Efficient estimation for the volatility of stochastic interest rate models

Abstract

The joint analysis of non-stationary and high frequency financial data poses theoretical challenges due to that such massive data varies with time and possesses no fixed density function. This paper proposes the local linear smoothing to estimate the unknown volatility function in scalar diffusion models based on Gamma asymmetric kernels for high frequency financial big data. Under the mild conditions, we obtain the asymptotic normality for the estimator at both interior and boundary design points. Besides the standard properties of the local linear estimator such as simple bias representation and boundary bias correction, the local linear smoothing using Gamma asymmetric kernels possesses some extra advantages such as variance reduction and resistance to sparse design, which is validated through finite sample simulation study and empirical analysis on 6-month Shanghai Interbank Offered Rate (abbreviated as Shibor) in China.

Appendices

Appendix

In this section, we present some technical lemmas and the proofs for the main theorems.

A. Technical lemmas

Lemma 2

(The occupation time formula) Let $X_{t}$ be a semimartingale with local time $(L_{X}(\cdot ,a))_{a \in \mathscr {D}}.$ Let g be a bounded Borel measurable function. Then

$$\begin{aligned} \int ^{\infty }_{-\infty } L_{X}(t, a)g(a)da = \int _{0}^{t}g(X_{s})d[X]_{s},~~a.s., \end{aligned}$$

(A.1)

where [X] is the quadratic variation of X.

Lemma 3

(Jacod’s stable convergence theorem) A sequence of $\mathbb {R}-$valued variables $\{\zeta _{n, i}: i \ge 1\}$ defined on the filtered probability space $(\Omega , \mathscr {F}, (\mathscr {F})_{t \ge 0}, {P})$ is $\mathscr {F}_{i \Delta _{n}}-$measurable for all n, i. Assume there exists a continuous adapted $\mathbb {R}-$valued process of finite variation $B_{t}$ and a continuous adapted and increasing process $C_{t}$, for any $t > 0,$ we have

$$\begin{aligned}&\sup _{0 \le s \le t}\big |\sum _{i = 1}^{[s/\Delta _{n}]}{E}\big [\zeta _{n, i} \mid \mathscr {F}_{(i - 1)\Delta _{n}}\big ] - B_{s}\big | {\mathop {\longrightarrow }\limits ^{P}}0, \end{aligned}$$

(A.2)

$$\begin{aligned}&\sum _{i = 1}^{[t/\Delta _{n}]}\big ({E}\big [\zeta _{n, i}^{2} \mid \mathscr {F}_{(i - 1)\Delta _{n}}\big ] - {E}^{2}\big [\zeta _{n, i} \mid \mathscr {F}_{(i - 1)\Delta _{n}}\big ]\big ) - C_{t} {\mathop {\longrightarrow }\limits ^{P}}0, \end{aligned}$$

(A.3)

$$\begin{aligned}&\sum _{i = 1}^{[t/\Delta _{n}]}{E}\big [\zeta _{n, i}^{4} \mid \mathscr {F}_{(i - 1)\Delta _{n}}\big ] {\mathop {\longrightarrow }\limits ^{P}}0. \end{aligned}$$

(A.4)

Assume also

$$\begin{aligned} \sum _{i = 1}^{[t/\Delta _{n}]}{E}\big [\zeta _{n, i} \Delta _{n}^{i}H \mid \mathscr {F}_{(i - 1)\Delta _{n}}\big ] {\mathop {\longrightarrow }\limits ^{P}}0, \end{aligned}$$

(A.5)

where either H is one of the components of Wiener process W or is any bounded martingale orthogonal (in the martingale sense) to W and $\Delta _{n}^{i}H = H_{i\Delta _{n}} - H_{(i - 1)\Delta _{n}}.$

