Second-order properties of thresholded realized power variations of FJA additive processes

Abstract

For a class of additive processes of finite jump activity (FJA), we give precise conditions for the mean-squared consistency and feasible Central Limit Theorems of thresholded realized power variation estimators (TRV). To justify that the proposed conditions are the “best possible”, we also show that these are necessary for FJA Lévy processes. Non-asymptotic upper bounds and asymptotic decompositions of the mean-squared errors of our estimators are also provided. For comparison purposes, we also obtain the analogous asymptotic decomposition for a general multi-power realized variation (MPV). These results theoretically justify the relatively large bias of MPV (when compared to TRV) observed numerically in earlier Monte Carlo studies.

Appendices

Appendix A: Proofs of Section 4.1

Throughout, Z denotes a standard normal random variable whose absolute moments are denoted by $m_{r}:={\mathbb {E}}|Z|^{r}$.

Proof of Theorem 4.1

Let us start by noting the following useful decomposition, valid for even positive integers k,

$$\begin{aligned} {\mathbb {E}}\left( |\Delta _{i}^{n}X|^{k}\mathbf{1}_{[|\Delta _{i}^{n}X|\le {}B_{n}]} \right) =\sigma _{n,i}^{k}m_{k}+{\mathcal {R}}_{n,i}^{(1,k)}+{\mathcal {R}}_{n,i}^{(2,k)}-{\mathcal {R}}_{n,i}^{(3,k)}, \end{aligned}$$

(A.1)

where the remainder terms are given by

$$\begin{aligned} {\mathcal {R}}_{n,i}^{(1,k)}&={\mathbb {E}}\left( |\Delta _{i}^{n}X|^{k}\mathbf{1}_{[\Delta _{i}^{n}N={}0]}\right) -\sigma _{n,i}^{k}m_{k}=\left( e^{-\lambda _{n,i}}-1\right) \sigma _{n,i}^{k}m_{k}\\&\quad +e^{-\lambda _{n,i}}\sum _{j=1}^{k}\left( {\begin{array}{c}k\\ j\end{array}}\right) \gamma _{n,i}^{j}\sigma _{n,i}^{k-j}{\mathbb {E}}\left( Z^{k-j}\right) ,\\ {\mathcal {R}}_{n,i}^{(2,k)}&={\mathbb {E}}\left( |\Delta _{i}^{n}X|^{k}\mathbf{1}_{[|\Delta _{i}^{n}X|\le {}B_{n}, \Delta _{i}^{n}N\ne {}0]}\right) ,\\ {\mathcal {R}}_{n,i}^{(3,k)}&={\mathbb {E}}\left( |\Delta _{i}^{n}X|^{k}\mathbf{1}_{[|\Delta _{i}^{n}X|>B_{n},\Delta _{i}^{n}N=0]}\right) . \end{aligned}$$

Using (2.2) and arguments alike those used in deriving (C.3) and (C.9), we have that

$$\begin{aligned} |{\mathcal {R}}^{(1,k)}_{n,i}|\le {}C^{(1)} h_{n}^{\frac{k}{2}+1}, \quad |{\mathcal {R}}_{n,i}^{(2,k)}|\le C^{(2)}B_{n}^{k+1}h_{n}, \quad |{\mathcal {R}}_{n,i}^{(3,k)}|\le C^{(3)} \sqrt{h_{n}}B_{n}^{k-1}\phi \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T}\sqrt{h_{n}}} \right) , \end{aligned}$$

(A.2)

for universal constants $C^{(\ell )}$, independent of i and n, and depending only on k, ${{\bar{\sigma }}}_{T}$, ${{\bar{\gamma }}}_{T}$, ${{\bar{\lambda }}}_{T}$, and $\Vert f\Vert _{\infty }$. For completeness, we check the bound for ${\mathcal {R}}_{n,i}^{(3,k)}$. Let us start by nothing that

$$\begin{aligned} |{\mathcal {R}}_{n,i}^{(3,k)}|&\le \sum _{j=0}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) |\gamma _{n,i}|^{j} \sigma _{n,i}^{k-j} {\mathbb {E}}\left[ |Z|^{k-j} \mathbf{1}_{[|\gamma _{n,i} + \sigma _{n,i} Z |> B_{n}]} \right] \\&\le 2 \sum _{j=0}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) {{\bar{\gamma }}}_{T}^{j} {{\bar{\sigma }}}_{T}^{k-j} h_{n}^{\frac{k+j}{2}}{\mathbb {E}}\left( Z^{k-j} \mathbf{1}_{\left[ Z>u_{n}\right] } \right) , \end{aligned}$$

where we set $u_{n}:=\frac{B_{n}-{{\bar{\gamma }}}_{T}h_{n}}{{{\bar{\sigma }}}_{T}\sqrt{h_{n}}}$ and assumed that n is large enough for $u_{n}>0$ (recall that $B_{n}\gg \sqrt{h_{n}}$). By making a change of variables, we have

$$\begin{aligned} {\mathbb {E}}\left( Z^{m} \mathbf{1}_{\left[ Z>u_{n}\right] } \right) = \frac{2^{m/2-1}}{\pi ^{1/2}} {\overline{\Gamma }} \left( \frac{m+1}{2}, {\frac{{u}_{n}^{2}}{2}} \right) \sim \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T} \sqrt{h_{n}}} \right) ^{m-1} \phi \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T}\sqrt{h_{n}}} \right) , \quad n\rightarrow \infty , \end{aligned}$$

where ${\overline{\Gamma }}(s,x):=\int _{x}^{\infty }u^{s-1}e^{-u}du$ is the upper incomplete gamma function and we used the well known asymptotic property ${\overline{\Gamma }}(s,x)\sim x^{s-1}e^{-x}$, as $x\rightarrow \infty$. Finally, for n large enough, independently of i,

$$\begin{aligned} |{\mathcal {R}}_{n,i}^{(3,k)}|&\le 4\sum _{j=0}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) {{\bar{\gamma }}}_{T}^{j} {{\bar{\sigma }}}_{T}^{k-j} h_{n}^{\frac{k+j}{2}} \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T} \sqrt{h_{n}}} \right) ^{k-j-1} \phi \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T}\sqrt{h_{n}}} \right) \\&=4{{\bar{\sigma }}}_{T}h_{n}^{\frac{1}{2}} B_{n}^{k-1} \phi \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T}\sqrt{h_{n}}} \right) \left( 1+o(1)\right) , \end{aligned}$$

where in the last equality we used that $B_{n}\gg \sqrt{h_{n}}$.

Now, we are ready to prove the result. Let us first analyze the bias of $TPV(X)[\mathbf{B}]_{T}^{n,k}$, which for simplicity is denoted by $TPV_{n}$. Note that, from (A.1),

$$\begin{aligned} Bias\left( TPV_{n}\right)&=\sum _{i=1}^{n}\frac{1}{m_{k}h_{n}^{k/2-1}}{\mathbb {E}}\left[ |\Delta _{i}^{n}X|^{k} \mathbf{{1}}_{[|\Delta _{i}^{n}X|\le B_{n}]} \right] - \int _{0}^{T}\sigma _{u}^{k}du\nonumber \\&=\sum _{i=1}^{n}\sigma _{t^{*}_{i}}^{k}h_{n}-\int _{0}^{T}\sigma _{u}^{k}du +\bar{{\mathcal {R}}}_{n}^{(1)}+\bar{{\mathcal {R}}}_{n}^{(2)}-\bar{{\mathcal {R}}}_{n}^{(3)}, \end{aligned}$$

(A.3)

where $t_{i}^{*}\in (t_{i-1},t_{i})$ is such that $\int _{t_{i-1}}^{t_{i}}\sigma _{u}^{2}du=\sigma _{t_{i}^{*}}^{2}h_{n}$ and, by (A.2), the reminder terms satisfy

$$\begin{aligned} |\bar{{\mathcal {R}}}^{(1)}_{n}|\le {}{\bar{C}}^{(1)} T h_{n}, \quad |\bar{{\mathcal {R}}}_{n}^{(2)}|\le {\bar{C}}^{(2)}T \frac{B_{n}^{k+1}}{h_{n}^{\frac{k}{2}-1}}, \quad |\bar{{\mathcal {R}}}_{n}^{(3)}|\le {\bar{C}}^{(3)} T \frac{B_{n}^{k-1}}{h_{n}^{\frac{k-1}{2}}}\phi \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T}\sqrt{h_{n}}} \right) , \end{aligned}$$

for some constants ${\bar{C}}^{\ell }$. It is now clear that the conditions given in (4.1) are sufficient for the bias to vanish. For $\varepsilon >0$, let $\delta >0$ be such that $|\sigma ^{k}_{v}-\sigma ^{k}_{u}|<\varepsilon$ whenever $|u-v|<\delta$. Then, for n large enough,

$$\begin{aligned} \left| \sum _{i=1}^{n}\sigma _{t^{*}_{i}}^{k}h_{n}-\int _{0}^{T}\sigma _{u}^{k}du\right| \le {}\sum _{i=1}^{n}\int _{t_{i-1}}^{t_{i}}|\sigma ^{k}_{u}-\sigma _{t_{i}^{*}}^{k}|du\le T \varepsilon , \end{aligned}$$

which shows that the first term on the right-hand side of (A.3) converges to 0. Now, we consider the variance of the TPV estimator. Clearly, by the independence of the increments of X and the decomposition (A.1),

$$\begin{aligned} \mathrm{Var}\left( TPV_{n}\right)&\le \frac{1}{m^{2}_{k}h_{n}^{k-2}}\sum _{i=1}^{n}{\mathbb {E}}\left[ |\Delta _{i}^{n}X|^{2k} \mathbf{{1}}_{[|\Delta _{i}^{n}X|\le B_{n}]} \right] \\&\le \widetilde{{\mathcal {R}}}^{(0)}_{n}+\widetilde{{\mathcal {R}}}_{n,i}^{(1)}+\widetilde{{\mathcal {R}}}_{n,i}^{(2)}, \end{aligned}$$

where the terms $\widetilde{{\mathcal {R}}}^{(\ell )}_{n}$ are given by

$$\begin{aligned} \widetilde{{\mathcal {R}}}^{(0)}_{n}= \frac{m_{2k}}{m^{2}_{k}h_{n}^{k-2}}\sum _{i=1}^{n}\left( \int _{t_{i-1}}^{t_{i}}\sigma ^{2}_{u}du\right) ^{k}, \quad \widetilde{{\mathcal {R}}}_{n,i}^{(\ell )}=\frac{1}{m^{2}_{k}h_{n}^{k-2}}\sum _{i=1}^{n}{\mathcal {R}}^{(\ell ,2k)}_{n,i},\quad \ell =1,2, \end{aligned}$$

and, thus, by (A.2) and the fact that $\left( \int _{t_{i-1}}^{t_{i}}\sigma ^{2}_{u}du\right) ^{k}\le {{\bar{\sigma }}}_{T}^{2k}h_{n}^{k}$, are such that

$$\begin{aligned} |\widetilde{{\mathcal {R}}}^{(0)}_{n}|\le {}{\widetilde{C}}^{(0)} T h_{n},\quad |\widetilde{{\mathcal {R}}}^{(1)}_{n}|\le {}{\widetilde{C}}^{(1)} T h^{2}_{n}, \quad |\widetilde{{\mathcal {R}}}_{n}^{(2)}|\le {\bar{C}}^{(2)}T \frac{B_{n}^{2k+1}}{h_{n}^{k-2}}, \end{aligned}$$

for some constants ${\widetilde{C}}^{\ell }$. Therefore, we conclude that the variance converges to 0 whenever the conditions in (4.1) are satisfied.

Let us now show that, for a FJA Lévy model, the conditions given in (4.1) are necessary. To this end, note that $\mathrm{Bias}_{n}:=\mathrm{Bias}\left( TPV(X)[\mathbf{B}]_{T}^{n,k};T\sigma ^{k}\right)$ is such that

$$\begin{aligned} \mathrm{Bias}_{n}&= T\left( \frac{1}{{m}_{k}h_{n}^{k/2}}{\mathbb {E}}\left[ |\Delta _{1}^{n}X|^{k} \mathbf{{1}}_{[|\Delta _{1}^{n}X|\le B_{n}, \Delta _{1}^{n}N= 0]} \right] - \sigma ^{k} \right) \nonumber \\&\quad + \frac{T}{{m_{k}}h_{n}^{k/2}}{\mathbb {E}}\left[ |\Delta _{1}^{n}X|^{k} \mathbf{{1}}_{[|\Delta _{1}^{n}X| \le B_{n}, \Delta _{1}^{n}N\ne 0]} \right] \nonumber \\&=: A_{n}^{(1)} + A_{n}^{(2)}. \end{aligned}$$

(A.4)

Lemma C.2 implies that, as far as $B_{n}\rightarrow {}0$,

$$\begin{aligned} A_{n}^{(2)} \sim \frac{2 T\lambda {\mathcal {C}}_{0}({f})}{(k+1){m}_{k}}\frac{B_{n}^{k+1}}{h_{n}^{k/2-1}}. \end{aligned}$$

(A.5)

Now, suppose that $L:=\liminf _{n\rightarrow \infty }B_{n}/\sqrt{h_{n}}\in [0,\infty )$. Then, as shown in the proof of Lemma C.5, for a subsequence $\{n_{j}\}_{j\ge {}1}$, $\lim _{j\rightarrow \infty } A_{n_{j}}^{(1)}$ exists and is different from 0. Moreover,

$$\begin{aligned} \lim _{j\rightarrow \infty } \frac{B_{n_{j}}^{k+1}}{h_{n_{j}}^{k/2-1}}=\lim _{j\rightarrow \infty } \left( \frac{B_{n_{j}}}{h_{n_{j}}^{1/2}}\right) ^{k} {B_{n_{j}}h_{n_{j}}}=0, \end{aligned}$$

which implies that $\lim _{j\rightarrow \infty }A_{n_{j}}^{(2)}=0$ and, hence, $\lim _{j\rightarrow \infty }\mathrm{Bias}_{n_{j}}=\lim _{j\rightarrow \infty }A_{n_{j}}^{(1)}\ne {}0$, which contradicts that the MSE vanishes as $n\rightarrow \infty$. So, hereafter, we assume that condition (4.1-i) holds and show the necessity of (4.1-ii). First, note that

$$\begin{aligned} {\mathrm{Var}}\left( TPV(X)[\mathbf{B}]_{T}^{n,k}\right)&=\frac{T}{{m}_{k}^{2}h_{n}^{k-1}}{\mathbb {E}}\left[ |\Delta _{1}^{n}X|^{2k} \mathbf{1}_{[|\Delta _{1}^{n}X| \le B_{n}]}\right] -\frac{T}{{m}_{k}^{2}h_{n}^{k-1}}{\mathbb {E}}\left[ |\Delta _{1}^{n}X|^{k} \mathbf{1}_{[|\Delta _{1}^{n}X| \le B_{n}]}\right] ^{2}\\&=:{\mathcal {D}}_{n}^{(1)}-\left( {\mathcal {D}}_{n}^{(2)}\right) ^{2}. \end{aligned}$$

Next, by decomposing $\mathbf{1}_{[|\Delta _{1}^{n}X| \le B_{n}]}$ as $\mathbf{1}_{[|\Delta _{1}^{n}X| \le B_{n},\Delta _{1}^{n}N\ne {}0]}+\mathbf{1}_{[|\Delta _{1}^{n}X| \le B_{n},\Delta _{1}^{n}N=0]}$ and applying Lemmas C.2 and C.5, it follows that

$$\begin{aligned} {\mathcal {D}}_{n}^{(1)}&=\frac{T}{{m}_{k}^{2}h_{n}^{k-1}}\left( h_{n} \lambda \frac{2B_{n}^{2k+1}}{2k+1} {\mathcal {C}}_{0}(f)\right) +\frac{T}{{m}_{k}^{2}h_{n}^{k-1}}\left( h_{n}^{k} \frac{\sigma ^{2k}2^{k}}{\pi ^{1/2}} \Gamma \left( \frac{2k+1}{2} \right) \right) +\mathrm{{h.o.t.}},\\ {\mathcal {D}}_{n}^{(2)}&=\frac{T^{1/2}}{{m}_{k}h_{n}^{\frac{k-1}{2}}}\left( h_{n} \lambda \frac{2B_{n}^{k+1}}{k+1} {\mathcal {C}}_{0}(f)\right) +\frac{T^{1/2}}{{m}_{k}h_{n}^{\frac{k-1}{2}}}\left( h_{n}^{k/2}\frac{\sigma ^{k}2^{k/2}}{\pi ^{1/2}} \Gamma \left( \frac{k+1}{2} \right) \right) +{\mathrm{{h.o.t.}}} \end{aligned}$$

Since the second terms in the decompositions of ${\mathcal {D}}_{n}^{(1)}$ and ${\mathcal {D}}_{n}^{(2)}$ converge to 0, regardless of $B_{n}$, and

$$\begin{aligned} \frac{B_{n}^{2k+2}}{h_{n}^{k-3}}=\frac{B_{n}^{2k+1}}{h_{n}^{k-2}}B_{n}h_{n}\ll \frac{B_{n}^{2k+1}}{h_{n}^{k-2}}, \end{aligned}$$

we conclude that, for the variance to vanish, it is necessary that condition (4.1-ii) holds true.

