2.1. Hypothesis and notation

Throughout the paper, the process $(Y_t,\, t\in\mathbb{R}_+)$ is a continuous semimartingale and $(Z_t,\,t\in\mathbb{R}_+)$ is a Lévy jump process starting at 0. Both processes take their values on $\mathbb{R}$ and are defined on the complete and right-continuous filtered probability space $(\Omega ,\mathcal F,(\mathcal F_t)_{t\geq0},\mathbb{P})$ .

We have the following hypothesis.

Hypothesis (H): The process Y is a weak solution of the equation

(5) \begin{equation} {\mathrm{d}} Y_t = a_t\,{\mathrm{d}} t + \sigma(Y_t)\,{\mathrm{d}} W_t, \end{equation}

where W is a standard Brownian motion and the function $\sigma\,:\, \mathbb{R}\to\mathbb{R}$ is of class $C^3$ with $\sigma(x)> 0$ for all $x\in\mathbb{R}$ . Moreover, the process a is progressively measurable, and for almost all $\omega$ we have $\int_0^t a^2_s(\omega)\,{\mathrm{d}} s< +\infty$ for any real $t>0$ . For each starting point $Y_0$ , the process Y is non-explosive.

Set $X\,:\!= Y+Z$ , where Z is a Lévy jump process given by

(6) \begin{equation} Z_t=\int_0^t\int_{\{|x|\le 1\}} x\,(\mu-\nu)({\mathrm{d}} s,{\mathrm{d}} x) + \int_0^t\int_{\{|x|> 1\}} x\,\mu({\mathrm{d}} s,{\mathrm{d}} x).\end{equation}

Here $\mu$ is a Poisson random measure independent of Y and with compensator $\nu({\mathrm{d}} s,{\mathrm{d}} x)={\mathrm{d}} s\,F({\mathrm{d}} x)$ , where F satisfies $\int_{\mathbb{R}} (1\wedge|x|^2)\,F({\mathrm{d}} x) < +\infty$ .

We will set ${\mathcal{A}}\,:\!= \sigma (\{ Y_t, \,t\ge 0\}).$

If (H) is satisfied and if the measure F satisfies

(7) \begin{equation}\int_{\mathbb{R}} (1\wedge|x|)\,F({\mathrm{d}} x) < +\infty, \end{equation}

then the process Z can be written as

(8) \begin{equation} Z_t\, =\, -\int_0^t\int_{\{|x|\le 1\}} x\,\nu({\mathrm{d}} s,{\mathrm{d}} x) + \int_0^t\int_{\mathbb{R}} x\,\mu({\mathrm{d}} s,{\mathrm{d}} x).\end{equation}

Hence, under (7), if we set

(9) \begin{equation} \gamma\,:\!=\,\gamma(X) = \inf\left\{p\ge 0\,: \int_{\mathbb{R}} (1\wedge|x|^p)\,F({\mathrm{d}} x) < +\infty\right\}\!,\end{equation}

then clearly $\gamma$ belongs to [0,1]. Specifying the role of $\gamma$ , we state the next hypothesis.

Hypothesis (H- $(\gamma)$ ): We have hypothesis (H), the process Z has finite variation, and (up to modification of the drift of Y) the process X is written as

\begin{align*}X_t = Y_t + \sum_{s\le t} \Delta X_s = Y_t + \sum_{s\le t} \Delta Z_s ,\end{align*}

where $\Delta X_s$ is the size of the jump of X at time s. Furthermore, the real $\gamma$ defined in (9) is the Blumenthal–Getoor index of the process X.

We denote by $\mathbb{P}_x$ the law of X starting at x on a canonical space.

Let us now proceed to giving the hypothesis on the functions.

Unless otherwise stated, the functions $(f_n)_{n\ge 1}$ and f are defined on $\mathbb{R}\times [0,1]\times\mathbb{R}\times\mathbb{R}$ and take their values on $\mathbb{R}$ .

For $r=1,\,2$ and 3, we have the following hypothesis.

Hypothesis (K-( r )): For any fixed $n\ge 1$ , $u\in[0,1]$ and $w\in\mathbb{R}$ , the function $(x,y)\to f_n(x,u,y,w)$ is ${\mathcal{C}}^r$ on $\mathbb{R}^2$ . Moreover, for $i,j\in \mathbb{N}$ such that $0\le i+j\le r$ and for all fixed real $M>0$ , there are positive constants $C_M$ and $p_M$ such that

(10) \begin{equation}{\sup}_{\{n\in\mathbb{N}^{\star},\,|x|,|y|\le M,\,u\in[0,1]\}}\left|\,\frac{\partial^{i+j} f_n}{\partial x^i\partial {y}^j}(x,u,y,w)\,\right|\le C_M \big(1+|w|^{p_M}\big).\end{equation}

Furthermore, there exists a function $f\,:\, \mathbb{R}\times[0,1]\times\mathbb{R}\times\mathbb{R}\to \mathbb{R}$ such that for all $(x,y)\in\mathbb{R}^2$ , $f_n(x,u,y,w)$ converges ${\mathrm{d}} u\otimes {\mathrm{d}} w$ -almost everywhere to f(x,u,y,w).

Note that (K-(3)) $\Rightarrow $ (K-(2)) $\Rightarrow$ (K-(1)).

We set $\beta_n\,:\!=\frac{\alpha_n}{\sqrt{\Delta_n}}$ , which throughout this paper we assume converges to $\beta\in[0,+\infty)$] . For $p,p' \ge 0$ , we state the following two hypotheses.

Hypothesis (L-( p )): For any fixed $K>0$ , there exists a constant $C_K\ge 0$ such that

\begin{align*}{\sup}_{\{n\in\mathbb{N}^{\star},\,|x|,|y|\le K,\,u\in[0,1], \,w\in\mathbb{R}\}} \bigl|f_n(x,u,y,w+z) - f_n(x,u,y,w)\bigr|< C_K|z|^p.\end{align*}

Hypothesis (L-( p , p ^′)): For any fixed $K>0$ , there exists a constant $C_K\ge 0$ such that

\begin{align*}{\sup}_{\{n\in\mathbb{N}^{\star},\,|x|,|y|\le K,\,u\in[0,1],\,w\in\mathbb{R}\}} \bigl|f_n(x,u,y,w+z) - f_n(x,u,y,w)\bigr|< C_K\big(|z|^p + \beta_n^{p'}\big).\end{align*}

For processes $(X^n)_{n\ge 1}$ and X, we say that $(X^n)$ converges uniformly on compacts in probability (u.c.p.) to X, denoted by $X^n \stackrel {\mathrm{u.c.p.}} {\longrightarrow} X$ , if for every $t,\varepsilon>0$ we have

\begin{align*}\lim_{n\to+\infty} \mathbb{P}\!\left({\sup}_{s\le t}|X_s-X^n_s|>\varepsilon\right) = 0.\end{align*}

If $(X^n)$ converges in law to X for the Skorokhod topology, we denote this by $X^n\stackrel{{\mathcal{L}}}{\longrightarrow} X$ . For the notions of convergence in law for the Skorokhod topology and ${\mathcal{A}}$ -stable convergence we refer to [Reference Jacod and Shiryaev12].

The fractional part of a real x will be denoted by $\{x\}=x-[x]\in[0,1)$] .

Recall that $h_{\sigma}$ is the density of a normal law ${\mathcal{N}}(0,\sigma^2)$ and set $h\,:\!=h_1$ . For any sequence of functions $(f_n)$ , we set

(11) \begin{equation} \xi_i^n = f_n\!\left(X_{(i-1)\Delta_n},\,\left\{\frac{X_{(i-1)\Delta_n}}{\alpha_n}\right\},\,Y_{(i-1)\Delta_n},\,\frac{\Delta_i^n X}{\sqrt{\Delta_n}}\right) \end{equation}

and

(12) \begin{equation} V^n(f_n,X, Y)_t\,:\!=\Delta_n\sum_{i=1}^{[t/\Delta_n]} \xi_i^n,\end{equation}

where $\Delta_i^n X= X_{i\Delta_n}-X_{(i-1)\Delta_n}$ , and $(\Delta_n)$ and $(\alpha_n)$ are sequences converging to 0.

Now we define the operators m and M in the following way: we denote by ${\mathcal{C}}(G)$ the class of measurable functions

\begin{align*}f\,:\,\begin{array}{lll}\mathbb{R} \times [0,1]\times \mathbb{R}\times \mathbb{R} & \to & \mathbb{R}\\(x,u,y,w) & \mapsto & f(x,u,y,w) \end{array} \end{align*}

such that

\begin{align*}\int_0^1\int_{\mathbb{R}} | f(x,u,y,w)|\, h_a(w)\,{\mathrm{d}} w\,{\mathrm{d}} u < +\infty\quad\text{for all real }\ x, \ y \ \text{and all $a>0$}.\end{align*}

Then, if $f\in{\mathcal{C}}(G)$ , we set

\begin{align*}mf\,:\,\begin{array}{lll}\mathbb{R} \times [0,1]\times \mathbb{R} & \to & \mathbb{R}\\(x,u,y) & \mapsto & mf(x,u,y) = \int_{\mathbb{R}} f(x,u,y,w) h_{\sigma(y)}(w)\,{\mathrm{d}} w \end{array} \end{align*}

and

\begin{align*}Mf\,:\,\begin{array}{lll}\mathbb{R} \times \mathbb{R} & \to & \mathbb{R}\\(x,y) & \mapsto & Mf(x,y) = \int_0^1 mf(x,u,y)\,{\mathrm{d}} u. \end{array} \end{align*}

Note that if f satisfies (K-(1)), then $f\in {\mathcal{C}}(G)$ .

2.2. Law of large numbers

Then

\begin{align*}V^n(f_n,X,Y)\stackrel{\mathrm{u.c.p.}}{\longrightarrow} \int_0^\cdot Mf(X_s,Y_s)\,{\mathrm{d}} s.\end{align*}

2.3. Central limit theorem

Before proceeding to the central limit theorem, we give some interesting auxiliary results.

Let $(f_n)_{n\ge 1}$ and (f) belong to ${\mathcal{C}}(G)$ ; then recall (11) and set

(14) \begin{equation} \overline{\zeta}_i^n \,:\!= \, \xi_i^n- \mathbb{E}\!\left\{ \left. \xi_i^n\,\right|{\mathcal{F}}_{(i-1)\Delta_n}\right\}\end{equation}

and

(15) \begin{equation} \overline{\zeta'}_i^n \,:\!=\, \xi_i^n - m f_n\!\left(X_{(i-1)\Delta_n},\,\left\{\frac{X_{(i-1)\Delta_n}}{\alpha_n}\right\}\!,\,Y_{(i-1)\Delta_n}\right)\!. \end{equation}

For $x,y,w\in\mathbb{R}$ , and $u\in[0,1]$ , we define the functions

(16) \begin{equation}\psi(x,u,y,w)\,:\!=\frac{w}{\sigma(y)}, \quad g(x,u,y,w)\,:\!=\frac{w\, \sigma'(y)}{2\sigma(y)}.\left(\frac{w^2}{\sigma^2(y)}-3\right)\!.\end{equation}

Then $\psi$ and g belong to ${\mathcal{C}}(G)$ .

Now let B be another one-dimensional Brownian motion defined on an auxiliary space; then we extend the original space and B will be independent of W and $\mu$ . So we define

(17) \begin{equation}L_t = \int_0^t\sqrt{M(f^2)(X_s,Y_s)-\int_0^1\left(m f(X_s,u,Y_s) \right)^2\,{\mathrm{d}} u - \left(M (f\psi)(X_s,Y_s)\right)^2}\;{\mathrm{d}} B_s,\end{equation}

which is, conditionally on ${\mathcal{F}}$ , a continuous Gaussian martingale; we will see below that it is well-defined.

We can now state the first results of this subsection.

Theorem 2. Assume (H) and (K-(1)), and further assume that the functions $(f_n)_{n\ge 1}$ satisfy one of the following two conditions.

