {N}}\left( 0, 36\nu _0\left( \frac{b^2(z;\sigma _0)}{b^4(z;\sigma )}(\partial _{\theta ^{(1)}}a_1)^2\left( 1+\frac{1}{(\partial _ya_1)^2}\right) \right) \right) \end{aligned}$$
For the second order derivative we split again the expression (27) in several parts and study their convergence:
$$\begin{aligned} T_1:= & {} \frac{\varDelta _n}{n} \sum _{i=1}^{n-1}-\frac{12\varDelta _n}{\varDelta _n^2 b^2(Z_i; \sigma ) (\partial _ya_1)^2}\left[ (\partial _{\theta ^{(1)}} a_1)^2 \right. \\&\left. + \,(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))\frac{(\partial ^2_{\theta ^{(1)}\theta ^{(1)}}a_1)(\partial _ya_1)}{(\partial _ya_1)^2} \right] \\ T_2:= & {} \frac{\varDelta _n}{n} \sum _{i=1}^{n-1} \frac{6(Y_{i+1} -\bar{A}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ))}{\varDelta _n b^2(Z_i;\sigma ) } \frac{(\partial _ya_1)^2\partial ^2_{ \theta ^{(1)} \theta ^{(1)}}a_1}{(\partial _ya_1)^4} \end{aligned}$$
It is easy to see that the terms \(T_2\) converges to 0 by Lemmas 6 and 5. \(T_1\), according to the Lemmas 4 and 6, converges to \(12\int \frac{(\partial _{\theta ^{(1)}}a_1)^2}{b^2(z; \sigma ) (\partial _ya_1)}\nu _0(dz) \). That gives the result. \(\square \)
Proof of Lemma 2
Note that we cannot infer the value of \(\theta ^{(2)}\) with the same scaling as the parameter of the smooth coordinate because the estimator for each variable converges with different speed. Thus, we fix the parameter \(\theta ^{(1)}\) to its estimated value \({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} \) and consider the same sum, but with a different scaling, namely :
$$\begin{aligned}&\lim _{n\rightarrow \infty , {\varDelta _n} \rightarrow 0} \frac{1}{n{\varDelta _n}} \left[ {\mathcal {L}}_{n, {\varDelta _n}}({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ,\theta ^{(2)}, \sigma ^2; Z_{0:n}) - {\mathcal {L}}_{n, {\varDelta _n}}({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} , \theta ^{(2)}_0, \sigma ^2; Z_{0:n}) \right] \\&= T_1 + T_2 + T_3 \end{aligned}$$
where the terms are given as follows:
$$\begin{aligned} T_1= & {} \frac{6}{n{\varDelta _n^4}}\sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )\left( \partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )\right) ^2}\left[ \left( X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)},\sigma )\right) ^2 \right. \\&\quad -\,\left. \left( X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0,\sigma )\right) ^2 \right] \\ T_2= & {} -\frac{6}{n{\varDelta _n^3}}\sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )}\left[ \frac{\left( X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)},\sigma ) \right) \left( Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}, \sigma )\right) }{\partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )} \right. \\&\quad \left. -\,\frac{\left( X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0,\sigma ) \right) \left( Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma )\right) }{\partial _ya_1(Z_i;{{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )} \right] \\ T_3= & {} \frac{2}{n{\varDelta _n^2}}\sum _{i=0}^{n-1} \frac{ \left[ (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}, \sigma ))^2 - (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))^2 \right] }{b^2(Z_i; \sigma )} \end{aligned}$$
We start with \(T_1\):
$$\begin{aligned} T_1= & {} \frac{6}{n{\varDelta _n^4}}\sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )\left( \partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )\right) ^2}\left[ \left( {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0,\sigma ) - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)},\sigma )\right) \right. \\&\times \left( X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0,\sigma )\right) \\&\left. - \left( {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0,\sigma ) - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)},\sigma )\right) ^2 \right] \\= & {} \frac{6}{n{\varDelta _n^4}}\sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )\left( \partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )\right) ^2} \\&\times \left[ \frac{\varDelta ^2_n}{2}\left( \partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )\right) \left( a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}) \right) \right. \\&\times \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}_0,\sigma ) \right) + \frac{\varDelta ^3_n}{2}\left( \partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )\right) \\&\times \left( a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}) \right) \\&\times \left( a_1(Z_i; \theta ^{(1)}_0)-a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n})\right) + \frac{\varDelta ^4_n}{4}\left( \partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )\right) ^2 \\&\times \left( a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}) \right) ^2 \\&\left. -\,\frac{\varDelta ^4_n}{4}\left( \partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )\right) ^2\left( a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}) \right) ^2 \right] \end{aligned}$$