Finally, from the previous decomposition for the variance, the asymptotic decomposition (4.3) holds. While (4.2) is deduced from the decomposition (A.4) and the asymptotic property (A.5) and the following easy consequence of Lemma C.5:

$$\begin{aligned} A_{n}^{(1)}= - \frac{T\pi ^{1/2}}{2^{k/2-1}\Gamma \left( \frac{k+1}{2}\right) } \frac{B_{n}^{k-1}}{h_{n}^{(k-1)/2}}\phi \left( \frac{B_{n}}{\sigma _{n}}\right) + T\Lambda _{n}^{(1)} h_{n}+\mathrm{h.o.t.} \end{aligned}$$

$\square$

Proof of Theorem 4.2

We prove the result through three steps:

Step 1. Here, we show that the estimator $MPV(X)_{{T}}^{n[\mathbf{r}]}$ is an asymptotically unbiased estimator for ${T}\sigma ^{\mathbf{r}^{+}}$ if and only if $r_{\max }<2$ and that the asymptotic behavior of the bias is given as in (4.12). Note that, by conditioning on $\Delta _{i}^{n}N$, ${\mathbb {E}}[|\Delta _{i}^{n}X|^{r_{i}}] = h_{n}^{r_{i}/2} e^{-\lambda _{n}} \sigma ^{r_{i}} m_{r_{i}} + R_{n}(r_{i})$, where

$$\begin{aligned} R_{n}(s):= & {} e^{-\lambda _{n}} {\mathbb {E}}\left[ |\gamma _{n} + \sigma \Delta _{i}^{n}W|^{s} \right] - e^{-\lambda _{n}} {\mathbb {E}}\left[ |\sigma \Delta _{i}^{n}W|^{s} \right] \nonumber \\&+ \,{\mathbb {E}}\left[ \left| \gamma _{n} + \sigma \Delta _{i}^{n}W+ \Delta _{i}^{n}J\right| ^{s} \mathbf{1}_{[\Delta _{i}^{n}N\ne 0]} \right] , \end{aligned}$$

(A.6)

and from Lemmas C.7 and C.8, $R_{n}(r) \sim \lambda h_{n} {\mathbb {E}}[|\zeta _{1}|^{r}]$, as $n \rightarrow {} \infty$. Let $a_{i}^{(n)} := h_{n}^{r_{i}/2} e^{-\lambda _{n}} \sigma ^{r_{i}} m_{r_{i}}$ and $b_{i}^{(n)} := R_{n}(r_{i})$. Using the fact that the increments are i.i.d., we may then express the first moment as follows:

$$\begin{aligned} {\mathbb {E}}[MPV(X)_{{T}}^{n [\mathbf{r}]} ]&= \frac{h_{n}^{1-\mathbf{r}^{+}/2}}{C(\mathbf{r})} (M_{n}-k+1) \prod _{i=1}^{k} \left( a_{i}^{(n)}+b_{i}^{(n)}\right) \nonumber \\&= \frac{h_{n}^{1-\mathbf{r}^{+}/2}}{C(\mathbf{r})} (M_{n}-k+1)\sum _{j=1}^{2^{k}} \prod _{i=1}^{k} c_{i,j}^{(n)}, \end{aligned}$$

(A.7)

where $c_{i,j}^{(n)} \in \lbrace a_{i}^{(n)},b_{i}^{(n)} \rbrace$ for each fixed j. It is then clear that there exists a term in the expansion (A.7) of the order $h_{n}^{1-r_{\max }/2}$, which would not converge to 0, as $n \rightarrow \infty$, if $r_{\max } \ge {} 2$. In turn, this shows the necessity of the condition $r_{\max }<2$ for the bias to converge to 0, since all of the terms in the expansion (A.7) are positive for n large enough. To show (4.12), let us observe that, under the condition $r_{\max } < 2$, the leading order term corresponds to $\prod _{i=1}^{k} a_{i}^{(n)}$, which can be written as

$$\begin{aligned} T\sigma ^{\mathbf{r}^{+}} + {T}(e^{-k\lambda _{n}}-1) \sigma ^{\mathbf{r}^{+}} - {T}\frac{(k-1)}{M_{n}}\sigma ^{\mathbf{r}^{+}} e^{-k\lambda _{n}}=T\sigma ^{\mathbf{r}^{+}}+O(h_{n}). \end{aligned}$$

To find the rate of convergence of the bias, consider $I_{\max } := \lbrace { i \in \lbrace {1,2,\ldots ,k \rbrace } : r_{i} = r_{\max } \rbrace }$ and let $|I_{\max }|$ denotes its cardinality. Then, the bias $H_{n}$ is such that

$$\begin{aligned} H_{n}&\sim T\sum _{i'\in I_{\max }} \frac{h_{n}^{-\frac{\mathbf{r}^{+}}{2}}}{C(\mathbf{r})} \frac{M_{n}-k+1}{M_{n}} b_{i^{'}}^{(n)} \times \prod _{i \ne i^{'}} a_{i}^{(n)} \sim T |I_{\max }|\lambda \sigma ^{(\mathbf{r}^{+}-r_{\max })} {\mathbb {E}}[|\zeta _{1}|^{r_{\max }}] m_{r_{\max }}^{-1}h_{n}^{1-\frac{r_{\max }}{2}}, \end{aligned}$$

(A.8)

as $n\rightarrow \infty$, which justifies (4.12).

Step 2. Throughout this step, we assume that $r_{\max }<2$, and show that the condition (4.10-ii) is necessary and sufficient for

$$\begin{aligned} \mathrm{Bias}\left( \left| MPV(X)_{T}^{n [\mathbf{r}]} \right| ^{2};T^{2}\sigma ^{2\mathbf{r}^{+}}\right) \rightarrow {}0. \end{aligned}$$

(A.9)

We begin by considering the following obvious decomposition:

$$\begin{aligned} \left| MPV(X)_{T}^{n [\mathbf{r}]} \right| ^{2} = S_{n}^{(1)} + 2 \sum _{\ell = 2}^{k} S_{n}^{(\ell )} + 2S_{n}^{(k+1)}, \end{aligned}$$

(A.10)

where, with the convention that $1=\prod _{j=k+1}^{k}$,

$$\begin{aligned} S_{n}^{(\ell )}&:= \frac{h_{n}^{2-\mathbf{r}^{+}}}{C(\mathbf{r})^{2}} \sum _{i=1}^{M_{n}-k-\ell +2}\, \prod _{j=1}^{\ell -1} |\Delta _{i+j-1} X|^{r_{j}} \prod _{j=\ell }^{k} |\Delta _{i+j-1}X|^{r_{j-\ell +1} + r_{j}} \nonumber \\&\qquad \prod _{j=k+1}^{k+\ell -1} |\Delta _{i+j-1}X|^{r_{j-\ell +1}}; \quad \ell = {1}, \dots , k, \end{aligned}$$

(A.11)

$$\begin{aligned} S_{n}^{(k+1)}&:= \frac{h_{n}^{2-\mathbf{r}^{+}}}{C(\mathbf{r})^{2}} \sum _{i=1}^{M_{n} - 2k + 1} \sum _{m = k + i}^{M_{n} - k + 1} \prod _{j=1}^{k}|\Delta _{i+j-1}X|^{r_{j}} \prod _{q=1}^{k}|\Delta _{m+q-1}X|^{r_{q}}. \end{aligned}$$

(A.12)

From here it is important to note that the right-hand side of (A.10) consists of two types of components or terms. Namely, those corresponding to overlapping blocks, as defined in (A.11), and a non-overlapping block term given in (A.12). As it turns out, under the condition $r_{\max } < 2$ the non-overlapping block component will converge in mean to the target value $T^{2} \sigma ^{2\mathbf{r}^{+}}/2$. Hence, in order to ensure the second moment bias of the sequence of MPV estimators converges to 0 in the limit the expected value of the overlapping block components necessarily needs to vanish.

Note that each overlapping block component, $S_{n}^{(\ell )}$, is the sum of $M_{n}-k+2-\ell$ identically distributed terms for $\ell = 1, \dots , k$, while the non-overlapping block component, $S_{n}^{(k+1)}$, contains $\left( {\begin{array}{c}M_{n}-2k+2\\ 2\end{array}}\right)$ such terms. Let $R_{n}(s)$ be defined as in (A.6) and recall that $R_{n}(s) \sim \lambda h_{n} {\mathbb {E}}[|\zeta _{1}|^{s}]$, as $n \rightarrow {} \infty$. An analysis of the expected value of the non-overlapping block term shows that

$$\begin{aligned} {\mathbb {E}}\left[ S_{n}^{(k+1)} \right]&= \frac{h_{n}^{2-\mathbf{r}^{+}}}{C(\mathbf{r})^{2}} \left( {\begin{array}{c}M_{n} - 2k + 2\\ 2\end{array}}\right) \prod _{j=1}^{k} \left[ e^{-\lambda _{n}} h_{n}^{r_{j}/2} \sigma ^{r_{j}} m_{r_{j}}+ R_{n}(r_{j}) \right] ^{2} \end{aligned}$$

(A.13)

$$\begin{aligned}&= \frac{T^{2}h_{n}^{-\mathbf{r}^{+}}}{C(\mathbf{r})^{2}} \frac{(M_{n}-2k+2)(M_{n}-2k+1)}{2M_{n}^{2}} \nonumber \\&\qquad \prod _{j=1}^{k} \left[ e^{-2\lambda _{n}} h_{n}^{r_{j}} \sigma ^{2r_{j}} m_{r_{j}}^{2} + 2e^{-\lambda _{n}} h_{n}^{r_{j}/2} \sigma ^{r_{j}} m_{r_{j}} R_{n}(r_{j}) + R_{n}(r_{j})^{2} \right] \nonumber \\&= \frac{T^{2}\sigma ^{2\mathbf{r}^{+}}}{2} + \frac{3-4k}{M_{n}}{\frac{T^{2}\sigma ^{2\mathbf{r}^{+}}}{2}} + {\mathcal {R}}_{n} \nonumber \\&\quad +\, o(h_{n}^{1-r_{\max }/2}), \quad \text {where} \quad {\mathcal {R}}_{n} = {\widehat{C}}_{\mathbf{r}}^{(k+1)} h_{n}^{1-r_{\max }/2}, \qquad (n \rightarrow {} \infty ), \end{aligned}$$

(A.14)

and ${\widehat{C}}_{\mathbf{r}}^{(k+1)} := 2T^{2}|I_{\max }| \lambda \sigma ^{(2\mathbf{r}^{+}-r_{\max })} {\mathbb {E}}\left[ |\zeta |^{r_{\max }}\right] m_{r_{\max }}^{-1}$ and $|I_{\max }| := \# \{ i \in \{1,2,\ldots ,k \} : r_{i} = r_{\max } \}$. We are now ready to show that (4.10-ii) is necessary and sufficient for (A.9) to hold. First, from (A.14), it transpires that if $r_{\max } < 2$, then $\lim _{n \rightarrow {} \infty } {\mathbb {E}}\left[ S_{n}^{(k+1)} \right] =T^{2} \sigma ^{2\mathbf{r}^{+}}/2$. Therefore, for (A.9) to be satisfied, it suffices to show that

$$\begin{aligned} \lim _{n \rightarrow {} \infty } {\mathbb {E}}\left[ S_{n}^{(\ell )} \right] = 0, \quad \ell = 1,2,\ldots ,k. \end{aligned}$$

(A.15)

The necessity of the above condition is also clear since each overlapping block term $S_{n}^{\ell }$ is nonnegative. To show that the condition (4.10-ii) is equivalent to (A.15), let us note that

$$\begin{aligned} {\mathbb {E}}\left[ S_{n}^{(\ell )} \right]&= T^{2}\frac{h_{n}^{-\mathbf{r}^{+}}}{C(\mathbf{r})^{2}} \frac{M_{n}-k-\ell +2}{M_{n}^{2}} \prod _{j=1}^{\ell -1} \left( e^{-\lambda _{n}} h_{n}^{r_{j}/2} \sigma ^{r_{j}} m_{r_{j}} + R_{n}(r_{j}) \right) \nonumber \\&\qquad \times \prod _{j=\ell }^{k} \left( e^{-\lambda _{n}} h_{n}^{(r_{j-\ell +1} + r_{j})/2} \sigma ^{(r_{j-\ell +1} + r_{j})} m_{(r_{j-\ell +1} + r_{j})} + R_{n}(r_{j-\ell +1} + r_{j}) \right) \nonumber \\&\qquad \times \prod _{j=k+1}^{k+\ell -1} \left( e^{-\lambda _{n}} h_{n}^{r_{j-\ell +1}/2} \sigma ^{r_{j-\ell +1}} m_{r_{j-\ell +1}} + R_{n}(r_{j-\ell +1}) \right) , \end{aligned}$$

(A.16)

whose slowest-converging term takes the form

$$\begin{aligned} T^{2}\frac{h_{n}^{-\mathbf{r}^{+}}}{C(\mathbf{r})^{2}} \frac{1}{M_{n}} {\prod _{j=1}^{\ell -1}h_{n}^{\frac{r_{j}}{2}} \sigma ^{r_{j}} m_{r_{j}} \prod _{j=\ell }^{k} h_{n}^{1\wedge \frac{r_{j-\ell +1} + r_{j}}{2}} k_{j,\ell }(\mathbf{r})} \prod _{j=k+1}^{k+\ell -1} h_{n}^{\frac{r_{j-\ell +1}}{2}} \sigma ^{r_{j-\ell +1}} m_{r_{j-\ell +1}}, \end{aligned}$$

where $k_{j,\ell }(\mathbf{r})=\sigma ^{(r_{j-\ell +1} + r_{j})} m_{(r_{j-\ell +1} + r_{j})}{} \mathbf{1}_{\{(r_{j-\ell +1} + r_{j})/2\le {}1\}}+\lambda {\mathbb {E}}|\zeta _{1}|^{r_{j-\ell +1} + r_{j}}\mathbf{1}_{\{(r_{j-\ell +1} + r_{j})/2\ge {}1\}}$. It is now clear that this term will vanish if and only if ${g}(\ell )>-1$, where

$$\begin{aligned} {g(\ell )}:= \sum _{j=1}^{\ell -1}\frac{r_{j}}{2} + \sum _{j=\ell }^{k} \left[ \frac{r_{j-\ell +1} + r_{j}}{2} \wedge 1 \right] + \sum _{j=k+1}^{k+\ell -1}\frac{r_{j-\ell +1}}{2} - \mathbf{r}^{+}; \quad \ell = 1, 2, \dots , k. \end{aligned}$$

The above function can be simplified to the expression given in (4.11). For future reference, let us also remark that, under the conditions given in (4.10),

$$\begin{aligned} {\mathbb {E}}\left[ S_{n}^{(\ell )} \right] \sim {T} {\widehat{C}}_{\mathbf{r}}^{(\ell )} h_{n}^{1+{g(\ell )}} \qquad (n \rightarrow {} \infty ), \qquad \ell = 1, \dots , k, \end{aligned}$$

(A.17)

where

$$\begin{aligned} {{\widehat{C}}_{\mathbf{r}}^{(\ell )}=\frac{1}{C(\mathbf{r})^{2}} \prod _{j=1}^{\ell -1}\sigma ^{r_{j}} m_{r_{j}} \prod _{j=\ell }^{k}k_{j,\ell }(\mathbf{r}) \prod _{j=k+1}^{k+\ell -1} \sigma ^{r_{j-\ell +1}} m_{r_{j-\ell +1}}.} \end{aligned}$$

(A.18)

Step 3. Let us first note that Step 1 and 2 above imply that first assertion of the theorem. Indeed, if both conditions in (4.10) are satisfied, then it was proved that

$$\begin{aligned} \mathrm{Bias}\left( \left| MPV(X)_{T}^{n [\mathbf{r}]} \right| ;T\sigma ^{\mathbf{r}^{+}}\right) \rightarrow {}0,\qquad \mathrm{Bias}\left( \left| MPV(X)_{T}^{n [\mathbf{r}]} \right| ^{2};T^{2}\sigma ^{2\mathbf{r}^{+}}\right) \rightarrow {}0,\qquad n\rightarrow \infty ,\nonumber \\ \end{aligned}$$

(A.19)

which, as it is well known, implies that $\mathrm{MSE}\left( \left| MPV(X)_{T}^{n [\mathbf{r}]} \right| ;T\sigma ^{\mathbf{r}^{+}}\right) \rightarrow {}0$. Reciprocally, if the latter limit holds, then the first limit in (A.19) must hold since $\mathrm{MSE}\ge {}\mathrm{Bias}^{2}$. As shown in Step 1, if the bias of the MPV estimator vanishes, then $r_{\max }<2$, and, as shown in Step 2 above, we will then have that the condition (4.10-ii) holds as well.