Then the process

$$\begin{aligned} \sum _{i = 1}^{[t/\Delta _{n}]}\zeta _{n, i} {\mathop {\longrightarrow }\limits ^{\mathcal {S} - \mathcal {L}}}B_{t} + M_{t}, \end{aligned}$$

where ${\mathop {\longrightarrow }\limits ^{\mathcal {S} - \mathcal {L}}}$ denotes stable convergence in law, $M_{t}$ is a continuous process defined on an extension $\big (\widetilde{\Omega }, \widetilde{P}, \widetilde{\mathscr {F}}\big )$ of the filtered probability space $\big ({\Omega }, {P}, {\mathscr {F}}\big )$ and which, conditionally on the the $\sigma -$filter $\mathscr {F}$, is a centered Gaussian $\mathbb {R}-$valued process with $\widetilde{E}\big [M_{t}^{2} \mid \mathscr {F}\big ] = C_{t}.$

Remark A.1

For lemma 3, one can refer to Jacod (2012) (Lemma 4.4) for more details. The stable convergence implies the following crucial property required in the detailed proof of Theorem 2.3.

If $Z_{n} {\mathop {\longrightarrow }\limits ^{\mathcal {S} - \mathcal {L}}}Z$ and if $Y_{n}$ and Y are variables defined on $(\Omega , \mathscr {F}, {P})$ and with values in the same Polish space F, then

$$\begin{aligned} Y_{n} {\mathop {\longrightarrow }\limits ^{P}}Y~~~~~\Rightarrow ~~~~~(Y_{n}, ~Z_{n}) {\mathop {\longrightarrow }\limits ^{\mathcal {S} - \mathcal {L}}}(Y, ~Z), \end{aligned}$$

(A.6)

which implies that $Y_{n} \times Z_{n} {\mathop {\longrightarrow }\limits ^{\mathcal {S} - \mathcal {L}}}Y \times Z$ through the continuous function $g(x, y) = x \times y.$

Lemma 4

Under Assumptions 1-3, we have

$$\begin{aligned} \Delta _{n}\sum _{j=1}^{n}K_{G(x/h_{n}+1, h_{n})}(x_{j-1})(x_{j-1}-x)^{k} {\mathop {\rightarrow }\limits ^{P}} \int _{0}^{T}K_{G(x/h_{n}+1,h_{n})}(x_{s})(x_{s}-x)^{k}ds.\nonumber \\ \end{aligned}$$

(A.7)

Remark A.2

Denote $\alpha _{k}(x):=E[(\xi -x)^{k}],$ where $\xi {\mathop {=}\limits ^{d}} K_{G(x/h_{n} + 1, h_{n})}({u}).$ For $\alpha _{k}(x),$ we have

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\alpha _{0}(x)=1,\\ &{}\alpha _{1}(x)=h_{n},\\ &{}\alpha _{2}(x)=xh_{n}+2h_{n}^{2},\\ &{}\alpha _{l}(x)=O_{p}(h_{n}^{2}), \end{array}\right. } \end{aligned}$$

(A.8)

for $l \ge 3,$ which implies that $\xi = x + o_{p}(\sqrt{h_{n}}).$ For more details of the results (A.8), one can sketch Sect. 4, especially equation (4.2) in Chen (2002).

Based on the equations above and Theorem 1.1 on the continuity of local times in Eisenbaum and Kaspi (2007), we can obtain that

$$\begin{aligned} \int _{0}^{T} K_{G(x/h_{n}+1, h_{n})}(x_{s})(x_{s}-x)^{k}ds&= \int _{0}^{T}K_{G(x/h_{n}+1, h_{n})}(x_{s})(x_{s}-x)^{k} \frac{\sigma ^{2}(x_{s})}{\sigma ^{2}(x_{s})}ds\nonumber \\&= \int _{0}^{T}K_{G(x/h_{n}+1, h_{n})}(x_{s})(x_{s}-x)^{k} \frac{d[X]_{s}}{\sigma ^{2}(x_{s})} \nonumber \\&= \int _{0}^{\infty }K_{G(x/h_{n}+1, h_{n})}(u)(u-x)^{k}\bar{L}_{X}(T,u)du \nonumber \\&= \int _{0}^{\infty }K_{G(x/h_{n}+1, h_{n})}(u)(u-x)^{k}[\bar{L}_{X}(T,u)-\bar{L}_{X}(T,x)]du \nonumber \\&\quad + \bar{L}_{x}(T,x)\int _{0}^{\infty }K_{G(x/h_{n}+1, h_{n})}(u)(u-x)^{k}du \nonumber \\&= E\left\{ (\xi -x)^{k}[\bar{L}_{X}(T,\xi )-\bar{L}_{X}(T,x)]\right\} + \bar{L}_{X}(T,x)E[(\xi -x)^{k}] \nonumber \\&= \bar{L}_{X}(T,x) \cdot \alpha _{k}(x) \cdot (1 + o_{p}(1)). \end{aligned}$$