1. (i) For any $K>0$ , there exists a constant $C_K>0$ such that

   \begin{align*} {\sup}_{\{n\in\mathbb{N}^{\star},\,|x|,|y|\le K,\,u\in[0,1],\, w\in\mathbb{R}\}} \left|\frac{\partial^i f_n}{\partial x^i}(x,u,y,w)\right|\,\le\, C_K \quad\mbox{for}\ i=0,1.\end{align*}

2. (ii) For any real $M>0$ , the constant $p_M$ in inequality (10) is always in [0,1), and for all real $\varepsilon,K,K$ ’ $>0$ there exist $\varepsilon'>0$ and $N\in\mathbb{N}^{*}$ such that

   \begin{align*}{\sup}_{\{|x|,|y|\le K,\,u\in[0,1],\,|w|\le K',\,|z|\le\varepsilon',\,n\ge N \}}\left|f_n(x,u,y,w+z) - f_n(x,u,y,w)\right|\, \le\,\varepsilon.\end{align*}

Then

\begin{align*}\sqrt{\Delta_n}\sum_{i=1}^{[\cdot/\Delta_n]} \overline{\zeta}_i^n\,\,\stackrel{{\mathcal{L}}}{\longrightarrow} \,\,L \, +\,\int_0^{\cdot} M (f\psi)(X_s,Y_s)\,{\mathrm{d}} W_s.\end{align*}

Moreover, the convergence also takes place ${\mathcal{A}}$ -stably in law.

Theorem 3. Assume (H- $(\gamma)$ ) for a $\gamma\in[0,1)$] and (K-(1)). Further, assume that the functions $(f_n)_{n\ge 1}$ satisfy one of the following three conditions.

1. (i) For any real $K>0$ , there exists a constant $C_K>0$ such that

   \begin{align*} {\sup}_{\{n\in\mathbb{N}^{\star},\,|x|,|y|\le K,\,u\in[0,1],\, w\in\mathbb{R}\}} \left|\frac{\partial^i f_n}{\partial x^i}(x,u,y,w)\right|\,\le\, C_K\quad\mbox{for}\ i=0,1.\end{align*}

2. (ii) For any real $M>0$ , the constant $p_M$ (in inequality (10)) is always in [0,1), and the functions $(f_n)_{n\ge 1}$ satisfy (L-(p,p^′)) for $p\in(\gamma,1)$ and $p'>0$ such that ${\beta^{p'}_n}/{\sqrt{\Delta_n}}\to0$ .

3. (iii) For any real $M>0$ , the constant $p_M$ (in inequality (10)) is always in [0,1), and the functions $(f_n)_{n\ge 1}$ satisfy (L-(p)) for $p\in(\gamma,1)$ .

Then the process $\sqrt{\Delta_n}\sum_{i=1}^{[\cdot/\Delta_n]} \overline{\zeta'}_i^n$ converges in law for the Skorokhod topology to

\begin{align*} L \, +\,\int_0^\cdot \left( \frac{M (f\psi)(X_s,Y_s)}{\sigma(Y_s)}\right)\,{\mathrm{d}} Y_s \,+ \,\int_0^\cdot M(fg)(X_s,Y_s)\,{\mathrm{d}} s.\end{align*}

Moreover, the convergence also takes place ${\mathcal{A}}$ -stably in law.

Now let us introduce some more notation.

For any integer $n\ge 1$ , functions $f_n,f\in {\mathcal{C}}(G)$ , real numbers x and y and $v\in[0,1]$ , let

(18) \begin{equation} \overline{m}f(x,v,y) = {m}f(x,v,y) - {M}f(x,y)\end{equation}

and

(19) \begin{equation} U_t^n\,:\!=\, \Delta_n\sum_{i=1}^{[t/\Delta_n]} M f_n(X_{(i-1)\Delta_n}, Y_{(i-1)\Delta_n}). \end{equation}

Next, for any real $\beta>0$ , set

(20) \begin{align}&\overline{\Lambda}(f,f)(X_s,Y_s,\beta) \nonumber\\& \quad = \int_0^1 \int_{\mathbb{R}}\biggl( f(X_{s},v,Y_{s},w) - Mf(X_{s},Y_{s})+\overline{m}f\!\left(X_{s},\left\{\frac{w}{\beta} + v\,\right\}Y_{s}\right) \nonumber\\& \qquad - \,\overline{m}f(X_{s},v,Y_{s}) + \sum_{j=1}^{+\infty} \int_{\mathbb{R}} \overline{m}f\!\left(X_{s},\left\{\frac{x+w}{\beta} + v\,\right\}, Y_{s}\right)h_{\sigma( Y_{s})\sqrt{j}}(x)\,{\mathrm{d}} x \nonumber\\ & \qquad - \sum_{j=1}^{+\infty} \int_{\mathbb{R}} \overline{m}f\!\left(X_{s},\left\{\frac{x}{\beta}+v\right\},Y_{s}\right)h_{\sigma(Y_{s})\sqrt{j}}(x)\,{\mathrm{d}} x \biggr)^2h_{\sigma(Y_{s})}(w)\,{\mathrm{d}} w\,{\mathrm{d}} v \end{align}

and

(21) \begin{equation} \overline{\Lambda}(f,f)(X_s,Y_s,0)\, = \,M(f^2)(X_s,Y_s) - (Mf(X_s,Y_s))^2.\end{equation}

If $(f_n)$ satisfies (K-(3)) and f is the limit function, then, using Lemma 6.1 of [Reference Delattre and Jacod6] (see also Lemma 4 of this paper), since $\sigma>0$ and is bounded, we see that

(22) \begin{align}&\left|\int_{\mathbb{R}} \overline{m}f\!\left(X_{s},\,\left\{\frac{x+w}{\beta} + v\right\}, Y_{s}\right)\,h_{\sigma( Y_{s})\sqrt{j}}(x)\,{\mathrm{d}} x \, \right| \nonumber \\&+ \left|\int_{\mathbb{R}} \overline{m}f\!\left(X_{s},\,\left\{\frac{x}{\beta}+v\right\},Y_{s}\right)\,h_{\sigma(Y_{s})\sqrt{j}}(x)\,{\mathrm{d}} x\, \right|\,\le\, C\, \frac{\beta^3}{j^{\frac{3}{2} }},\end{align}

where C may depend on $\omega$ and T, with $s\le T$ . Then $\overline{\Lambda}(f,f)$ is well-defined.

We also set

\begin{align*}B'_t\,:\!=\int_0^t\sqrt{\overline{\Lambda}(f,f)(X_s,Y_s,\beta) - (M (f\psi)(X_s,Y_s))^2}\;{\mathrm{d}} B_s.\end{align*}

We will see below that B ^′ is well-defined.

Now we give the central limit theorem associated with Theorem 1.

Theorem 4. Assume that the process X satisfies (H- $(\gamma)$ ) for $\gamma<1$ and that the functions $(f_n)_{n\ge 1}$ fulfil hypothesis (K-(3)). For all x and y, ${\mathrm{d}} u\otimes {\mathrm{d}} w$ -almost all (u,w) and all sequences $(u_n)_{n\ge 1}$ converging to u, the sequence $f_n(x,u_n,y,w)$ converges to f(x,u,y,w). Furthermore, $(f_n)$ satisfies one of the following three conditions.

1. (i) For any real $K>0$ , there exists a constant $C_K>0$ such that

   \begin{align*} {\sup}_{\{n\in\mathbb{N}^{\star},\,|x|,|y|\le K,\,u\in[0,1],\, w\in\mathbb{R}\}} \left|\frac{\partial^{i+j} f_n}{\partial x^i\partial y^j}(x,u,y,w)\right|\,\le\, C_K\quad\mbox{for}\ 0\le i+j \le 3.\end{align*}

2. (ii) For any real $M>0$ , the constant $p_M$ (in inequality (10)) always satisfies $p_M\in [0,1)$] , and the functions $(f_n)_{n\ge 1}$ fulfil hypothesis (L-(p,p^′)) with $p\in(\gamma,1)$ and $p'>0$ such that ${\beta^{p'}_n}/{\sqrt{\Delta_n}}\to0$ .

3. (iii) For any real $M>0$ , the constant $p_M$ always satisfies $p_M\in [0,1)$] and the functions $(f_n)$ fulfil hypothesis (L-(p)) with $p\in(\gamma,1)$ .

Then each of the two processes

\begin{align*}\frac{1}{\sqrt{\Delta_n}}\left(V^n(f_n,X,Y) - \int_0^\cdot M f_n(X_s,Y_s)\,{\mathrm{d}} s\right)\end{align*}

and

\begin{align*}\frac{1}{\sqrt{\Delta_n}}\left(V^n(f_n,X,Y) - U^n\right)\end{align*}

converges in law, for the Skorokhod topology, to the process

\begin{align*} B' + \int_0^\cdot \left(\frac{ M (f\psi)(X_s,Y_s)}{\sigma(Y_s)}\right)\,{\mathrm{d}} Y_s + \int_0^\cdot M (fg)(X_s,Y_s)\,{\mathrm{d}} s.\end{align*}

Moreover, the convergence also takes place ${\mathcal{A}}$ -stably in law.

Remark 3.

1. (i) The central limit theorems obtained here generalise those of [Reference Delattre and Jacod6], and if X is a continuous diffusion, we recover the same limit.

2. (ii) There are two other aspects that we can examine further:

   1. (a) the case of a more general semimartingale, such as Itô’s semimartingale;

   2. (b) a central limit theorem in the case where $\beta_n \to + \infty $ .

4.2. Passing to the canonical process and change of probability

Under $(\Omega, \mathcal F,\mathbb{P})$ , if the process Y is as in (LH), there exists (see Chapter I, part 4.B, of [Reference Delattre7]) a probability measure $\overline{\mathbb{P}}$ on $(\Omega,\mathcal F)$ that is absolutely continuous with respect to $\mathbb{P}$ and under which Y satisfies the equation

(36) \begin{equation}{\mathrm{d}} Y_t = b(Y_t)\,{\mathrm{d}} t + \sigma(Y_t)\,{\mathrm{d}} W'_t, \end{equation}

where $b(y)\,:\!=\frac{\sigma(y)\sigma'(y)}{2}$ and $W'=:W_t-\int_0^t \frac{b(Y_s)-a_s}{\sigma(Y_s)}\,{\mathrm{d}} s$ is a standard Brownian motion under $\overline{\mathbb{P}}$ . Since it is well known that changing the probability measure (when $\overline{\mathbb{P}} \sim \mathbb{P}$ ) does not have repercussions for the convergences used here (see part I.4.B of [Reference Delattre7]), by the localization procedure it will suffice to prove the theorem under $(\Omega,\mathcal F,(\mathcal F_t)_{t\in [0,T_0] },\overline{\mathbb{P}})$ for a fixed and large enough real $T_0>0$ .

From now on we will also suppose that X satisfies the following two hypotheses, which replace (LH) and (LH- $(\gamma)$ ).

Hypothesis (LH^′): The function $\sigma$ is $C^2_b$ with $\inf_x \sigma(x)>0$ , and Y is the solution of

(37) \begin{equation} {\mathrm{d}} Y_t = b(Y_t)\,{\mathrm{d}} t + \sigma(Y_t)\,{\mathrm{d}} W_t,\end{equation}

where $b(y)=\frac{\sigma(y)\sigma'(y)}{2}$ and W is a standard Brownian motion. Then $X_t=Y_t+Z_t$ , where Z is as in (LH).

Hypothesis (LH^′- $(\gamma)$ ): We have (LH^′) and (H- $(\gamma)$ ).

Consider now the canonical space $\Omega'=C([0,T_0],\mathbb{R})$ and its canonical filtration $(\mathcal F'_t)_{t\in [0,T_0] }$ . There exists a probability $\mathbb{P}'$ such that the coordinate mapping is the solution of (36). From now on, we can assume that Y is the canonical process on $(\Omega',\mathcal F'_{T_0},(\mathcal F')_{t\in [0,T_0] },\mathbb{P}')$ .

Let us again take another auxiliary space $(\Omega'', {\mathcal{F}}'',({\mathcal{F}}'')_{t\in [0,T_0]},\mathbb{P}'')$ supporting a Lévy jump process Z starting at 0 and satisying (LH^′).

Finally, we set

(38) \begin{equation} {\Omega}=\Omega'\times\Omega'',\quad{\mathbb{P}}=\mathbb{P}'\otimes\mathbb{P}'',\quad{\mathcal{F}}={\mathcal{F}}'_{T_0}\otimes{\mathcal{F}}_{T_0}'', \quad{\mathcal{F}}_t={\mathcal{F}}'_t\otimes{\mathcal{F}}''_t\quad\forall t\in [0,T_0].\end{equation}

In the proof we will sometimes omit $T_0$ . For any $x\in\mathbb{R}$ , we denote by $\mathbb{P}_x$ the probability measure such that $\mathbb{P}_x(X_0=x)=1$ . Then, for any $A\in\mathcal F$ , we have $\mathbb{P}(A)=\int_{\mathbb{R}} \mathbb{P}_x(A)\,\delta({\mathrm{d}} x)$ , where $\delta$ is the law of $X_0=Y_0$ .