The asymptotics (4.12) was shown in the Step 1 above. Therefore, the only remaining assertion to prove is the asymptotics (4.14). To that end, let us first recall from (A.7), (A.10), and (A.13) that the first and second moment of the MPV estimator are given by

$$\begin{aligned} {\mathbb {E}}\left[ MPV(X)_{T}^{n [\mathbf{r}]}\right]&= \frac{h_{n}^{1-\mathbf{r}^{+}/2}}{C(\mathbf{r})} (M_{n}-k+1) \prod _{j=1}^{k} \left( e^{-\lambda _{n}} h_{n}^{r_{j}/2} \sigma ^{r_{j}} m_{r_{j}} + R_{n}(r_{j}) \right) \\ {\mathbb {E}}\left[ \left| MPV(X)_{T}^{n [\mathbf{r}]} \right| ^{2} \right]&= \frac{2h_{n}^{2-\mathbf{r}^{+}}}{C(\mathbf{r})^{2}} \left( {\begin{array}{c}M_{n} - 2k + 2\\ 2\end{array}}\right) \prod _{j=1}^{k} \left[ e^{-\lambda _{n}} h_{n}^{r_{j}/2} \sigma ^{r_{j}} m_{r_{j}} + R_{n}(r_{j}) \right] ^{2}\\&\quad +{\mathbb {E}}\left[ S_{n}^{(1)} \right] + 2 \sum _{\ell =2}^{k} {\mathbb {E}}\left[ S_{n}^{(\ell )} \right] , \end{aligned}$$

where $\{S_{n}^{(\ell )}\}_{\ell = 1,2,\dots ,k}$ are given as in (A.10). From these representations, it follows that

$$\begin{aligned} Var \left( MPV(X)_{T}^{n[\mathbf{r}]} \right)&={T^{2}}D_{n,k} \frac{h_{n}^{-\mathbf{r}^{+}}}{C(\mathbf{r})^{2}} \prod _{j=1}^{k} \left( e^{-\lambda _{n}} \sigma ^{r_{j}} h_{n}^{r_{j}/2} m_{r_{j}} + R_{n}(r_{j}) \right) ^{2} \nonumber \\&\quad + {\mathbb {E}}\left[ S_{n}^{(1)} \right] + 2 \sum _{\ell =2}^{k} {\mathbb {E}}\left[ S_{n}^{(\ell )} \right] , \end{aligned}$$

(A.20)

where

$$\begin{aligned} D_{n,k} := \frac{(M_{n}-2k+2)(M_{n}-2k+1)}{M_{n}^{2}} - \frac{(M_{n}-k+1)^{2}}{M_{n}^{2}} \sim \frac{(1-2k)}{M_{n}}, \quad \text {as} \quad n \rightarrow {} \infty . \end{aligned}$$

The asymptotics of ${\mathbb {E}}\left[ S_{n}^{(\ell )} \right]$ for $\ell =1,\dots ,k$ are stated in (A.17), while the asymptotics of the first term in (A.20) is given by [see also (A.13)]

$$\begin{aligned} {T^{2}}D_{n,k} \frac{h_{n}^{-\mathbf{r}^{+}}}{C(\mathbf{r})^{2}} \prod _{j=1}^{k} \left( e^{-\lambda _{n}} \sigma ^{r_{j}} h_{n}^{r_{j}/2} m_{r_{j}} + R_{n}(r_{j}) \right) ^{2}&= {T}(1-2k)\sigma ^{2\mathbf{r}^{+}}h_{n} + o(h_{n}). \end{aligned}$$

Therefore, we deduce the asymptotic behavior (4.14) for $\mathrm{Var} \left( MPV(X)_{T}^{n[\mathbf{r}]} \right)$ with a constant ${\overline{L}}(\mathbf{r})$ given by

$$\begin{aligned} {\overline{L}}(\mathbf{r}) := \mathbf{1}_{0<r_{\max }\le {}1}\sigma ^{2\mathbf{r}^{+}}(1-2k) + {\widehat{C}}_{\mathbf{r}}^{(1)} + 2\sum _{\ell =2}^{k} {\widehat{C}}_{\mathbf{r}}^{(\ell )}, \end{aligned}$$

(A.21)

where the constants ${\widehat{C}}_{\mathbf{r}}^{(\ell )}, \; \ell = 1,2,\dots ,k$, are defined by (A.18). $\square$

Proof of Theorem 4.4

Let us begin by noting that

$$\begin{aligned} h_{n}^{-1/2} \sum _{i=1}^{M_{n}}\left( |\Delta _{i}^{n}X|^{2}{} \mathbf{1}_{[\Delta _{i}^{n}N= 0]} - \sigma _{n,i}^{2}\right) {\mathop {\Longrightarrow }\limits ^{n \rightarrow {} \infty }}_{D} N\left( 0,2\int _{0}^{T} \sigma ^{4}_{s}ds\right) . \end{aligned}$$

(A.22)

The limit (A.22) can be derived from Theorem 1 in Barndorff-Nielsen and Shephard (2007), which states that

$$\begin{aligned} h_{n}^{-1/2} \sum _{i=1}^{M_{n}}\left( |\Delta _{i}^{n}X^{c}|^{2} - \sigma _{n,i}^{2}\right) {\mathop {\Longrightarrow }\limits ^{n \rightarrow {} \infty }}_{D} N\left( 0,2\int _{0}^{T} \sigma ^{4}_{s}ds\right) , \end{aligned}$$

(A.23)

and the fact that ${\mathcal {R}}_{n}:=h_{n}^{-1/2} \sum _{i=1}^{M_{n}}|\Delta _{i}^{n}X^{c}|^{2}{} \mathbf{1}_{[\Delta _{i}^{n}N\ne 0]} \overset{n \rightarrow {} \infty }{\rightarrow _{{\mathbb {P}}}} 0$ since

$$\begin{aligned} {\mathbb {E}}\left[ {\mathcal {R}}_{n}\right] =h_{n}^{-1/2} \sum _{i=1}^{M_{n}}\left( 1-e^{-\lambda _{n,i}}\right) \left( \gamma _{n,i}^{2}+\sigma _{n,i}^{2}\right) \le h_{n}^{1/2}T{{\bar{\lambda }}}_{T}\left( {{\bar{\gamma }}}_{T}^{2}h_{n}+{{\bar{\sigma }}}_{T}^{2}\right) \,{\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }}\,0. \end{aligned}$$

Next, we show that, under the conditions given in (4.17),

$$\begin{aligned} h_{n}^{-1/2} \left( \sum _{i=1}^{M_{n}} |\Delta _{i}^{n}X|^{2} \mathbf{1}_{[|\Delta _{i}^{n}X|\le B_{n}, \Delta _{i}^{n}N= 0]} - \int _{0}^{T} \sigma ^{2}_{s}ds \right) {\mathop {\Longrightarrow }\limits ^{n \rightarrow {} \infty }}_{D} N\left( 0,2\int _{0}^{T} \sigma ^{4}_{s}ds\right) .\nonumber \\ \end{aligned}$$

(A.24)

To this end, we show that

$$\begin{aligned} T_{n}:= \sum _{i=1}^{M_{n}} |\Delta _{i}^{n}X|^{2} \mathbf{1}_{[|\Delta _{i}^{n}X| > B_{n}, \Delta _{i}^{n}N= 0]}=o_{P}(h_{n}^{1/2}). \end{aligned}$$

Since $T_{n}$ is nonnegative, it suffices to show that $h_{n}^{-1/2} {\mathbb {E}}T_{n}\rightarrow {}0$, as $n\rightarrow \infty$. Using (2.7),

$$\begin{aligned} {\mathbb {E}}\left[ {T}_{n}\right]&=\sum _{i=1}^{M_{n}}e^{-\lambda _{n,i}}\left( {\sigma }_{n,i}^{2}+{\gamma }_{n,i}^{2}\right) \left( {{\bar{\Phi }}}\left( \frac{B_{n}+{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) +{{\bar{\Phi }}}\left( \frac{B_{n}-{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) \right) \nonumber \\&\quad +\sum _{i=1}^{M_{n}}e^{-\lambda _{n,i}}{\sigma }_{n,i}\left( \left( B_{n}-{\gamma }_{n,i}\right) \phi \left( \frac{B_{n}+{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) +\left( B_{n}+{\gamma }_{n,i}\right) \phi \left( \frac{B_{n}-{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) \right) . \end{aligned}$$

(A.25)

Applying (2.6) and after some simplifications, for large enough n,

$$\begin{aligned} {\mathbb {E}}\left[ {T}_{n}\right]&\le {} \sum _{i=1}^{M_{n}}{\sigma }_{n,i}\left( B_{n}^{2}+{\sigma }_{n,i}^{2}\right) \left( \frac{1}{B_{n}+{\gamma }_{n,i}}\phi \left( \frac{B_{n}+{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) +\frac{1}{B_{n}-{\gamma }_{n,i}}\phi \left( \frac{B_{n}-{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) \right) . \end{aligned}$$

(A.26)

Finally, using that $|{\gamma }_{n,i}|\le {}{\bar{\gamma }}_{T}h_{n}$ and ${\underline{\sigma }}_{T}^{2}h_{n}\le {}{\sigma }^{2}_{n,i}\le {}{{\bar{\sigma }}}_{T}^{2}h_{n}$, one can see that the right-hand side of (A.26) multiplied by $h_{n}^{-1/2}$ converges to 0 if and only if (4.17) holds.

We are now ready to show (4.16). We begin with the decomposition

$$\begin{aligned} TQV(X)[{\mathbf{B}}]_{{T}}^{n}= & {} \sum _{i=1}^{M_{n}}|\Delta _{i}^{n}X|^{2} \mathbf{1}_{[|\Delta _{i}^{n}X| \le B_{n}, \Delta _{i}^{n}N= 0]} \\&+ \sum _{i=1}^{M_{n}}|\Delta _{i}^{n}X|^{2} \mathbf{1}_{[|\Delta _{i}^{n}X| \le B_{n}, \Delta _{i}^{n}N\ne 0]} =: D_{n}^{(1)} + D_{n}^{(2)}. \end{aligned}$$

From (A.24) and Slutsky’s Theorem, it is clear that it suffices to show that $h_{n}^{-1/2} D_{n}^{(2)} \overset{n \rightarrow {} \infty \;\;}{\rightarrow _{{\mathbb {P}}}} 0$. For the latter identity, let us consider the following:

$$\begin{aligned}&{\mathbb {P}}\left( \frac{1}{h_{n}^{1/2}} \sum _{i=1}^{M_{n}}|\Delta _{i}^{n}X|^{2} \mathbf{1}_{[|\Delta _{i}^{n}X| \le B_{n}, \Delta _{i}^{n}N\ne 0]} \ne 0 \right) \nonumber \\&\quad \le {\mathbb {P}}\left( \bigcup _{i=1}^{M_{n}} \left[ |\Delta _{i}^{n}X|\le B_{n}, \Delta _{i}^{n}N\ne 0 \right] \right) \nonumber \\&\quad = 1 - \exp \left( \sum _{i=1}^{M_{n}}\ln \left( 1- {\mathbb {P}}\left( |\Delta _{i}^{n}X|\le B_{n}, \Delta _{i}^{n}N\ne 0 \right) \right) \right) \nonumber \\&\quad \le {} 1 - \exp \left( \sum _{i=1}^{M_{n}}\ln \left( 1- 2\Vert f\Vert _{\infty }B_{n}h_{n}{\bar{\lambda }}_{T} \right) \right) \end{aligned}$$

(A.27)

where the last inequality follows from the next easy implications of (C.9):

$$\begin{aligned} {\mathbb {P}}\left( |\Delta _{i}^{n}X|\le B_{n}, \Delta _{i}^{n}N\ne 0 \right)&=e^{-{\lambda }_{n,i}}\sum _{k=1}^{\infty }{\mathbb {P}}\left( \left| {\sigma }_{n,i}Z+{\gamma }_{n,i}+\sum _{j=1}^{k}\zeta _{j}\right| \le B_{n} \right) \frac{{\lambda }_{n,i}^{k}}{k!}\\&\le 2B_{n}e^{-{\lambda }_{n,i}}\sum _{k=1}^{\infty }\Vert f^{*k}\Vert _{\infty }\frac{{\lambda }_{n,i}^{k}}{k!}\\&\le {}2\Vert f\Vert _{\infty }B_{n}{\lambda }_{n,i}. \end{aligned}$$

It is now clear that, provided that $B_{n}\rightarrow {}0$, (A.27) tends to 0, which in turn implies that $\displaystyle {\lim _{n \rightarrow {} \infty }}{\mathbb {P}}\left( h_{n}^{-1/2}D_{n}^{(2)} \ne 0 \right) = 0$. The proof is now complete. $\square$

Appendix B: Proofs of Section 4.2

Proof of Proposition 4.5

Throughout, Z denotes a standard normal variable. We fix $A_{i} := \mathbf{1}_{[|\Delta _{i}^{n}X| > B_{n}]} - \Delta _{i}^{n}N$, for $i = 1,2,\dots ,M_{n}$, and note that $(A_{i})_{i=1}^{M_{n}}$ form a collection of independent random variables. We begin by recording the first and second moments of each $A_{i}$. By conditioning on $\Delta _{i}^{n}N$,

$$\begin{aligned} {\mathbb {E}}[A_{i}]&= e^{-\lambda _{n,i}} {\mathbb {P}}\left( |\gamma _{n,i} + \sigma _{n,i} Z |> B_{n} \right) - \lambda _{n,i} {\mathbb {P}}\left( |\gamma _{n,i} + \sigma _{n,i} Z + \zeta _{1}| \le B_{n} \right) + {R}_{n,i}^{(1)},\\ {\mathbb {E}}[A_{i}^{2}]&= e^{-\lambda _{n,i}} {\mathbb {P}}\left( |\gamma _{n,i} + \sigma _{n,i} Z | > B_{n} \right) + \lambda _{n,i} {\mathbb {P}}\left( |\gamma _{n,i} + \sigma _{n,i} Z + \zeta _{1}| \le B_{n} \right) + {R}^{(2)}_{n,i}, \end{aligned}$$

where

$$\begin{aligned} {R}_{n,i}^{(1)}&:=\lambda _{n,i}{\mathbb {P}}\left( \left| \gamma _{n,i} + \sigma _{n,i} Z + \zeta _{1} \right|> {B_{n}} \right) (e^{-\lambda _{n,i}}-1)\\&\quad +\sum _{j=2}^{\infty } e^{-\lambda _{n,i}} \frac{\lambda _{n,i}^{j}}{j!} {\mathbb {P}}\left( \left| \lambda _{n,i} + \sigma _{n,i} Z + \zeta _{1} + \cdots + \zeta _{j} \right|> B_{n} \right) \\ {R}_{n,i}^{(2)}&:=\lambda _{n,i} {\mathbb {P}}\left( \left| \gamma _{n,i} + \sigma _{n,i} Z + \zeta _{1} \right|> B_{n} \right) (1-e^{-\lambda _{n,i}}) +\lambda _{n,i}^{2}\\&\quad +\sum _{j=2}^{\infty } e^{-\lambda _{n,i}} \frac{\lambda _{n,i}^{j}}{j!}(1-2j){\mathbb {P}}\left( \left| \gamma _{n,i} + \sigma _{n,i}Z + \zeta _{1} + \cdots + \zeta _{j} \right| > B_{n} \right) . \end{aligned}$$

Note that $\left| {R}_{n,i}^{(1)}\right| \le 2\lambda _{n,i}^{2}$ and $\left| {R}_{n,i}^{(2)}\right| \le 4\lambda _{n,i}^{2}$. Next, using the mutual independence of the $A_{i}$’s,

$$\begin{aligned} {\mathbb {E}}\left[ \left| {\widehat{N}}[\mathbf B ]_{T_{n}}^{n} - N_{T_{n}} \right| ^{2} \right]&= \sum _{i=1}^{M_{n}} {\mathbb {E}}[A_{i}^{2}] + \sum _{i \ne j}{\mathbb {E}}[A_{i}]{\mathbb {E}}[A_{j}] \\&\le {} \sum _{i=1}^{M_{n}} {\mathbb {E}}[A_{i}^{2}] + \left( \sum _{i=1}^{M_{n}}{\mathbb {E}}[A_{i}]\right) ^{2} \\&={\bar{V}}_{n}^{(1)}+{\bar{V}}_{n}^{(2)}+\left( {\bar{V}}_{n}^{(1)}+{\bar{V}}_{n}^{(2)}+{\bar{R}}_{n}^{(1)}\right) ^{2}+{\bar{R}}^{(2)}_{n}, \end{aligned}$$

where we have fixed

$$\begin{aligned}&{\bar{V}}^{(1)}_{n}:=\sum _{i=1}^{M_{n}}e^{-\lambda _{n,i}} {\mathbb {P}}\left( |\gamma _{n,i} + \sigma _{n,i}Z | > B_{n} \right) , \quad {\bar{V}}^{(2)}_{n}:=\sum _{i=1}^{M_{n}}\lambda _{n,i} {\mathbb {P}}\left( |\gamma _{n,i} + \sigma _{n,i}Z + {\zeta _{1}} | \le B_{n} \right) ,\\&{\bar{R}}_{n}^{(1)}:=\sum _{i=1}^{M_{n}}R^{(1)}_{n,i}, \qquad {\bar{R}}_{n}^{(2)}:= \sum _{i=1}^{M_{n}} {R}_{n,i}^{(2)}. \end{aligned}$$

