Appendix A: Some More Proofs
Proof of Lemma 2.23
Introduce \( \Phi ^{\ell }_p(u, x, v) = {\chi ^{\ell }(u, x)\chi ^{-\ell }(x, v)}/{p(x)}\) for all \(x \in {\mathcal {S}}(p)\) and 0 elsewhere, which allows to perform “probabilistic integration” as follows: if \(f \in \mathrm {dom}(\Delta ^{-\ell })\) is such that \((\Delta ^{-\ell }f)\) is integrable on \([x_1, x_2] \cap {\mathcal {S}}(p)\), then
$$\begin{aligned} f(x_2)-f(x_1) = {\mathbb {E}}\left[ \Phi ^{\ell }_p(x_1, X, x_2) \Delta ^{-\ell }f(X)\right] \end{aligned}$$
(A.1)
for all \(x_1 < x_2 \in {\mathcal {S}}(p)\). We can use this function to obtain
$$\begin{aligned} \bar{h}(x)&={\mathbb {E}}\left[ (h(x)-h(X))(\chi ^\ell (X,x)+\chi ^{-\ell }(x,X)) \right] \\&= {\mathbb {E}}\left[ \Delta ^{-\ell } h(X_2) {\mathbb {E}}\left[ \Phi _p^\ell (X,X_2,x)-\Phi _p^\ell (x,X_2,X) | X_2\right] \right] \end{aligned}$$
(we use the fact that \(\chi ^\ell (x,y)+\chi ^{-\ell }(y,x) =1+{\mathbb {I}}[\ell =0]{\mathbb {I}}[x=y]\)) and it only remains to reorganize the integrand to obtain the claim. To this end, we note how, by definition,
$$\begin{aligned} {\mathbb {E}}\left[ \Phi _p^\ell (X,y,x)-\Phi _p^\ell (x,y,X) \right]&=\frac{\chi ^{-\ell }(y,x)}{p(y)} {\mathbb {E}}[\chi ^\ell (X,y)] - \frac{\chi ^{\ell }(x,y)}{p(y)}{\mathbb {E}}[\chi ^{-\ell }(y,X)] \\&= \chi ^{-\ell }(y, x) \frac{P(y - {\mathbb {I}}[\ell = 1]<)}{p(y)}\\&\quad - \chi ^{\ell }(x, y)\frac{\bar{P}(y+{\mathbb {I}}[\ell = -1]) }{p(y)} \end{aligned}$$
where the first identity is immediate by definition of \(\Phi _p^\ell \) and the last identity follows from the definition of the generalized indicator \(\chi ^{\ell }\). \(\square \)
Proof of Lemma 2.26
The expressions (2.21) and (2.22) of the solution g are direct from the definition of \({\mathcal {L}}_p^\ell \) and its representation (2.19). The first expression (2.26) of the derivative is direct from the expression (2.8). For the second claim, we shall first prove the following results:
$$\begin{aligned}&\Delta ^{-\ell }g(x) \nonumber \\&\quad = \frac{{\mathbb {E}} \left[ \tilde{K}_p^{\ell }(X_1, x+\ell ) R_p^{\ell }(x, X_2) \bigg ( \Delta ^{-\ell }\eta (X_2) \Delta ^{-\ell }h(X_1) - \Delta ^{-\ell } h(X_2) \Delta ^{-\ell } \eta (X_1) \bigg ) \right] }{\big (-{\mathcal {L}}_p^\ell \eta (x)\big )\big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big )} \end{aligned}$$
(A.2)
$$\begin{aligned}&\quad = \frac{{\mathbb {E}} \left[ \bigg (\tilde{K}_p^{\ell }(X_1, x+\ell ) R_p^{\ell }(x, X_2)- R_p^{\ell }(x, X_1)\tilde{K}_p^{\ell }(X_2, x+\ell ) \bigg ) \Delta ^{-\ell }h(X_1) \Delta ^{-\ell }\eta (X_2) \right] }{\big (-{\mathcal {L}}_p^\ell \eta (x)\big )\big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big )} \end{aligned}$$