4.4. Preliminary results

In this subsection we give a series of lemmas which generalize lemmas in [Reference Delattre and Jacod6]. The following one will be used again and again and corresponds to Lemma 6.1 of [Reference Delattre and Jacod6]. We will use a constant C for the boundary and its value may change from line to line.

Lemma 1. Let $k\,:\,\mathbb{R}\to\mathbb{R}$ and $f\,:\,\mathbb{R}\times[0,1)^2\to \mathbb{R}$] be two measurable functions. For all $u,v\in[0,1)$] , we suppose that the mapping $x\to g(x,u,v)\,:\!=k(x)f(x,u,v)$ is of class $C^1$ . Furthermore, assume that for all real z and $\alpha>0 $ ,

\begin{align*} \int_{\mathbb{R}} \int_0^1 \left(|g|+\left|\frac{\partial g}{\partial x} \right| \right)\left(x,u,\left\{u+\frac{z}{\alpha}\right\}\right)\,{\mathrm{d}} u\,{\mathrm{d}} x < +\infty.\end{align*}

Then

(58) \begin{align}&\left|\int_{\mathbb{R}} k(x)f\!\left(x,\left\{\frac{x}{\alpha}\right\},\left\{\frac{x+z}{\alpha}\right\}\right)\,{\mathrm{d}} x - \int_{\mathbb{R}}k(x)\int_0^1 f\!\left(x,u,\left\{u+\frac{z}{\alpha}\right\}\right)\,{\mathrm{d}} u\,{\mathrm{d}} x\right|\nonumber\\& \quad \le\,\alpha \int_{\mathbb{R}}\left(\int_0^1 \left|\frac{\partial g}{\partial x}\left(x,u, \left\{u+\frac{z}{\alpha}\right\}\right)\right|\,{\mathrm{d}} u\right)\,{\mathrm{d}} x.\end{align}

Proof. The proof proceeds similarly to the proof of Proposition 5.1 in [Reference Delattre7]. In fact, integration by parts gives

(59) \begin{align} g\left(\alpha k+\alpha u, u, \left\{u+\frac{z}{\alpha}\right\}\right) & = \int_0^1 g\left(\alpha k+\alpha v, u, \left\{u+\frac{z}{\alpha}\right\}\right)\,{\mathrm{d}} v \nonumber\\ & \quad + \alpha\int_0^1 \frac{\partial g}{\partial x}\left(\alpha k+\alpha v, u, \left\{u+\frac{z}{\alpha}\right\}\right)\chi(u,v)\,{\mathrm{d}} v \end{align}

for almost all $u\in[0,1)$] . Here $\chi (u,v)=v-1_{\{u < v\}}$ . Then, since for $j\in \mathbb{Z}$ and $u\in\mathbb{R}$ we have $\{u+j\}=\{u\}$ , it follows that

\begin{eqnarray*}\int_{\mathbb{R}} g\left(x,\left\{\frac{x}{\alpha}\right\},\left\{\frac{x+z}{\alpha}\right\}\right)\,{\mathrm{d}} x & = & \alpha \sum_{j\in\mathbb{Z}} \int_0^1g\left(\alpha j+\alpha u, u, \left\{u+\frac{z}{\alpha}\right\}\right)\,{\mathrm{d}} u\\ & = & \alpha \sum_{j\in\mathbb{Z}}\int_0^1\int_0^1 g\left(\alpha j+\alpha v, u, \left\{u+\frac{z}{\alpha}\right\}\right)\,{\mathrm{d}} u\,{\mathrm{d}} v\\ & & +\, \alpha^2\sum_{j\in\mathbb{Z}} \int_0^1\int_0^1\frac{\partial g}{\partial x}\left(\alpha j+\alpha v, u, \left\{u+\frac{z}{\alpha}\right\}\right)\chi(u,v)\,{\mathrm{d}} u\,{\mathrm{d}} v\end{eqnarray*}

\begin{eqnarray*} & = & \sum_{j\in\mathbb{Z}}\int_{\alpha j}^{\alpha (j+1)}\int_0^1 g\left(x, u, \left\{u+\frac{z}{\alpha}\right\}\right)\,{\mathrm{d}} u\,{\mathrm{d}} x\\ & & +\; \alpha\sum_{j\in\mathbb{Z}} \int_{\alpha j}^{\alpha (j+1)}\int_0^1\frac{\partial g}{\partial x}\left(x, u, \left\{u+\frac{z}{\alpha}\right\}\right)\chi\!\left(u,\frac{x}{\alpha}-j\right){\mathrm{d}} u\,{\mathrm{d}} x\\ & = & \int_{\mathbb{R}}\int_0^1 g\left(x, u, \left\{u+\frac{z}{\alpha}\right\}\right)\,{\mathrm{d}} u\,{\mathrm{d}} x\\ & & +\; \alpha\sum_{j\in\mathbb{Z}} \int_{\alpha j}^{\alpha (j+1)}\int_0^1\frac{\partial g}{\partial x}\left(x, u, \left\{u+\frac{z}{\alpha}\right\}\right)\chi\!\left(u,\frac{x}{\alpha}-j\right){\mathrm{d}} u\,{\mathrm{d}} x.\end{eqnarray*}

Since $|\chi|\le 1$ , we have

\begin{align*}&\left|\, \alpha\sum_{j\in\mathbb{Z}} \int_{\alpha j}^{\alpha (j+1)}\int_0^1\frac{\partial g}{\partial x}\left(x, u, \left\{u+\frac{z}{\alpha}\right\}\right)\chi\!\left(u,\frac{x}{\alpha}-j\right)\,{\mathrm{d}} u\,{\mathrm{d}} x\, \right|\\& \quad\le \alpha \int_{\mathbb{R}}\int_0^1\left|\frac{\partial g}{\partial x}\left(x, u, \left\{u+\frac{z}{\alpha}\right\}\right)\right|\,{\mathrm{d}} u \,{\mathrm{d}} x,\end{align*}

and thus we get the result.

Lemma 2. Assume (LH^′), and for any $n\ge 1$ let $g_n:\mathbb{R}^2\to\mathbb{R}$ be a measurable function such that there exists a constant $C >0$ and an integer p such that

\begin{align*}{\sup}_n |g_n(x,y)|\,\,\le\, C(1+|x|^2+|y|^p)\quad\mbox{for all real}\ x,y.\end{align*}

1. 1. If, further, there exists a function $g\,:\,\mathbb{R}^2\to\mathbb{R}$ such that for all real x and y and for any sequence $(x_n,y_n)_{n\ge1}$ tending to (x, y), the sequence $(g_n(x_n,y_n))$ tends to g(x,,y), then

   (60) \begin{equation}\mathbb{E}\!\left(\,{\sup}_{t\le T}\,\left|\,\Delta_n\sum_{i=1}^{[t/\Delta_n]} g_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n})-\int_0^tg(X_s,Y_s)\,{\mathrm{d}} s\right|\,\right)\,\to\,0\quad\mbox{as}\ n\to+\infty.\end{equation}

2. 2. Moreover, if the functions $(g_n)$ are of class $C^2$ on $\mathbb{R}^2$ and if there exists a constant $C >0$ and an integer p such that

   \begin{align*} {\sup}_n\,\left|\frac{\partial^{i+j} g_n(x,y)}{\partial x^i \partial y^j}\right|\,\le\, C(1+|x|^{1}+|y|^p)\quad\mbox{ for }1\le i+j\le 2 \mbox{ and }x,y\in\mathbb{R},\end{align*}
   then
   (61) \begin{equation}\mathbb{E}\!\left\{\,{\sup}_{t\le T}\,\left|\,\Delta_n\sum_{i=1}^{[t/\Delta_n]} g_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n})-\int_0^t g_n(X_s,Y_s)\,{\mathrm{d}} s\right|^2\,\right\}\,\le\,{C_t}\Delta_n^2, \end{equation}
   where $C_t$ is a constant depending on t.

Let $q, T\in\mathbb{R}_+$ and $r\in\{1,2,3\}$ . Then we have the following lemma.

Lemma 3. Under (LH^′) and (LK-(r,q)), the functions $(m_nf_n)$ and $(mf_n)$ are $C^r$ for the first and third variables and we have

(62) \begin{equation} {\sup}_{\{0\le i+j\le r,\,x,y\in\mathbb{R},\,u\in[0,1]\}}\,\left\{\left|\frac{\partial^{i+j}\, m f_n}{\partial x^i\,\partial y^i}(x,u,y)\right| + \left|\frac{\partial^{i+j} \,m_nf_n}{\partial x^i\,\partial y^j}(x,u,y)\right| \right\} \,\le\, C.\end{equation}

Furthermore,

(63) \begin{equation} {\sup}_{\{0\le i\le r,\,x,y\in\mathbb{R},\,u\in[0,1]\}}\!\left\{\left|{m}_n f_n(x,u, y) - {m} f_n(x,u,y)\right| + \left|\overline{m}_n f_n(x, y) - \overline{m}' f_n(x, y)\right|\right\}\le C\sqrt{\Delta_n}\end{equation}

and

(64) \begin{equation}{\sup}_{\{0\le i\le r,\,x,y\in\mathbb{R},\,u\in[0,1]\}}\left\{\left|{m}_n f_n(x,u, y) - {m} f_n(x,u,y) - \sqrt{\Delta_n} m\tilde{f_n}(x,u,y)\right|\right\}\le C{\Delta_n}. \end{equation}

Lemma 4. Under (LH^′) and (LK-(r,q)), for $j\le \frac{T}{\Delta_n}$ and for all real a,b, c and y, we have the two inequalities

(65) \begin{equation} \left| \int_{\mathbb{R}} \overline{m}_nf_n(a+x,x)p_{j\Delta_n}(y,x)\,{\mathrm{d}} x\,\right|\,\le\,C\,\frac{\beta_n^r}{j^{\frac{r}{2}}}\,(1+T^{\frac{r}{2}}) \end{equation}

and

(66) \begin{equation} \left|\,\int_{\mathbb{R}}\overline{m}' f_n(x\sqrt{\Delta_n}+a,x\sqrt{\Delta_n}+b,x\sqrt{\Delta_n}+c)h_{\sigma( y)\sqrt{j}}(x)\,{\mathrm{d}} x\,\right|\,\le\,C\,\frac{\beta_n^r}{j^{\frac{r}{2}}}\,(1+T^{\frac{r}{2}}).\end{equation}

For all $q\ge 0$ , using all the above results (in particular Lemma 3 and the inequality (52)), we deduce the following inequalities, which have already been obtained as (8-15), (8-16) and (8-17) in [Reference Delattre and Jacod6].

Lemma 5. Assume (LH^′) and (LK-(1,q)), and suppose that one of the following conditions is fulfilled.

1. (i) There exists a constant $C>0$ such that

   \begin{align*} {\sup}_{\{n\in\mathbb{N}^{\star},\,x,y,y'\in\mathbb{R},\,u\in[0,1]\}}| f_n(x,u,y',y)|\,\le\, C . \end{align*}

2. (ii) We have $q\in[0,2)$] , and for all $\varepsilon, K,K'>0$ there exists an integer $N>0$ and a real $\varepsilon'>0$ such that

   \begin{align*}{\sup}_{\{|x|,|y'|\le K,\,u\in[0,1],\,|y|\le K',\,|z|\le\varepsilon',\,n\ge N \}}|f_n(x,u,y',y+z) - f_n(x,u,y',y)|\, \le\,\varepsilon.\end{align*}

Then

\begin{align*} V^n(f_n,Y,X)-\overline{V}^n(f_n,Y,X)\,\stackrel{\mathrm{u.c.p.}}{\longrightarrow}\, 0\quad\mbox{ as } n\to+\infty.\end{align*}

Proof. (i) Recall the notation (55). To simplify, we temporarily write

(71) \begin{align} f_n\!\left(X_{(i-1)\Delta_n},\left\{\frac{X_{(i-1)\Delta_n}}{\alpha_n}\right\},Y_{(i-1)\Delta_n}, \cdot\right) \ \mbox{as}\ f_n^i(\!\cdot\!). \end{align}

Then

\begin{align*}V^n(f_n,Y,X)-\overline{V}^n(f_n,Y,X) = \Delta_n\sum_{i=1}^{[t/\Delta_n]}\left(L^n_i-\mathbb{E}(L^n_i|\mathcal F_{(i-1)\Delta_n})\right) +\Delta_n\sum_{i=1}^{[t/\Delta_n]}\mathbb{E}(L^n_i|\mathcal F_{(i-1)\Delta_n}).\end{align*}

The process $\left(\Delta_n\sum_{i=1}^{[t/\Delta_n]}\left(L^n_i-\mathbb{E}(L^n_i|\mathcal F_{(i-1)\Delta_n})\right)\right) $ is a square-integrable martingale with respect to the filtration $(\mathcal F_{[t/\Delta_n]\Delta_n})$ . Its bracket is less than $\Delta_n^2\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}((L^n_i)^2|\mathcal F_{(i-1)\Delta_n})$ . And since $(f_n)$ is uniformly bounded, we easily have by the Doob inequality the local uniform convergence in probability.

