Appendix
A modified stochastic differential equation
In Remark 5, we claim the existence of a modified stochastic differential equation (SDE) whose solution is well approximated by the RTS-RK method. Let us denote by \({\widetilde{f}}\) the function defining the modified equation corresponding to the numerical flow \(\Psi _h\) truncated after l terms, i.e.
$$\begin{aligned} {\widetilde{f}}(y) = f(y) + h^q f_{q+1}(y) + h^{q+1} f_{q+2}(y) + \cdots + h^l f_{l+1}(y). \end{aligned}$$
Details about the construction of such a function can be found in Sect. 7.2. In particular, analyticity of the function f is needed for a rigorous backward error analysis to hold. Therefore, we will refer in this section to Assumption 5 (see Sect. 7.2). For the additive noise method presented in Conrad et al. (2017), the authors consider the SDE
$$\begin{aligned} \,\mathrm {d}Y = {\widetilde{f}}(Y) \,\mathrm {d}t + \sqrt{Q h^{2p}} \,\mathrm {d}W, \end{aligned}$$
(51)
where W is a d-dimensional standard Brownian motion. It is possible to show (Conrad et al. 2017, Theorem 2.4) that the solution of (51) satisfies
$$\begin{aligned} {|}{{\mathbb {E}}}\big (\Phi (Y_N) - \Phi (Y(T)) \mid Y_0 = y \big ){|} \le Ch^{2p}, \end{aligned}$$
where \(T = Nh\) and \(Y_N\) is the numerical solution given by the additive noise method after N steps. Here, we present a similar construction for the RTS-RK method. In particular, let us consider the modified SDE
$$\begin{aligned} \,\mathrm {d}{\widetilde{Y}}= & {} \Big ({\widetilde{f}}({\widetilde{Y}}) + \frac{1}{2}Ch^{2p}\partial _{tt}\Psi _h({\widetilde{Y}})\Big ) \,\mathrm {d}t \nonumber \\&\quad +\, \sqrt{Ch^{2p}\partial _t \Psi _h({\widetilde{Y}})\partial _t \Psi _h({\widetilde{Y}})^\top } \,\mathrm {d}W, \end{aligned}$$
(52)
where C is given in Assumption 1(iii). Let us denote by \({\widetilde{{\mathcal {L}}}}\) the generator of (52), which can be written explicitly as
$$\begin{aligned} {\widetilde{{\mathcal {L}}}} = \Big ({\widetilde{f}} + \frac{1}{2}Ch^{2p}\partial _{tt}\Psi _h\Big ) \cdot \nabla + \frac{1}{2}Ch^{2p}\partial _t \Psi _h \partial _t \Psi _h^\top : \nabla ^2, \end{aligned}$$
and, adopting the semi-group notation, it satisfies
$$\begin{aligned} {{\mathbb {E}}}\big (\Phi ({\widetilde{Y}}(h))\mid {\widetilde{Y}}(0) = y\big ) = \hbox {e}^{h{\widetilde{{\mathcal {L}}}}}\Phi (y). \end{aligned}$$
In the following lemma, we consider the error over one step between the numerical solution given by the RTS-RK method and the solution of (52) in the weak sense. The proof is inspired by the calculations presented in Conrad et al. (2017, Section 2.4).
Lemma 9
Under the assumptions of Lemma 1 and if Assumption 5 holds, then
$$\begin{aligned} {|}{{\mathbb {E}}}\big (\Phi (Y_1)- \Phi ({\widetilde{Y}}(h))\mid Y_0 = y\big ){|} \le C h^{2p+1}, \end{aligned}$$
where C is a positive constant independent of h and of y, \({\widetilde{Y}}\) is the solution of (52) and \(Y_1\) is the numerical solution given by the RTS-RK method after one step.