(A.9)

Proof

One can refer to Bandi and Phillips (2003) for similar proof procedure. For brevity, here we omit the technical details. $\square$

B. Detailed proof for theorem 2.3

Proof

For $(x_{i}-x_{i-1})^{2},$ based on Itô formula, we have

$$\begin{aligned}&(x_{i}-x_{i-1})^{2} \nonumber \\&\quad = \int _{(i-1)\Delta _{n}}^{i \Delta _{n}} \sigma ^{2}(x_{s})ds + 2\int _{(i-1)\Delta _{n}}^{i \Delta _{n}}(x_{s} - x_{i-1}) \mu (x_{s})ds + 2\int _{(i-1)\Delta _{n}}^{i \Delta _{n}}(x_{s} - x_{i-1}) \sigma (x_{s})dW_{s}.\nonumber \\ \end{aligned}$$

(B.1)

For simplicity, we set $T = 1,$ and also based on formula (B.1),

$$\begin{aligned} \frac{R(h_{n})}{\sqrt{\Delta _{n}}}\left( \hat{\sigma }^{2}_{n}(x) - \sigma ^{2}(x)\right)&= \sqrt{n}R(h_{n}) \left( \hat{\sigma }^{2}_{n}(x) - \sigma ^{2}(x)\right) \\&= \frac{\sqrt{n}R(h_{n})\Delta _{n}^{2}\sum \nolimits _{i=1}^{n}\omega _{i-1} \left( \frac{(x_{i}-x_{i-1})^{2}}{\Delta _{n}}-\sigma ^{2}(x)\right) }{\Delta _{n}^{2}\sum \nolimits _{i=1}^{n}\omega _{i-1}}\\&= \frac{\sqrt{n}R(h_{n})\Delta _{n}\sum \nolimits _{i=1}^{n}\omega _{i-1}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}} (\sigma ^{2}(x_{s})-\sigma ^{2}(x))ds}{\Delta _{n}^{2}\sum \nolimits _{i=1}^{n}\omega _{i-1}}\\&\quad + \frac{2\sqrt{n}R(h_{n})\Delta _{n}\sum \limits _{i=1}^{n}\omega _{i-1}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}} (x_{s}-x_{i-1})\mu (x_{s})ds}{\Delta _{n}^{2}\sum \limits _{i=1}^{n}\omega _{i-1}}\\&\quad + \frac{2\sqrt{n}R(h_{n})\Delta _{n}\sum \limits _{i=1}^{n}\omega _{i-1}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}} (x_{s}-x_{i-1})\sigma (x_{s})dW_{s}}{\Delta _{n}^{2}\sum \limits _{i=1}^{n}\omega _{i-1}}\\&= A_{1}+A_{2}+A_{3}, \end{aligned}$$

where the regularization coefficient $R(h_{n}) = \Big \{\begin{array}{ll} \sqrt{h_{n}^{1/2}},\quad interior~x, \\ \sqrt{h_{n}},\quad boundary~x. \end{array}$ According to Lemma 4 and the expression for $\omega _{i-1},$ one can get that

$$\begin{aligned} \Delta _{n}^{2}\sum \limits _{i=1}^{n}\omega _{i-1} {\mathop {\rightarrow }\limits ^{P}} \bar{L}^{2}_{X}(T,x) \cdot [\alpha _{0}(x) \cdot \alpha _{2}(x) - \alpha ^{2}_{1}(x)] \cdot (1 + o_{p}(1)) = \bar{L}^{2}_{X}(T,x) \cdot [x h_{n} + h^{2}_{n}].\nonumber \\ \end{aligned}$$