Recall that \(\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}_0,\sigma )\right) \) is of \(O(\varDelta ^3_n)\) by Proposition 3. Thus, the first summand of the \(T_1\) is of order \(\varDelta ^5_n\) and converges to 0. The second summand, under assumption (A5), can be rewritten as
$$\begin{aligned}&\frac{3}{n{\varDelta _n^4}}\sum _{i=0}^{n-1} \frac{\varDelta ^3_n\left( \theta ^{(1)}_0 - {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}\right) ^T g(X_i)\left( a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}) \right) }{b^2(Z_i; \sigma )\left( \partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )\right) } \\&\quad =\frac{3}{\sqrt{n\varDelta _n} n}\sum _{i=0}^{n-1} \frac{\sqrt{\frac{n}{\varDelta _n}}\left( \theta ^{(1)}_0 - {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}\right) ^T g(X_i)\left( a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}) \right) }{b^2(Z_i; \sigma )\left( \partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )\right) }. \end{aligned}$$
\(\sqrt{\frac{n}{\varDelta _n}}\left( \theta ^{(1)}_0 - {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}\right) ^T\) converges to a normal variable with zero mean due to Theorem 1, and the whole expression converges to 0, because \(n\varDelta _n\rightarrow \infty \). Thus \(T_1\) converges to 0. Consider \(T_2\):
$$\begin{aligned} T_2= & {} -\frac{6}{n{\varDelta _n^3}}\sum _{i=0}^{n-1} \left[ \frac{\varDelta _n\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) (a_2(Z_i; \theta ^{(2)}) - a_2(Z_i; \theta ^{(2)}_0))}{\partial _ya_1(Z_i;{{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )b^2(Z_i; \sigma )} \right. \\&\left. +\,\frac{\varDelta _n^2 (a_1(Z_i; \theta ^{(1)}_0) -a_1(Z_i;{{\hat{\theta }}}^{(1)}_{n,\varDelta _n})) (a_2(Z_i; \theta ^{(2)}) - a_2(Z_i; \theta ^{(2)}_0))}{\partial _ya_1(Z_i;{{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )b^2(Z_i; \sigma )} \right] \end{aligned}$$
Then, the first part of the sum converges to zero in probability after applying Lemma 6. The second part of the sum also converges to zero because \(n\varDelta _n\rightarrow \infty \) by design, and \({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} \overset{\mathbb {P}_0}{\longrightarrow }\theta ^{(1)}_0\). So that, recalling (A5), and applying the arguments used above for \(T_1\) to \(a_1(Z_i; \theta ^{(1)}_0) - a_1(Z_i;{{\hat{\theta }}}^{(1)}_{n,\varDelta _n}) = \left( \theta ^{(1)}_0 - {{\hat{\theta }}}^{(1)}_{n,\varDelta _n} \right) g(X_i) \), we prove that \(T_2\) also converges to 0. So we just have to consider the remaining term \(T_3\):
$$\begin{aligned} T_3= & {} \frac{2}{n\varDelta _n^2} \sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )} \left[ (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))^2 \right. \\&\quad \quad \left. +\,(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))({\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ) - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}, \sigma )) \right. \\&\quad \quad \left. +\,({\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ) - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}, \sigma ))^2 - (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))^2 \right] \\&\quad =\frac{2}{n\varDelta _n^2} \sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )} \left[ {\varDelta _n}(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))(a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)})) \right. \\&\quad \quad +\left. {\varDelta _n^2}(a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}))^2 \right] \end{aligned}$$
The first part of the sum converges to 0 due to Lemma 6. Then we apply Lemma 5 and get the convergence:
$$\begin{aligned}&\lim _{n\rightarrow \infty , {\varDelta _n} \rightarrow 0} \frac{1}{n{\varDelta _n}} \left[ {\mathcal {L}}_{n, {\varDelta _n}}({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ,\theta ^{(2)}, \sigma ^2; Z_{0:n}) - {\mathcal {L}}_{n, {\varDelta _n}}({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} , \theta ^{(2)}_0, \sigma ^2_0; Z_{0:n}) \right] \overset{\mathbb {P}_0}{\longrightarrow } \\&\quad 2 \int \frac{(a_2(z; \theta ^{(2)}_0) - a_2(z; \theta ^{(2)}))^2}{b^2(z; \sigma )} \nu _0(dz) \end{aligned}$$
\(\square \)
Proof of Lemma 3
We can split the contrast in the following sum:
$$\begin{aligned} \lim _{n\rightarrow \infty , {\varDelta _n} \rightarrow 0} \frac{1}{2n} {\mathcal {L}}_{n, {\varDelta _n}}({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} , \theta ^{(2)}, \sigma ^2; Z_{0:n}) = \lim _{n\rightarrow \infty , {\varDelta _n} \rightarrow 0}\left[ 3T_1 - 3T_2 + T_3 + T_4 \right] \end{aligned}$$
where terms are given by follows:
$$\begin{aligned} T_1&= \frac{1}{n}\sum _{i=0}^{n-1}\frac{(X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)},\sigma ))^2}{{\varDelta _n^3} b^2(Z_i; \sigma )(\partial _{y} a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))^2} \\ T_2&= \frac{1}{n}\sum _{i=0}^{n-1}\frac{(X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)},\sigma ))(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}, \sigma ))}{{\varDelta _n^2} b^2(Z_i; \sigma )(\partial _{y} a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))}\\ T_3&= \frac{1}{n}\sum _{i=0}^{n-1}\frac{(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}, \sigma ))^2}{{\varDelta _n} b^2(Z_i; \sigma ) } \\ T_4&= \frac{1}{n}\sum _{i=0}^{n-1} \log b^2(Z_i; \sigma ) \end{aligned}$$
For the term \(T_1\) we have:
$$\begin{aligned} T_1= & {} \frac{1}{n{\varDelta _n^3}}\sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )} \frac{ \left( X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)},\sigma ) \right) ^2}{(\partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))^2}\\= & {} \frac{1}{n}\sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )} \frac{ \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) +{\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)},\sigma ) \right) ^2}{{\varDelta _n^3}(\partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))^2} \\= & {} \frac{1}{n} \sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )} \left[ \frac{ \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) ^2}{{\varDelta _n^3}(\partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))^2} \frac{(\partial _ya_1(Z_i; \theta ^{(1)}_0))^2}{(\partial _ya_1(Z_i; \theta ^{(1)}_0))^2} \right. \\&+\left. \frac{2\varDelta _n\left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma )\right) \left( a_1(Z_i;\theta ^{(1)}_0)-a_1(Z_i;{{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ) \right) }{\varDelta _n^3 (\partial _{y} a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))^2 } \right. \\&\left. + \frac{\varDelta _n^2}{\varDelta _n^3}\frac{\left( a_1(Z_i;\theta ^{(1)}_0)-a_1(Z_i;{{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ) \right) ^2}{b^2(Z_i; \sigma )(\partial _{y} a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))^2} \right] \end{aligned}$$
Thanks to the Lemmas 5 and 6, we know that the second term of the sum converges to 0 in probability, and for the first one we have:
$$\begin{aligned}&\frac{1}{n} \sum _{i=0}^{n-1} \frac{1}{b^2(Z_i; \sigma )} \frac{ \left( X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right) ^2}{{\varDelta _n^3}(\partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))^2} \frac{(\partial _ya_1(Z_i; \theta ^{(1)}_0))^2}{(\partial _ya_1(Z_i; \theta ^{(1)}_0))^2}\\&\quad \overset{\mathbb {P}_0}{\longrightarrow } \int \frac{b^2(z; \sigma _0)}{b^2(z; \sigma )}\frac{(\partial _ya_1(z; \theta ^{(1)}_0))^2}{(\partial _ya_1(z; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))^2} \nu _0(dz) \end{aligned}$$
For the third term, we use the assumption (A5), and then obtain the convergence to 0 in probability thanks to Theorem 1, the continuous mapping theorem and Lemma 4:
$$\begin{aligned} \frac{2}{n} \sum _{i=0}^{n-1} \frac{\varDelta _n^2}{\varDelta _n^3}\frac{\left( a_1(Z_i;\theta ^{(1)}_0)-a_1(Z_i;{{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ) \right) ^2}{b^2(Z_i; \sigma ) (\partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))^2} = \frac{2}{n^2} \sum _{i=0}^{n-1} \frac{\left( \sqrt{\frac{n}{\varDelta _n}}(\theta ^{(1)}_0-{{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )\right) ^2 g^2(X_i)}{b^2(Z_i; \sigma ) (\partial _ya_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))^2} \overset{\mathbb {P}_0}{\longrightarrow } 0. \end{aligned}$$
Then, \(T_2\) decomposes as:
$$\begin{aligned}&\frac{1}{n}\sum _{i=0}^{n-1}\frac{(X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}, \sigma ))(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}, \sigma ))}{{\varDelta _n^2} b^2(Z_i; \sigma )(\partial _{y} a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))} \\&\quad = \frac{1}{n}\sum _{i=0}^{n-1}\frac{(\partial _{y} a_1(Z_i; \theta ^{(1)}_0))}{b^2(Z_i; \sigma )(\partial _{y} a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))} \\&\quad \quad \times \left[ \frac{(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))}{{\varDelta _n^2} (\partial _{y} a_1(Z_i; \theta ^{(1)}_0))} \right. \\&\quad \quad +\,\left. \frac{\varDelta _n(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(a_2(Z_i; \theta ^{(2)}_0)-a_2(Z_i; \theta ^{(2)}))}{{\varDelta _n^2} (\partial _{y} a_1(Z_i; \theta ^{(1)}_0))} \right. \\&\quad \quad +\,\left. \frac{\varDelta _n(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))(a_1(Z_i; \theta ^{(1)}_0)-a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))}{{\varDelta _n^2} (\partial _{y} a_1(Z_i; \theta ^{(1)}_0))} \right. \\&\quad \quad \left. +\,\frac{\varDelta _n^2(a_1(Z_i; \theta ^{(1)}_0)-a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))(a_2(Z_i; \theta ^{(2)}_0)-a_2(Z_i; \theta ^{(2)}))}{{\varDelta _n^2} (\partial _{y} a_1(Z_i; \theta ^{(1)}_0))} \right] \end{aligned}$$