Next, concerning $\Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}(L^n_i|\mathcal F_{(i-1)\Delta_n})$ , we have, after a variable change,

(72) \begin{align}& \Delta_n\sum_{i=1}^{[t/\Delta_n]} |\mathbb{E}(L^n_i|\mathcal F_{(i-1)\Delta_n})| \nonumber\\& = \Delta_n\sum_{i=1}^{[t/\Delta_n]} \left|\mathbb{E}\Big( \Big(\int_{\mathbb{R}} f_n^i\Big(\frac{y}{\sqrt{\Delta_n}}\Big) \big(q_{\Delta_n}(Y_{(i-1)\Delta_n},y-\Delta_i^n Z) - q_{\Delta_n}(Y_{(i-1)\Delta_n},y)\big)\,{\mathrm{d}} y\,\Big)\,\Big|\,\mathcal F_{(i-1)\Delta_n}\,\Big)\right| \nonumber \\&= \Delta_n\sum_{i=1}^{[t/\Delta_n]}\left|\mathbb{E}\Big( \Big(\int_{\mathbb{R}} f_n^i\Big(\frac{y}{\sqrt{\Delta_n}}\Big)\,\times\, \Big(\int_{y\wedge (y-\Delta_i^n Z)}^{y\vee (y-\Delta_i^n Z)} \frac{\partial q_{\Delta_n}}{\partial z}(Y_{(i-1)\Delta_n},z)\,dz\Big)\,{\mathrm{d}} y\,\Big)\,\Big|\,\mathcal F_{(i-1)\Delta_n}\,\Big) \right|,\end{align}

and the last quantity is smaller than

\begin{align*} \Delta_n\sum_{i=1}^{[t/\Delta_n]}\left|\mathbb{E}\Big( \Big(\int_{\mathbb{R}} \big|\Delta_n^i Z\big| \times \Big|\frac{\partial q_{\Delta_n}}{\partial z}(Y_{(i-1)\Delta_n},z)\Big|\,{\mathrm{d}} z\,\Big)\,\Big|\,\mathcal F_{(i-1)\Delta_n}\,\Big) \right|.\end{align*}

Hence, using the inequality (5.7) of [Reference Delattre and Jacod6] and the fact that f is bounded, we easily see that

\begin{eqnarray*} \Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}(|L^n_i|\,|\,\mathcal F_{(i-1)\Delta_n}) &\le & C \Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\!\left(\left(\frac{\big|\Delta_n^i Z\big|}{\sqrt{\Delta_n}}\wedge 1\right)\right). \end{eqnarray*}

Finally, Lemma 4.1 of [Reference Jacod10] allows us to conclude the convergence to 0 of the last quantity as $n\to+\infty$ .

(ii) The proof of this part is inspired by the proof of Lemma 4.4(ii) of [Reference Jacod10], in which we replace $\beta_i^n\,:\!=\sigma_{(i-1)\Delta_n}\frac{\Delta_i^n W}{\sqrt{\Delta_n}}$ by $\frac{\Delta_i^n Y}{\sqrt{\Delta_n}}$ , which has no consequences.

Recall the notation (56).

Lemma 6. Assume (LH^′) and (LK-(1,q)), and suppose that one of the following conditions is fulfilled.

1. (i) For each n, $(f_n)$ and its first derivatives are uniformly bounded.

2. (ii) We have $q\in[0,1)$] , and for all $\varepsilon, K, K'>0$ there exists an integer $N>0$ and a real $\varepsilon'>0$ such that

   \begin{align*}{\sup}_{\{x,y\in [0,K],\,u\in[0,1],\, w\in[0,K'],\,|z|\le\varepsilon',\,n\ge N \}}|f_n(x,u,y,w+z) - f_n(x,u,y,w)|\,\le\,\varepsilon.\end{align*}

Then

\begin{align*} R^n\,\stackrel{\mathrm{u.c.p.}}{\to} \,0 \quad \mbox{as}\ n\to+\infty.\end{align*}

Proof. (i) Recall (55) and (71). For any fixed integer $n\ge 1$ , the process $R^n$ is a square-integrable martingale with respect to the filtration $(\mathcal F_{[t/\Delta_n]\Delta_n})_{t\ge 0}$ , and its bracket satisfies

\begin{align*}\langle R^n, R^n\rangle_t \, \le \, \Delta_n\sum_{i=1}^{[t/\Delta_n]}\mathbb{E}\bigl\{ (L^n_i )^2\,\big|\,\mathcal F_{(i-1)\Delta_n}\bigr\},\end{align*}

which is less than

\begin{align*} & \Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\!\left\{\left(\int_{\mathbb{R}} f^{i\,2}_n\left(\frac{y}{\sqrt{\Delta_n}}\right)\, \times\,\left(q_{\Delta_n}(Y_{(i-1)\Delta_n},y-\Delta_i^n Z) - q_{\Delta_n}(Y_{(i-1)\Delta_n},y )\right){\mathrm{d}} y\right)\bigg|\,{\mathcal F}_{(i-1)\Delta_n}\right\}\\ &+ 2{\Delta_n}\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\!\left\{\left(\int_{\mathbb{R}} f^{i\,2}_n\left(\frac{y}{\sqrt{\Delta_n}}\right)\,\left(q_{\Delta_n}(Y_{(i-1)\Delta_n},y) - q_{\Delta_n}(Y_{(i-1)\Delta_n},y-\Delta_i^n Z )\right){\mathrm{d}} y\right)\bigg|\,{\mathcal F}_{(i-1)\Delta_n}\right\}\\ & + 2 {\Delta_n}\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\Biggl\{\biggl(\int_{\mathbb{R}} f_n^i\left(\frac{y+\Delta_i^n Z}{\sqrt{\Delta_n}}\right)\left(f_n^i\left(\frac{y+\Delta_i^n Z}{\sqrt{\Delta_n}}\right) - f_n^i\left(\frac{y}{\sqrt{\Delta_n}}\right)\right)\\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \times q_{\Delta_n}(Y_{(i-1)\Delta_n},y)\,{\mathrm{d}} y\biggr)\,\bigg|\,{\mathcal F}_{(i-1)\Delta_n}\,\Biggr\}.\end{align*}

The two first terms are treated as in the proof of Lemma 5, where $f_n$ is replaced by $f_n^2$ . For the third one, which we call $R^n_t(3)$ , we easily obtain

\begin{align*}|R^n_t(3)| & \le 2 K {\Delta_n}\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\!\left\{\left(\int_{\mathbb{R}}\Big|f_n^i\left(\frac{y+\Delta_i^n Z}{\sqrt{\Delta_n}}\right) - f_n^i\left(\frac{y}{\sqrt{\Delta_n}}\right)\Big|\,q_{\Delta_n}(Y_{(i-1)\Delta_n},y)\,{\mathrm{d}} y\right)\bigg|\,{\mathcal F}_{(i-1)\Delta_n}\right\}\\ & \le K {\Delta_n}\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\!\left\{\left. \left(1\wedge \left|\frac{\Delta_i^n Z}{\sqrt{\Delta_n}}\right|\right)\,\right|\,{\mathcal F}_{(i-1)\Delta_n}\right\}\\ & \le K\left(\sqrt{\Delta_n}\Big( 1+\varepsilon_n^{-1}\Big)+\big(\gamma(\varepsilon_n)\big)^{\frac{1}{2}}+\Delta_n\varepsilon_n^{-2}\Big)\right), \end{align*}

where $\gamma(\varepsilon)=\int_{\{|x|\le \varepsilon\}} x^2\, F({\mathrm{d}} x)$ and we have used Lemma 5.3 of [Reference Jacod10]. Choosing $\varepsilon_n=\Delta_n^{\frac{1}{4}}$ , we get $ R^n_t(3)\stackrel{\mathrm{u.c.p.}}{\to} 0$ . Finally, we conclude that

\begin{align*} R^n_t\,\stackrel{\mathrm{u.c.p.}}{\to} \,0.\end{align*}

(ii) Noting again that $\langle R^n, R^n\rangle_t \le \Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\big((L^n_i)^2\,|\,\mathcal F_{(i-1)\Delta_n}\big)$ , we conclude as in part (ii) of Lemma 5.

Recalling (57), we have the following result.

Lemma 7. Suppose that X satisfies (LH^′- $(\gamma)$ ) with $\gamma\in[0,1)$] and that the functions $(f_n)_{n\ge 1}$ satisfy (LK-(1,q)) and one of the following hypotheses:

1. (i) We have $q\equiv 0$ .

2. (ii) Hypothesis (LL-(p,p^′)) is satisfied with $p\in(\gamma,1)$ and $p'>0$ such that $\frac{\beta^{p'}_n}{\sqrt{\Delta_n}}\to0$ .

3. (iii) We have (LL-(p)) for $p\in(\gamma,1)$ .

Then

(73) \begin{equation} G^n\,\stackrel{\mathrm{u.c.p.}}{\to}\,0\quad\mbox{as}\ n\to+\infty.\end{equation}

Proof. Let $(\varepsilon_n)$ be a sequence converging to zero. We have the decomposition

\begin{align*}\frac{\Delta_i^nX}{\sqrt{\Delta_n}} = \frac{\Delta_i^n Y}{\sqrt{\Delta_n}} +\frac{{Z'}^n_i}{\sqrt{\Delta_n}}+\frac{{Z''}^n_i}{\sqrt{\Delta_n}}\,,\end{align*}

where

(74) \begin{equation} {Z'}^n_i = \frac1{\sqrt{\Delta_n}}\int_{(i-1)\Delta_n}^{i\Delta_n} \int_{\{|x|>\varepsilon_n\}} |x|\,\mu({\mathrm{d}} x,{\mathrm{d}} s)\end{equation}

and

(75) \begin{equation} {Z''}^n_i = \frac1{\sqrt{\Delta_n}}\int_{(i-1)\Delta_n}^{i\Delta_n} \int_{\{|x|\le\varepsilon_n\}} |x|\,\mu({\mathrm{d}} x,{\mathrm{d}} s).\end{equation}

1. (i) Since the $f_n$ are uniformly bounded and the mapping $y\to q_t(x,y)$ is differentiable, we can copy the proofs of Lemmas 6 and 5 to get

   (76) \begin{equation} |G^n_t| \,\le\, K{\Delta_n}\sum_{i=1}^{[t/\Delta_n]}\left(\mathbb{E}\!\left\{\left.\!\bigl[1\wedge|{Z'}^n_i|\big]\,\right|\,{\mathcal F}_{(i-1)\Delta_n}\right\} + \mathbb{E}\!\left\{ |{Z''}_i^n|\:|\,{\mathcal F''}_{(i-1)\Delta_n}\right\} \right). \end{equation}
   Then
   \begin{align*}&\mathbb{E}\!\left\{\int_{(i-1)\Delta_n}^{i\Delta_n} \int_{\{|x|\le\varepsilon_n\}} |x|\,\mu({\mathrm{d}} x,{\mathrm{d}} s) \,\bigg|\,\mathcal F_{(i-1)\Delta_n}\,\right\} = \\ &\quad \mathbb{E}\!\left\{\int_{(i-1)\Delta_n}^{i\Delta_n} \int_{\{|x|\le\varepsilon_n\}} |x|\,\nu({\mathrm{d}} x,{\mathrm{d}} s) \,\bigg|\,\mathcal F_{(i-1)\Delta_n}\,\right\} \\ & \le C\Delta_n \varepsilon_n^{1-\gamma'},\end{align*}
   where $\gamma<\gamma'<1$ . Using inequality (5.9) in Lemma 5.3 of [Reference Jacod10], we have
   (77) \begin{equation} |G^n_t| \,\le\, Kt\bigl( \varepsilon_n^{1-\gamma'} + \sqrt{\Delta_n}\varepsilon_n^{-\gamma'}\bigr).\end{equation}
   Choosing $\varepsilon_n=\Delta_n^a$ for $0 < a < \frac{1}{2\gamma'}$ , we get the result.