Proof
Let us consider the modified ODE
$$\begin{aligned} {\widehat{y}}'(t) = {\widetilde{f}}({\widehat{y}}), \end{aligned}$$
(53)
and denote its flow as \({\widehat{\varphi }}_t\). The generator \({\widehat{{\mathcal {L}}}} = {\widetilde{f}} \cdot \nabla \) satisfies, adopting the semi-group notation,
$$\begin{aligned} \Phi ({\widehat{\varphi }}_h(y)) = \hbox {e}^{h{\widehat{{\mathcal {L}}}}}\Phi (y). \end{aligned}$$
We can now compute the distance between the solution to (52) and (53) as
$$\begin{aligned} \begin{aligned}&\hbox {e}^{h {\widetilde{{\mathcal {L}}}}}\Phi (y)- \hbox {e}^{h {\widehat{{\mathcal {L}}}}}\Phi (y) = \hbox {e}^{h{\widetilde{f}} \cdot \nabla }\\&\quad \Big (\hbox {e}^{\frac{1}{2}Ch^{{2p+1}}\partial _{tt}\Psi _h \cdot \nabla + \frac{1}{2}Ch^{{2p+1}}\partial _t \Psi _h \partial _t \Psi _h^\top : \nabla ^2} - I\Big ) \Phi (y)\\&\quad = \big (1 + {\mathcal {O}}(h)\big )\Big (\frac{1}{2}Ch^{{2p+1}}\partial _{tt}\Psi _h \cdot \nabla + \frac{1}{2}Ch^{{2p+1}}\partial _t \Psi _h \partial _t \Psi _h^\top : \nabla ^2 \\&\qquad + {\mathcal {O}}\big (h^{{4p+1}}\big )\Big )\Phi (y) \\&\quad = \frac{1}{2}Ch^{2p+1}\partial _{tt}\Psi _h \cdot \nabla \Phi (y) + \frac{1}{2}Ch^{2p+1}\partial _t \Psi _h \partial _t \Psi _h^\top : \nabla ^2 \Phi (y) \\&\qquad + {\mathcal {O}}\big (h^{{4p+1}}\big ). \end{aligned} \end{aligned}$$
Let us recall that equation (10) gives
$$\begin{aligned} \begin{aligned}&\hbox {e}^{h{\mathcal {L}}_h}\Phi (y) - \Phi (\Psi _h(y)) = \frac{1}{2} Ch^{2p+1}\partial _{tt}\Psi _h(y) \cdot \nabla \Phi (y)\\&\quad +\frac{1}{2}Ch^{2p+1}\partial _t \Psi _h(y) \partial _t \Psi _h(y)^\top :\nabla ^2\Phi (y) + {\mathcal {O}}(h^{2p+1}), \end{aligned} \end{aligned}$$
which implies that
$$\begin{aligned} \hbox {e}^{h{\widetilde{{\mathcal {L}}}}}\Phi (y)- \hbox {e}^{h{\mathcal {L}}_h}\Phi (y) = \hbox {e}^{h{\widehat{{\mathcal {L}}}}}\Phi (y) - \Phi (\Psi _h(y)) + {\mathcal {O}}(h^{2p+1}). \end{aligned}$$
Now, the theory of backward error analysis (see Sect. 7.2 or e.g. Hairer et al. 2006, Chapter IX) guarantees that
$$\begin{aligned} \hbox {e}^{h{\widehat{{\mathcal {L}}}}}\Phi (y) - \Phi (\Psi _h(y)) = {\mathcal {O}}(h^{q+l+2}). \end{aligned}$$
Choosing \(l = 2p - q - 1\), we have therefore
$$\begin{aligned} \hbox {e}^{h{\widetilde{{\mathcal {L}}}}}\Phi (y)- \hbox {e}^{h{\mathcal {L}}_h}\Phi (y) = {\mathcal {O}}(h^{2p+1}), \end{aligned}$$
which is the desired result. \(\square \)
The error can be then propagated to final time as in Theorem 1, as presented in the following theorem.
Theorem 7
Under the assumptions of Lemma 9 and Theorem 1, and if there exists a constant \(L > 0\) independent of h such that for all \(\Phi \in {\mathcal {C}}^\infty _b({\mathbb {R}}^d, {\mathbb {R}})\)
$$\begin{aligned} \sup _{u\in {\mathbb {R}}^d} {|}\hbox {e}^{h{\widetilde{{\mathcal {L}}}}}\Phi (u){|} \le (1 + Lh)\sup _{u\in {\mathbb {R}}^d}{|}\Phi (u){|}, \end{aligned}$$
then it holds
$$\begin{aligned} {|}{{\mathbb {E}}}\big (\Phi (Y_N)- \Phi ({\widetilde{Y}}(T))\mid Y_0 = y\big ){|} \le Ch^{2p}, \end{aligned}$$
where \(T = Nh\) and C is a positive constant independent of h and of y, \({\widetilde{Y}}\) is the solution of (52) and \(Y_N\) is the numerical solution given by the RTS-RK method after N steps.