Again, using Lemma 6, we know that the second and the third terms are converging to 0 in probability. For the first term, thanks to Lemma 5 we have the following convergence:
$$\begin{aligned}&\frac{1}{n}\sum _{i=0}^{n-1} \frac{(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))(\partial _{y} a_1(Z_i; \theta ^{(1)}_0))}{{\varDelta _n^2} b^2(Z_i; \sigma )(\partial _{y} a_1(Z_i; \theta ^{(1)}_0))(\partial _{y} a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))} \overset{\mathbb {P}_0}{\longrightarrow } \\&\quad \int \frac{b^2(z; \sigma _0)}{b^2(z; \sigma )}\frac{\partial _ya_1(z; \theta ^{(1)}_0) }{\partial _ya_1(z; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )} \nu _0(dz) \end{aligned}$$
Finally, we treat the last term:
$$\begin{aligned} \frac{1}{n}\sum _{i=0}^{n-1} \frac{\varDelta _n^2(a_1(Z_i; \theta ^{(1)}_0)-a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))(a_2(Z_i; \theta ^{(2)}_0)-a_2(Z_i; \theta ^{(2)}))}{\varDelta _n^2b^2(Z_i; \sigma )(\partial _{y} a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} ))} \end{aligned}$$
Using again the Lipschitz continuity of \(a_1\), Theorem 1 and the Slutsky’s theorem, we obtain a convergence to zero in probability for this term. \(T_4\) converges in probability to \(\int \log b^2(z; \sigma ) \nu _0(dz)\) due to Lemma 4. Consider \(T_3\):
$$\begin{aligned} T_3= & {} \frac{1}{n \varDelta _n}\sum _{i=0}^{n-1}\frac{1}{b^2(Z_i; \sigma )}\left[ (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))^2 \right. \\&+\left. 2(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))({\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ) - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}, \sigma )) \right. \\&+\left. ({\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ) - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}, \sigma ))^2\right] \\= & {} \frac{1}{n \varDelta _n }\sum _{i=0}^{n-1} \frac{(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))^2}{b^2(Z_i; \sigma )} \\&+2\frac{{\varDelta _n}}{n {\varDelta _n} }\sum _{i=0}^{n-1} \frac{(Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n}, \theta ^{(2)}_0, \sigma ))(a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}))}{b^2(Z_i; \sigma )}\\&+ \frac{{\varDelta _n^2}}{n {\varDelta _n}}\sum _{i=0}^{n-1} \frac{(a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}))^2}{b^2(Z_i; \sigma )} \end{aligned}$$
Thanks to Lemmas 5 and 6 we conclude that
$$\begin{aligned} T_3 \overset{\mathbb {P}_0}{\longrightarrow } \int \frac{b^2(z; \sigma _0)}{b^2(z; \sigma )} \nu _0(dz) +0 +0 \end{aligned}$$
Finally, we obtain
$$\begin{aligned}&\frac{1}{n} {\mathcal {L}}_{n, \varDelta _n}(\theta , \sigma ^2; Z_{0:n}) \overset{\mathbb {P}_0}{\longrightarrow } \\&\quad \int \left( \frac{b^2(z;\sigma _0)}{b^2(z; \sigma )}\left[ 3\left( \frac{\partial _ya_1(z; \theta ^{(1)}_0)}{\partial _ya_1(z; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )}\right) ^2 - 3\frac{\partial _ya_1(z; \theta ^{(1)}_0) }{\partial _ya_1(z; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )} + 1 \right] + \log b^2(z; \sigma ) \right) \nu _0(dz) \end{aligned}$$
By assumption (A5) \(\partial _ya_1(\cdot )\) does not depend on \(\theta ^{(1)}\), thus \(\frac{\partial _ya_1(z; \theta ^{(1)}_0) }{\partial _ya_1(z; {{\hat{\theta }}}^{(1)}_{n, \varDelta _n} )} = 1 \). It gives the Lemma. \(\square \)
Proof of Theorem 2
Consistency The consistency of the estimator for the parameter \(\theta ^{(2)}\) is based on Lemma 2, with the arguments analogous to the proof of Theorem 1. For the diffusion parameter \(\sigma \), the result follows from Lemma 3. Denote \({\mathcal {I}}(\sigma , \sigma _0): =\frac{b^2(z;\sigma _0)}{b^2(z; \sigma )} + \log b^2(z; \sigma )\). We can choose some subsequence \(n_k\) such that \({{\hat{\sigma }}}_{n, \varDelta _{n}}\) converges to some \(\sigma _{\infty }\). By the definition of the estimator we know that \({\mathcal {I}}(\sigma _{\infty }, \sigma _0) \le {\mathcal {I}}(\sigma _{0}, \sigma _0)\). But we also know that \(\frac{b^2(z;\sigma _0)}{b^2(z; \sigma )} + \log b^2(z; \sigma ) \ge 1 + \log b^2(z;\sigma _0)\) and thus \({\mathcal {I}}(\sigma _{\infty }, \sigma _0) \ge {\mathcal {I}}(\sigma _{0}, \sigma _0)\), and by the identifiability assumption \(\sigma _{\infty } \equiv \sigma _0\). It proves the consistency of \({\hat{\sigma }}\).
Asymptotic normality The proof follows the standard pattern. Throughout the proof we assume that \(\theta ^{(2)} \text { and } \sigma \in \mathbb {R}\) in order to simplify the notations. We write the Taylor expansion of the contrast function defined in (17) and apply an appropriate scaling