(B.2)

In what follows, we only need to calculate the numerators of $A_{1},$ $A_{2}$ and $A_{3}.$

We firstly introduce the uniform boundedness for the increments of diffusion process (more details seen in Karatzas and Shreve (2000)), that is

$$\begin{aligned} \lim \sup _{\Delta _{n} \rightarrow 0} \frac{\max _{i \le n} \sup _{(i-1) \Delta _{n} \le s \le i \Delta _{n}} \mid x_{s} - x_{i-1} \mid }{(\Delta _{n} \log (1/\Delta _{n}))^{1/2}} \end{aligned}$$

where C is a suitable constant.

For the numerator of $A_{1},$ we obtain

$$\begin{aligned} \sqrt{n}R(h_{n})\Delta _{n}&\sum \limits _{i=1}^{n}\omega _{i-1}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(\sigma ^{2}(x_{s})-\sigma ^{2}(x))ds\\&= \sqrt{n}R(h_{n})\Delta _{n}\sum \limits _{i=1}^{n}\omega _{i-1}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(\sigma ^{2}(x_{s}) - \sigma ^{2}(x_{i-1}) + \sigma ^{2}(x_{i-1}) - \sigma ^{2}(x))ds\\&= \sqrt{n}R(h_{n})\Delta _{n}\sum \limits _{i=1}^{n}\omega _{i-1}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(\sigma ^{2}(x_{s}) - \sigma ^{2}(x_{i-1}))ds\\&\quad + \sqrt{n}R(h_{n})\Delta _{n}\sum \limits _{i=1}^{n}\omega _{i-1}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(\sigma ^{2}(x_{i-1}) - \sigma ^{2}(x))ds\\&= A_{11} + A_{12}. \end{aligned}$$

According to uniform boundedness for the increments of diffusion process (hereafter indicated as the UBI) and Assumption 1(ii), $A_{11}=o(A_{12})$ and similarly $A_{2}=o(A_{12})$ based on Assumption 1(i), so we only need to consider $A_{12}.$

As for $A_{12},$ firstly through Taylor expansion, we can get

$$\begin{aligned} \sigma ^{2}(x_{i-1})-\sigma ^{2}(x)&=(\sigma ^{2})^{'}(x)(x_{i-1}-x) + \frac{1}{2}(\sigma ^{2})^{''}(x+\theta (x_{i-1}-x))(x_{i-1}-x)^{2}, \end{aligned}$$

where $\theta \in [0,1].$

It can be obviously deduced that

$$\begin{aligned} \sum ^{n}_{i=1}\omega _{i-1}(x_{i-1}-x) \equiv 0. \end{aligned}$$

Then, according to Lemma 4 and the expression for $\omega _{i-1},$

$$\begin{aligned} \frac{A_{12}}{\sqrt{n} R(h_{n})}&= \Delta _{n}\sum \limits _{i=1}^{n}\omega _{i-1}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}} \left[ \frac{1}{2}(\sigma ^{2})^{''}(x+\theta (x_{i-1}-x))(x_{i-1}-x)^{2}\right] ds\\&\quad \rightarrow \frac{1}{2} (\sigma ^{2})''(x)\left[ \alpha _{2}^{2}(x) - \alpha _{1}(x) \cdot \alpha _{3}(x)\right] \cdot \bar{L}_{X}^{2}(T, x)\\&= \frac{1}{2} (\sigma ^{2})''(x) \cdot \bar{L}_{X}^{2}(T, x) \left[ (x h_{n} + h_{n}^{2})^{2} - O_{p}(h_{n}^{3})\right] \\&= \frac{x^{2}}{2} (\sigma ^{2})''(x) \cdot \bar{L}_{X}^{2}(T, x) \cdot h_{n}^{2}. \end{aligned}$$