Proof
The proof follows by replacing \({\mathcal {L}}\) with \(\widetilde{{\mathcal {L}}}\) and Lemma 1 with Lemma 9 in the proof of Theorem 1. \(\square \)
Proof of Lemma 6
In the following, we denote by \(\llbracket a, b \rrbracket \) the interval \(\llbracket a, b \rrbracket = [a, b]\) if \(a < b\) and \(\llbracket a, b \rrbracket = [b, a]\) if \(a \ge b\). Let us first consider \(r \ge 2\) and the function \(\gamma _r(x) = x^r \hbox {e}^{-r\kappa /x}\), whose first derivative is given by
$$\begin{aligned} \gamma _r'(x) = rx^{r-2}(x + \kappa ) \hbox {e}^{-r\kappa /x}. \end{aligned}$$
Under Assumption 6, we have that \(H_j \le Mh\) almost surely, and hence for any \(t \in \llbracket h, H_j\rrbracket \)
$$\begin{aligned} {|}\gamma _r'(t){|} \le r (Mh)^{r-2} (Mh + \kappa )\hbox {e}^{-r\kappa /(Mh)}, \end{aligned}$$
where we exploited that \(\hbox {e}^{-r\kappa /x}\) is a growing function of x. The fundamental theorem of calculus gives
$$\begin{aligned} \begin{aligned} {|}\gamma _r(H_j){|}&= \Big |\gamma _r(h) + \int _{h}^{H_j} \gamma _r'(t) \,\mathrm {d}t \,\Big |\\&\le \gamma _r(h) + r (Mh)^{r-2} (Mh + \kappa )\hbox {e}^{-r\kappa /(Mh)} {|}H_j - h{|}, \quad \text {almost surely}. \end{aligned} \end{aligned}$$
Taking expectation on both sides and since by (33) it holds \({|}\eta _j{|}^r \le C\gamma _r(H_j)\), we obtain
$$\begin{aligned} {{\mathbb {E}}}{|}\eta _j{|}^r \le C\left( \gamma _r(h) + r M^{r-2} h^{p+r-3/2} (Mh + \kappa )\hbox {e}^{-r\kappa /(Mh)}\right) , \end{aligned}$$
which proves the desired inequality. This is because Assumption 6 and Assumption 1(ii) imply that \(M \ge 1\), and because Mh can be bounded by M. Let us now consider \(r = 1\). In this case, we have for \(t \in \llbracket h, H_j\rrbracket \)
$$\begin{aligned} {|}\gamma _1'(t){|} \le (mh)^{-1} (Mh + \kappa )\hbox {e}^{-\kappa /(Mh)}, \quad \text {almost surely.} \end{aligned}$$
Hence, we apply the same reasoning as above and obtain almost surely
$$\begin{aligned} {|}\gamma _1(H_j){|} \le \gamma _1(h) + (mh)^{-1} (Mh + \kappa ) \hbox {e}^{-\kappa /(Mh)} {|}H_j - h{|}, \end{aligned}$$
which implies the desired result by proceeding as above. \(\square \)
Proof of Lemma 7
We first expand the square as
$$\begin{aligned} \begin{aligned}&\left( \sum _{j=0}^{n-1} \left( \sum _{k=q}^{N-1} a_{jk} + b_j\right) \right) ^2 = \sum _{j=0}^{n-1} \left( \sum _{k=q}^{N-1} a_{jk} + b_j\right) ^2 \\&\quad + 2 \sum _{j=1}^{n-1} \sum _{i=0}^{j-1}\left( \sum _{k=q}^{N-1} a_{jk} + b_j\right) \left( \sum _{k=q}^{N-1} a_{ik} + b_i\right) . \end{aligned} \end{aligned}$$
(54)
Then, we expand the square in the first sum and obtain