$$\begin{aligned}&\int C_{n, \varDelta _n}\left( \varphi _0 + u({{\hat{\varphi }}}_{n, \varDelta _n} - \varphi _0); z \right) du \, E_{n, \varDelta _n} \\&\quad = - D_{n, \varDelta _n}(\varphi _0), \end{aligned}$$
where by \(\varphi \) we now denote \((\theta ^{(2)}, \sigma )\) and the parameter \(\theta ^{(1)}\) is fixed to its estimate \({{\hat{\theta }}}^{(1)}_{n,\varDelta _n}\) throughout the proof, and
$$\begin{aligned} C_{n, \varDelta _n}(\theta )&:= \left[ \begin{matrix} {\frac{1}{n\varDelta _n}} \frac{\partial ^2}{\partial \theta ^{(2)} \partial \theta ^{(2)}} {\mathcal {L}}_{n, \varDelta _n}({{\hat{\theta }}}^{(1)}_{n,\varDelta _n},\theta ^{(2)}, \sigma ; Z_{0:n}) &{}\quad \frac{1}{n\sqrt{\varDelta _n}}\frac{\partial ^2}{\partial \sigma \partial \theta ^{(2)}} {\mathcal {L}}({{\hat{\theta }}}^{(1)}_{n,\varDelta _n},\theta ^{(2)}, \sigma ; Z_{0:n}) \\ {\frac{1}{n\sqrt{\varDelta _n}}}\frac{\partial ^2}{\partial \theta ^{(2)}\partial \sigma } {\mathcal {L}}_{n, \varDelta _n}({{\hat{\theta }}}^{(1)}_{n,\varDelta _n},\theta ^{(2)}, \sigma ; Z_{0:n}) &{}\quad \frac{1}{n} \frac{\partial ^2}{\partial \sigma \partial \sigma }{\mathcal {L}}_{n, \varDelta _n}({{\hat{\theta }}}^{(1)}_{n,\varDelta _n},\theta ^{(2)}, \sigma ; Z_{0:n})\end{matrix}\right] , \\ E_{n, \varDelta _n}&:= \left[ \begin{matrix} {\sqrt{n\varDelta _n}} ({{\hat{\theta }}}^{(2)}_n - \theta ^{(2)}_0) \\ {\sqrt{n}} ({{\hat{\sigma }}}_n - \sigma _0) \end{matrix} \right] , \quad D_{n, \varDelta _n} = \left[ \begin{matrix} {\frac{1}{\sqrt{n\varDelta _n}}} \frac{\partial }{\partial \theta ^{(2)} } {\mathcal {L}}_{n, \varDelta _n}({{\hat{\theta }}}^{(1)}_{n,\varDelta _n},\theta ^{(2)}, \sigma ; Z_{0:n}) \\ {\frac{1}{\sqrt{n}}} \frac{\partial }{\partial \sigma } {\mathcal {L}}_{n, \varDelta _n}({{\hat{\theta }}}^{(1)}_{n,\varDelta _n},\theta ^{(2)}, \sigma ; Z_{0:n}) \end{matrix}\right] . \end{aligned}$$
First, we compute the higher-order terms of the partial derivatives of first and second order with respect to \(\theta ^{(2)}\) and \(\sigma \):
$$\begin{aligned} \frac{\partial }{\partial \theta ^{(2)}}{\mathcal {L}}_{n, \varDelta _n}(\cdot )= & {} \sum _{i=1}^{n-1} \left[ - 6 \frac{\varDelta _n(\partial _{\theta ^{(2)}}a_2) (X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n^2b^2(Z_i; \sigma )(\partial _{{{\hat{\theta }}}^{(1)}_{n,\varDelta _n}}a_1)} \right. \\&+\left. 2\frac{\varDelta _n(\partial _{\theta ^{(2)}}a_2) (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n b^2(Z_i; \sigma )} \right] =: D^{1}_{n, \varDelta _n}\\ \frac{\partial }{\partial \sigma }{\mathcal {L}}_{n, \varDelta _n}(\cdot )= & {} - \sum _{i=1}^{n-1}\frac{\partial _\sigma b}{b^3(Z_i; \sigma )} \left[ 6 \frac{ (X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))^2}{\varDelta _n^3 (\partial _{{{\hat{\theta }}}^{(1)}_{n,\varDelta _n}}a_1)^2} \right. \\&-\left. 6\frac{ (X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma )( Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n^2 (\partial _{{{\hat{\theta }}}^{(1)}_{n,\varDelta _n}}a_1)} \right. \\&+ \left. 2\frac{ (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))^2}{\varDelta _n} \right] + \frac{\partial _\sigma b}{b(Z_i; \sigma )} =: D^2_{n, \varDelta _n}\\ \frac{\partial ^2}{\partial \theta ^{(2)}\partial \theta ^{(2)}}{\mathcal {L}}_{n, \varDelta _n}(\cdot )= & {} \sum _{i=1}^{n-1} \left[ - 6 \frac{\varDelta _n(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2) (X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n^2b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \right. \\&\left. + 2\frac{\varDelta _n(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2) (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n b^2(Z_i; \sigma )} \right. \\&\left. + \frac{\varDelta _n^2(\partial _{\theta ^{(2)}}a_2)^2}{\varDelta _n b^2(Z_i; \sigma )} \right] =: C^{11}_{n, \varDelta _n} \\ \frac{\partial ^2}{\partial \theta ^{(2)}\partial \sigma }{\mathcal {L}}_{n, \varDelta _n}(\cdot )= & {} \sum _{i=1}^{n-1}\frac{\partial _\sigma b}{b^2(Z_i; \sigma )} \\&\quad \left[ 12 \frac{\varDelta _n(\partial _{\theta ^{(2)}}a_2) (X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n^2b(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \right. \\&+\left. 4\frac{\varDelta _n(\partial _{\theta ^{(2)}}a_2) (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n b(Z_i; \sigma )} \right] =: C^{12}_{n, \varDelta _n}=C^{21}_{n, \varDelta _n}\\&\quad \frac{\partial ^2}{\partial \sigma ^2}{\mathcal {L}}_{n, \varDelta _n}(\cdot ) = - \sum _{i=1}^{n-1}\frac{6(\partial _\sigma b)^2 - 2b(Z_i; \sigma )(\partial ^2_{\sigma \sigma } b) }{b^4(Z_i; \sigma )}\\&\quad \left[ 6 \frac{ (X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))^2}{\varDelta _n^3 (\partial _{\theta ^{(1)}}a_1)^2} \right. \\&-\left. 6\frac{ (X_{i+1} - {\bar{A}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))( Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n^2 (\partial _{\theta ^{(1)}}a_1)} \right. \\&\left. + 2\frac{ (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))^2}{\varDelta _n} \right] \\&+ 2\frac{b(Z_i;\sigma )(\partial ^2_{\sigma \sigma } b)-(\partial _\sigma b)^2}{b^2(Z_i; \sigma )} =: C^{22}_{n, \varDelta _n} \end{aligned}$$