(A.3)
We first prove (A.2). Starting from (2.7) and applying repeatedly (2.19) then (2.20) (once to h and once to \(\eta \)), we obtain
$$\begin{aligned}&\Delta ^{-\ell } g(x)\\&\quad =\frac{{\mathbb {E}} \left[ \tilde{K}_p^{\ell }(X_1, x+\ell ) \bigg ( \bar{\eta }(x) \Delta ^{-\ell }h(X_1)\big ) - \bar{h}(x) \Delta ^{-\ell } \eta (X_1) \bigg ) \right] }{\big (-{\mathcal {L}}_p^\ell \eta (x)\big )\big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big )} \\&\quad = \frac{{\mathbb {E}} \left[ \tilde{K}_p^{\ell }(X_1, x+\ell ) R_p^{\ell }(x, X_2) \bigg ( \Delta ^{-\ell }\eta (X_2) \Delta ^{-\ell }h(X_1)\big ) - \Delta ^{-\ell } h(X_2) \Delta ^{-\ell } \eta (X_1) \bigg ) \right] }{\big (-{\mathcal {L}}_p^\ell \eta (x)\big )\big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big )}. \end{aligned}$$
We now prove (A.3). By similar arguments as above, this follows from
$$\begin{aligned}&\Delta ^{-\ell } g(x) \\&\quad =\frac{{\mathbb {E}} \left[ \tilde{K}_p^{\ell }(X_1, x+\ell ) \bar{\eta }(x) \Delta ^{-\ell }h(X_1) \right] -\big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big ) {\mathbb {E}}\left[ R_p^\ell (x,X_1)\Delta ^{-\ell }h(X_1)\right] }{\big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big )\big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big )} \\&\quad = \frac{{\mathbb {E}} \left[ \bigg (\tilde{K}_p^{\ell }(X_1, x+\ell ) \bar{\eta }(x) - R_p^{\ell }(x, X_1)\big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big ) \bigg ) \Delta ^{-\ell }h(X_1) \right] }{\big (-{\mathcal {L}}_p^\ell \eta (x)\big )\big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big )}\\&\quad = \frac{{\mathbb {E}} \left[ \bigg (\tilde{K}_p^{\ell }(X_1, x+\ell ) R_p^{\ell }(x, X_2)- R_p^{\ell }(x, X_1)\tilde{K}_p^{\ell }(X_2, x+\ell ) \bigg ) \Delta ^{-\ell }h(X_1)\Delta ^{-\ell }\eta (X_2) \right] }{\big (-{\mathcal {L}}_p^\ell \eta (x)\big ) \big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big )} . \end{aligned}$$
To conclude, we decompose the above expectation into four parts with: \(X_i<x+{\mathbb {I}}[\ell = 1]\) and/or \(X_i\ge x+{\mathbb {I}}[\ell = 1]\), for \(i=1,2\) (i.e., using either \(\chi ^{-\ell }(X_i,x)\) or \(\chi ^{\ell }(x,X_i)\)). Therefore, by considering separately \(\ell \in \{0,-1,1\}\), we can easily verify that
$$\begin{aligned} \tilde{K}_p^\ell (y,x+\ell ) = {\left\{ \begin{array}{ll} \dfrac{P(y-{\mathbb {I}}[\ell = 1])\bar{P}(x+{\mathbb {I}}[\ell = 1])}{p(y)p(x+\ell )}&{} \text{ if }\quad y<x+{\mathbb {I}}[\ell = 1] \\ \dfrac{P(x-{\mathbb {I}}[\ell = -1])\bar{P}(y+{\mathbb {I}}[\ell = -1])}{p(y)p(x+\ell )}&{} \text{ if }\quad y\ge x+{\mathbb {I}}[\ell = 1] \end{array}\right. } \end{aligned}$$