$$\begin{aligned} \begin{aligned}&\left( \sum _{k=q}^{N-1} a_{jk} + b_j\right) ^2 = \left( \sum _{k=q}^{N-1} a_{jk}\right) ^2 + b_j^2 + 2 b_j \sum _{k=q}^{N-1} a_{jk}\\&\qquad = \sum _{k=q}^{N-1} a_{jk}^2 + 2\sum _{k=q+1}^{N-1}\sum _{l=q}^{k-1} a_{jk} a_{jl} + b_j^2 + 2 b_j \sum _{k=q}^{N-1} a_{jk}\\&\qquad = a_{jq}^2 + \sum _{k=q+1}^{N-1} a_{jk}^2 + 2\sum _{k=q+1}^{N-1}\sum _{l=q}^{k-1} a_{jk} a_{jl} + b_j^2 + 2 b_j \sum _{k=q}^{N-1} a_{jk}. \end{aligned} \end{aligned}$$
(55)
We then rewrite the term appearing in the double sum in (54) as
$$\begin{aligned} \begin{aligned}&\left( \sum _{k=q}^{N-1} a_{jk} + b_j\right) \left( \sum _{k=q}^{N-1} a_{ik} + b_i\right) = a_{jq}a_{iq} + \sum _{k=q}^{N-1}\sum _{\begin{array}{c} l=q \\ l+k > 2q \end{array}}^{N-1} a_{jk}a_{il} \\&\qquad + b_j \sum _{k=q}^{N-1} a_{ik} + b_i \sum _{k=q}^{N-1} a_{jk} + b_i b_j. \end{aligned} \end{aligned}$$
(56)
Substituting expressions (55) and (56) in (54), we finally get
$$\begin{aligned}&\left( \sum _{j=0}^{n-1} \left( \sum _{k=q}^{N-1} a_{jk} + b_j\right) \right) ^2 = \sum _{j=0}^{n-1} a_{jq}^2\\&\qquad + 2 \sum _{j=1}^{n-1} \sum _{i=0}^{j-1} a_{jq}a_{iq} + R(a) + S(a, b), \end{aligned}$$
where the remainder R(a) can be written as \(R = R_1 + R_2 + R_3\) where
$$\begin{aligned} \begin{aligned}&R_1(a) = \sum _{j=0}^{n-1}\sum _{k=q+1}^{N-1} a_{jk}^2, \\&R_2(a) = 2 \sum _{j=0}^{n-1}\sum _{k=q+1}^{N-1}\sum _{l=q}^{k-1} a_{jk} a_{jl},\\&R_3(a) = 2 \sum _{j=1}^{n-1} \sum _{i=0}^{j-1} \sum _{k=q}^{N-1}\sum _{\begin{array}{c} l=q \\ l+k > 2q \end{array}}^{N-1} a_{jk}a_{il}, \end{aligned} \end{aligned}$$
and the remainder S(a, b) can be written as \(S = S_1 + S_2 + S_3 + S_4\) where
$$\begin{aligned}&S_1(a, b) = \sum _{j=0}^{n-1} b_j^2, \\&S_2(a, b) = 2\sum _{j=1}^{n-1} \sum _{i=0}^{j-1} b_i b_j, \\&S_3(a, b) = 2\sum _{j=1}^{n-1} \sum _{k=q}^{N-1} b_j a_{jk}, \\&S_4(a, b) = 2\sum _{j=1}^{n-1}\sum _{i=0}^{n-1}\left( b_j \sum _{k=q}^{N-1} a_{ik} + b_i \sum _{k=q}^{N-1} a_{jk}\right) .\\ \end{aligned}$$
which proves the desired result. \(\square \)
Proof of Lemma 8
In the following, all the constants are independent of h and n, but can depend on N and q. Moreover, since \(h < 1\), we often apply \(h^r \le h^s\) for \(r \ge s\). We first notice that, under Assumption 2 and Assumption 5, we get for all \(j = 0, \ldots , n-1\) and \(k = q, \ldots , N-1\)
$$\begin{aligned} \begin{aligned} {|}\Delta _{j,k}{|}&= {|}Q_{k+1}(Y_j) - Q_{k+1}(Y_{j+1}){|}\\&\le C {{\Vert }\Psi _0(Y_j) - \Psi _{H_j}(Y_j){\Vert }}\\&\le C_\Delta {|}H_j{|}, \end{aligned} \end{aligned}$$
(57)
almost surely and where \(C_\Delta \) is independent of h. Above, we exploited that \(Q_{k+1}\) is Lipschitz continuous for all \(k=q, \ldots , N+1\) due to Assumption 5. Let us now consider \(R(\Delta )\). Due to (57) and to Assumption 6, we have
$$\begin{aligned} \begin{aligned} {{\mathbb {E}}}(H_j^k - h^k)^2 \Delta _{j,k}^2&\le {C_\Delta ^2}{{\mathbb {E}}}(H_j^{k+1} - H_j h^k)^2\\&= {C_\Delta ^2}\big (h^{2(k+1)} + C_{2(k+1)}h^{2p+2(k+1)-1} \\&+ h^{2(k+1)} + C_2 h^{2p+2k+1} \\&\quad - 2h^{2k+2} - 2C_{k+2}h^{2p+2k+1}\big )\\&= {C_\Delta ^2}\big ((C_{2(k+1)} + C_2 - 2C_{k+2})h^{2p+2k+1}\big )\\&\le C h^{2p+2k+1}, \end{aligned} \end{aligned}$$