We start with proving the convergence for the terms \(C_{n, \varDelta _n}\). Then we can obtain a convergence in probability after few technical steps. We start with \(C^{11}_{n, \varDelta _n}\):
$$\begin{aligned} \frac{1}{n\varDelta _n}C^{11}_{n, \varDelta _n}= & {} \frac{1}{n\varDelta _n}\sum _{i=1}^{n-1} \left[ - 6 \frac{\varDelta _n(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2) (X_{i+1} - {{\bar{A}}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n^2b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \right. \\&+\left. 2\frac{\varDelta _n(\partial ^2_{\theta ^{(2)}\theta ^{(2)}_0}a_2) (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n b^2(Z_i; \sigma )} + \frac{\varDelta _n^2(\partial _{\theta ^{(2)}}a_2)^2}{\varDelta _n b^2(Z_i; \sigma )} \right] \\= & {} \frac{1}{n\varDelta _n}\sum _{i=1}^{n-1} \left[ - 6 \frac{\varDelta _n(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2) (X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))}{\varDelta _n^2b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \right. \\&-\left. 6 \frac{\varDelta _n^2(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2) (a_1(Z_i;\theta ^{(1)}_0) - a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}))}{\varDelta _n^2b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \right. \\&+ \left. 2\frac{\varDelta _n(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2) (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n b^2(Z_i; \sigma )} + \frac{\varDelta _n^2(\partial _{\theta ^{(2)}}a_2)^2}{\varDelta _n b^2(Z_i; \sigma )} \right] \end{aligned}$$
Note that thanks to Lemma 6 we know that
$$\begin{aligned}&\frac{1}{n\varDelta _n}\sum _{i=1}^{n-1} \left[ - 6 \frac{\varDelta _n(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2) (X_{i+1} - {\bar{A}}_1(Z_i;\theta ^{(1)}_0, \theta ^{(2)}, \sigma ))}{\varDelta _n^2b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \right. \\&\quad +\left. 2\frac{\varDelta _n(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2) (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n b^2(Z_i; \sigma )} \right. \\&\quad +\left. \frac{\varDelta _n^2(\partial _{\theta ^{(2)}}a_2)^2}{\varDelta _n b^2(Z_i; \sigma )} \right] \overset{\mathbb {P}_0}{\rightarrow } \int \frac{(\partial _{\theta ^{(2)}}a_2)^2}{b^2(z; \sigma )}\nu _0(dz) \end{aligned}$$
What about the remaining term, thanks to the assumption (A5) we have:
$$\begin{aligned}&-\frac{6}{n\varDelta _n}\sum _{i=1}^{n-1}\frac{(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2) (a_1(Z_i;\theta ^{(1)}_0) - a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}))}{b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \\&\quad =- \frac{6 }{\sqrt{n\varDelta _n}}\frac{1}{n}\sum _{i=1}^{n-1}\frac{(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2)a_1(Z_i;\sqrt{\frac{n}{\varDelta _n}}\left( \theta ^{(1)}_0 - {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}\right) ) }{b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \end{aligned}$$
We know that \(\left( \theta ^{(1)}_0 - {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}\right) \sqrt{\frac{n}{\varDelta _n}}\) is normally distributed by Theorem 1, and \( \frac{1}{n}\sum _{i=1}^{n-1}\frac{(\partial ^2_{\theta ^{(2)}\theta ^{(2)}}a_2) }{b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \) converges to its invariant density by Lemma 4. Then by Slutsky’s and the continuous mapping theorem the product also converges in distribution to a normal variable, which is, divided by \(\sqrt{n\varDelta _n}\) converges to zero since \(n\varDelta _n\rightarrow \infty \) by design. However, as \(n\varDelta _n\rightarrow \infty \), this term converges to 0 in probability. As a result,
$$\begin{aligned} \frac{1}{n\varDelta _n}C^{11}_{n, \varDelta _n} \overset{\mathbb {P}_0}{\rightarrow } \int \frac{(\partial _{\theta ^{(2)}}a_2)^2}{b^2(z; \sigma )}\nu _0(dz) \end{aligned}$$
With the same arguments we prove that \(\frac{1}{n\sqrt{\varDelta _n}}C^{12}_{n, \varDelta _n}=\frac{1}{n\sqrt{\varDelta _n}}C^{21}_{n, \varDelta _n} \overset{\mathbb {P}_0}{\rightarrow } 0\) and that
$$\begin{aligned} \frac{1}{n}C^{22}_{n, \varDelta _n}\overset{\mathbb {P}_0}{\rightarrow } - 4 \int \frac{(\partial _\sigma b)^2 }{b^2(z; \sigma _0)} \nu _0(dz) \end{aligned}$$
Then we consider the remaining term, recalling the assumption (A5): We start with the term