Note that, for $\ell =1,2$ and some constants $K_{\ell }$,

$$\begin{aligned} \left| {\bar{R}}_{n}^{(\ell )}\right| \le K_{\ell } \sum _{i=1}^{M_{n}}\lambda _{n,i}^{2}\le K_{\ell }h_{n}T_{n}{\bar{\lambda }}_{T_{n}}^{2}\rightarrow {}0,\quad n\rightarrow {}0, \end{aligned}$$

which converges to 0 due to the last limit in (4.20). Thus, for the MSE of ${\widehat{N}}[\mathbf B ]_{t}^{n}$ to converges to 0, it suffices that $\lim _{n\rightarrow {} \infty }{\bar{V}}_{n}^{(1)}=\lim _{n\rightarrow \infty }{\bar{V}}_{n}^{(2)} =0$. From (C.9),

$$\begin{aligned} {\bar{V}}_{n}^{(2)}\le {} {2} \Vert f\Vert _{\infty }B_{n}\sum _{i=1}^{M_{n}}\lambda _{n,i}\le {}2\Vert f\Vert _{\infty }B_{n}{{\bar{\lambda }}}_{T_{n}}T_{n}, \end{aligned}$$

which converges to 0 by the second limit in (4.20). Now, from the third condition in (4.20) and the Gaussian tail estimate (2.6),

$$\begin{aligned} {\bar{V}}_{n}^{(1)}&= \sum _{i=1}^{M_{n}}e^{-\lambda _{n,i}}\left( {{\bar{\Phi }}}\left( \frac{B_{n}+\gamma _{n,i}}{\sigma _{n,i}}\right) +{{\bar{\Phi }}}\left( \frac{B_{n}-\gamma _{n,i}}{\sigma _{n,i}}\right) \right) \\&\le {}2 M_{n}{{\bar{\Phi }}}\left( \frac{B_{n}-{\bar{\gamma }}_{T_{n}}h_{n}}{{\bar{\sigma }}_{T_{n}}\sqrt{h_{n}}}\right) \\&\le {}2\frac{M_{n}{{\bar{\sigma }}}_{T_{n}}\sqrt{h_{n}}}{B_{n}-{{\bar{\gamma }}}_{T_{n}}h_{n}}\phi \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T_{n}}\sqrt{h_{n}}}\right) \exp \left( {\frac{B_{n}{{\bar{\gamma }}}_{T_{n}}}{{{\bar{\sigma }}}_{T_{n}}^{2}}}\right) , \end{aligned}$$

which converges to 0 by conditions in (4.20).

Now, we consider the second assertion dealing with a FJA Lévy model. In that case, it is easy to see that ${\bar{R}}_{n}^{(2)}=M_{n} {R}_{n}^{(2)} \sim M_{n}h_{n}^{2} \lambda ^{2}/2>0$ whenever $h_{n},B_{n}\rightarrow {}0$. Therefore, for large enough n,

$$\begin{aligned} {\mathbb {E}}\left[ \left| {\widehat{N}}[\mathbf B ]_{T_{n}}^{n} - N_{T_{n}} \right| ^{2} \right] \ge {\bar{V}}_{n}^{(1)} + {\bar{V}}_{n}^{(2)}\ge {}0. \end{aligned}$$

Thus, if the mean-squared error vanishes, then $\lim _{n\rightarrow {} \infty }{\bar{V}}_{n}^{(1)}=\lim _{n\rightarrow \infty }{\bar{V}}_{n}^{(2)} =0$. Now, by (C.10) with $k=0$, ${\mathbb {P}}\left( |\gamma h_{n} + \sigma \sqrt{h_{n}}Z + \zeta _{1}| \le B_{n} \right) \sim {2}B_{n}{\mathcal {C}}_{0}(f)$ when $n \rightarrow {} \infty$ and, hence, ${\bar{V}}_{n}^{(2)} \sim {2}\lambda M_{n} h_{n} B_{n}{\mathcal {C}}_{0}(f)$, as $n \rightarrow {} \infty$. In particular, if ${\bar{V}}_{n}^{(2)}$ converges to 0, the second condition in (4.20) must hold. On the other hand, if ${\bar{V}}_{n}^{(1)}$ converges to 0, then necessarily $\lim _{n\rightarrow \infty } B_{n}/h_{n}^{1/2} =\infty$ and, by (2.6),

$$\begin{aligned} {\bar{V}}_{n}^{(1)} \sim \frac{2 M_{n} \sigma \sqrt{h_{n}}}{B_{n}} \phi \left( \frac{B_{n}}{\sigma \sqrt{h_{n}}} \right) , \qquad (n\rightarrow \infty ), \end{aligned}$$

(B.1)

which further implies that the first condition in (4.20) must hold. Finally, the decomposition (4.22) directly follows from the asymptotic rates for ${\bar{V}}_{n}^{(1)}$, ${\bar{V}}_{n}^{(2)}$, ${\bar{R}}_{n}^{(1)}$, and ${\bar{R}}_{n}^{(2)}$ stated above and the next decomposition:

$$\begin{aligned} {\mathbb {E}}\left[ \left| {\widehat{N}}[\mathbf B ]_{T_{n}}^{n} - N_{T_{n}} \right| ^{2} \right] ={\bar{V}}_{n}^{(1)}+{\bar{V}}_{n}^{(2)}+\frac{M_{n}(M_{n}-1)}{M_{n}^{2}}\left( {\bar{V}}_{n}^{(1)}+{\bar{V}}_{n}^{(2)}+{\bar{R}}_{n}^{(1)}\right) ^{2}+{\bar{R}}^{(2)}_{n}. \end{aligned}$$

$\square$

Proof of Proposition 4.6

Throughout we will assume $\gamma \ne 0$. The proof is similar and simpler if $\gamma = 0$. Let $Z_{1},Z_{2},\dots$ denote independent standard Gaussian variables and let $Y_{i} := \Delta _{i}^{n}X\mathbf{1}_{[|\Delta _{i}^{n}X| > B_{n}]} - \Delta _{i}^{n}J$ so that ${\widehat{J}}[\mathbf B ]_{T_{n}}^{n} - J_{T_{n}} = \sum _{i=1}^{M_{n}} Y_{i}.$ Note that $\Delta _{i}^{n}J= \Delta _{i}^{n}J\mathbf{1}_{[\Delta _{i}^{n}N\ne 0]}$ and, thus,

$$\begin{aligned} Y_{i}&= (\Delta _{i}^{n}X) \mathbf{1}_{[|\Delta _{i}^{n}X| > B_{n}, \Delta _{i}^{n}N= 0]} - (\Delta _{i}^{n}X) \mathbf{1}_{[\Delta _{i}^{n}N\ne 0, |\Delta _{i}^{n}X| \le B_{n}]}+\left( \Delta _{i}^{n}X^{c}\right) \mathbf{1}_{[ \Delta _{i}^{n}N\ne 0]} \\&=: {\bar{T}}_{{n,i}}^{(1)}+ {\bar{T}}_{{n,i}}^{(2)}+{\bar{T}}_{{n,i}}^{(3)}. \end{aligned}$$

Hence,

$$\begin{aligned} {\mathbb {E}}\left[ \left| {{\widehat{J}}[\mathbf B ]_{T_{n}}^{n}} - J_{T_{n}}\right| ^{2} \right]&= \sum _{i=1}^{M_{n}} {\mathbb {E}}[Y_{i}^{2}] + \sum _{i \ne j} {\mathbb {E}}[Y_{i}]{\mathbb {E}}[Y_{j}]\le {} \sum _{i=1}^{M_{n}} {\mathbb {E}}[Y_{i}^{2}] + \left( \sum _{i =1}^{M_{n}} {\mathbb {E}}[Y_{i}]\right) ^{2} \end{aligned}$$

(B.2)

$$\begin{aligned}&\le {}{3}\sum _{\ell =1}^{3} \sum _{i=1}^{M_{n}} {\mathbb {E}}\left[ \left( {\bar{T}}_{{n,i}}^{(\ell )}\right) ^{2}\right] +\left( \sum _{\ell =1}^{3}\sum _{i=1}^{M_{n}}{\mathbb {E}}\left[ {\bar{T}}_{{n,i}}^{(\ell )}\right] \right) ^{2}. \end{aligned}$$

(B.3)

Therefore, for the above expression to converge to 0, it suffices to show that

$$\begin{aligned} \lim _{n\rightarrow {}\infty }\sum _{i=1}^{M_{n}} {\mathbb {E}}\left[ \left( {\bar{T}}_{{n,i}}^{(\ell )}\right) ^{2}\right] =0, \qquad \lim _{n\rightarrow \infty }\sum _{i=1}^{M_{n}} {\mathbb {E}}\left[ {\bar{T}}_{{n,i}}^{(\ell )}\right] =0,\quad \ell =1,2,3. \end{aligned}$$

First, using (2.7), we can find the first two moments of ${\bar{T}}_{n,i}^{(1)}$ as

$$\begin{aligned} {\mathbb {E}}\left[ {\bar{T}}_{n,i}^{(1)}\right]&=e^{-\lambda _{n,i}}{\sigma }_{n,i}\left( \phi \left( \frac{B_{n}-{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) -\phi \left( \frac{B_{n}+{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) \right) \nonumber \\&\quad +e^{-\lambda _{n,i}}\gamma _{n,i}\left( {{\bar{\Phi }}}\left( \frac{B_{n}+{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) +{{\bar{\Phi }}}\left( \frac{B_{n}-{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) \right) \nonumber \\ {\mathbb {E}}\left[ \left( {\bar{T}}_{n,i}^{(1)}\right) ^{2}\right]&=e^{-\lambda _{n,i}}\left( {\sigma }_{n,i}^{2}+{\gamma }_{n,i}^{2}\right) \left( {{\bar{\Phi }}}\left( \frac{B_{n}+{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) +{{\bar{\Phi }}}\left( \frac{B_{n}-{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) \right) \nonumber \\&\quad +e^{-\lambda _{n,i}}{\sigma }_{n,i}\left( \left( B_{n}-{\gamma }_{n,i}\right) \phi \left( \frac{B_{n}+{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) +\left( B_{n}+{\gamma }_{n,i}\right) \phi \left( \frac{B_{n}-{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) \right) . \end{aligned}$$

(B.4)

Next, using (2.6), which can be applied in light of the second condition in (4.26), and some simplifications,

$$\begin{aligned} \left| {\mathbb {E}}\left[ {\bar{T}}_{n,i}^{(1)}\right] \right|\le & {} \phi \left( \frac{B_{n}}{\sigma _{n,i}}\right) e^{\frac{B_{n}|\gamma _{n,i}|}{\sigma _{n,i}^{2}}}\frac{B_{n}|\gamma _{n,i}|}{\sigma _{n,i}}\\&+ |\gamma _{n,i}|{\sigma }_{n,i}\left( \frac{1}{B_{n}+{\gamma }_{n,i}}\phi \left( \frac{B_{n}+{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) +\frac{1}{B_{n}-{\gamma }_{n,i}}\phi \left( \frac{B_{n}-{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) \right) . \end{aligned}$$

Thus, using (2.2),

$$\begin{aligned} \sum _{i=1}^{M_{n}}\left| {\mathbb {E}}\left[ {\bar{T}}_{n,i}^{(1)}\right] \right|&\le M_{n}\sqrt{h_{n}}\phi \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T_{n}}\sqrt{h_{n}}}\right) \exp \left( \frac{B_{n}{\bar{\gamma }}_{T_{n}}}{{\underline{\sigma }}^{2}_{T_{n}}}\right) \frac{B_{n}{\bar{\gamma }}_{T_{n}}}{{\underline{\sigma }}_{T_{n}}}\\&\quad +2M_{n}\frac{{{\bar{\gamma }}}_{T_{n}}{{\bar{\sigma }}}_{T_{n}}h_{n}^{3/2}}{B_{n}-{{\bar{\gamma }}}_{T_{n}}h_{n}}\phi \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T_{n}}\sqrt{h_{n}}}\right) \exp \left( \frac{B_{n}{\bar{\gamma }}_{T_{n}}}{{\bar{\sigma }}_{T_{n}}^{2}}\right) , \end{aligned}$$

which vanishes as $n\rightarrow {}\infty$ in light of the conditions in (4.25)–(4.26). Similarly, applying (2.6),

$$\begin{aligned} \sum _{i=1}^{M_{n}}{\mathbb {E}}\left[ \left( {\bar{T}}_{n,i}^{(1)}\right) ^{2}\right]&\le {} \sum _{i=1}^{M_{n}}{\sigma }_{n,i}\left( B_{n}^{2}+{\sigma }_{n,i}^{2}\right) \left( \frac{1}{B_{n}+{\gamma }_{n,i}}\phi \left( \frac{B_{n}+{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) +\frac{1}{B_{n}-{\gamma }_{n,i}}\phi \left( \frac{B_{n}-{\gamma }_{n,i}}{{\sigma }_{n,i}}\right) \right) \nonumber \\&\le 2M_{n}{{\bar{\sigma }}}_{T_{n}}\sqrt{h_{n}}\frac{B_{n}^{2}+{{\bar{\sigma }}}^{2}_{T_{n}}h_{n}}{B_{n}-{{\bar{\gamma }}}_{T_{n}}h_{n}}\phi \left( \frac{B_{n}}{{{\bar{\sigma }}}_{T_{n}}\sqrt{h_{n}}}\right) \exp \left( \frac{B_{n}{\bar{\gamma }}_{T_{n}}}{{\bar{\sigma }}_{T_{n}}^{2}}\right) , \end{aligned}$$

(B.5)

which again converges to 0 due to the conditions in (4.25)–(4.26). To handle ${\bar{T}}^{(2)}$, let us fix ${{\bar{\zeta }}}_{k}:=\sum _{j=1}^{k}\zeta _{j}$ and note that, by (C.9), for $m=1,2$,

$$\begin{aligned} \left| {\mathbb {E}}[({\bar{T}}_{{n,i}}^{(2)})^{m}]\right|&\le e^{-{\lambda }_{n,i}}\sum _{k=1}^{\infty }{\mathbb {E}}\left( \left| {\sigma }_{n,i}Z+{\gamma }_{n,i}+{{\bar{\zeta }}}_{k}\right| ^{m} \mathbf{1}_{\left[ \left| {\sigma }_{n,i}Z+{\gamma }_{n,i}+{{\bar{\zeta }}}_{k}\right| \le B_{n}\right] } \right) \frac{{\lambda }_{n,i}^{k}}{k!}\nonumber \\&\le 2\frac{B_{n}^{m+1}}{m+1}e^{-{\lambda }_{n,i}}\sum _{k=1}^{\infty }\Vert f^{*k}\Vert _{\infty }\frac{{\lambda }_{n,i}^{k}}{k!} \le {}2\Vert f\Vert _{\infty }\frac{B_{n}^{m+1}}{m+1}{\lambda }_{n,i}. \end{aligned}$$

(B.6)

Therefore, using (2.2),

$$\begin{aligned} \sum _{i=1}^{M_{n}}\left| {\mathbb {E}}[({\bar{T}}_{{n,i}}^{(2)})^{m}]\right| \le 2\Vert f\Vert _{\infty }\frac{B_{n}^{m+1}}{m+1}\sum _{i=1}^{M_{n}}\lambda _{n,i}\le 2\Vert f\Vert _{\infty }\frac{B_{n}^{m+1}}{m+1}T_{n}{{\bar{\lambda }}}_{T_{n}}, \end{aligned}$$

which vanishes due to the last condition in (4.26). Finally, we can directly compute

$$\begin{aligned} {\mathbb {E}}[{\bar{T}}_{n,i}^{(3)}]&= \gamma _{n,i}\left( 1-e^{-\lambda _{n,i}}\right) ,\quad {\mathbb {E}}[|{\bar{T}}_{n}^{(3)}|^{2}] =\left( 1-e^{-\lambda _{n,i}}\right) \left( \gamma _{n,i}^{2}+\sigma _{n,i}^{2}\right) . \end{aligned}$$

(B.7)

Thus,

$$\begin{aligned} \sum _{i=1}^{M_{n}}\left| {\mathbb {E}}[{\bar{T}}_{n,i}^{(3)}]\right| \le {}T_{n}{{\bar{\gamma }}}_{T_{n}}{{\bar{\lambda }}}_{T_{n}}h_{n},\quad \sum _{i=1}^{M_{n}}{\mathbb {E}}[({\bar{T}}_{n}^{(3)})^{2}]\le T_{n}{{\bar{\lambda }}}_{T_{n}}h_{n}\left( {{\bar{\gamma }}}_{T_{n}}^{2}h_{n}+{{\bar{\sigma }}}_{T_{n}}^{2}\right) , \end{aligned}$$

which vanishes due to the last three conditions in (4.24).