and
$$\begin{aligned} R_p^\ell (x,y) = {\left\{ \begin{array}{ll} \dfrac{P(y-{\mathbb {I}}[\ell = 1])}{p(y)}&{} \text{ if }\quad y<x+{\mathbb {I}}[\ell = 1] \\ \dfrac{-\bar{P}(y+{\mathbb {I}}[\ell = -1])}{p(y)}&{} \text{ if }\quad y\ge x+{\mathbb {I}}[\ell = 1] \end{array}\right. } \end{aligned}$$
Basic manipulations then give
$$\begin{aligned}&\Delta ^{-\ell } g(x) \big (-{\mathcal {L}}_p^\ell \eta (x)\big )\big (-{\mathcal {L}}_p^\ell \eta (x+\ell )\big ) =\frac{\bar{P}(x+{\mathbb {I}}[\ell = 1])+P(x-{\mathbb {I}}[\ell = -1])}{p(x+\ell )}\\&\quad \Bigg ({\mathbb {E}}\left[ \Delta ^{-\ell } h(X_1) \frac{\bar{P}(X_1+{\mathbb {I}}[\ell =-1])}{p(X_1)}\chi ^{\ell }(x,X_1)\right] \\&\qquad \times {\mathbb {E}}\left[ \Delta ^{-\ell } \eta (X_2) \frac{P(X_2-{\mathbb {I}}[\ell = 1])}{p(X_2)}\chi ^{-\ell }(X_2,x)\right] \\&\qquad - {\mathbb {E}}\left[ \Delta ^{-\ell } h(X_1) \frac{P(X_1-{\mathbb {I}}[\ell = 1])}{p(X_1)}\chi ^{-\ell }(X_1,x)\right] \\&\qquad \times {\mathbb {E}}\left[ \Delta ^{-\ell } \eta (X_2) \frac{\bar{P}(X_2+{\mathbb {I}}[\ell = -1])}{p(X_2)}\chi ^{\ell }(x,X_2)\right] \Bigg ) \end{aligned}$$
which leads to the claim as \(\bar{P}(x+{\mathbb {I}}[\ell =1]) +P(x-{\mathbb {I}}[\ell = -1])=1\) and \(\ell ={\mathbb {I}}[\ell =1] -{\mathbb {I}}[\ell = -1]\). \(\square \)
Proof of Lemma 2.27
The condition implies that \(g^-\) is nondecreasing and nonnegative over \({\mathcal {S}}(p) \cap (-\infty , \xi ]\) and nondecreasing and nonpositive over \({\mathcal {S}}(p) \cap (\xi , \infty )\). Therefore, the absolute value of the solution for point mass Eq. (2.28) reaches his supremum at \(\xi \) or \(\xi +1\), which gives the bound (2.29). Moreover, the supremum of the difference is observed between \(\xi \) and \(\xi +1\). Using the explicit expression (2.17) and the relation \(\tau _p^\ell (x+\ell )p(x+\ell )=\tau _p^{-\ell }(x)p(x)\), we have
$$\begin{aligned} \sup _x |\Delta g(x)|&= g^-(\xi ) - g^-(\xi +1) =\frac{P(\xi -1)}{\tau _p^+(\xi )} +\frac{(1-P(\xi ))p(\xi )}{\tau _p^+(\xi +1)p(\xi +1)} \\&= \frac{P(\xi -1)}{\tau _p^+(\xi )} + \frac{1-P(\xi )}{\tau _p^-(\xi )}. \end{aligned}$$
Furthermore, as \(x-{\mathbb {E}}[X] = \tau _p^+(x)-\tau _p^-(x)\), we have \(\tau ^-_p(\xi ) \ge \tau ^+_p(\xi )\) if \(\xi \le {\mathbb {E}}[X]\) (resp. \(\tau ^-_p(\xi ) \le \tau ^+_p(\xi )\) if \(\xi \ge {\mathbb {E}}[X]\)). Therefore, the supremum is bounded by \(\frac{P(\xi -1)+1-P(\xi )}{\tau _p^+(\xi )} =\frac{1-p(\xi )}{\tau _p^+(\xi )}\) if \(\xi \le {\mathbb {E}}[X]\) and otherwise by \(\frac{1-p(\xi )}{\tau _p^-(\xi )}\).