2. (ii) Now, if (L(p, p ^′)) is satisfied, we have

   (78) \begin{equation} \sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\!\left\{\left| L^n_i\right|\right\} \le K\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\big\{\,|{Z'}^n_i|^p+|{Z''}^n_i|^p+ \beta_n^{p'}\big\}. \end{equation}
   Next, by inequality (5.8) in Lemma 5.3 of [Reference Jacod10] and the fact that $\gamma<\gamma'$ , we have
   \begin{align*}\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\!\left\{|{Z'}^n_i|^p\right\}\, \le\, Ct\Delta_n^{\frac{1-p}{2}}\varepsilon^{-(\gamma'-p)^+}_n, \end{align*}
   and thus we get the convergence of the last expression to 0 as $n\to+\infty$ provided that $p<1$ and $\gamma'\le p$ .

We also have

\begin{align*}\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]}\left(\mathbb{E}\!\left\{\left|{Z''}^n_i\right|^p\,\right\}\right) \, \le \,Ct \Delta_n^{\frac{p-1}2} \varepsilon_n^{p(1-\gamma')} .\end{align*}

We choose $\varepsilon_n=\Delta_n^a$ with $\frac{1-p}{2p(1-\gamma')} < a < \frac{1-p}{2(\gamma'-p)}$ , which is possible only if $\gamma'<\frac{2p}{1+p}$ . Since if $p\in(0,1)$ then $p<\frac{2p}{1+p}$ , we obtain the convergence to 0.

(iii) The proof of this part is similar to that of (ii).

Lemma 8. Assume (LH^′- $(\gamma)$ ) with $\gamma < 1$ , and define the sequence of measurable functions $g_n\,:\, \mathbb{R}\times [0,1]\to\mathbb{R}$ , for $n\ge 1$ , that are uniformly bounded and converge pointwise to a function g. Finally, fix $a\in \mathbb{R}\setminus \mathbb{Z}$ and let $(a_n)_{n\ge 1}$ be a sequence converging to a. Then

\begin{align*} \mathbb{E}\!\left(\,\left| g_n\left( Z_{\Delta_n}, \left\{ a_n+\frac{Z_{\Delta_n}}{\alpha_n}\right\}\right) - g(0,\{a\}) \right|\, \right)\end{align*}

converges to 0 as $n\to+\infty$ .

Proof. By Lemma 4.1 of [Reference Genon-Catalot14], the process $U_t\,:\!=\frac{Z_t}{t}$ converges in probability to 0 as $t\to 0$ . Next,

(79) \begin{equation}\mathbb{E}\!\left(\,\left| g_n\left( Z_{\Delta_n}, \left\{ a_n+\frac{Z_{\Delta_n}}{\alpha_n}\right\}\right) - g(0,\{a\}) \right|\,\right) = \sum_{i=1}^2 G^n_i,\end{equation}

where

\begin{align*}G^n_1=\!: \,\mathbb{E}\!\left(\left|(g_n(\Delta_n U_{\Delta_n},\{a_n+\sqrt{\Delta_n}U_{\Delta_n}/\beta_n\}) - g(0,\{a\})\, \right|\,1_{\{|U_{\Delta_n}|>\varepsilon\}}\right)\,\le\, C \mathbb{P}(|U_{\Delta_n}|>\varepsilon),\end{align*}

which converges to 0 as $n\to+\infty$ . Then

\begin{align*}G^n_2=\!: \, \mathbb{E}\!\left(\left|g_n(\Delta_n U_{\Delta_n},\{a_n+\sqrt{\Delta_n}U_{\Delta_n}/\beta_n\}) - g(0,\{a\})\,\right|\,1_{\{|U_{\Delta_n}|<\varepsilon\}}\,\right)\end{align*}

clearly converges to 0 for every $\varepsilon>0$ . Thus, taking in (79) the superior limit in n and then the limit as $\varepsilon\to 0$ , we get the result.

5.1. Proof of the law of large numbers

The proof is based on that of Theorem 2.1 in [Reference Delattre and Jacod6] and on an approximation of $V^n(f_n,X,Y)$ by $\overline{V}^n(f_n,X,Y)$ . Using the notation of the previous subsection, we have

\begin{align*}\overline{V}^n(f_n,X, Y) - U^n = \overline{U}^n - U^n + \sqrt{\Delta_n}\; M^n(k_n) - \sqrt{\Delta_n}\;H^n(k_n)\end{align*}

for any $k_n\in\mathbb{N}^*$ . Using, respectively, part (i) of Lemma 2 and the inequality (63) in Lemma 3, we get

\begin{align*}U^n\,\stackrel{\mathrm{u.c.p.}}{\longrightarrow} \,\int_0^\cdot Mf(X_s,Y_s,Y_s)\,{\mathrm{d}} s&\quad\mbox{as }n\to+\infty,\\{\sup}_{s\le t} \left|\overline{U}^n_s - U^n_s\right|\,\longrightarrow\,0&\quad \mbox{as }n\to+\infty.\end{align*}

Now we set $k_n=\bigl[\Delta_n^{\frac{-1}3}\bigr]$ . Then by (70) we have

\begin{align*} \sqrt{\Delta_n}\;|H^n(k_n)_t|\,\le\, \frac{Ct}{\sqrt{k_n}}\,\to\,0\quad\mbox{since}\ k_n\to+\infty\ \mbox{as}\ n\to+\infty.\end{align*}

Next, the processes $(M^n_t(k_n))_{n\ge 1}$ are martingales with respect to the filtration $(\mathcal F_{[t/\Delta_n]\Delta_n})$ . Then, using (67), we have

\begin{align*}\mathbb{E}\!\left({\sup}_{s\le t}|M^n_s(k_n)|^2\right)\,\le\,4\mathbb{E}\!\left(|M^n_t(k_n)|^2\right)\,\le\,4\Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\!\left(|\mu_i^n(k_n)|^2\right)\,\le\,Ct\Delta_n^{\frac{-1}3}.\end{align*}

Hence, for $t\in[0,T_0]$ we have

\begin{align*} \sqrt{\Delta_n} M^n(k_n)\,\stackrel{\mathrm{u.c.p.}}{\longrightarrow}\,0\quad\mbox{as}\ n\to+\infty.\end{align*}

Finally, we end the proof by applying Lemma 5.

5.2. Proof of the central limit theorems

The proofs of the central limit theorems are based on control of the jump parts and on classical theorems on stable convergence of discretized processes to a semimartingale (see [Reference Delattre and Jacod6, Reference Jacod and Shiryaev12] or [Reference Jacod and Protter9] for more details).

5.2.1. Proof of Theorem 4

We always take $t\le T_0$ . Recall the notation of the previous section; then we have

\begin{align*}\frac1{\sqrt{\Delta_n}}\left(V^n(f_n,X,Y)-U^n\right) = \frac1{\sqrt{\Delta_n}}\left(\overline{V}^n(f,X,Y) - U^n\right) + R^n + G^n.\end{align*}

By Lemmas 6 and 7, we have the convergence u.c.p. to 0 of the processes $R^n$ and $G^n$ .

Now we study the processes $ \frac1{\sqrt{\Delta_n}}\bigl(\overline{V}^n(f,X,Y) - U^n\bigr)$ . Recall (71). We remark that for all $k\ge 0$ ,

(80) \begin{equation} \frac1{\sqrt{\Delta_n}}\left(\overline{V}^n(f,X,Y) - U^n\right) = M^n(k) + \tilde{U}^n + \frac{1}{\sqrt{\Delta_n}}\left( \overline{U}^n - U^n - \sqrt{\Delta_n}\tilde{U}^n\right) - H^n(k).\end{equation}

By the inequality (64), we have

(81) \begin{eqnarray}{\sup}_{s\le t} \,\frac1{\sqrt{\Delta_n}}\left|\,\overline{U}^n_s - U^n_s - \sqrt{\Delta_n}\tilde{U}^n_s\,\right| & \le & Ct \sqrt{\Delta_n}, \nonumber\\ & {\rightarrow}& 0\quad\mbox{as}\ n\to+\infty.\end{eqnarray}

Next, from part (i) of Lemma 2 we deduce that

\begin{align*}{\sup}_{s\le t} \,\left|\,\tilde{U}^n_s - \int_0^s\int_0^1\int_{\mathbb{R}} \tilde{f}(X_{s},u, Y_{s},y) h_{\sigma(Y_{s})}(y)\,{\mathrm{d}} y\,{\mathrm{d}} u\,{\mathrm{d}} s \,\right|\,\rightarrow\,0\quad\mbox{as }n\to+\infty.\end{align*}

Now, using (70) under (LK-(3,q)), if $k_n=1\vee[\Delta_n^{-\frac{1}{2}}]$ , then we have ${\sup}_{t\le t} |H^n_t(k_n)|\rightarrow 0 $ as $n\to+\infty$ .

Let us now concentrate on $M^n_t(k_n)$ . Following Section 9 of [Reference Delattre and Jacod6], we divide the rest of the proof into three steps.

Step 1: We consider the process

(82) \begin{equation} C^n_t = \langle M^n_t(k_n), M^n_t(k_n)\rangle = \Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\big\{(\mu_i^n(k_n))^2\,\big|\,\mathcal F_{(i-1)\Delta_n}\big\}.\end{equation}

Let $(k_n')_{n\ge 1}$ be a sequence of integers such that $2\le k_n'\le k_n$ and $k'_n\to+\infty$ . Then

(83) \begin{equation}({\mu}_i^n(k_n))^2-({\mu}_i^n(k'_n))^2=-({\mu}_i^n(k_n)-{\mu}_i^n(k'_n))^2 +2{\mu}_i^n(k_n)({\mu}_i^n(k_n)-{\mu}_i^n(k'_n)).\end{equation}

Then, under (LK-3) and applying (67) and (68), the two sums $\Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\{({\mu}_i^n(k_n))^2|\mathcal F_{(i-1)\Delta_n}\}$ and $\Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\{({\mu}_i^n(k'_n))^2|\mathcal F_{(i-1)\Delta_n}\}$ are $\mathbb{P}$ -equivalent, in the sense that

(84) \begin{equation}\Delta_n\sum_{i=1}^{[t/\Delta_n]}\mathbb{E}\!\left\{\left.\left(({\mu}_i^n(k_n))^2-({\mu}_i^n(k'_n))^2\right)\,\right|\,\mathcal F_{(i-1)\Delta_n}\,\right\}\,\stackrel{\mathbb{P}}{\longrightarrow} \,0\quad\mbox{as}\ n\to+\infty.\end{equation}

Likewise, using Lemma 3, we can replace all the $\overline{m}_nf_n$ by $\overline{m}' f_n$ in the expression of ${\mu}_i^n(k'_n)$ . Next, we replace $p_t(y,x)$ by $h_{\sigma(y)\sqrt{t}}(x-y)$ and $\eta_i^n$ by

(85) \begin{equation} {\eta'}_i^n \, = \, {\xi'}^n_i - \int_{\mathbb{R}}\int_0^1 f_n\!\left(X_{(i-1)\Delta_n},u,Y_{(i-1)\Delta_n},\frac{y}{\sqrt{\Delta_n}}\right)h_{\sigma(Y_{(i-1)\Delta_n})\sqrt{\Delta_n}}(y)\,{\mathrm{d}} y\,{\mathrm{d}} u.\end{equation}

Thus, if we set

(86) \begin{align} {\mu'}_i^n(k'_n) & = {\eta'}_i^n + \overline{m}'f_n(X_{i\Delta_n},Y_{i\Delta_n}) - \overline{m}'f_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n}) \nonumber\\ & \quad +\sum_{j=1}^{k'_n-2} \int_{\mathbb{R}} \int_{\mathbb{R}}\overline{m}'f_n(x+Z_{i\Delta_n}+z,x)h_{\sigma(Y_{i\Delta_n})\sqrt{j\Delta_n}}(x-Y_{i\Delta_n})\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x \nonumber\\ & \quad -\ \sum_{j=1}^{k'_n-1} \int_{\mathbb{R}}\int_{\mathbb{R}} \overline{m}'f_n(x+Z_{(i-1)\Delta_n}+z,x)h_{\sigma(Y_{(i-1)\Delta_n})\sqrt{j\Delta_n}}(x-Y_{(i-1)\Delta_n})\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x, \end{align}

then

(87) \begin{equation} \Delta_n\sum_{i=1}^{[t/\Delta_n]}\left|\,\mathbb{E}\!\left\{\left.\left(({\mu}_i^n(k'_n))^2-({\mu'}_i^n(k'_n))^2\right)\right|\mathcal F_{(i-1)\Delta_n}\right\}\,\right|\,\le\, CT (k'_n)^{\frac{3}{2}}\sqrt{\Delta_n}\left(1+ (k'_n)^{\frac{3}{2}}\sqrt{\Delta_n}\right). \end{equation}

Hence the expressions $\Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\{({\mu}_i^n(k'_n))^2|\mathcal F_{(i-1)\Delta_n}\}$ and $\Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\{({\mu'}_i^n(k'_n))^2|\mathcal F_{(i-1)\Delta_n}\}$ are $\mathbb{P}$ -equivalent provided that $(k'_n)^{\frac{3}{2}}\sqrt{\Delta_n}\to0$ . We can then set $k'_n=1\vee[\Delta_n^{-\frac{1}{4}}]$ for the rest of the proof.