We now show the second assertion of the proposition and, thus, hereafter we assume a FJA Lévy model and fix $\gamma _{n}=\gamma h_{n}$, $\sigma _{n}=\sigma \sqrt{h_{n}}$, and $\lambda _{n}=\lambda h_{n}$. In that case, it is easy to check that the three conditions in (4.28) imply all the conditions in (4.24)–(4.26). To show the necessity of (4.28), it is useful to note that, in the FJA case, (B.2) can further be simplified as follows:

$$\begin{aligned} {\mathbb {E}}\left[ |{{\widehat{J}}[\mathbf B ]_{T_{n}}^{n}} - J_{T_{n}}|^{2} \right]&= \sum _{i=1}^{M_{n}} {\mathbb {E}}[Y_{i}^{2}] + \sum _{i \ne j} {\mathbb {E}}[Y_{i}]{\mathbb {E}}[Y_{j}] = {M_{n} {\mathbb {E}}[Y_{1}^{2}] + M_{n}(M_{n}-1) {\mathbb {E}}[Y_{1}]^{2}}. \end{aligned}$$

(B.8)

By writing ${\bar{\zeta }}_{k}=\sum _{i=1}^{k}\zeta _{i}$ and conditioning on $N_{h_{n}}$,

$$\begin{aligned} {\mathbb {E}}[Y_{1}^{2}]=\sum _{k=0}^{\infty }e^{-\lambda _{n}}\frac{\lambda _{n}^{k}}{k!}\,{\mathbb {E}}\left[ \left\{ \left( \gamma _{n} + \sigma _{n} Z +{\bar{\zeta }}_{k}\right) \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z +{\bar{\zeta }}_{k}| > B_{n}]}-{\bar{\zeta }}_{k}\right\} ^{2} \right] \end{aligned}$$

(B.9)

Note that the first term in (B.9) (corresponding to $k=0$) is precisely ${\mathbb {E}}[|{\bar{T}}_{n,1}^{(1)}|^{2}]$. Therefore,

$$\begin{aligned} {\mathbb {E}}\left[ |{{\widehat{J}}[\mathbf B ]_{T_{n}}^{n}} - J_{T_{n}}|^{2} \right] \ge M_{n}{\mathbb {E}}[|{\bar{T}}_{n}^{(1)}|^{2}]= M_{n} {\mathbb {E}}\left[ \left| \gamma _{n} + \sigma _{n} Z \right| ^{2} \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z | > B_{n}, \Delta _{i}^{n}N= 0]} \right] \ge {}0. \end{aligned}$$

Observe that

$$\begin{aligned} {\mathbb {E}}[|{\bar{T}}_{n}^{(1)}|^{2}]&= e^{-\lambda _{n}} \sigma ^{2} h_{n} \left[ \int _{(B_{n}-\gamma _{n})/\sigma _{n}}^{\infty } |\gamma h_{n}^{1/2}/\sigma + z|^{2} \phi (z)dz \right. \\&\left. \quad + \int ^{-(B_{n}+\gamma _{n})/\sigma _{n}}_{-\infty } |\gamma h_{n}^{1/2}/\sigma + z|^{2} \phi (z)dz \right] . \end{aligned}$$

Suppose that ${L} := \liminf _{n \rightarrow {} \infty }B_{n}/h_{n}^{1/2}<\infty$ and let $n_{k}$ be an increasing subsequence such that $\lim _{k \rightarrow {} \infty }B_{n_{k}}/\sqrt{h_{n_{k}}}={L}$. Then, from the Dominated Convergence Theorem,

$$\begin{aligned} \lim _{k\rightarrow \infty } \frac{{\mathbb {E}}[|{\bar{T}}_{n_{k}}^{(1)}|^{2}]}{h_{n_{k}}} = \sigma ^{2} \int _{(-L/\sigma ,L/\sigma )^{c}} z^{2} \phi (z)dz, \end{aligned}$$

which implies that $\lim _{k \rightarrow {} \infty } M_{n_{k}}{\mathbb {E}}[|{\bar{T}}_{n_{k}}^{(1)}|^{2}] >0$, since, by the assumption of the sampling design, $\lim _{n \rightarrow {} \infty } M_{n}h_{n} \in (0,\infty ]$. Therefore, $\lim _{n \rightarrow {} \infty } MSE \left( {\widehat{J}}[B]_{T_{n}}^{n}; J_{T_{n}} \right) = 0$ implies that $\lim _{n \rightarrow {} \infty }B_{n}/\sqrt{h_{n}}=\infty$. Furthermore, from Lemma C.1,

$$\begin{aligned} {\mathbb {E}}[|{\bar{T}}_{n}^{(1)}|^{2}] \sim 2\sigma B_{n}h_{n}^{1/2} \phi \left( \frac{B_{n}}{\sigma _{n}} \right) , \end{aligned}$$

and, thus, the second limit in (4.28) must hold. To show that the last limit therein holds, consider the second term on the right-hand side of (B.8) and note that we must have that $M_{n}{\mathbb {E}}[Y_{1}]\rightarrow {}0$. Next, applying Lemmas C.1 and C.2 to the decomposition

$$\begin{aligned} {\mathbb {E}}[Y_{1}]&={\mathbb {E}}[ (\Delta _{1}^{n}X \mathbf{1}_{[|\Delta _{1}^{n}X |> B_{n}]} - \Delta _{1}^{n}J)\mathbf{1}_{[\Delta _{1}^{n}N=0]} ] +{\mathbb {E}}[ (\Delta _{1}^{n}X \mathbf{1}_{[|\Delta _{1}^{n}X | > B_{n}]} - \Delta _{1}^{n}J)\mathbf{1}_{[\Delta _{1}^{n}N\ne {}0]} ]\\&={\mathbb {E}}[ {\bar{T}}_{n,1}^{(1)}]+{\mathbb {E}}[ \Delta _{1}^{n}X^{c}\mathbf{1}_{[\Delta _{1}^{n}N\ne {}0]}]- {\mathbb {E}}[ (\Delta _{1}^{n}X) \mathbf{1}_{[|\Delta _{1}^{n}X |\le {} B_{n},\Delta _{1}^{n}N\ne {}0]}], \end{aligned}$$

we get:

$$\begin{aligned} {\mathbb {E}}[Y_{1}]= 2\frac{\gamma }{\sigma } B_{n}h_{n}^{1/2} \phi \left( \frac{B_{n}}{\sigma _{n}} \right) +\gamma h_{n}(1-e^{-\lambda h_{n}}) -\frac{\lambda h_{n} B_{n}^{2}}{2} \overline{{\mathcal {C}}}(f)+\mathrm{h.o.t.} \end{aligned}$$

Since $B_{n}\gg \sqrt{h_{n}}$ and $T_{n}B_{n}h_{n}^{-1/2} \phi \left( \frac{B_{n}}{\sigma _{n}} \right) \rightarrow {}0$, we must have that the last limit in (4.28) holds. $\square$

Appendix C: Supporting technical lemmas

Throughout this section, X denotes a FJA Lévy process with constant drift $\gamma$, volatility $\sigma$, and jump intensity $\lambda$. Let us recall the notation $\gamma _{n} := \gamma h_{n}$, $\sigma ^{2}_{n} :=\sigma ^{2}h_{n}$, and $\lambda _{n}=\lambda h_{n}$. In order to unify the presentation, we shall use the following terminology for a given threshold sequence $\mathbf B = (B_{n})_{n}$:

$$\begin{aligned}&{\mathcal {L}}_{n}^{I}[\mathbf{B}](k) := {\mathbb {E}}\left[ (\Delta _{1}^{n}X)^{k} \mathbf{{1}}_{[|\Delta _{1}^{n}X| > B_{n}, \Delta _{1}^{n}N= 0]} \right] ,\quad {\mathcal {L}}_{n}^{II}[\mathbf{B}](k) := {\mathbb {E}}\left[ (\Delta _{1}^{n}X)^{k} \mathbf{{1}}_{[|\Delta _{1}^{n}X| \le B_{n}, \Delta _{1}^{n}N\ne 0]} \right] , \end{aligned}$$

(C.1)

$$\begin{aligned}&{\mathcal {L}}_{n}^{II,Abs}[\mathbf{B}](k) := {\mathbb {E}}\left[ |\Delta _{1}^{n}X|^{k} \mathbf{{1}}_{[|\Delta _{1}^{n}X| \le B_{n}, \Delta _{1}^{n}N\ne 0]} \right] ,\quad {\mathcal {G}}^{I}_{n}[\mathbf{B}](k) := {\mathbb {E}}\left[ |\Delta _{1}^{n}X|^{k} \mathbf{{1}}_{[|\Delta _{1}^{n}X|\le B_{n}, \Delta _{1}^{n}N= 0]} \right] . \end{aligned}$$

(C.2)

Lemma C.1

Let $\mathbf B = (B_{n})_{n}$ be a threshold sequence such that $\lim _{n \rightarrow {} \infty } B_{n}h_{n}^{-1/2} = \infty$. Then, as $n \rightarrow {} \infty$,

$$\begin{aligned} {\mathcal {L}}_{n}^{I}[\mathbf B ](k) \sim \left\{ \begin{array}{cc} 2 \sigma _{n} B_{n}^{k-1} \phi \left( \frac{B_{n}}{\sigma _{n}} \right) ; &{} \quad k \in \lbrace {0,2,4,\dots \rbrace },\\ 2 \sigma _{n} B_{n}^{k} \frac{\gamma }{\sigma ^{2}}\phi \left( \frac{B_{n}}{\sigma _{n}} \right) ; &{}\quad k \in \lbrace {1,3,5,\dots \rbrace }, \\ \end{array} \right. \end{aligned}$$

(C.3)

Furthermore, if $\gamma = 0$ and k is an odd positive integer, then ${\mathcal {L}}^{I}_{n}[\mathbf B ](k) = 0$.

Proof

Throughout, Z denotes a standard normal random variable. We begin by assuming $\gamma \ne 0$ and by noticing that

$$\begin{aligned} {\mathcal {L}}_{n}^{I}[\mathbf B ](k)&= e^{-\lambda _{n}} \sum _{j=0}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) \gamma _{n}^{j} \sigma _{n}^{k-j} {\mathbb {E}}\left[ Z^{k-j} \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z | > B_{n}]} \right] . \end{aligned}$$

(C.4)

Let ${\mathcal {G}}_{n}^{(m)}({\mathbf{B}}) := {\mathbb {E}}\left[ Z^{m} \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z | > B_{n}]} \right]$ and note that

$$\begin{aligned} {\mathcal {G}}_{n}^{(m)}(\mathbf B ) = \int _{\frac{B_{n}-\gamma _{n}}{\sigma _{n}}}^{\infty } z^{m} \phi (z) dz + \int ^{-\frac{(B_{n}+\gamma _{n})}{\sigma _{n}}}_{-\infty } z^{m} \phi (z) dz. \end{aligned}$$

Let ${\overline{u}}_{n} := (B_{n}-\gamma _{n})/\sqrt{2\sigma _{n}^{2}}$ and ${\underline{u}}_{n} := (B_{n}+\gamma _{n})/\sqrt{2\sigma _{n}^{2}}$ and observe that, since $\lim _{n \rightarrow {} \infty } {\overline{u}}_{n} = \lim _{n \rightarrow {} \infty } {\underline{u}}_{n} = +\infty$, for n sufficiently large, $\min \lbrace {{\overline{u}}_{n},{\underline{u}}_{n}\rbrace } > 0$. In that case, by making a change of variables, we have

$$\begin{aligned} {\mathcal {G}}_{n}^{(m)}(\mathbf B ) = \frac{2^{m/2-1}}{\pi ^{1/2}} {\overline{\Gamma }} \left( \frac{m+1}{2}, {{\overline{u}}_{n}^{2}} \right) + (-1)^{m} \frac{2^{m/2-1}}{\pi ^{1/2}} {\overline{\Gamma }} \left( \frac{m+1}{2},{ {\underline{u}}_{n}^{2}} \right) , \end{aligned}$$

where ${\overline{\Gamma }}(s,x):=\int _{x}^{\infty }u^{s-1}e^{-u}du$ is the upper incomplete gamma function. In the case that $m \in \lbrace {0,2,4,\dots \rbrace }$, the well known asymptotic property ${\overline{\Gamma }}(s,x)\sim x^{s-1}e^{-x}$, as $x\rightarrow \infty$, implies that, as $n \rightarrow {} \infty$,

$$\begin{aligned} {\mathcal {G}}_{n}^{(m)}(\mathbf B ) \sim 2 \left( \frac{B_{n}}{\sigma _{n}} \right) ^{m-1} \phi \left( \frac{B_{n}}{\sigma _{n}} \right) . \end{aligned}$$

(C.5)

For $m \in \lbrace {1,3,5,\dots \rbrace }$, note that

$$\begin{aligned} {\mathcal {G}}_{n}^{(m)}(\mathbf B )= & {} \frac{2^{m/2-1}}{\pi ^{1/2}} \left[ \int _{{{\overline{u}}^{2}_{n}}}^{\infty }w^{(m-1)/2} e^{-w}dw - \int _{{{\underline{u}}^{2}_{n}}}^{\infty }w^{(m-1)/2} e^{-w}dw \right] \\= & {} sign(\gamma ) \frac{2^{m/2-1}}{\pi ^{1/2}} \int _{{{\overline{u}}^{2}_{n} \wedge {\underline{u}}^{2}_{n}}}^{{{\overline{u}}^{2}_{n} \vee {\underline{u}}^{2}_{n}}}w^{(m-1)/2} e^{-w}dw, \end{aligned}$$

and so, without loss of generality, we suppose that $\gamma > 0$, which implies that ${\underline{u}}_{n} > {\overline{u}}_{n}$. Now, if $m=1$, then

$$\begin{aligned} \int _{{{\overline{u}}^{2}_{n}}}^{{{\underline{u}}^{2}_{n}}} w^{(m-1)/2} e^{-w}dw= & {} e^{-{{\overline{u}}^{2}_{n}}}-e^{-{{\underline{u}}^{2}_{n}}}=e^{-(B_{n}^{2}+\gamma _{n}^{2})/2\sigma _{n}^{2}}\left( e^{B_{n}\gamma /\sigma ^{2}}- e^{-B_{n}\gamma /\sigma ^{2}} \right) \\&\sim \pi ^{1/2} \frac{B_{n}}{2^{-3/2}} \frac{\gamma }{\sigma ^{2}} \phi \left( \frac{B_{n}}{\sigma _{n}} \right) . \end{aligned}$$

For general m, we can proceed by induction to show that

$$\begin{aligned} \int _{{{\overline{u}}_{n}^{2}}}^{{{\underline{u}}^{2}_{n}}} w^{(m-1)/2} e^{-w}dw\sim \pi ^{1/2} \frac{B_{n}^{m}}{2^{m/2-2}\sigma _{n}^{m-1}} \frac{\gamma }{\sigma ^{2}} \phi \left( \frac{B_{n}}{\sigma _{n}} \right) . \end{aligned}$$

(C.6)

Indeed, suppose that (C.6) holds for $m=M-2$ for an odd positive M. Now, for $m=M$, integration by parts yields

$$\begin{aligned} \int _{{{\overline{u}}^{2}_{n}}}^{{{\underline{u}}^{2}_{n}}} w^{\frac{M-1}{2}} e^{-w}dw&= \frac{e^{-(B_{n}^{2}+\gamma _{n}^{2})/2\sigma _{n}^{2}}}{2^{\frac{M-1}{2}} \sigma _{n}^{M-1}} \left[ e^{\frac{B_{n}\gamma }{\sigma ^{2}}}(B_{n}-\gamma _{n})^{M-1} - e^{-\frac{B_{n}\gamma }{\sigma ^{2}}}(B_{n} + \gamma _{n})^{M-1} \right] \\&\quad + \frac{M-1}{2} \int _{{{\overline{u}}^{2}_{n}}}^{{{\underline{u}}^{2}_{n}}} w^{(M-3)/2} e^{-w} dw. \end{aligned}$$

The expression in the square brackets above is such that

$$\begin{aligned} e^{B_{n}\gamma /\sigma ^{2}}(B_{n}-\gamma _{n})^{M-1} - e^{-B_{n}\gamma /\sigma ^{2}} (B_{n} + \gamma _{n})^{M-1} = 2\frac{B_{n}^{M}\gamma }{\sigma ^{2}} + o(B_{n}^{M}), \qquad (n \rightarrow {} \infty ). \end{aligned}$$

Therefore, in light of the induction step, as $n \rightarrow {} \infty$,

$$\begin{aligned} \int _{{{\overline{u}}^{2}_{n}}}^{{{\underline{u}}^{2}_{n}}} w^{(M-1)/2} e^{-w}dw = \pi ^{1/2} \frac{B_{n}^{M}}{2^{M/2-2}\sigma _{n}^{M-1}} \frac{\gamma }{\sigma ^{2}} \phi \left( \frac{B_{n}}{\sigma _{n}} \right) + O \left( \frac{B_{n}^{M-2}}{\sigma _{n}^{M-3}}\phi \left( \frac{B_{n}}{\sigma _{n}}\right) \right) ,\nonumber \\ \end{aligned}$$