By remark 2.21, the solution \(g_A^\ell (x)\) is explicit and defined by \(g_\xi \) for \(\xi \in A\). The sign of \(g_\xi \) changes according to the relative position of \(\xi \) and x. Then, combined with the hypotheses, the maximal value of \(|g_A^-(x)|\) is either observed at \(x=\min _{\xi \in A}\{\xi \}=:\xi _{1}\) or \(x=\max _{\xi \in A}\{\xi \}+1=:\xi _{2}+1\). Then,
$$\begin{aligned} \sup _x |g^-_A(x)|&= \max \left\{ \frac{P(\xi _1-1)}{p(\xi _1)\tau _p^+(\xi _1)}\sum _{j\in A}p(j), \frac{1-P(\xi _2)}{p(\xi _2)\tau _p^-(\xi _2)}\sum _{j\in A}p(j) \right\} \\&\le \left( \sum _{j\in A}p(j) \right) \sup _{\xi \in A} \left\{ \frac{1}{\tau _p^+(\xi )p(\xi )}, \frac{1}{\tau _p^-(\xi )p(\xi )}\right\} . \end{aligned}$$
Finally, due to the monotonicity of each \(g_\xi (x)\) function, the maximal difference \(|\Delta g_A(x)|\) is bounded by the supremum of \(|\Delta g_\xi (x)|\) for \(\xi \in A\), which is enough to conclude. \(\square \)
Proof of Theorem 3.2
First take \(c_1(x)= c_2(x) = 1 \) in (3.4). Without any further assumptions on h, the solution \(g_h^*\) of (1.5) with \(c(x)=1\) can be represented as
$$\begin{aligned} g_h^*(x) = \frac{{\mathcal {L}}_\infty ^\ell h(x+\ell )}{c_1(x+\ell )} = {\mathcal {L}}_\infty ^\ell h(x+\ell ) \end{aligned}$$
Hence, we obtain (3.5).
Next take \(\eta _1 = \eta _2 = \mathrm {Id}\) in (3.3). Then, \(-{\mathcal {L}}_{\infty }^\ell \eta _1(x) = \tau _\infty ^\ell (x)\) and \(-{\mathcal {L}}_n^\ell \eta _2(x) = \tau _n^\ell (x)\), the Stein kernels of \(p_\infty \) and \(p_n\). Without any further assumptions on h, the solution \(g_h(x)\) of (1.4) with \(\eta =\mathrm {Id}\) can be represented as
$$\begin{aligned} g_h(x) = \frac{-{\mathcal {L}}_\infty ^\ell h(x+\ell )}{\tau _\infty (x+\ell )} \end{aligned}$$
Hence we get (3.6).
Appendix B: Some more inequalities
Corollary B.1
(Identity (3.6), Stein kernels and \(\ell = 0\)) Under the same assumptions and with exactly the same notations as in Corollary 3.4, the following results hold true.
-
1.