$$\begin{aligned} \frac{1}{\sqrt{n\varDelta _n}} D^{1}_{n, \varDelta _n}= & {} \frac{1}{\sqrt{n\varDelta _n}}\sum _{i=1}^{n-1} \left[ - 6 \frac{\varDelta _n(\partial _{\theta ^{(2)}}a_2) (X_{i+1} - {{\bar{A}}}_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n^2b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \right. \\&+\left. 2\frac{\varDelta _n(\partial _{\theta ^{(2)}}a_2) (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{\varDelta _n b^2(Z_i; \sigma )} \right] \\= & {} \frac{1}{\sqrt{n\varDelta _n}}\sum _{i=1}^{n-1} \left[ - 6 \frac{(\partial _{\theta ^{(2)}}a_2) (X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))}{\varDelta _n b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \right. \\&-\left. 6 \frac{(\partial _{\theta ^{(2)}}a_2) a_1(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}- \theta ^{(1)}_0) }{b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \right. \\&\left. + 2\frac{(\partial _{\theta ^{(2)}}a_2) (Y_{i+1} - {\bar{A}}_2(Z_i; {{\hat{\theta }}}^{(1)}_{n,\varDelta _n}, \theta ^{(2)}, \sigma ))}{b^2(Z_i; \sigma )} \right] \end{aligned}$$
For the first and the third term we simply apply Lemma 7 and obtain convergence in distribution to \({\mathcal {N}}\left( 0, \nu _0\left( \frac{(\partial _{\theta ^{(2)}}a_2)^2}{b^2(z; \sigma _0)} \right) \right) \). For the second term we apply the result of Theorem 1, as well as the continuous mapping and Slutsky’s theorem we may state that:
$$\begin{aligned} -6\sum _{i=1}^{n-1} \frac{(\partial _{\theta ^{(2)}}a_2) \, a_1(Z_i; \sqrt{\frac{n}{\varDelta _n}}({{\hat{\theta }}}^{(1)}_{n,\varDelta _n}- \theta ^{(1)}_0)) }{b^2(Z_i; \sigma )(\partial _{\theta ^{(1)}}a_1)} \overset{{\mathcal {D}}}{\longrightarrow } -6 \int \frac{(\partial _{\theta ^{(2)}}a_2) }{b^2(z; \sigma )(\partial _{\theta ^{(1)}}a_1)}a_1(z; {{\tilde{\eta }}})\nu _0(dz), \end{aligned}$$
where \({{\tilde{\eta }}}\) is distributed as stated in Theorem 1. Then as \(n\rightarrow 0\),
$$\begin{aligned} -\frac{6}{n}\sum _{i=1}^{n-1} \frac{(\partial _{\theta ^{(2)}}a_2) \, a_1(Z_i; \sqrt{\frac{n}{\varDelta _n}}({{\hat{\theta }}}^{(1)}_{n,\varDelta _n}- \theta ^{(1)}_0)) }{b^2(Z_i; \sigma )(\partial _{{{\hat{\theta }}}^{(1)}_{n,\varDelta _n}}a_1)} \overset{\mathbb {P}_0}{\longrightarrow } 0 \end{aligned}$$
By analogy, we prove the convergence for the term \(D^2_{n, \varDelta _n}\), obtaining:
$$\begin{aligned} \frac{1}{\sqrt{n}}D^2_{n, \varDelta _n} \overset{{\mathcal {D}}}{\longrightarrow } {\mathcal {N}}\left( 0, 32\nu _0\left( \frac{(\partial _\sigma b)^2}{b^2(z; \sigma _0)} \right) \right) \end{aligned}$$
That gives the result. \(\square \)
1.4 Consistency and normality of the least squares contrast
Proof of Theorem 3
The proof will follow the one of the classical contrast. First, we define the following quantities:
$$\begin{aligned}&{\mathcal {L}}^{(1),LSE}_{n,\varDelta _n}(\theta ^{(1)}, \theta ^{(2)}, \sigma ;Z_{0:n}) = \frac{1}{n} \sum _{i=0}^{n-1} \frac{(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ))^2}{\varDelta ^3_n}\\&{\mathcal {L}}^{(2),LSE}_{n,\varDelta _n}(\theta ^{(1)}, \theta ^{(2)}, \sigma ;Z_{0:n}) = \frac{1}{n} \sum _{i=0}^{n-1} \frac{(Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ))^2}{\varDelta _n} \end{aligned}$$
Consistency of \({{\hat{\theta }}}^{(1)}\). First, consider:
$$\begin{aligned}&\varDelta _n\left[ {\mathcal {L}}^{(1),LSE}_{n,\varDelta _n}(\theta ^{(1)}, \theta ^{(2)}, \sigma ; Z_{0:n})-{\mathcal {L}}^{(1),LSE}_{n,\varDelta _n}(\theta ^{(1)}_0, \theta ^{(2)}, \sigma ; Z_{0:n})\right] \\&\quad = \frac{\varDelta _n}{n\varDelta ^3_n}\sum _{i=0}^{n-1} \left[ (X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) \right. \\&\quad \quad \left. + {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ) - {\bar{A}}_1(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ))^2 \right. \\&\quad \quad -\left. (X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))^2 \right] = \frac{\varDelta ^2_n}{n\varDelta ^3_n}\sum _{i=0}^{n-1} \\&\quad \quad \times \left[ 2(X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(a_1(Z_i; \theta ^{(1)}_0) -a_1(Z_i; \theta ^{(1)})) \right. \\&\quad \quad +\left. \varDelta _n (a_1(Z_i; \theta ^{(1)}_0) - a_1(Z_i; \theta ^{(1)}))^2 + O(\varDelta ^2_n) \right] \end{aligned}$$
Then we have from Lemmas 5 and 6:
$$\begin{aligned}&\frac{2 }{n\varDelta _n} \sum _{i=0}^{n-1} (X_{i+1} - {\bar{A}}_1(Z_i; \theta ^{(1)}_0, \theta ^{(2)}, \sigma ))(a_1(Z_i; \theta ^{(1)}_0) - a_1(Z_i; \theta ^{(1)})) \overset{\mathbb {P}_0}{\longrightarrow } 0 \\&\quad \frac{1}{n} \sum _{i=0}^{n-1} (a_1(Z_i; \theta ^{(1)}_0) - a_1(Z_i; \theta ^{(1)}))^2 \overset{\mathbb {P}_0}{\longrightarrow }\int (a_1(z; \theta ^{(1)}_0) - a_1(z; \theta ^{(1)}))^2 \nu _0(dz) \end{aligned}$$
We conclude that there exists a subsequence \({{\hat{\theta }}}^{(1)}_{n,{\varDelta _n}} = \underset{\theta }{\arg \min } \, {\mathcal {L}}^{LSE}_{n,\varDelta _n}(\theta ;Z_{0:n}) \) that tends to \(\theta _\infty \). Since the minimum is attained at the point \(\theta _0\) and from (A4), we conclude that \(\theta _\infty = \theta _0\). Hence the estimator is consistent.