Now we define

(88) \begin{align} & \overline{\mu}_i^n(k'_n) = {\eta'}_i^n + \overline{m}'f_n(X_{(i-1)\Delta_n},X_{i\Delta_n}, Y_{(i-1)\Delta_n}) - \overline{m}'f_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n}) \nonumber \\ &\quad +\sum_{j=1}^{k'_n-2}\int_{\mathbb{R}} \int_{\mathbb{R}} \overline{m}'f_n(x+Z_{(i-1)\Delta_n}+z,x+Z_{i\Delta_n}+z,x)h_{\sigma(Y_{(i-1)\Delta_n})\sqrt{j\Delta_n}}(x-Y_{i\Delta_n})\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x \nonumber \\ &\quad - \sum_{j=1}^{k'_n-1} \int_{\mathbb{R}}\int_{\mathbb{R}} \overline{m}'f_n(x+Z_{(i-1)\Delta_n}+z,x)h_{\sigma(Y_{(i-1)\Delta_n})\sqrt{j\Delta_n}}(x-Y_{(i-1)\Delta_n})\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x. \end{align}

Using the properties of $\sigma$ and of the density of a normal law, we easily see that

\begin{align*}\left|{\mu'}_i^n(k'_n)-\overline{\mu}_i^n(k'_n)\right| \,\le\, Ck'_n\left(|Y_{i\Delta_n}-Y_{(i-1)\Delta_n}| + |Z_{i\Delta_n}-Z_{(i-1)\Delta_n}|\right).\end{align*}

Thus we have

\begin{align*} \Delta_n\sum_{i=1}^{[t/\Delta_n]}\left|\,\mathbb{E}\!\left\{\left({\mu'}_i^n(k'_n))^2-(\overline{\mu}_i^n(k'_n))^2\right)\big|\,\mathcal F_{(i-1)\Delta_n}\right\}\right|\,\le\, Ct k'_n\sqrt{\Delta_n}\left(1+ k'_n\sqrt{\Delta_n}\right). \end{align*}

Hence, we can continue with

(89) \begin{align}&\Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\{(\overline{\mu}_i^n(k'_n))^2|\mathcal F_{(i-1)\Delta_n}\} \nonumber \\& \quad = \Delta_n\,\sum_{i=1}^{[t/\Delta_n]}\int_{\mathbb{R}} \int_{\mathbb{R}} \,\Biggl\{ f_n^i\left(\frac{w}{\sqrt{\Delta_n}}\right)\, -\,Mf_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n})\nonumber \\ & \qquad\qquad\qquad + \overline{m}'f_n(X_{(i-1)\Delta_n},w+z'+X_{(i-1)\Delta_n},w+Y_{(i-1)\Delta_n})\, -\,\overline{m}'f_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n})\nonumber \\ &\qquad\qquad + \sum_{j=1}^{k'_n-2} \int_{\mathbb{R}}\int_{\mathbb{R}}\Big( \overline{m}'f_n(x+z+Z_{(i-1)\Delta_n},x+z+z'+Z_{(i-1)\Delta_n},x) \nonumber \\[-7pt] &\qquad\qquad\qquad\qquad\qquad\times h_{\sigma( Y_{(i-1)\Delta_n})\sqrt{j\Delta_n}}(x-w-Y_{(i-1)\Delta_n})\Big)\,\theta_{j\Delta_n}({\mathrm{d}} z){\mathrm{d}} x \nonumber \\ & \qquad - \sum_{j=1}^{k'_n-1} \int_{\mathbb{R}}\int_{\mathbb{R}} \Big(\overline{m}'f_n(x+Z_{(i-1)\Delta_n}+z,x)\, \times\,h_{\sigma(Y_{(i-1)\Delta_n})\sqrt{j\Delta_n}}(x-Y_{(i-1)\Delta_n})\,\Big)\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x \Bigg\}^2 \nonumber\\ &\qquad\qquad\qquad\qquad\qquad\times q_{\Delta_n}(Y_{(i-1)\Delta_n},w)\,{\mathrm{d}} w\,\theta_{\Delta_n}(dz'). \end{align}

By (4.18) of [Reference Delattre7] and as below, we have the $\mathbb{P}$ -equivalency

\begin{align*}& \Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\{(\overline{\mu}_i^n(k'_n))^2|\mathcal F_{(i-1)\Delta_n}\} \\& \quad \stackrel{\mathbb{P}}{\sim} \, \Delta_n\,\sum_{i=1}^{[t/\Delta_n]}\, \int_{\mathbb{R}} \int_{\mathbb{R}}\,\Biggl\{ f_n^i\left(\frac{w}{\sqrt{\Delta_n}}\right)\, -\,Mf_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n})\\&\qquad\qquad\qquad + \overline{m}'f_n(X_{(i-1)\Delta_n},w+z'+X_{(i-1)\Delta_n},w+Y_{(i-1)\Delta_n})\, -\,\overline{m}'f_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n}) \\& \qquad\qquad + \sum_{j=1}^{k'_n-2} \int_{\mathbb{R}}\int_{\mathbb{R}}\Big( \overline{m}'f_n(x+z+Z_{(i-1)\Delta_n},x+z+z'+Z_{(i-1)\Delta_n},x)\\[-7pt]& \qquad\qquad\qquad\qquad\qquad\times h_{\sigma( Y_{(i-1)\Delta_n})\sqrt{j\Delta_n}}(x-w-Y_{(i-1)\Delta_n})\,\Big)\,\theta_{j\Delta_n}({\mathrm{d}} z) \,{\mathrm{d}} x \\ & \qquad- \sum_{j=1}^{k'_n-1} \int_{\mathbb{R}}\int_{\mathbb{R}}\Big( \overline{m}'f_n(x+Z_{(i-1)\Delta_n}+z,x)\,\times\,h_{\sigma(Y_{(i-1)\Delta_n})\sqrt{j\Delta_n}}(x-Y_{(i-1)\Delta_n})\,\Big)\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x \Biggr\}^2 \\&\qquad\qquad\qquad\qquad\qquad\times h_{\sigma(Y_{(i-1)\Delta_n})\sqrt{\Delta_n}}(w)\,{\mathrm{d}} w\,\theta_{\Delta_n}({\mathrm{d}} z').\end{align*}

Next, we have

\begin{align*}\Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\bigl\{(\overline{\mu}_i^n(k'_n))^2\,\big|\,\mathcal F_{(i-1)\Delta_n}\bigr\} \,\stackrel{\mathbb{P}}{\sim} \, \Delta_n\, \sum_{i=1}^{[t/\Delta_n]} {\Lambda}_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n})\,,\end{align*}

where

(90) \begin{align} & {\Lambda}_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n}) = \int_{\mathbb{R}} \int_{\mathbb{R}}\,\Biggl\{\,f_n^i(w)-Mf_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n}) \nonumber\\ & +\,\overline{m}'f_n(X_{(i-1)\Delta_n},w\sqrt{\Delta_n}+z'+X_{(i-1)\Delta_n},w\sqrt{\Delta_n}+Y_{(i-1)\Delta_n}) - \overline{m}'f_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n})\nonumber\\ & + \sum_{j=1}^{k'_n-2} \int_{\mathbb{R}}\int_{\mathbb{R}}\,\Big(\, \overline{m}'f_n(x\sqrt{\Delta_n}+w\sqrt{\Delta_n}+z+X_{(i-1)\Delta_n},x\sqrt{\Delta_n}+w\sqrt{\Delta_n}+z+z'+X_{(i-1)\Delta_n},\nonumber\\[-7pt] & \qquad\qquad\qquad\qquad\qquad x\sqrt{\Delta_n}+w\sqrt{\Delta_n}+ Y_{(i-1)\Delta_n})\,h_{\sigma( Y_{(i-1)\Delta_n})\sqrt{j}}(x)\,\Big)\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x \nonumber\\ & - \sum_{j=1}^{k'_n-1} \int_{\mathbb{R}}\int_{\mathbb{R}}\,\Big(\, \overline{m}'f_n(x\sqrt{\Delta_n}+X_{(i-1)\Delta_n}+z,x\sqrt{\Delta_n}+Y_{(i-1)\Delta_n})\,h_{\sigma(Y_{(i-1)\Delta_n})\sqrt{j}}(x)\,\Big)\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x \Biggr\}^2\nonumber\\ & \qquad\qquad\qquad\qquad\qquad\times h_{\sigma(Y_{(i-1)\Delta_n})}(w)\,{\mathrm{d}} w\,\theta_{\Delta_n}({\mathrm{d}} z'). \end{align}

Then ${\Lambda}_n$ is a function of $X_{(i-1)\Delta_n}$ , $Y_{(i-1)\Delta_n}$ and fractional parts of the form $\left\{\frac{X_{(i-1)\Delta_n} + a_m^n }{\alpha_n} \right\}$ with $m=1,2, \ldots, k'_n$ . Thus, as in Remark 9.1 of [Reference Delattre and Jacod6], we can use Theorem 1 for the adequate function. Then, for any T such that $0\le T\le T_0$ , we have

\begin{align*}{\sup}_{t\le T}\, \left| \Delta_n \sum_{i=1}^{[t/\Delta_n]} {\Lambda}_n(X_{(i-1)\Delta_n},Y_{(i-1)\Delta_n}) - \int_0^t \Lambda'_n(X_{s},Y_{s} )\,{\mathrm{d}} s\right|\,\,\to\,\,0\end{align*}

in probability as $n\to+\infty,$ where

(91) \begin{align} &\Lambda'_n(X_{s},Y_{s} ) \,:\!=\int_0^1 \int_{\mathbb{R}} \int_{\mathbb{R}}\,\Biggl[ \, f_n\!\left(X_{s},v,Y_{s},w\right) - Mf_n(X_{s},Y_{s})\nonumber\\ &\qquad\qquad +\, \overline{m}f_n\!\left(X_{s},\left\{\frac{w\sqrt{\Delta_n}+z'}{\alpha_n} + v\,\right\},w\sqrt{\Delta_n}+Y_{s}\right) - \overline{m}f_n(X_{s},v,Y_{s}) \nonumber\\ &\quad +\, \sum_{j=1}^{k'_n-2} \int_{\mathbb{R}}\int_{\mathbb{R}} \, \bigg(\,\overline{m}f_n\biggl(x\sqrt{\Delta_n}+w\sqrt{\Delta_n}+z+X_{s},\left\{\frac{x\sqrt{\Delta_n}+w\sqrt{\Delta_n}+z+z'}{\alpha_n} + v\,\right\}, \nonumber\\[-7pt] &\qquad\qquad\qquad\qquad\qquad\quad x\sqrt{\Delta_n}+w\sqrt{\Delta_n}+ Y_{s})\,h_{\sigma( Y_{s})\sqrt{j}}(x)\,\bigg)\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x \nonumber\\ & \quad - \sum_{j=1}^{k'_n-1} \int_{\mathbb{R}}\int_{\mathbb{R}}\bigg(\, \overline{m}f_n\bigg(x\sqrt{\Delta_n}+X_{s}+z,\left\{\frac{x\sqrt{\Delta_n}+z}{\alpha_n}+v\right\},x\sqrt{\Delta_n}+Y_{s}\bigg)h_{\sigma(Y_{s})\sqrt{j}}(x)\bigg)\nonumber\\ &\qquad\qquad\qquad\qquad\qquad \times \theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x \Biggr]^2\,h_{\sigma(Y_{s})}(w)\,{\mathrm{d}} w\,\theta_{\Delta_n}(dz')\,{\mathrm{d}} v. \end{align}

First, assume that $\beta >0$ . Owing to (66), the series in j,

\begin{align*}\left(\int_{\mathbb{R}}\overline{m}' f_n(x\sqrt{\Delta_n}+a,x\sqrt{\Delta_n}+b,x\sqrt{\Delta_n}+c)h_{\sigma( Y_{(i-1)\Delta_n})\sqrt{j}}(x)\right),\end{align*}

converges. Since the rest of a converging series tends to 0, by the Lebesgue dominated convergence theorem, the quantity $\int_0^t|\Lambda'_n(X_{s},Y_{s} ) - \Lambda''_n(X_{s},Y_{s} )|\,{\mathrm{d}} s$ converges in $L^1(\mathbb{P})$ to 0, where