The Kolmogorov distance between the random variables \(X_n\) and \(X_\infty \) is
$$\begin{aligned}&\mathrm {Kol}(X_n,X_\infty ) \nonumber \\&\quad = \sup _z \Bigg | {\mathbb {E}}\Bigg [ \frac{\tau _n(X_n) -\tau _\infty (X_n)}{\tau _{\infty }(X_n)}{\mathbb {I}}_{{\mathcal {S}}_{\infty }}(X_n) \end{aligned}$$
(B.1)
$$\begin{aligned}&\qquad \times \Bigg ( {P_{\infty }(z) - {\mathbb {I}}[X_n \le z]} + \frac{X_n -{\mathbb {E}}[X_{\infty }]}{\tau _{\infty }(X_n)} \frac{P_{\infty }(X_n \wedge z) \bar{P}_{\infty }(X_n\vee z)}{p_{\infty }(X_n)} \Bigg ) \Bigg ] + \kappa _{\mathrm {Id}}(z) \Bigg | \nonumber \\&\quad \le {\mathbb {E}}\left[ \left| \frac{\tau _n(X_n)}{\tau _{\infty }(X_n)}-1 \right| \left( 1 + \frac{|X_n -{\mathbb {E}}[X_{\infty }]|}{\tau _{\infty }(X_n)} \frac{P_\infty (X_n)\bar{P}_\infty (X_n)}{p_\infty (X_n)} \right) {\mathbb {I}}_{{\mathcal {S}}_{\infty }}(X_n) \right] + \sup _z |\kappa _{\mathrm {Id}}(z)| \end{aligned}$$
(B.2)
where
$$\begin{aligned} \kappa _{\mathrm {Id}}(z)&= (\mu _n - \mu _{\infty }) {\mathbb {E}} \left[ \frac{P_{\infty }(X_n \wedge z) \bar{P}_{\infty }(X_n\vee z)}{\tau _{\infty }(X_n)p_{\infty }(X_n)} \right] \\&\quad + \lim _{x \nearrow b_n \wedge b_{\infty }}\frac{\tau _n(x)}{\tau _\infty (x)} \frac{p_n(x)}{p_\infty (x)}P_\infty (x\wedge z)\bar{P}_\infty (x\vee z)\\&\quad -\lim _{x \searrow a_n \vee a_{\infty }}\frac{\tau _n(x)}{\tau _\infty (x)} \frac{p_n(x)}{p_\infty (x)}P_\infty (x\wedge z)\bar{P}_\infty (x\vee z). \end{aligned}$$
-
2.
The total variation distance between \(X_n\) and \(X_\infty \) is
$$\begin{aligned}&\mathrm {TV}(X_n, X_{\infty })\nonumber \\&\quad = \kappa _{\mathrm {Id}}({\mathbb {I}}_{A_{n}^{\infty }}) +{\mathbb {E}}\Bigg [ \frac{\tau _n(X_n) - \tau _\infty (X_n)}{\tau _{\infty }(X_n)}{\mathbb {I}}_{{\mathcal {S}}_{\infty }}(X_n) \end{aligned}$$
(B.3)
$$\begin{aligned}&\qquad \times \Bigg ({P}_{\infty }(A_n^{\infty }) - {\mathbb {I}}_{ A_n^{\infty }}(X_n)\nonumber \\&\qquad +\frac{X_n -{\mathbb {E}}[X_{\infty }]}{\tau _{\infty }(X_n)} \frac{ {P}_\infty (A_{n}^{\infty }\cap (-\infty , X_n])-{P}_\infty (A_{n}^{\infty }) P_{\infty }(X_n)}{p_\infty (X_n)} \Bigg ) \Bigg ] \nonumber \\&\quad \le {\mathbb {E}}\left[ \left| \frac{\tau _n(X_n)}{\tau _{\infty }(X_n)}-1 \right| \left( 1 + \frac{|X_n -{\mathbb {E}}[X_{\infty }]|}{\tau _{\infty }(X_n)} \frac{P_\infty (X_n)\bar{P}_\infty (X_n)}{p_\infty (X_n)} \right) {\mathbb {I}}_{{\mathcal {S}}_{\infty }}(X_n)\right] + \kappa _{\mathrm {Id}}({\mathbb {I}}_{A_{n}^{\infty }}) \end{aligned}$$
(B.4)
with
$$\begin{aligned} \kappa _{\mathrm {Id}}({\mathbb {I}}_{A_{n}^{\infty }})&= \lim _{x \nearrow b_n \wedge b_{\infty }}\frac{\tau _n(x)}{\tau _\infty (x)} \frac{p_n(x)}{p_\infty (x)} \left( {P}_\infty (A_{n}^{\infty }\cap (-\infty ,x]) -{P}_\infty (A_{n}^{\infty }) P_{\infty }(x)\right) \\&\quad - \lim _{x \searrow a_n \vee b_n}\frac{\tau _n(x)}{\tau _\infty (x)} \frac{p_n(x)}{p_\infty (x)} \left( {P}_\infty (A_{n}^{\infty }\cap (-\infty ,x]) -{P}_\infty (A_{n}^{\infty }) P_{\infty }(x)\right) \\&\quad + (\mu _n-\mu _{\infty }) {\mathbb {E}} \left[ \frac{ {P}_\infty (A_{n}^{\infty } \cap (-\infty , X_n])-{P}_\infty (A_{n}^{\infty }) P_{\infty }(X_n)}{{\tau _{\infty }(X_n)} p_\infty (X_n)} \right] \end{aligned}$$
-
3.