(58)
where \(C > 0\) is a positive constant. Now, since \(k \ge q+1\), we get
$$\begin{aligned} {{\mathbb {E}}}(H_j^k - h^k)^2 \Delta _{j,k}^2 \le C h^{2(p+q+1)}. \end{aligned}$$
Hence, for \(R_1(\Delta )\) there exists a constant \({\widetilde{C}}_1\) such that
$$\begin{aligned} {{\mathbb {E}}}R_1(\Delta ) \le {\widetilde{C}}_1 n h^{2(p+q+1)}. \end{aligned}$$
We now proceed to the second remainder \(R_2(\Delta )\). Applying the Cauchy–Schwarz inequality and (58), we get
$$\begin{aligned} \begin{aligned}&{{\mathbb {E}}}\left( (H_j^k - h^k)\Delta _{j,k}(H_j^l - h^l) \Delta _{j,l}\right) \le \Big ({{\mathbb {E}}}\big ((H_j^k - h^k)^2\Delta _{j,k}^2\Big )^{1/2}\\&\qquad \Big ({{\mathbb {E}}}\big ((H_j^l - h^l)^2\Delta _{j,l}^2\Big )^{1/2}\\&\quad \le C h^{2p+k+l+1}, \end{aligned} \end{aligned}$$
where \(C > 0\) is a positive constant. Now, since in the definition of \(R_2(a)\) in (57) we have \(k \ge q+1\) and \(l \ge q\), we have here \(k+l \ge 2q+1\). Therefore, there exists a constant \({\widetilde{C}}_2\) such that
$$\begin{aligned} {{\mathbb {E}}}R_2(\Delta ) \le {\widetilde{C}}_2 n h^{2(p+q+1)}. \end{aligned}$$
We now consider the term \(R_3(\Delta )\). Since \(H_i\) and \(H_j\) are independent for \(i \ne j\), we have
$$\begin{aligned} \begin{aligned}&{{\mathbb {E}}}\big ((H_j^k - h^k)\Delta _{j,k}(H_i^l - h^l) \Delta _{i,l}\big ) \\&\quad = {{\mathbb {E}}}(H_j^k - h^k)\Delta _{j,k} {{\mathbb {E}}}(H_i^l - h^l) \Delta _{i,l}. \end{aligned} \end{aligned}$$
Computing the two factors singularly, we have due to (57) and to Assumption 6
$$\begin{aligned} \begin{aligned} {{\mathbb {E}}}(H_j^k - h^k)\Delta _{j,k}&\le C_\Delta {{\mathbb {E}}}(H_j^{k+1} - H_jh^k) \\&= C_\Delta C_{k+1} h^{2p+k}, \end{aligned} \end{aligned}$$
(59)
and analogously for \({{\mathbb {E}}}(H_i^l - h^l) \Delta _{i,l}\). Then, since \(k+l \ge 2q+1\)
$$\begin{aligned}&{{\mathbb {E}}}\big ((H_j^k - h^k)\Delta _{j,k}(H_i^l - h^l) \Delta _{i,l}\big )\nonumber \\&\quad \le C_\Delta ^2 C_{k+1} C_{l+1} h^{2(2p+q+1/2)}. \end{aligned}$$
(60)
Hence, we have for a constant \({\widetilde{C}}_3 > 0\)
$$\begin{aligned} {{\mathbb {E}}}R_3(\Delta ) \le {\widetilde{C}}_3 n^2 h^{2(2p+q+1/2)}. \end{aligned}$$
Finally, replacing \(t_n = nh\), we can write for a constant \(C > 0\)
$$\begin{aligned} \begin{aligned} {{\mathbb {E}}}R(\Delta )&\le ({\widetilde{C}}_1 + {\widetilde{C}}_2) n h^{2(p+q+1)} + {\widetilde{C}}_3 n^2 h^{2(2p+q+1/2)}\\&= ({\widetilde{C}}_1 + {\widetilde{C}}_2) t_n h^{2(p + q + 1/2)} + {\widetilde{C}}_3 t_n^2 h^{2(2p + q - 1/2)}. \end{aligned} \end{aligned}$$
Let us now consider \(S(\Delta , \eta )\). First, we notice that under the assumption \(p \ge 3/2\) we have for any \(r \ge 1\), \(\min \{r, p+r-3/2\} = r\), and therefore Lemma 6 simplifies to