Consistency of \({{\hat{\theta }}}^{(2)}\) Consider:
$$\begin{aligned}&\frac{1}{\varDelta _n}\left[ {\mathcal {L}}^{(2),LSE}_{n,\varDelta _n}(\theta ^{(1)}, \theta ^{(2)},\sigma ; Z_{0:n})-{\mathcal {L}}^{(2),LSE}_{n,\varDelta _n}(\theta ^{(1)}, \theta ^{(2)}_0,\sigma ; Z_{0:n})\right] \\&\quad = \left[ \frac{1}{n\varDelta ^2_n}\sum _{i=0}^{n-1} (Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}_0, \sigma ) \right. \\&\quad \quad + {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}_0, \sigma ) - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}, \sigma ))^2 \\&\quad \quad -\left( Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}_0, \sigma ))^2 \right] = \frac{\varDelta _n}{n\varDelta ^2_n}\sum _{i=0}^{n-1} \\&\quad \quad \times \left[ 2(Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}_0, \sigma ))(a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)})) \right. \\&\quad \quad +\left. \varDelta _n(a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}))^2 + O(\varDelta ^2_n) \right] \end{aligned}$$
Thanks to Lemmas 4 and 5:
$$\begin{aligned}&\frac{2 {\varDelta _n}}{n{\varDelta _n^2}} \sum _{i=0}^{n-1} (Y_{i+1} - {\bar{A}}_2(Z_i; \theta ^{(1)}, \theta ^{(2)}_0, \sigma ))(a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)})) \overset{\mathbb {P}_0}{\longrightarrow } 0\\&\frac{{\varDelta _n^2}}{n{\varDelta _n^2}} \sum _{i=0}^{n-1} (a_2(Z_i; \theta ^{(2)}_0) - a_2(Z_i; \theta ^{(2)}))^2 \overset{\mathbb {P}_0}{\longrightarrow } \int (a_2(z; \theta ^{(2)}_0) - a_2(z; \theta ^{(2)}))^2 \nu _0(dz) \end{aligned}$$
The consistency is concluded following the same arguments as in the case of \(\theta ^{(1)}\).
Asymptotic normality We apply again a Taylor formula for a function (22):
$$\begin{aligned} \int C_n\left( \theta _0 + u({{\hat{\theta }}}_n - \theta _0 \right) )du\, E_n = D_n(\theta _0), \end{aligned}$$
where we define
$$\begin{aligned} C_n(\theta )&:= \left[ \begin{matrix} {\frac{\varDelta _n}{n}}\frac{\partial ^2}{\partial \theta ^{(1)} \partial \theta ^{(1)}} {\mathcal {L}}^{LSE}_{n,\varDelta _n}(\theta ;Z_{0:n})&{}\quad \frac{1}{n}\frac{\partial ^2}{\partial \theta ^{(1)} \partial \theta ^{(2)}} {\mathcal {L}}^{LSE}_{n,\varDelta _n}(\theta ;Z_{0:n}) \\ {\frac{1}{n }}\frac{\partial ^2}{\partial \theta ^{(1)} \partial \theta ^{(2)}} {\mathcal {L}}^{LSE}_{n,\varDelta _n}(\theta ;Z_{0:n}) &{}\quad \frac{1}{n \varDelta _n}\frac{\partial ^2}{\partial \theta ^{(2)} \partial \theta ^{(2)}} {\mathcal {L}}^{LSE}_{n,\varDelta _n}(\theta ;Z_{0:n})\end{matrix}\right] ,\\ E_n&:= \left[ \begin{matrix} {\sqrt{\frac{n}{\varDelta _n}}} ({{\hat{\theta }}}^{(1)}_{n, \varDelta _n} - \theta ^{(1)}_0) \\ {\sqrt{n\varDelta _n}} ({{\hat{\theta }}}^{(2)}_n - \theta ^{(2)}_0) \end{matrix} \right] , \quad D_n (\theta ) = \left[ \begin{matrix} {\frac{\sqrt{\varDelta _n}}{n}}\frac{\partial }{\partial \theta ^{(1)} } {\mathcal {L}}^{LSE}_{n,\varDelta _n}(\theta ;Z_{0:n}) \\ {\frac{1}{n\sqrt{\varDelta _n}}}\frac{\partial }{\partial \theta ^{(2)} } {\mathcal {L}}^{LSE}_{n,\varDelta _n}(\theta ;Z_{0:n}) \end{matrix}\right] . \end{aligned}$$
Using Lemma 7 we get:
$$\begin{aligned} D_n(\theta _0) \overset{{\mathcal {D}}}{\longrightarrow } -2 {\mathcal {N}}\left( 0,I_{2}\cdot \left[ \begin{matrix} {1\over 3}\int b^2(z;\sigma _0) (\partial _ya_1(z; \theta ^{(1)}_0))^2 (\partial _{\theta ^{(1)}}a_1(z; \theta ^{(1)}_0))^2 \nu _0(dz) \\ \int b^2(z;\sigma _0) (\partial _{\theta ^{(2)}}a_2(z; \theta ^{(2)}_0))^2 \nu _0(dz) \end{matrix} \right] \right) , \end{aligned}$$
where \(I_{2}\) is \(2\times 2\) identity matrix. And by Lemmas 5 and 4 we have the result for \(C_n(\theta )\):
$$\begin{aligned} C_n(\theta _0) \overset{\mathbb {P}_0}{\longrightarrow } -2 \left[ \begin{matrix} \int (\partial _{\theta ^{(1)}}a_1(z; \theta ^{(1)}_0))^2 \nu _0(dz) &{} 0 \\ 0 &{} \int (\partial _{\theta ^{(2)}}a_2(z; \theta ^{(2)}_0))^2 \nu _0(dz) \end{matrix} \right] . \end{aligned}$$
That, in the combination with the consistency result, gives the theorem. \(\square \)