The Wasserstein distance between \(X_n\) and \(X_\infty \) is
$$\begin{aligned} \mathrm {Wass}(X_n, X_{\infty })&= \sup _{h \in \mathrm {Lip}(1)} \Bigg | \kappa _{\mathrm {Id}}(h) \end{aligned}$$
(B.5)
$$\begin{aligned}&\quad + {\mathbb {E}}\left[ \frac{\tau _n(X_n) -\tau _\infty (X_n)}{\tau _{\infty }(X_n)} h'(X_\infty ) \right. \nonumber \\&\quad \left. \left( R_{\infty }(X_n, X_{\infty }) +\frac{X_n-{\mathbb {E}}[X_{\infty }]}{\tau _{\infty }(X_n)} \tilde{K}_\infty (X_n,X_\infty )\right) {\mathbb {I}}_{{\mathcal {S}}_{\infty }}(X_n)\right] \Bigg | \nonumber \\&\quad \le 2 {\mathbb {E}}\left[ \left| \frac{\tau _n(X_n)}{\tau _{\infty }(X_n)}-1 \right| |X_n - {\mathbb {E}}[X_{\infty }]|{\mathbb {I}}_{{\mathcal {S}}_{\infty }}(X_n)\right] + \sup _{h \in \mathrm {Lip}(1)} \kappa _{\mathrm {Id}}(h) \end{aligned}$$
(B.6)
where
$$\begin{aligned} \kappa _{\mathrm {Id}}(h)&= \lim _{x \searrow a_n \vee a_{\infty }} \frac{\tau _n(x)}{\tau _\infty (x)}\frac{p_n(x)}{p_{\infty }(x)} \int _{a_{\infty }}^{b_{\infty }} h'(u) {P_{\infty }(x \wedge u) \bar{P}_{\infty }(x \vee u) } \mathrm {d}u\\&\quad - \lim _{x \nearrow b_n \wedge b_{\infty }}\frac{\tau _n(x)}{\tau _\infty (x)} \frac{p_n(x)}{p_{\infty }(x)} \int _{a_{\infty }}^{b_{\infty }}h'(u) {P_{\infty }(x \wedge u) \bar{P}_{\infty }(x \vee u) } \mathrm {d}u\\&\quad + (\mu _n - \mu _{\infty }) {\mathbb {E}}\left[ \frac{h'(X_\infty )}{\tau _{\infty }(X_{n})} \left( R_{\infty }(X_n, X_{\infty }) + \frac{X_n - {\mathbb {E}}[X_{\infty }]}{\tau _{\infty }(X_n)} \tilde{K}_\infty (X_n, X_\infty )\right) \right] \end{aligned}$$
Corollary B.2
(Identity (3.6), Stein kernels, \(\ell = \pm 1\)) Under the same assumptions and with exactly the same notations as in Corollary 3.6, the following results hold true.