(C.7)

which proves (C.6) for $m=M$. From (C.6), we deduce that for $m \in \lbrace {1,3,5,\dots \rbrace }$

$$\begin{aligned} {\mathcal {G}}_{n}^{(m)}(\mathbf B ) \sim 2 \frac{B_{n}^{m}}{\sigma _{n}^{m-1}} \frac{\gamma }{\sigma ^{2}}\phi \left( \frac{B_{n}}{\sigma _{n}} \right) . \end{aligned}$$

(C.8)

Finally, from (C.4), (C.5), and (C.8), we conclude the validity of (C.3). In the case that k is odd and $\gamma = 0$, from (C.4) and the symmetry of the integral term, it is clear that ${\mathcal {L}}^{I}_{n}[\mathbf B ](k) = \sigma _{n}^{k} {\mathbb {E}}\left[ Z^{k} \mathbf{1}_{[|\sigma _{n} Z | >B_{n}]} \right] = 0$. $\square$

Lemma C.2

Let f be of the mixture form given in (2.3) such that $f_{+}$ and $f_{-}$ are bounded and continuous at 0. Then, for any threshold sequence $\mathbf{B}:=(B_{n})_{n}$ such that $B_{n}\rightarrow {}0$, and any non-negative integer k, as $n \rightarrow {} \infty$,

$$\begin{aligned} {\mathcal {L}}_{n}^{II,Abs}[\mathbf{B}](k) \sim h_{n} \lambda \frac{{2}B_{n}^{k+1}}{k+1} {{\mathcal {C}}_{0}(f)}. \end{aligned}$$

Furthermore, if k is odd and $\lim _{n \rightarrow {} \infty } B_{n}/h_{n}^{1/2} = \infty$, then

$$\begin{aligned} {\mathcal {L}}_{n}^{II}[\mathbf{B}](k) \sim h_{n} \lambda \frac{B_{n}^{k+1}}{k+1} \overline{{\mathcal {C}}}(f). \end{aligned}$$

Proof

The results follows directly from conditioning on how many jumps occur over the time interval $(h_{n}i,h_{n}(i+1)]$ and applying the results of Lemma C.3 below. $\square$

Lemma C.3

Let f be a bounded density function, let $\mathbf B =(B_{n})_{n}$ be a positive sequence such that $B_{n}\rightarrow {}0$, as $n\rightarrow \infty$, and $k=0,1,\dots$. Then, the following assertions hold:

• 1. For arbitrary constants $\sigma \in {\mathbb {R}}_{+}$ and $\gamma \in {\mathbb {R}}$,

  $$\begin{aligned} {\mathbb {E}}\left[ \left| \gamma + \sigma Z +\zeta _{1} \right| ^{k} \mathbf{1}_{[|\gamma + \sigma Z + \zeta _{1}| \le B_{n}]} \right] \le {}2\Vert f\Vert _{\infty }\frac{B_{n}^{k+1}}{k+1}. \end{aligned}$$

  (C.9)

• 2. If, additionally, f has the mixture form given in (2.3) such that $f_{+}:[0,\infty )\rightarrow [0,\infty )$ and $f_{-}:(-\infty ,0]\rightarrow [0,\infty )$ are left- and right-continuous at 0, respectively, then, as $n \rightarrow {} \infty$,

  $$\begin{aligned} \lim _{n\rightarrow \infty }\frac{k+1}{B_{n}^{k+1}}{\mathbb {E}}\left[ \left| \gamma _{n} + \sigma _{n} Z+\zeta _{1} \right| ^{k} \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z+ \zeta _{1}| \le B_{n}]} \right]&=pf_{+}(0)+qf_{-}(0), \end{aligned}$$

  (C.10)

  where $\sigma _{n}=\sigma \sqrt{h_{n}}$ and $\gamma _{n}=\gamma h_{n}$. Furthermore, if k is odd, and $B_{n}h_{n}^{-1/2} \rightarrow \infty$, then

  $$\begin{aligned} \lim _{n\rightarrow {}\infty }\frac{k+1}{B_{n}^{k+1}}{\mathbb {E}}\left[ \left( \gamma _{n} + \sigma _{n} Z +\zeta _{1} \right) ^{k} \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z+ \zeta _{1} | \le B_{n}]} \right] =pf_{+}(0)-qf_{-}(0).\qquad \end{aligned}$$

  (C.11)

Proof of Lemma C.3

For the first assertion, note that

$$\begin{aligned} {\mathbb {E}}\left[ \left| \gamma + \sigma Z +\zeta _{1} \right| ^{k} \mathbf{1}_{[|\gamma + \sigma Z + \zeta _{1}| \le B_{n}]} \right]&=\int _{{\mathbb {R}}}\int _{-B_{n}}^{B_{n}}|w|^{k}\sigma ^{-1}\phi \left( \frac{w-x-\gamma }{\sigma }\right) dwf(x)dx\\&\le {}\Vert f\Vert _{\infty }\int _{-B_{n}}^{B_{n}}|w|^{k}\int _{{\mathbb {R}}}\sigma ^{-1}\phi \left( \frac{w-x-\gamma }{\sigma }\right) dxdw\\&={}\Vert f\Vert _{\infty }\int _{-B_{n}}^{B_{n}}|w|^{k}dw, \end{aligned}$$

which implies (C.9). To show the second assertion, let us start by introducing some notation needed in the sequel. Let $\phi _{n}(z):=\phi (z/\sigma _{n})/\sigma _{n}$ denote the density of $\sigma W_{h_{n}}$ and, for $a,b\in \{-1,1\}$, let

$$\begin{aligned} I_{n}^{a,b}(w) := \int _{0}^{\infty } \left( \phi _{n}(w-b x-\gamma _{n}) +a \phi _{n}(-w-b x-\gamma _{n}) \right) dx. \end{aligned}$$

Note that, for $w > 0$,

$$\begin{aligned} {I_{n}^{1,b}(w)} = 1 - b \cdot sgn(\gamma ) \int ^{(w+|\gamma _{n}|)/\sigma _{n}}_{(w-|\gamma _{n}|)/\sigma _{n}} \phi (z)dz, \quad {I_{n}^{-1,b}(w)} = b \int ^{(w+\gamma _{n})/\sigma _{n}}_{(-w+\gamma _{n})/\sigma _{n}} \phi (z)dz.\nonumber \\ \end{aligned}$$

(C.12)

Without loss of generality, we assume hereafter that $\gamma > 0$ and, for future reference, let us also remark that, by the mean value theorem, $I_{n}^{1,b}(w) = 1 -2 \gamma b \sigma ^{-1} h_{n}^{1/2}\phi (\zeta _{n}^{b}(w))$, for some $\zeta _{n}^{b}(w)\in {\mathbb {R}}$.

Let us denote the expectations appearing in (C.10) and (C.11) by $C_{n,k}$ and $D_{n,k}$, respectively, and note that

$$\begin{aligned} C_{n,k}&=\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}|z+x+\gamma _{n}|^{k}{\mathbf{1}_{[|z+x+\gamma _{n}| \le B_{n}]}}f(x)\phi _{n}(z)dxdz= T_{n,k}^{(1)}+T_{n,k}^{(-1)},\\ D_{n,k}&=\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}(z+x+\gamma _{n})^{k}{\mathbf{1}_{[|z+x+\gamma _{n}| \le B_{n}]}}f(x)\phi _{n}(z)dxdz= T_{n,k}^{(1)}+(-1)^{k}T_{n,k}^{(-1)}, \end{aligned}$$

where, for $b\in \{-1,1\}$,

$$\begin{aligned} T_{n,k}^{(b)}&:=\int _{0}^{B_{n}}\int _{{\mathbb {R}}} f(x) \phi _{n}(b w-x-\gamma _{n}) dx w^{k}dw \\&=\int _{0}^{B_{n}}\int _{{\mathbb {R}}} \left[ pf_{+}(x) \mathbf{1}_{[x \ge 0]} + qf_{-}(x)\mathbf{1}_{[x< 0]} \right] \phi _{n}(b w-x-\gamma _{n}) dx w^{k}dw. \end{aligned}$$

Let us analyze the two expressions $C_{n,k}$ and $D_{n,k}$ via the generalized sum

$$\begin{aligned} H_{n,k}^{(a)} := \frac{T_{n,k}^{(1)} + a T_{n,k}^{(-1)}}{{{\bar{m}}}_{n}}, \qquad a\in \{-1,1\}, \end{aligned}$$

where ${{\bar{m}}}_{n}:=\int _{0}^{B_{n}} w^{k}dw=B_{n}^{k+1}/(k+1)$. To this end, consider the decomposition

$$\begin{aligned} H_{n,k}^{(a)}&= \frac{p}{{{\bar{m}}}_{n}} \int _{0}^{B_{n}} \int _{0}^{\infty } \left[ f_{+}(x)-f_{+}(0)\right] \left[ \phi _{n}(w-x-\gamma _{n})+a\phi _{n}(-w-x-\gamma _{n}\right] dx w^{k}dw \nonumber \\&\quad + \frac{q}{{{\bar{m}}}_{n}} \int _{0}^{B_{n}} \int _{-\infty }^{0} \left[ f_{-}(x)-f_{-}(0)\right] \left[ \phi _{n}(w-x-\gamma _{n})+a\phi _{n}(-w-x-\gamma _{n})\right] dx w^{k}dw \nonumber \\&\quad + \frac{p}{{{\bar{m}}}_{n}} f_{+}(0) \int _{0}^{B_{n}} I_{n}^{a,1}(w) w^{k}dw + \frac{q}{{{\bar{m}}}_{n}} f_{-}(0) \int _{0}^{B_{n}} I_{n}^{a,-1}(w) w^{k}dw \nonumber \\&=: {\overline{C}}^{(a)}_{n,k} + {\underline{C}}^{(a)}_{n,k} + C^{(a)}_{n,k}. \end{aligned}$$

(C.13)

Depending on the sign of a, the last term on the right-hand side of (C.13) equals

$$\begin{aligned} C_{n,k}^{(a)} = \left\{ \begin{array}{ll} pf_{+}(0) + qf_{-}(0) + \gamma O(h_{n}^{1/2}){,} &{}\quad \text {if }a=1,\\ \frac{pf_{+}(0)-qf_{-}(0)}{{{\bar{m}}}_{n}} \int _{0}^{B_{n}} \int ^{\frac{w+\gamma _{n}}{\sigma _{n}}}_{\frac{(-w+\gamma _{n})}{\sigma _{n}}} \phi (v) dvw^{k} dw{,} &{}\quad \text {if }a=-1. \end{array} \right. \end{aligned}$$

(C.14)

In order to show the asymptotic behavior of $C_{n,k}^{(-1)}$, we next show that

$$\begin{aligned} J_{n,k}:=\frac{1}{{{\bar{m}}}_{n}} \int _{0}^{B_{n}} \int _{\frac{-w+\gamma _{n}}{\sigma _{n}}}^{\frac{w+\gamma _{n}}{\sigma _{n}}} \phi (v) dv w^{k} dw{\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }}{}1. \end{aligned}$$

(C.15)

First note that, by a change of variables,

$$\begin{aligned} {J_{n,k}} =\frac{\sigma _{n}^{k+1}}{{{\bar{m}}}_{n}} \int _{0}^{\frac{B_{n}}{\sigma _{n}}} \int _{-u+\frac{\gamma _{n}}{\sigma _{n}}}^{u+\frac{\gamma _{n}}{\sigma _{n}}} \phi (v) dv u^{k} du. \end{aligned}$$

Let $u_{0}>0$ be large enough that $\int _{(-u_{0},u_{0})^{c}}\phi (v)dv<\varepsilon$. Let N large enough that $B_{n}/\sigma _{n}>u_{0}+1$ and $|\gamma _{n}/\sigma _{n}|<1$ for all $n\ge {}N$. Then, for $n\ge {}N$,

$$\begin{aligned} 0\le 1-J_{n,k}&=\frac{\sigma _{n}^{k+1}}{{{\bar{m}}}_{n}} \int _{0}^{\frac{B_{n}}{\sigma _{n}}} \int _{(-u+\frac{\gamma _{n}}{\sigma _{n}},u+\frac{\gamma _{n}}{\sigma _{n}})^{c}} \phi (v) dv u^{k} du\\&\le \frac{\sigma _{n}^{k+1}}{{{\bar{m}}}_{n}} \int _{0}^{u_{0}+1} u^{k} du+\frac{\sigma _{n}^{k+1}}{{{\bar{m}}}_{n}} \int _{u_{0}+1}^{\frac{B_{n}}{\sigma _{n}}} \int _{(-u_{0},u_{0})^{c}} \phi (v) dv u^{k} du\\&\le \frac{\sigma _{n}^{k+1}}{{{\bar{m}}}_{n}} \int _{0}^{u_{0}+1} u^{k} du+\varepsilon \frac{\sigma _{n}^{k+1}}{{{\bar{m}}}_{n}} \int _{0}^{\frac{B_{n}}{\sigma _{n}}}u^{k} du{\mathop {\longrightarrow }\limits ^{n\rightarrow {}\infty }}\varepsilon , \end{aligned}$$

because $\sigma _{n}^{k+1}/{{\bar{m}}}_{n}\rightarrow {}0$ and both $u+\gamma _{n}/\sigma _{n}>u_{0}$ and $-u+\gamma _{n}/\sigma _{n}<-u_{0}$ whenever $u>u_{0}+1$. Since $\varepsilon$ is arbitrary, we conclude the first limit in (C.15). Together (C.14) and (C.15) imply that

$$\begin{aligned} C_{n,k}^{(a)} {\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }} \left\{ \begin{array}{ll} pf_{+}(0) + qf_{-}(0) ; &{}\quad \text {if }a=1,\\ pf_{+}(0)- qf_{-}(0); &{} \quad \text {if }a=-1. \end{array} \right. \end{aligned}$$

(C.16)

We will now show that the first two terms appearing on the right-hand side of (C.13) are such that

$$\begin{aligned} \lim _{n\rightarrow {}\infty } {\overline{C}}^{(a)}_{n,k} = \lim _{n\rightarrow {}\infty }{\underline{C}}^{(a)}_{n,k} =0. \end{aligned}$$

(C.17)

Given $\varepsilon >0$, by the continuity of $f_{+}$ at 0, there exists a $\delta >0$ such that $|f_{+}(x)-f_{+}(0)|<\varepsilon /2$, for all $x\in (0,\delta )$. The ensuing upper bound follows:

$$\begin{aligned} |{\overline{C}}^{(a)}_{n,k}|&\le \frac{p}{{{\bar{m}}}_{n}}\int _{0}^{B_{n}}\int _{0}^{\delta } |f_{+}(x)-f_{+}(0)| \left[ \phi _{n}(w-x-\gamma _{n})+\phi _{n}(w+x+\gamma _{n})\right] dx w^{k}dw\\&\quad + \frac{p}{{{\bar{m}}}_{n}}\int _{0}^{B_{n}}\int _{\delta }^{\infty } f_{+}(x) \left[ \phi _{n}(w-x-\gamma _{n})+\phi _{n}(w+x+\gamma _{n}) \right] dx w^{k}dw\\&\quad + f_{+}(0)\frac{p}{{{\bar{m}}}_{n}}\int _{0}^{B_{n}}\int _{\delta }^{\infty } \left[ \phi _{n}(w-x-\gamma _{n})+\phi _{n}(w+x+\gamma _{n}) \right] dx w^{k}dw. \end{aligned}$$

Let us denote each of the three terms in the right-hand side of the last inequality by ${\overline{C}}_{n,k,1}$,${\overline{C}}_{n,k,2}$, and ${\overline{C}}_{n,k,3}$, respectively. Clearly,

$$\begin{aligned} {\overline{C}}_{n,k,1}&\le \frac{\varepsilon }{2} \frac{p}{{{\bar{m}}}_{n}}\int _{0}^{B_{n}}\int _{0}^{\delta } \left[ \phi _{n}(w-x-\gamma _{n}) + \phi _{n}(-w-x-\gamma _{n})\right] dx w^{k}dw\le \varepsilon ,\\ {\overline{C}}_{n,k,2}&\le \sup _{{\mathop {{x\in (\delta ,\infty )}}\limits ^{w\in (0,B_{n})}}} \phi _{n}(w-x-\gamma _{n}) + \sup _{{\mathop {{x\in (\delta ,\infty )}}\limits ^{w\in (0,B_{n})}}} \phi _{n}(w+x+\gamma _{n}). \end{aligned}$$

Note that, for n large enough, $\min \lbrace |w-x-\gamma _{n}|, |w+x+\gamma _{n}| \rbrace >\delta /2$ for all $w\in (0,B_{n})$ and ${x\in (\delta ,\infty )}$. Thus,

$$\begin{aligned} {\overline{C}}_{n,k,2} \le 2\sup _{|z|>\delta /2}\phi _{n}(z) \rightarrow 0, \quad (n\rightarrow \infty ). \end{aligned}$$