(92) \begin{align} \Lambda''_n(X_{s},Y_{s} ) &\,:\! =\int_0^1 \int_{\mathbb{R}} \int_{\mathbb{R}}\,\Biggl[\, f_n\!\left(X_{s},v,Y_{s},w\right) - Mf_n(X_{s},Y_{s})\nonumber\\ & \qquad\qquad +\, \overline{m}f_n\!\left(X_{s},\left\{\frac{w\sqrt{\Delta_n}+z'}{\alpha_n} + v\,\right\},w\sqrt{\Delta_n}+Y_{s}\right) - \overline{m}f_n(X_{s},v,Y_{s}) \nonumber\\ & \quad +\, \sum_{j=1}^{+\infty} \int_{\mathbb{R}}\int_{\mathbb{R}}\bigg( \overline{m}f_n\bigg(x\sqrt{\Delta_n}+w\sqrt{\Delta_n}+z+X_{s},\left\{\frac{x\sqrt{\Delta_n}+w\sqrt{\Delta_n}+z+z'}{\alpha_n} + v\,\right\},\nonumber\\[-7pt] & \qquad\qquad\qquad\qquad\qquad x\sqrt{\Delta_n}+w\sqrt{\Delta_n}+ Y_{s}\bigg)\,h_{\sigma( Y_{s})\sqrt{j}}(x)\,\Big)\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x \nonumber\\ & \quad - \sum_{j=1}^{+\infty} \int_{\mathbb{R}}\int_{\mathbb{R}}\bigg( \overline{m}f_n\bigg(x\sqrt{\Delta_n}+X_{s}+z,\left\{\frac{x\sqrt{\Delta_n}+z}{\alpha_n}+v\right\},x\sqrt{\Delta_n}+Y_{s}\bigg) \nonumber\\ & \qquad\qquad\qquad\quad\times h_{\sigma(Y_{s})\sqrt{j}}(x)\bigg)\,\theta_{j\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} x \Biggr]^2\,h_{\sigma(Y_{s})}(w)\,{\mathrm{d}} w\,\theta_{\Delta_n}(dz')\,{\mathrm{d}} v. \end{align}

Next, using Lemma 8, we have that $\int_0^t|\Lambda''_n(X_{s},Y_{s} ) - \overline{\Lambda}(f,f)(X_{s},Y_{s} )|\,{\mathrm{d}} s$ converges in $L^1(\mathbb{P})$ to 0, where

\begin{align*}\overline{\Lambda}(f,f)(X_s,Y_s) &\,:\!=\int_0^1 \int_{\mathbb{R}}\,\Biggl[ \,f\!\left(X_{s},v,Y_{s},w\right) - Mf(X_{s},Y_{s})\\ & \qquad\qquad\quad +\, \overline{m}f\bigg(X_{s},\left\{\frac{w}{\beta} + v\,\right\},Y_{s}\bigg) - \overline{m}f(X_{s},v,Y_{s}) \\ & \qquad +\, \sum_{j=1}^{+\infty} \int_{\mathbb{R}}\bigg( \overline{m}f\bigg(X_{s},\left\{\frac{x+w}{\beta} + v\,\right\}, Y_{s}\bigg)h_{\sigma( Y_{s})\sqrt{j}}(x)\bigg)\,{\mathrm{d}} x \\ &\qquad -\, \sum_{j=1}^{+\infty} \int_{\mathbb{R}}\bigg(\overline{m}f\bigg(X_{s},\left\{\frac{x}{\beta}+v\right\},Y_{s}\bigg)h_{\sigma(Y_{s})\sqrt{j}}(x)\bigg)\,{\mathrm{d}} x \Biggr]^2 h_{\sigma(Y_{s})}(w)\,{\mathrm{d}} w\,{\mathrm{d}} v, \end{align*}

Finally, we have that for all $t\in [0,T_0]$ ,

(93) \begin{equation} \Delta_n\sum_{i=1}^{[t/\Delta_n]} \mathbb{E}\big\{(\mu_i^n(k_n))^2\,\big|\,\mathcal F_{(i-1)\Delta_n}\big\}\,\stackrel{\mathrm{ u.c.p.}}{\rightarrow}\,\int_0^t\overline{\Lambda}(f,f)(X_s,Y_s)\,{\mathrm{d}} s\quad\mbox{as}\ n\to +\infty.\end{equation}

Next, suppose that $\beta=0$ . Recall (91). Because of (66) and since $(f_n)$ satisfies (LK-(3,q)), we have that

\begin{align*}\int_0^t|\Lambda'_n(X_{s},Y_{s} ) - \Lambda'''_n(X_{s},Y_{s} )|\,{\mathrm{d}} s\end{align*}

converges in $L^1(\mathbb{P})$ to 0, where

(94) \begin{align} {\Lambda'''}_n(X_{s},Y_{s} ) & \,:\! =\int_0^1 \int_{\mathbb{R}} \int_{\mathbb{R}}\,\Bigg[ f_n\!\left(X_{s},v,Y_{s},w\right) - Mf_n(X_{s},Y_{s})\nonumber\\ & \quad +\, \overline{m}f_n\bigg(X_{s},\left\{\frac{w\sqrt{\Delta_n}+z}{\alpha_n} + v\,\right\},w\sqrt{\Delta_n}+Y_{s}\bigg) - \overline{m}f_n(X_{s},v,Y_{s}) \Bigg]^2 \nonumber\\ & \qquad\qquad\qquad\quad\times h_{\sigma(Y_{s})}(w)\,{\mathrm{d}} w\,\theta_{\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} v. \end{align}

Now, since $\beta_n =\frac{\alpha_n}{\sqrt{\Delta_n}}$ converges to 0, we can once again use Lemma 1 to get

\begin{align*}\int_0^t |\Lambda'''_n(X_{s},Y_{s} ) - \Lambda^{(4)}_n(X_{s},Y_{s} )|\,{\mathrm{d}} s\,\to\, 0\quad \mbox{as } n\to +\infty,\end{align*}

where

(95) \begin{align} \Lambda_n^{(4)}(X_{s},Y_{s} ) & \,:\!=\int_0^1\int_0^1 \int_{\mathbb{R}} \int_{\mathbb{R}}\Bigg[ f_n\!\left(X_{s},v,Y_{s},w\right) - Mf_n(X_{s},Y_{s})\nonumber\\ & \quad +\, \overline{m}f_n\bigg(X_{s},\left\{\frac{z}{\alpha_n} + v + v'\,\right\},w\sqrt{\Delta_n}+Y_{s}\bigg) - \overline{m}f_n(X_{s},v,Y_{s}) \Bigg]^2\nonumber\\ & \qquad\qquad\qquad\quad\times h_{\sigma(Y_{s})}(w)\,{\mathrm{d}} w\,\theta_{\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} v\,{\mathrm{d}} v'. \end{align}

Next, owing to (LK(3,r)) and (LH- $(\gamma)$ ), we still have

\begin{align*}\int_0^t |\Lambda^{(4)}_n(X_{s},Y_{s} ) - \Lambda^{(5)}_n(X_{s},Y_{s} )|\,{\mathrm{d}} s\,\to\, 0\quad\mbox{as }n\to +\infty,\end{align*}

where

(96) \begin{align} \Lambda_n^{(5)}(X_{s},Y_{s} ) & =\int_0^1\int_0^1 \int_{\mathbb{R}} \int_{\mathbb{R}}\Bigg[ f_n\!\left(X_{s},v,Y_{s},w\right) - Mf_n(X_{s},Y_{s})\nonumber\\ & \quad +\, \overline{m}f_n\bigg(X_{s},\left\{\frac{z}{\alpha_n} + v + v'\,\right\}\!,Y_{s}\bigg) - \overline{m}f_n(X_{s},v,Y_{s})\Bigg]^2 \nonumber\\ & \qquad\qquad\qquad\quad\times h_{\sigma(Y_{s})}(w)\,{\mathrm{d}} w\,\theta_{\Delta_n}({\mathrm{d}} z)\,{\mathrm{d}} v\,{\mathrm{d}} v'. \end{align}

After some easy calculations, we find that

\begin{align*} \Lambda_n^{(5)}(X_{s},Y_{s} ) & =M(f^2_n)(X_s,Y_s) + \left(Mf_n(X_s,Y_s)\right)^2 - \int_{[0,1]}\left(mf_n(X_s,v,Y_s)\right)^2\,{\mathrm{d}} v\\ & \quad +\int_{\mathbb{R}\times[0,1]^2}\left(mf_n\!\left(X_s,\left\{v+v'+\frac{z}{\alpha_n}\right\}\!,Y_s\right)\right)^2\,{\mathrm{d}} v\,{\mathrm{d}} v'\,\theta_{\Delta_n}({\mathrm{d}} z)\\ & \quad -\, 2 Mf_n(X_s,Y_s)\int_{\mathbb{R}\times[0,1]^2}\,mf_n\!\left(X_s,\left\{v+v'+\frac{z}{\alpha_n}\right\}\!,Y_s\right)\,{\mathrm{d}} v\,{\mathrm{d}} v'\,\theta_{\Delta_n}({\mathrm{d}} z).\end{align*}

Next, one easily shows that if U is a uniform law on [0,1], then for any other random variable V that is independent of U, we have $\{U+V\}\,\stackrel{\mathcal{L}}{=}\, U$ (see also Lemma 5.2(i) of [Reference Wilms16]). Then, clearly,

\begin{align*}\Lambda_n^{(5)}(X_{s},Y_{s} ) = M(f^2_n)(X_s,Y_s) + \left(Mf_n(X_s,Y_s)\right)^2,\end{align*}

and this converges pointwise to the process

(97) \begin{equation}\overline{\Lambda}'(f,f)(X_{s},Y_{s} ) \, = \, M(f^2)(X_s,Y_s) - \big(Mf(X_s,Y_s)\big)^2.\end{equation}

Finally, we conclude that

(98) \begin{equation}\Delta_n\sum_{i=1}^{[\cdot/\Delta_n]} \mathbb{E}\{(\mu_i^n(k_n))^2|\mathcal F_{(i-1)\Delta_n}\}\,\stackrel{\mathrm{u.c.p.}}{\to}\,\int_0^\cdot \overline{\Lambda}'(f,f)(X_s,Y_s)\,{\mathrm{d}} s\quad\mbox{as}\ n\to +\infty.\end{equation}

Step 2: We study the quantity $ \Delta_n^2 \sum_{i=1}^{[t/\Delta_n]}\mathbb{E}\{({\mu}_i^n(k_n))^4\,|\,\mathcal F_{(i-1)\Delta_n}\}$ . By Corollary 4 or, more precisely, the inequality (67), we easily see that under hypothesis (Lk-(r, q)) with $r\ge 2$ and $q\ge 0$ , the last expression converges pointwise to 0 as $n\to+\infty$ .

Step 3: Now we consider

(99) \begin{equation} B^n_t = \Delta_n \sum_{i=1}^{[t/\Delta_n]}\mathbb{E}\!\left({\mu}_i^n(k_n) \frac{\Delta_i^n W}{\sqrt{\Delta_n}}\,\bigg|\,\mathcal F_{(i-1)\Delta_n}\right).\end{equation}

We set

(100) \begin{equation} B'^n_t = \Delta_n \sum_{i=1}^{[t/\Delta_n]}\mathbb{E}\!\left({\mu}_i^n(k_n) \frac{\Delta_i^n Y}{\sigma(Y_{(i-1)\Delta_n})\sqrt{\Delta_n}}\,\Bigg|\,\mathcal F_{(i-1)\Delta_n}\right).\end{equation}

By the Cauchy–Schwarz inequality and the properties of $\sigma$ , we have that

\begin{align*}& B^n_t-B'^n_t \\ & \le \Delta_n\sum_{i=1}^{[t/\Delta_n]}\left(\mathbb{E}\!\left(\left.({\mu}_i^n(k_n))^2\right|\mathcal F_{(i-1)\Delta_n}\right)\right)^{\frac{1}{2}}\left(\mathbb{E}\!\left(\left.\left(\frac{\Delta_i^n W}{\sqrt{\Delta_n}} - \frac{\Delta_i^n Y}{\sigma(Y_{(i-1)\Delta_n})\sqrt{\Delta_n}} \right)^2\right|\mathcal F_{(i-1)\Delta_n}\right)\right)^{\frac{1}{2}}\\& \le Ct\Delta_n^{\frac{1}{2}}. \end{align*}

Next,

\begin{align*}B'^n_t=\Delta_n\sum_{i=1}^{[t/\Delta_n]}\mathbb{E}\!\left(\left.{\mu}_i^n(k_n)\psi\left(Y_{(i-1)\Delta_n},\frac{\Delta_i^n Y}{\sqrt{\Delta_n}}\right)\,\right|\, \mathcal F_{(i-1)\Delta_n}\right)\!,\end{align*}

where we recall that $\psi(x,u,y,w)=\psi(y,w)=\frac{w}{\sigma(y)}$ .