$$\begin{aligned}&\mathrm {TV}(X_n, X_{\infty }) \\&\quad = \kappa _{\mathrm {Id}}^{\ell }({\mathbb {I}}_{A_{n}^{\infty }}) +{\mathbb {E}}\Bigg [ \frac{\tau _n^{\ell }(X_n) -\tau _\infty ^{\ell }(X_n)}{\tau _{\infty }^{\ell } (X_n)} {\mathbb {I}}_{{\mathcal {S}}_{\infty }}(X_n+\ell ) \\&\qquad \times \Bigg ({P}_{\infty }(A_n^{\infty }) - {\mathbb {I}}_{ A_n^{\infty }}(X_n) + \frac{X_n -{\mathbb {E}}[X_{\infty }]}{\tau _{\infty }^{\ell }(X_n+\ell )}\\&\qquad \frac{ {P}_\infty (A_{n}^{\infty }\cap (-\infty , X_n-{\mathbb {I}}[\ell = -1]])-{P}_\infty (A_{n}^{\infty }) P_{\infty }(X_n-{\mathbb {I}}[\ell = -1])}{p_\infty (X_n+\ell )} \Bigg ) \Bigg ] \\&\quad \le {\mathbb {E}}\left[ \left| \frac{\tau _n^{\ell }(X_n)}{\tau _{\infty }^{\ell }(X_n)}-1 \right| \right. \\&\qquad \left. \left( 1 + \frac{|X_n -{\mathbb {E}}[X_{\infty }]|}{\tau _{\infty }^{\ell }(X_n+\ell )} \frac{P_\infty (X_n-{\mathbb {I}}[\ell = -1])\bar{P}_\infty (X_n-{\mathbb {I}}[\ell = -1])}{p_\infty (X_n+\ell )} \right) {\mathbb {I}}_{{\mathcal {S}}_{\infty }}(X_n+\ell ) \right] \\&\qquad + \kappa _{\mathrm {Id}}^{\ell }({\mathbb {I}}_{A_{n}^{\infty }}) \end{aligned}$$
with
$$\begin{aligned} \kappa _{\mathrm {Id}}^{+}({\mathbb {I}}_{A_{n}^{\infty }})&=-\lim _{x \searrow a_n \vee a_{\infty }}\frac{\tau _n^{+} (x)}{\tau _{\infty }^{+}(x)}\frac{p_n(x)}{p_\infty (x)} \left( {P}_\infty (A_{n}^{\infty }\cap (-\infty ,x-1]) -{P}_\infty (A_{n}^{\infty }) P_{\infty }(x-1)\right) \\&\quad + (\mu _\infty -\mu _n) {\mathbb {E}} \left[ \frac{ {P}_\infty (A_{n}^{\infty }\cap (-\infty , X_n])-{P}_\infty (A_{n}^{\infty }) P_{\infty }(X_n)}{{\tau _{\infty }^{+}(X_n+1)} p_\infty (X_n+1)} {\mathbb {I}}_{{\mathcal {S}}_{\infty }}(X_n+1)\right] \end{aligned}$$
and
$$\begin{aligned} \kappa _{\mathrm {Id}}^{-}({\mathbb {I}}_{A_{n}^{\infty }})&=\lim _{x \nearrow b_n \wedge b_{\infty }}\frac{\tau _n^{-}(x)}{\tau _{\infty }^{-}(x)} \frac{p_n(x)}{p_\infty (x)} \left( {P}_\infty (A_{n}^{\infty }\cap (-\infty ,x])-{P}_\infty (A_{n}^{\infty }) P_{\infty }(x)\right) \\&\quad + (\mu _\infty -\mu _{n}) {\mathbb {E}} \left[ \frac{ {P}_\infty (A_{n}^{\infty }\cap (-\infty ,X_n]) -{P}_\infty (A_{n}^{\infty }) P_{\infty }(X_n)}{{\tau _{\infty }^{-}(X_n-1)} p_\infty (X_n-1)} {\mathbb {I}}_{{\mathcal {S}}_{\infty }}(X_n-1)\right] \end{aligned}$$