Similarly,

$$\begin{aligned} {\overline{C}}_{n,k,3} \le f_{+}(0)\frac{2p}{{{\bar{m}}_{n}}}\int _{0}^{B_{n}}\int _{|z|\ge {}\delta /2}\phi _{n}(z) dz w^{k}dw\le 2f_{+}(0)\int _{|z|\ge {}\delta /2}\phi _{n}(z) dz\rightarrow {}0, \quad (n\rightarrow {}\infty ). \end{aligned}$$

Therefore,

$$\begin{aligned} \displaystyle {\limsup _{n \rightarrow {} \infty }}\left| {\overline{C}}^{(a)}_{n,k}\right| \le \limsup _{n\rightarrow \infty }{\overline{C}}_{n,k,1} +\limsup _{n\rightarrow \infty }{\overline{C}}_{n,k,2}+\limsup _{n\rightarrow \infty }{\overline{C}}_{n,k,3}\le \varepsilon , \end{aligned}$$

(C.18)

and, since $\varepsilon$ is arbitrary, we conclude that $\lim _{n\rightarrow \infty }{\overline{C}}^{(a)}_{n,k}=0$. One can similarly show that $\lim _{n\rightarrow \infty }{\underline{C}}^{(a)}_{n,k}=0$. Together (C.13), (C.16), and (C.17) yield (C.10) and (C.11). $\square$

Lemma C.4

Let f be of the mixture form given in (2.3) such that $f_{+}:[0,\infty )\rightarrow [0,\infty )$ and $f_{-}:(-\infty ,0]\rightarrow [0,\infty )$ are bounded. Then, for any nonnegative sequence $(B_{n})_{n}$ such that $B_{n}\rightarrow {}0$, any nonnegative integers k, and any $m\ge {}2$, we have:

$$\begin{aligned} \lim _{n\rightarrow {}0}\frac{k+1}{B_{n}^{k+1}}{\mathbb {E}}\left[ \left| \gamma _{n} + \sigma \Delta _{i}^{n}W+\zeta _{1}+\dots +\zeta _{m} \right| ^{k} \mathbf{1}_{[|\gamma _{n} + \sigma \Delta _{i}^{n}W+ \zeta _{1}+\dots +\zeta _{m} | \le B_{n}]} \right]&= 2f^{*m}(0), \end{aligned}$$

(C.19)

$$\begin{aligned} \lim _{n\rightarrow {}0}\frac{k+1}{B_{n}^{k+1}}{\mathbb {E}}\left[ \left( \gamma _{n} + \sigma \Delta _{i}^{n}W+\zeta _{1}+\dots +\zeta _{m} \right) ^{k} \mathbf{1}_{[|\gamma _{n} + \sigma \Delta _{i}^{n}W+ \zeta _{1}+\dots +\zeta _{m} | \le B_{n}]} \right]&=f^{*m}(0)(1+(-1)^{k}), \end{aligned}$$

(C.20)

as $n\rightarrow {}\infty$. Furthermore, if k is odd, $f^{*m}\in C_{b}^{1}$, and $B_{n}/h_{n}^{1/2}\rightarrow \infty$, then

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{k+2}{B_{n}^{k+2}}{\mathbb {E}}\left[ \left( \gamma _{n} + \sigma \Delta _{i}^{n}W+\zeta _{1}+\dots +\zeta _{m} \right) ^{k} \mathbf{1}_{[|\gamma _{n} + \sigma \Delta _{i}^{n}W+ \zeta _{1}+\dots +\zeta _{m} | \le B_{n}]} \right] = 2 (f^{*m})'(0).\nonumber \\ \end{aligned}$$

(C.21)

Proof of Lemma C.4

The first two relationships (C.19) and (C.20) follow from Lemma C.3 since $\zeta _{1}+\dots +\zeta _{m}$ can be seen as a random variable with a density satisfying the conditions of the Lemma. Indeed, for $m\ge {}2$, the density of $\zeta _{1}+\dots +\zeta _{m}$ is the m-fold convolution $f^{*m}$, which is actually continuous everywhere under the condition $f\in {\mathbb {L}}^{1}({\mathbb {R}})\cap {\mathbb {L}}^{\infty }({\mathbb {R}})$. So, we only need to show the second assertion and, as with (C.19) and (C.20), it suffices to consider $m=1$. Let us recall from the proof of Lemma C.3 that the expectation appearing in (C.21), denoted therein by $D_{n,k}$, can be expressed as

$$\begin{aligned} D_{n,k}=\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}(z+x+\gamma _{n})^{k}{\mathbf{1}_{[|z+x+\gamma _{n}| \le B_{n}]}}f(x)\phi _{n}(z)dxdz= T_{n,k}^{(1)}-T_{n,k}^{(-1)}, \end{aligned}$$

where, for $b\in \{-1,1\}$,

$$\begin{aligned} T_{n,k}^{(b)} :=\int _{0}^{B_{n}}\int _{{\mathbb {R}}} f(x) \phi _{n}(b u-x-\gamma _{n}) dx u^{k}du. \end{aligned}$$

Next, changing variables into $y=x+\gamma _{n}$ for $T_{n,k}^{(1)}$ and into $y = -x - \gamma _{n}$ for $T_{n,k}^{(-1)}$,

$$\begin{aligned} T_{n}^{(1)}&=\int _{0}^{B_{n}}\int _{{\mathbb {R}}}f(y-\gamma _{n})\phi _{n}(u-y) dy u^{k}du, \\ T_{n}^{(-1)}&=\int _{0}^{B_{n}}\int _{{\mathbb {R}}}f(-y-\gamma _{n})\phi _{n}(-u+y) dy u^{k}du \end{aligned}$$

Therefore, invoking the symmetry of $\phi _{n}$,

$$\begin{aligned} \frac{D_{n,k}}{B_{n}^{k+2}}&=\frac{1}{B_{n}^{k+2}}\int _{0}^{B_{n}}\int _{{\mathbb {R}}}(f(y-\gamma _{n})-f(-y-\gamma _{n}))\phi _{n}(u-y) dx u^{k}du\\&=\frac{1}{B_{n}^{k+2}}\int _{0}^{B_{n}}\int _{{\mathbb {R}}}\int _{-1}^{1}f'(y\beta -\gamma _{n})d\beta y \phi _{n}(u-y) dy u^{k}du. \end{aligned}$$

Denoting a standard Gaussian variable by Z and changing variables first from u to $v=u/B_{n}$, then from y to $z=B_{n}v-y$, and finally from $\beta$ to $\alpha =B_{n} \beta$, we can write

$$\begin{aligned} \frac{D_{n,k}}{B_{n}^{k+2}}&=\frac{1}{B_{n}}\int _{0}^{1}\int _{{\mathbb {R}}}\int _{-1}^{1}f'(y\beta -\gamma _{n})d\beta y \phi _{n}(B_{n}v-y) dy v^{k}dv\\&=\int _{0}^{1}{\mathbb {E}}\left( \frac{1}{B_{n}}\int _{-B_{n}}^{B_{n}}f'\left( \left( u-B_{n}^{-1}\sigma _{n}Z\right) \alpha -\gamma _{n}\right) d\alpha \left( u-B_{n}^{-1}\sigma _{n}Z\right) \right) u^{k}du. \end{aligned}$$

By the Dominated Convergence Theorem and the assumption that $h_{n}^{1/2}B_{n}^{-1}\rightarrow {}0$, we conclude that $\lim _{n\rightarrow {}\infty }D_{n,k}/B_{n}^{k+2}=2f'(0)/(k+2)$. $\square$

Lemma C.5

Let $\mathbf{B}:=(B_{n})_{n\ge {}1}$ be an arbitrary threshold sequence and $k=0,1,\dots$. Then, $\lim _{n \rightarrow {} \infty } B_{n}/h_{n}^{1/2} = +\infty$ if and only if

$$\begin{aligned} \displaystyle {\lim _{n \rightarrow {} \infty }}\frac{{\mathcal {G}}_{n}^{I}[\mathbf{B}](k)}{h_{n}^{k/2}} = \frac{\sigma ^{k}2^{k/2}}{\pi ^{1/2}} \Gamma \left( \frac{k+1}{2} \right) . \end{aligned}$$

(C.22)

Furthermore, when $\lim _{n \rightarrow {} \infty } B_{n}/h_{n}^{1/2} = +\infty$, we have, as $n \rightarrow {} \infty$,

$$\begin{aligned}&\frac{{\mathcal {G}}_{n}^{I}[\mathbf{B}](k)}{h_{n}^{k/2} } - \frac{\sigma ^{k} 2^{k/2} \Gamma (\frac{k+1}{2})}{\pi ^{1/2}} \nonumber \\&\quad \sim {\left\{ \begin{array}{ll} A_{n,k}, &{} \quad \text {if k is even or } \; \gamma = 0, \\ A_{n,k} + \frac{\gamma ^{k+1}}{\sigma }h_{n}^{(k+1)/2} \left( \frac{2}{\pi }\right) ^{1/2} \sum _{j=0}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) \frac{(-1)^{k-j}}{k-j+1}, &{} \quad \text {if k is odd and} \; \gamma > 0, \\ A_{n,k} - \frac{\gamma ^{k+1}}{\sigma }h_{n}^{(k+1)/2} \left( \frac{2}{\pi }\right) ^{1/2} \sum _{j=0}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) \frac{1}{k-j+1},&\quad \text {if k is odd and} \; \gamma < 0, \end{array}\right. }\nonumber \\ \end{aligned}$$

(C.23)

where

$$\begin{aligned} A_{n,k}:= & {} -2 \sigma \phi \left( \frac{B_{n}}{\sigma _{n}} \right) \left( \frac{B_{n}}{h_{n}^{1/2}} \right) ^{k-1} + \frac{h_{n}}{\pi ^{1/2}} \sigma ^{k-2} 2^{k/2-1} \\&\quad \times \left[ -2\lambda \sigma ^{2} \Gamma \left( \frac{k+1}{2} \right) + \gamma ^{2} {\left( {\begin{array}{c}k\\ 2\end{array}}\right) }\Gamma \left( \frac{k-1}{2} \right) \right] . \end{aligned}$$

Proof

Let us first assume that $\liminf _{n\rightarrow \infty }B_{n}/\sqrt{h_{n}}=:L\in [0,\infty )$ and let $\{n_{j}\}_{j\ge {}1}$ be a subsequence such that $\lim _{j\rightarrow \infty }B_{n_{j}}/h^{1/2}_{n_{j}}=L$. Then, by the dominated convergence theorem,

$$\begin{aligned} \frac{{\mathcal {G}}_{n_{j}}^{I}[\mathbf{B}](k)}{h_{n_{j}}^{k/2}}&= {e^{-\lambda _{n}}}{\mathbb {E}}\left[ |\gamma h^{1/2}_{n_{j}} + \sigma Z|^{k} \mathbf{{1}}_{[|\gamma h^{1/2}_{n_{j}} + \sigma Z| \le \frac{B_{n_{j}}}{h^{1/2}_{n_{j}}}]} \right] {\mathop {\longrightarrow }\limits ^{j\rightarrow \infty }} {\mathbb {E}}\left[ |\sigma Z|^{k} \mathbf{{1}}_{[|\sigma Z| \le L]} \right] . \end{aligned}$$

(C.24)

The limit value above can be expressed in terms of the lower incomplete gamma function, ${\underline{\Gamma }}(s,x):=\int _{0}^{x} u^{s-1} e^{-u}du$, as follows:

$$\begin{aligned} {\mathbb {E}}\left[ |\sigma Z|^{k} \mathbf{{1}}_{[|\sigma Z| \le L]} \right] =\mathbf{1}_{L\ne {}0} \frac{\sigma ^{k}2^{k/2}}{\sqrt{\pi }} {\underline{\Gamma }} \left( \frac{k+1}{2}, {\frac{L^{2}}{2\sigma ^{2}}} \right) . \end{aligned}$$

This proves that, for the validity of (C.22), it is necessary that $\lim _{n \rightarrow {} \infty } B_{n}/h_{n}^{1/2} = +\infty$. Hereafter, we assume the later condition. In that case, using again dominated convergence, ${\mathcal {G}}_{n}^{I}[\mathbf{B}](k)/h_{n}^{k/2}\rightarrow {\mathbb {E}}\left[ |\sigma Z|^{k} \right]$, and (C.22) follows from the well-known formula for the centered moments of a normal r.v. To show the second assertion, we assume that $\gamma >0$ (the cases $\gamma <0$ and $\gamma =0$ are proved similarly). Let us begin by noting that

$$\begin{aligned} D_{n,k}&:= \frac{{\mathcal {G}}_{n}^{I}[\mathbf{B}](k)}{h_{n}^{k/2}} - \frac{\sigma ^{k} 2^{k/2} \Gamma (\frac{k+1}{2})}{\pi ^{1/2}}= {e^{-\lambda _{n}}}h_{n}^{-k/2} \int _{-(B_{n}+\gamma _{n})/\sigma _{n}}^{(B_{n}-\gamma _{n})/\sigma _{n}} |\gamma _{n} + \sigma _{n}z|^{k} \phi (z)dz\nonumber \\&\quad - \frac{\sigma ^{k} 2^{k/2} \Gamma (\frac{k+1}{2})}{\pi ^{1/2}}. \end{aligned}$$

(C.25)

Since $-(B_{n}+\gamma _{n})/\sigma _{n})< -\gamma _{n}/\sigma _{n} < (B_{n}-\gamma _{n})/\sigma _{n}$, we may further decompose the integral term appearing on the right-hand side above as follows:

$$\begin{aligned} \int _{-(B_{n}+\gamma _{n})/\sigma _{n}}^{(B_{n}-\gamma _{n})/\sigma _{n}} |\gamma _{n} + \sigma _{n}z|^{k} \phi (z)dz&= \int _{-\gamma _{n}/\sigma _{n}}^{(B_{n}-\gamma _{n})/\sigma _{n}} \left( \gamma _{n} + \sigma _{n} z \right) ^{k} \phi (z)dz \nonumber \\&\quad + (-1)^{k} \int _{-(B_{n}+\gamma _{n})/\sigma _{n}}^{-\gamma _{n}/\sigma _{n}} \left( \gamma _{n} + \sigma _{n} z \right) ^{k} \phi (z)dz \nonumber \\&= \sum _{j=0}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) \gamma ^{j} \sigma ^{k-j} h_{n}^{(k+j)/2} \int _{-\gamma _{n}/\sigma _{n}}^{(B_{n}-\gamma _{n})/\sigma _{n}} z^{k-j} \phi (z)dz \nonumber \\&\quad + (-1)^{k} \sum _{j=0}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) \gamma ^{j} \sigma ^{k-j} h_{n}^{(k+j)/2} \int ^{-\gamma _{n}/\sigma _{n}}_{-(B_{n}+\gamma _{n})/\sigma _{n}} z^{k-j} \phi (z)dz. \end{aligned}$$

(C.26)

Next, by a change of variable,

$$\begin{aligned} D_{n,k}&= \frac{\sigma ^{k}2^{{k/2-1}}}{\pi ^{1/2}} \left[ e^{-\lambda _{n}} {\underline{\Gamma }} \left( \frac{k+1}{2}, \frac{({B_{n}+ \gamma _{n})^{2}}}{2\sigma _{n}^{2}} \right) -\Gamma \left( \frac{k+1}{2} \right) \right] \end{aligned}$$

(C.27)

$$\begin{aligned}&\quad +\frac{\sigma ^{k}2^{{k/2-1}}}{\pi ^{1/2}} \left[ e^{-\lambda _{n}} {\underline{\Gamma }} \left( \frac{k+1}{2}, \frac{({B_{n}- \gamma _{n})^{2}}}{2\sigma _{n}^{2}} \right) - \Gamma \left( \frac{k+1}{2} \right) \right] \end{aligned}$$

(C.28)

$$\begin{aligned}&\quad + e^{-\lambda _{n}} \sigma ^{k} \frac{2^{k/2-1}}{\pi ^{1/2}} {\underline{\Gamma }} \left( \frac{k+1}{2}, \frac{\gamma ^{2}h_{n}}{2\sigma ^{2}} \right) \left[ (-1)^{k}-1\right] \end{aligned}$$

(C.29)

$$\begin{aligned}&\quad + \sum _{j=1}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) \gamma ^{j} \sigma ^{k-j} h_{n}^{j/2} \frac{2^{(k-j)/2-1}}{\pi ^{1/2}} \left[ {\underline{\Gamma }} \left( \frac{k-j+1}{2}, \frac{(B_{n}-\gamma _{n})^{2}}{2\sigma _{n}^{2}} \right) \nonumber \right. \\&\quad +{(-1)^{2k-j}{\underline{\Gamma }} \left( \frac{k-j+1}{2}, \frac{(B_{n}+\gamma _{n})^{2}}{2\sigma _{n}^{2}} \right) } \nonumber \\&\quad \left. +\, {\underline{\Gamma }} \left( \frac{k-j+1}{2}, \frac{\gamma ^{2}h_{n}}{2\sigma ^{2}} \right) (-1)^{k-j}[1-(-1)^{k}] \right] . \end{aligned}$$

(C.30)