$$\begin{aligned} {{\mathbb {E}}}{|}\eta _j{|}^r \le Ch^r \hbox {e}^{-r\kappa /(Mh)}. \end{aligned}$$
We first consider \(S_1(\Delta , \eta )\). Applying Lemma 6 with \(r = 2\), we obtain for a constant \({\widehat{C}}_1 > 0\)
$$\begin{aligned} {{\mathbb {E}}}S_1(\Delta , \eta ) \le {\widehat{C}}_1 n h^2 \hbox {e}^{-2\kappa /(Mh)}. \end{aligned}$$
For the second term \(S_2(\Delta , \eta )\), we have by (33) that \({|}\eta _i{|}\le CH^i \hbox {e}^{-\kappa /H_i}\) and \(\eta _j\le CH^j \hbox {e}^{-\kappa /H_j}\) almost surely. These two bounds are independent for \(i \ne j\), and therefore, applying Lemma 6 with \(r = 1\), we have for a constant \({\widehat{C}}_2 > 0\)
$$\begin{aligned} {{\mathbb {E}}}S_2(\Delta , \eta ) \le {\widehat{C}}_2 n^2 h^2 \hbox {e}^{-2\kappa /(Mh)}. \end{aligned}$$
We now consider the third remainder \(S_3(\Delta , \eta )\). Applying the Cauchy–Schwarz inequality, we obtain
$$\begin{aligned} {{\mathbb {E}}}\eta _j (H_j^k - h^k) \Delta _{j,k} \le ({{\mathbb {E}}}\eta _j^2)^{1/2} ({{\mathbb {E}}}(H_j^k - h^k)^2 \Delta _{j,k}^2)^{1/2}. \end{aligned}$$
Applying Lemma 6 with \(r = 2\) to the first factor and (58) to the second, we get
$$\begin{aligned} \begin{aligned} {{\mathbb {E}}}\eta _j (H_j^k - h^k) \Delta _{j,k}&\le C h \hbox {e}^{-\kappa /(Mh)}h^{p+k+1/2}\\&= C h^{p+k+3/2}\hbox {e}^{-\kappa /(Mh)} \end{aligned} \end{aligned}$$
Now, since \(k \ge q\), we have for a constant \({\widehat{C}}_3 > 0\)
$$\begin{aligned} {{\mathbb {E}}}S_3(\Delta , \eta ) \le {\widehat{C}}_3 n h^{p+q+3/2}\hbox {e}^{-\kappa /(Mh)}. \end{aligned}$$
Finally, we consider the last term \(S_4(\Delta , \eta )\). Since by (33) it holds \({|}\eta _j{|} \le CH_j \hbox {e}^{-\kappa /H_j}\) almost surely, and this bound is independent of \(H_i\) for \(i \ne j\), applying (59) and Lemma 6 we have
$$\begin{aligned} \begin{aligned} {{\mathbb {E}}}\eta _j(H_i^k - h^k)\Delta _{i,k}&= {{\mathbb {E}}}\eta _j {{\mathbb {E}}}(H_i^k - h^k)\Delta _{i,k}\\&\le Ch \hbox {e}^{-\kappa /(Mh)}h^{2p+k}, \end{aligned} \end{aligned}$$
which, since \(k \ge q\), implies that there exists a constant \({\widehat{C}}_4 > 0\) such that
$$\begin{aligned} {{\mathbb {E}}}S_4(\Delta , \eta ) \le {\widehat{C}}_4 n^2 h^{2p+q+1}\hbox {e}^{-\kappa /(Mh)}. \end{aligned}$$
Finally, replacing \(t_n = nh\), we can write
$$\begin{aligned} \begin{aligned}&{{\mathbb {E}}}S(\Delta , \eta ) \le ({\widehat{C}}_1 n h^2 + {\widehat{C}}_2 n^2 h^2) \hbox {e}^{-2\kappa /(Mh)} \\&\qquad + ({\widehat{C}}_3 n h^{p+q+3/2} + {\widehat{C}}_4 n^2 h^{2p+q+1})\hbox {e}^{-\kappa /(Mh)}\\&\quad = ({\widehat{C}}_1 t_n h + {\widehat{C}}_2 t_n^2) \hbox {e}^{-2\kappa /(Mh)} \\&\qquad + ({\widehat{C}}_3 t_n h^{p+q+1/2} + {\widehat{C}}_4 t_n^2 h^{2p+q-1})\hbox {e}^{-\kappa /(Mh)}, \end{aligned} \end{aligned}$$
which completes the proof. \(\square \)