Following Step 1, we can easily see that $(B'^n_t)$ converges u.c.p. to $\int_0^t \overline{\Lambda}''(f,\psi)(X_s,Y_s)\,{\mathrm{d}} s$ on $[0,T_0]$ , where

(101) \begin{align}\overline{\Lambda}''(f,\psi)(X_s,Y_s) &\,:\! =\int_0^1 \int_{\mathbb{R}}\,\Bigg[\,f\!\left(X_{s},v,Y_{s},w\right) - Mf(X_{s},Y_{s})\nonumber\\ & \qquad\qquad +\, \overline{m}f\bigg(X_{s},\left\{\frac{w}{\beta} + v\,\right\},Y_{s}\bigg) - \overline{m}f(X_{s},v,Y_{s}) \nonumber\\ & \qquad \qquad+ \sum_{j=1}^{+\infty} \int_{\mathbb{R}} \overline{m}f\bigg(X_{s},\left\{\frac{x+w}{\beta} + v\,\right\}, Y_{s}\bigg)h_{\sigma( Y_{s})\sqrt{j}}(x)\,{\mathrm{d}} x \nonumber\\ & \qquad\qquad - \sum_{j=1}^{+\infty} \int_{\mathbb{R}} \overline{m}f\bigg(X_{s},\left\{\frac{x}{\beta}+v\right\},Y_{s}\bigg)h_{\sigma(Y_{s})\sqrt{j}}(x)\,{\mathrm{d}} x \Bigg]\nonumber\\ & \qquad\qquad\qquad\quad\times \left(\frac{w}{\sigma(Y_s)}\right)\,h_{\sigma(Y_{s})}(w)\,{\mathrm{d}} w\,{\mathrm{d}} v \end{align}

when $\beta>0$ .

Next, it is easy to see that if (W, U) are two independent random variables such that $W\sim{\mathcal{N}}(m,a^2)$ and $U\sim{\mathcal{U}}(0,1)$ , then $(W,\{U+W\})\stackrel{\mathcal{L}}{=} (W,U)$ . In addition, since the mapping $w\to w$ is odd, an easy development of $\overline{\Lambda}''(f,\psi)(X_s,Y_s)$ gives

\begin{align*}\overline{\Lambda}''(f,\psi)(X_s,Y_s) = \int_0^1 \int_{\mathbb{R}}\,\Big(\,f\!\left(X_{s},v,Y_{s},w\right)\Big)\Big(\frac{w}{\sigma(Y_s)}\Big)\,h_{\sigma(Y_{s})}(w)\,{\mathrm{d}} w\,{\mathrm{d}} v = M(f\psi)(X_s,Y_s).\end{align*}

When $\beta=0$ , we arrive at the same limit without difficulty.

To prove the conclusions of the theorem, we consider the two parts in turn.

First part: Following [Reference Delattre and Jacod6] (the proof of Theorem 2.2, to be precise), we conclude that the process $M^n(k_n)$ converges in law to

\begin{align*}\int_0^\cdot \sqrt{\overline{\Lambda}(f,f)(X_s,Y_s) - (M(f\psi)(X_s,Y_s,Y_s))^2}\; {\mathrm{d}} B_s + \int_0^\cdot M(f\psi)(X_s,Y_s,Y_s)\,{\mathrm{d}} W_s\end{align*}

when $\beta>0$ , and if $\beta=0$ we replace $\overline{\Lambda}(f,f)(X_s,Y_s)$ by $\overline{\Lambda}'(f,f)(X_s,Y_s)$ .

According to Proposition 3.17 and Lemma 3.31 from Chapter VI of [Reference Jacod and Shiryaev12], both of the processes $\frac1{\sqrt{\Delta_n}}\big(\overline{V}^n(f,X,Y) - U^n\big)$ and $\frac1{\sqrt{\Delta_n}}\left(V(f_n,X,Y) - U^n\right)$ converge in law when $\beta>0$ to

\begin{align*}&\int_0^\cdot \sqrt{\overline{\Lambda}(f,f)(X_s,Y_s) - (M(f\psi)(X_s,Y_s))^2}\;{\mathrm{d}} B_s + \int_0^\cdot \! M(f\psi)(X_s,Y_s)\,{\mathrm{d}} W_s + \int_0^\cdot \! M(\tilde{f})(X_s,Y_s)\,{\mathrm{d}} s\\&=\int_0^\cdot \sqrt{\overline{\Lambda}(f,f)(X_s,Y_s) - (M(f\psi)(X_s,Y_s))^2}\;{\mathrm{d}} B_s \\& \quad + \int_0^\cdot \frac{M(f\psi)(X_s,Y_s)}{\sigma(Y_s)}\,{\mathrm{d}} Y_s + \int_0^\cdot M(fg)(X_s,Y_s)\,{\mathrm{d}} s,\end{align*}

where g is defined in (16). If $\beta=0$ , in the above process we replace $\overline{\Lambda}(f,f)(X_s,Y_s)$ by $\overline{\Lambda}'(f,f)(X_s,Y_s)$ . This completes the proof.

For the process $\frac{1}{\sqrt{\Delta_n}}\big(V^n(f_n,X,Y)_t - \int_0^t M f_n(X_s,Y_s)\,{\mathrm{d}} s\big)$ , we have

\begin{align*} \frac{1}{\sqrt{\Delta_n}}\left(V^n(f_n,X,Y)_t - \int_0^t M f_n(X_s,Y_s)\,{\mathrm{d}} s\right) & = \frac{1}{\sqrt{\Delta_n}}\left(V^n(f_n,X,Y)_t - U^n_t\right)\\ & \quad + \frac{1}{\sqrt{\Delta_n}}\left( U^n_t - \int_0^t M f_n(X_s,Y_s)\,{\mathrm{d}} s\right)\!,\end{align*}

and by Lemmas 2 and 3, we conclude that it converges to the same limit.

Second part: Hence we have the convergence in law. For the ${\mathcal{A}}$ -stable convergence, the proof is a copy of that of Theorem 2.2 in [Reference Delattre and Jacod6].

5.2.2. Proof of Theorem 2

We remark that

(102) \begin{equation} \overline{\zeta}^n_i = \zeta^n_i + L^n_i - \mathbb{E}(L^n_i\,|\,\mathcal F_{(i-1)\Delta_n}).\end{equation}

By Lemma 6,

\begin{align*}\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]} L^n_i - \mathbb{E}(L^n_i|\mathcal F_{(i-1)\Delta_n})\stackrel{\mathrm{u.c.p.}}{\to} 0.\end{align*}

We divide the rest of the proof into three steps.

Step 1: First, the processes $(\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]}\zeta^n_i)_{n\ge 1}$ are martingales with respect to the filtrations $(\mathcal F_{[t/\Delta_n]\Delta_n})$ , and their brackets are given by

\begin{eqnarray*}\left\langle\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]}\zeta^n_i,\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]}\zeta^n_i\right\rangle & = & \Delta_n\sum_{i=1}^{[t/\Delta_n]}\Bigg( m_nf_n^2\left(X_{(i-1)\Delta_n},\left\{\frac{X_{(i-1)\Delta_n}}{\alpha_n}\right\},Y_{(i-1)\Delta_n}\right)\\ & & -\, \left(m_nf_n\!\left(X_{(i-1)\Delta_n},\left\{\frac{X_{(i-1)\Delta_n}}{\alpha_n}\right\},Y_{(i-1)\Delta_n}\right)\right)^2\Bigg).\end{eqnarray*}

By inequality (63) of Lemma 3, as $n\to+\infty$ they are $\mathbb{P}$ -equivalent to

\begin{align*}&\Delta_n\sum_{i=1}^{[t/\Delta_n]}\!\left(mf_n^2\left(X_{(i-1)\Delta_n},\left\{\frac{X_{(i-1)\Delta_n}}{\alpha_n}\right\},Y_{(i-1)\Delta_n}\right) \right.\\ &\quad- \left. \left(mf_n\!\left(X_{(i-1)\Delta_n},\left\{\frac{X_{(i-1)\Delta_n}}{\alpha_n}\right\},Y_{(i-1)\Delta_n}\right)\right)^2\right). \end{align*}

Then, using Theorem 1, we conclude that

\begin{align*}\left\langle\sqrt{\Delta_n}\sum_{i=1}^{[\cdot/\Delta_n]}\zeta^n_i,\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]}\zeta^n_i\right\rangle\,\stackrel{\mathrm{u.c.p.}}{\rightarrow}\,\int_0^\cdot \int_0^1 \left(mf^2\left(X_{s},u,Y_{s}\right) - \left(mf\!\left(X_s,u,Y_s\right)\right)^2\right){\mathrm{d}} u\,{\mathrm{d}} s.\end{align*}

Step 2: Consider the process $\Delta_n^2\sum_{i=1}^{[t/\Delta_n]}\mathbb{E}((\zeta_i^n)^4\,|\mathcal F_{(i-1)\Delta_n})$ . It is clearly smaller than $Ct\Delta_n$ and thus converges u.c.p. to 0.

Step 3: We study the process

\begin{align*}\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]}\mathbb{E}\!\left(\left.{\xi'}_i^n\,\right|\mathcal F_{(i-1)\Delta_n}\right),\end{align*}

which is equal to

(103) \begin{equation}\begin{split} & \sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]}\mathbb{E}\!\left(\left.{\xi'}_i^n\times\left(\frac{\Delta_i^n Y}{\sigma(Y_{(i-1)\Delta_n})}\right)\,\right|\mathcal F_{(i-1)\Delta_n}\right)\\& + \sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]}\mathbb{E}\!\left(\left.{\xi'}_i^n\times\left(\Delta_i^n W-\frac{\Delta_i^n Y}{\sigma(Y_{(i-1)\Delta_n})}\right)\,\right|\mathcal F_{(i-1)\Delta_n}\right).\end{split}\end{equation}

Note that

\begin{align*}\Delta_i^n W-\frac{\Delta_i^n Y}{\sigma(Y_{(i-1)\Delta_n}} = \frac{1}{\sigma(Y_{(i-1)\Delta_n})}\left(\,-\int_{(i-1)\Delta_n}^{i\Delta_n}\!\! b(Y_s)\,{\mathrm{d}} s +\! \int_{(i-1)\Delta_n}^{i\Delta_n} \!\!(\sigma(Y_{(i-1)\Delta_n})-\sigma(Y_s))\,{\mathrm{d}} W_s\right).\end{align*}

By the Cauchy–Schwarz inequality, and since $\sigma$ is $C^1_b$ , we have

\begin{align*}\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]}\mathbb{E}\!\left(\left. \left|\,{\xi'}_i^n\times\left(\Delta_i^n W-\frac{\Delta_i^n Y}{\sigma(Y_{(i-1)\Delta_n})}\right)\!\right|\;\right|\mathcal F_{(i-1)\Delta_n}\right)\,\le\,\sqrt{\Delta_n}\sum_{i=1}^{[t/\Delta_n]} C\Delta_n\,\le\,Ct\sqrt{\Delta_n}. \end{align*}

Thus, we get the convergence u.c.p. to 0 of the last process in (103).

Concerning the first process in (103), as in Step 1, when $n\to+\infty$ it is $\mathbb{P}$ -equivalent to

\begin{align*}\Delta_n\sum_{i=1}^{[t/\Delta_n]}\int_{\mathbb{R}} f_n^i\left(y\right)\times\frac{y}{\sigma(Y_{(i-1)\Delta_n})}\,h_{\sigma(Y_{(i-1)\Delta_n})}(y)\,{\mathrm{d}} y.\end{align*}

Thus, by Theorem 1, this process converges u.c.p. to

\begin{align*}\int_0^\cdot \,\int_0^1\,\int_{\mathbb{R}} \left(f(X_s,u,Y_s,y)\times \frac{y}{\sigma(Y_{s})}\times h_{\sigma(Y_{s})}(y)\right)\,{\mathrm{d}} y\,{\mathrm{d}} u\,{\mathrm{d}} s.\end{align*}

The conclusion of the proof is identical to that of the proof of Theorem 4.