We denote the previous four terms $P_{n}^{(1,+)}$, $P_{n}^{(1,-)}$, $P_{n}^{(2)}$, and $P_{n}^{(3)}$. From here we can use the asymptotic properties of the lower and upper incomplete gamma functions in order to analyze the previous four terms on the right-hand side of (C.30). Indeed, for the first two terms,

$$\begin{aligned} P_{n}^{(1,\pm )}&= -\frac{\sigma ^{k}2^{k/2}}{\pi ^{1/2}} {\overline{\Gamma }} \left( \frac{k+1}{2}, \frac{(B_{n}\pm \gamma _{n})^{2}}{2\sigma _{n}^{2}} \right) - \lambda h_{n} \frac{\sigma ^{k}2^{k/2-1}}{\pi ^{1/2}} {\underline{\Gamma }} \left( \frac{k+1}{2},\frac{(B_{n}\pm \gamma _{n})^{2}}{2\sigma _{n}^{2}}\right) + o(h_{n}) \nonumber \\&{=} -\sigma \phi \left( \frac{B_{n}}{\sigma _{n}} \right) \left( \frac{B_{n}}{h_{n}^{1/2}}\right) ^{k-1} - \lambda h_{n} \sigma ^{k} \frac{2^{k/2-1}}{\pi ^{1/2}} \Gamma \left( \frac{k+1}{2}\right) +{\mathrm{h.o.t.}}, \qquad (n \rightarrow {} \infty ), \end{aligned}$$

(C.31)

where ${\overline{\Gamma }}(s,x):=\int _{x}^{\infty } u^{s-1} e^{-u}du$ denotes the upper incomplete gamma function and we have used the asymptotics ${\overline{\Gamma }}(s,x)\sim x^{s-1}e^{-x}$ as $x\rightarrow \infty$. For the second term, note that $P_{n}^{(2)}=0$ is k if even, while, for odd k,

$$\begin{aligned} P_{n}^{(2)} \sim -\frac{\gamma ^{k+1}}{(2\pi \sigma ^{2})^{1/2}} \frac{h_{n}^{(k+1)/2}}{k+1} \end{aligned}$$

(C.32)

Lastly, for the last term in line (C.30), we have, as $n \rightarrow {} \infty$,

$$\begin{aligned} P_{n}^{(3)} \sim \; {\left\{ \begin{array}{ll} h_{n} { \left( {\begin{array}{c}k\\ 2\end{array}}\right) } \gamma ^{2} \sigma ^{k-2} \frac{2^{k/2-1}}{\pi ^{1/2}} \Gamma \left( \frac{k-1}{2} \right) ; &{}\quad \text {when} \; k \; \text {is even}, \\ h_{n} { \left( {\begin{array}{c}k\\ 2\end{array}}\right) } \gamma ^{2} \sigma ^{k-2} \frac{2^{k/2-1}}{\pi ^{1/2}} \Gamma \left( \frac{k-1}{2} \right) + \frac{\gamma ^{k+1}}{\sigma } h_{n}^{(k+1)/2} \left( { \frac{2}{\pi }}\right) ^{1/2} \sum _{j=1}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) \frac{(-1)^{k-j}}{k-j+1};&\quad \text {when} \; k \; \text {is odd}. \end{array}\right. }\nonumber \\ \end{aligned}$$

(C.33)

Indeed, for odd k, we make use of the fact that ${\underline{\Gamma }}(s,x)\sim x^{s}/s$ as $x\rightarrow {}0$ to conclude that

$$\begin{aligned} \gamma ^{j} \sigma ^{k-j} h_{n}^{j/2} \frac{2^{(kj)/ 2-1}}{\pi ^{1/2}} \underline{\Gamma } \left( \frac{k-j+1}{2}, \frac{\gamma ^{2}h_{n}}{2\sigma ^{2}} \right) \sim \frac{\gamma ^{k+1}}{(2\pi \sigma ^{2})^{1/2}} \frac{h_{n}^{(k+1)/2}}{kj+ 1}, \qquad (n \rightarrow {} \infty ), \end{aligned}$$

for each $j = 1,2, \ldots , k$. For even k, using that ${\underline{\Gamma }}(x,s)=\Gamma (s)-{\overline{\Gamma }}(s,x)$ and the asymptotics ${\overline{\Gamma }}(s,x)\sim x^{s-1}e^{-x}$, as $x\rightarrow \infty$, the term for odd j can be proved to be of order

$$\begin{aligned} h_{n}^{j-\frac{k-1}{2}}B_{n}^{k-j} \phi \left( \frac{B_{n}}{\sigma _{n}} \right) , \end{aligned}$$

which is negligible compared to the first term in (C.31). The term in (C.33) corresponds to $j=2$, which is the dominant among those terms with even j. This concludes the proof. $\square$

Lemma C.6

Let f be of the mixture form given in (2.3) such that $f_{+}:[0,\infty )\rightarrow [0,\infty )$ and $f_{-}:(-\infty ,0]\rightarrow [0,\infty )$ are bounded functions and let $(B_{n})_{n\ge {}1}$ be a sequence such that $\lim _{n \rightarrow {} \infty } B_{n}/h_{n}^{1/2} = \infty$. Then, as $n \rightarrow {} \infty$,

$$\begin{aligned} {\mathbb {E}}\left[ (\gamma _{n} + \sigma _{n}Z)^{2} \mathbf{1}_{[|\Delta _{i}^{n}X|>B_{n}, \Delta _{i}^{n}N\ne 0]} \right]&\sim \lambda \sigma ^{2}h_{n}^{2}, \end{aligned}$$

(C.34)

$$\begin{aligned} {\mathbb {E}}\left[ (\gamma _{n} + \sigma _{n}Z) \mathbf{1}_{[|\Delta _{i}^{n}X|>B_{n}, \Delta _{i}^{n}N\ne 0]} \right]&= \gamma \lambda h_{n}^{2}+o( h_{n}^{2}) + o(B_{n}h_{n}^{3/2}). \end{aligned}$$

(C.35)

Proof

To show (C.34), note that, for any nonnegative integer k,

$$\begin{aligned} {\mathbb {E}}\left[ |Z|^{k} \mathbf{1}_{[|\Delta _{i}^{n}X|>B_{n}, \Delta _{i}^{n}N\ne 0]} \right]= & {} e^{-\lambda _{n}} \sum _{j=1}^{\infty } \frac{\lambda _{n}^{j}}{j!} {\mathbb {E}}\left[ |Z|^{k} \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z + \zeta _{1} + \cdots + \zeta _{j} | > {B_{n}}]} \right] \\&\sim \lambda h_{n}{\mathbb {E}}\left[ |Z|^{k}\right] , \quad (n\rightarrow \infty ), \end{aligned}$$

since, by dominated convergence theorem, ${\mathbb {E}}\left[ |Z|^{k} \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z + \zeta _{1} + \cdots + \zeta _{j} | > {B_{n}}]} \right] \rightarrow {\mathbb {E}}\left[ |Z|^{k}\right]$, as $n\rightarrow \infty$. It is now clear that, as $n\rightarrow \infty$,

$$\begin{aligned} {\mathbb {E}}\left[ (\gamma _{n} + \sigma _{n}Z)^{2} \mathbf{1}_{[|\Delta _{i}^{n}X|>B_{n}, \Delta _{i}^{n}N\ne 0]} \right]&=\gamma ^{2}\lambda h_{n}^{3}+2\lambda \sigma h_{n}^{3/2}O(h_{n})+\sigma ^{2}\lambda h_{n}^{2}\sim \sigma ^{2}\lambda h_{n}^{2}. \end{aligned}$$

Let us denote the left-hand side of (C.35) by $A_{n}$, and then note that

$$\begin{aligned} A_{n}&= \gamma _{n} e^{-\lambda _{n}} \sum _{j=1}^{\infty } \frac{\lambda _{n}^{j}}{j!} {\mathbb {P}}\left( |\gamma _{n} + \sigma _{n} Z + \zeta _{1} + \cdots + \zeta _{j} |> B_{n} \right) \\&\quad + \sigma _{n} e^{-\lambda _{n}} \sum _{j=1}^{\infty } \frac{\lambda _{n}^{j}}{j!} {\mathbb {E}}\left[ Z \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z + \zeta _{1} + \cdots + \zeta _{j} | > {B_{n}}]} \right] \\&= \gamma \lambda h_{n}^{2} + o(h_{n}^{2} ) +T_{n} + R_{n}. \end{aligned}$$

where

$$\begin{aligned} T_{n}:=- \sigma \lambda e^{-\lambda _{n}} h_{n}^{3/2} {\mathbb {E}}\left[ Z \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z + \zeta _{1}| \le {B_{n}}]} \right] , \quad R_{n} := \sigma _{n} e^{-\lambda _{n}} \sum _{j=2}^{\infty } \frac{\lambda _{n}^{j}}{j!} {\mathbb {E}}\left[ Z \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z + \zeta _{1} + \cdots + \zeta _{j} | \le {B_{n}}]} \right] . \end{aligned}$$

Now, take $n \in {\mathbb {N}}$ large enough such that $-B_{n}< \gamma _{n} < B_{n}$ and consider the following decomposition:

$$\begin{aligned} {\mathbb {E}}\left[ Z \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z + \zeta _{1}| \le {B_{n}}]} \right]&= \int _{(B_{n}-\gamma _{n})/\sigma _{n}}^{\infty } \int _{-(B_{n}+\gamma _{n}+\sigma _{n}z)}^{B_{n}-\gamma _{n}-\sigma _{n}z} q f_{-}(x) dx z \phi (z) dz \\&\quad + \int ^{-(B_{n}+\gamma _{n})/\sigma _{n}}_{-\infty } \int _{-(B_{n}+\gamma _{n}+\sigma _{n}z)}^{B_{n}-\gamma _{n}-\sigma _{n}z} p f_{+}(x) dx z \phi (z) dz \\&\quad + \int ^{(B_{n}-\gamma _{n})/\sigma _{n}}_{-(B_{n}-\gamma _{n})/\sigma _{n}} \left[ \int _{-(B_{n}+\gamma _{n}+\sigma _{n}z)}^{0} q f_{-}(x) dx \right. \\&\left. \quad + \int ^{(B_{n}+\gamma _{n}+\sigma _{n}z)}_{0} p f_{+}(x) dx \right] z \phi (z) dz \\&=: H_{n}^{(1)} + H_{n}^{(2)} + H_{n}^{(3)}. \end{aligned}$$

In light of the uniform bounds on $f_{+}$ and $f_{-}$, we have

$$\begin{aligned} \frac{|H_{n}^{(1)}|}{2B_{n}}&\le q \Vert f_{-}\Vert _{\infty } \int _{(B_{n}-\gamma _{n})/\sigma _{n}}^{\infty } z \phi (z) dz = q \Vert f_{-}\Vert _{\infty } \phi \left( \frac{B_{n}-\gamma _{n}}{\sigma _{n}} \right) , \end{aligned}$$

(C.36)

$$\begin{aligned} \frac{|H_{n}^{(2)}|}{2B_{n}}&\le p \Vert f_{+}\Vert _{\infty } \int ^{-(B_{n}-\gamma _{n})/\sigma _{n}}_{-\infty } z \phi (z) dz = p \Vert f_{+}\Vert _{\infty } \phi \left( \frac{B_{n}+\gamma _{n}}{\sigma _{n}} \right) . \end{aligned}$$

(C.37)

For $H_{n}^{(3)}$, an application of the dominate convergence theorem yields

$$\begin{aligned} \displaystyle {\lim _{n \rightarrow {} \infty }}\frac{H_{n}^{(3)}}{B_{n}} = [pf_{+}(0) + qf_{-}(0)] \int _{-\infty }^{\infty } z \phi (z)dz = 0. \end{aligned}$$

(C.38)

From (C.36) to (C.38), it is clear that $T_{n} = o(B_{n}h_{n}^{3/2})$, as $n \rightarrow {} \infty$. On the other hand, the analysis used above also shows that, for any $j \in {\mathbb {N}}$, we have ${\mathbb {E}}\left[ Z \mathbf{1}_{[|\gamma _{n} + \sigma _{n} Z + \zeta _{1} + \cdots + \zeta _{j}| \le {B_{n}}]} \right] = o(B_{n})$ and, thus, $R_{n} = o(B_{n}h_{n}^{5/2})$. $\square$

Lemma C.7

Given $\gamma \ne 0$, $\sigma > 0$, $h > 0$, and $r > 0$, the quantity given by

$$\begin{aligned} D_{h,r} := {\mathbb {E}}\left[ \left| \gamma h + \sigma h^{1/2} Z \right| ^{r} \right] - {\mathbb {E}}\left[ \left| \sigma h^{1/2} Z \right| ^{r} \right] , \end{aligned}$$

is such that $D_{h,r} \sim {r \sigma ^{r-2} h^{r/2 + 1} \gamma ^{2} \frac{{2^{r/2-1}}}{\pi ^{1/2}} \Gamma \left( \frac{r+1}{2}\right) }$, as $h \rightarrow {} 0^{+}$.

Proof

From Winkelbauer (2012), for any $r > -1$,

$$\begin{aligned} {\mathbb {E}}\left[ \left| \gamma h + \sigma h^{1/2} Z \right| ^{r} \right] = \sigma ^{r} h^{r/2} \frac{2^{r/2}}{\pi ^{1/2}} \Gamma \left( \frac{{r+1}}{2} \right) \left. _{1}F_{1}\right. \left( -\frac{r}{2},\frac{1}{2},-\frac{\gamma ^{2}h}{{2}\sigma ^{2}} \right) , \end{aligned}$$

where $\left. _{1}F_{1}\right. (a,b,z)$ denotes Kummer’s confluent hypergeometric function, which is defined by the expansion $\sum _{i=0}^{\infty } \frac{a^{(i)}}{b^{(i)}} \frac{z^{i}}{i!}$ and $a^{(i)} := a(a+1) \cdots (a+i-1)$ and $a^{(0)}=1$. The stated asymptotics follow directly from here. $\square$

Lemma C.8

Let $r \in [0,\infty )$ and suppose that ${\mathbb {E}}[|\zeta _{1}|^{r}]<\infty$. Then, as $n \rightarrow {} \infty$,

$$\begin{aligned} {\mathbb {E}}\left[ \left| \gamma _{n} + \sigma \Delta _{1}^{n}W + \Delta _{1}^{n}J \right| ^{r}I_{[\Delta _{1}^{n}N \ne 0]} \right] = \lambda h_{n} {\mathbb {E}}[|\zeta _{1}|^{r}] + o(h_{n}). \end{aligned}$$

(C.39)

Proof

Let Z be a standard normal variable. For future reference, note that ${\mathbb {E}}[|\zeta _{1}|^{r}]<\infty$ implies that ${\mathbb {E}}[|\zeta _{1} + \dots + \zeta _{m}|^{r}]<\infty$, for all $m\ge {}1$, since $|\zeta _{1} + \dots + \zeta _{m}|^{r}\le {}m^{r}\max _{i\le {}m}\{|\zeta _{i}|\}^{r}\le m^{r}(|\zeta _{1}|^{r}+\dots +|\zeta _{m}|^{r})$. Now, fix $m \in {\mathbb {N}}$ and let

$$\begin{aligned} a_{n}^{r}(m)&:= {\mathbb {E}}\left[ \left| \gamma _{n} + \sigma \Delta _{1}^{n} W + (\zeta _{1} + \dots + \zeta _{m}) \right| ^{r} \right] e^{-\lambda _{n}} \frac{\lambda _{n}^{m}}{m!}. \end{aligned}$$

Clearly, for n large enough so that $h_{n}\le {}1$, $\left| \gamma _{n} + \sigma \Delta _{1}^{n} W + (\zeta _{1} + \dots + \zeta _{m}) \right| ^{r}\le 3^{r}(|\gamma |^{r}+{\sigma }|Z|^{r}+|\zeta _{1}+\dots +\zeta _{m}|^{r})$, which has finite expectation. Thus, from the dominated convergence theorem,

$$\begin{aligned} \displaystyle {\lim _{n \rightarrow \infty }} h_{n}^{-m}a_{n}^{r}(m) = \frac{\lambda ^{m}}{m!}{\mathbb {E}}\left[ \left| \zeta _{1} + \dots + \zeta _{m} \right| ^{r} \right] . \end{aligned}$$

(C.40)

Finally, by conditioning on the total number of jumps over the time interval $[(i-1)h_{n},ih_{n})$,

$$\begin{aligned}&{\mathbb {E}}\left[ \left| \gamma _{n} + \sigma \Delta _{i}^{n}W+ \Delta _{i}^{n}J\right| ^{r}I_{[\Delta _{i}^{n}N\ne 0]} \right] \\&\quad = \sum _{m \ge 1} {\mathbb {E}}\left[ \left| \gamma _{n} + \sigma \Delta _{i}^{n}W+ \zeta _{1} + \dots + \zeta _{m} \right| ^{r}\right] \frac{e^{-\lambda _n}}{m!} \lambda _n^{m}= a_{n}^{r}(1) + o(h_{n}), \end{aligned}$$

from which the results directly follows. $\square$