Thus, the bias for $\sqrt{n}R(h_{n}) \left( \hat{\sigma }^{2}_{n}(x) - \sigma ^{2}(x)\right)$ is

$$\begin{aligned} \frac{A_{12}}{\sqrt{n}R(h_{n}) \Delta _{n}^{2}\sum \limits _{i=1}^{n}\omega _{i-1}}&\rightarrow \frac{\frac{x^{2}}{2}(\sigma ^{2})''(x)\bar{L}_{X}^{2}(T,x)h_{n}^{2}}{\bar{L}_{X}^{2}(T,x)(xh+h^{2})} \\&\quad =\frac{x}{2}(\sigma ^{2})''(x)h_{n}. \end{aligned}$$

In what follows, we calculate the asymptotic variance for

$$\begin{aligned} \sqrt{n}R(h_{n}) \left( \hat{\sigma }^{2}_{n}(x) - \sigma ^{2}(x) - \frac{x}{2}(\sigma ^{2})''(x)h_{n}\right) . \end{aligned}$$

By Burkholder-Davis-Gundy (BDG) inequality, Hölder inequality and UBI property, we can conclude that $A_{2}=o_{a.s}(A_{3}).$ So here we only need to deal with $A_{3}.$ However, the $\omega _{i-1}$ in $A_{3}$ is not $\mathscr {F}_{i\Delta _{n}}$-measurable, so we can not directly use Jacod’s stable convergence theorem in Lemma 3. To solve this problem, we firstly introduce its convergence limit in probability divided by $\bar{L}_{X}(T,x) \cdot x h_{n},$ that is,

$$\begin{aligned} \frac{\Delta _{n} \cdot \omega _{i}}{\bar{L}_{X}(T,x) \cdot x h_{n}} {\mathop {\rightarrow }\limits ^{P}}K_{G(x/h_{n}+1,h_{n})}(x_{i})\left[ 1-(x_{i}-x)\frac{1}{x}\right] =: \omega ^{+}_{i}. \end{aligned}$$

So

$$\begin{aligned} \omega ^{+}_{i-1}=K_{G(x/h_{n}+1,h_{n})}(x_{i-1})\left[ 1-(x_{i-1}-x)\frac{1}{x}\right] \end{aligned}$$

is obviously $\mathscr {F}_{i\Delta _{n}}$-measurable, and we can utilize the Jacod’s stable convergence theorem for

$$\begin{aligned} 2\sqrt{n}R(h_{n})\sum \limits _{i=1}^{n}\omega _{i-1}^{+}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}:=\sum ^{n}_{i=1}q_{i} \end{aligned}$$

where

$$\begin{aligned} q_{i}=2\sqrt{n}R(h_{n})\omega _{i-1}^{+}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}. \end{aligned}$$

Jacod’s sable convergence theorem in Lemma 3 tells us that the following arguments,

$$\begin{aligned}&S_{1}=\sup _{1\le i\le n}|\sum ^{n}_{i=1}E_{i-1}[q_{i}]-0|{\mathop {\rightarrow }\limits ^{P}}0, \\&S_{2}=\sum ^{n}_{i=1}(E_{i-1}[q_{i}^{2}]-E_{i-1}^{2}[q_{i}]){\mathop {\rightarrow }\limits ^{P}} \left\{ \begin{aligned}&\frac{\sigma ^{4}(x)}{\sqrt{\pi }\sqrt{x}}\bar{L}_{x}(T,x),\qquad interior~x,\\&\frac{\sigma ^{4}(x)\Gamma (2k+1)}{2^{2k}\Gamma ^{2}(k+1)}\bar{L}_{X}(T,x),\qquad boundary~x,\\ \end{aligned} \right. \\&S_{3}=\sum ^{n}_{i=1}E_{i-1}[q_{i}^{4}]{\mathop {\rightarrow }\limits ^{P}}0, \nonumber \\&S_{4}=\sum ^{n}_{i=1}E_{i-1}[q_{i}\Delta _{i}^{n}H]{\mathop {\rightarrow }\limits ^{P}}0 \end{aligned}$$

implies $\sum ^{n}_{i=1}q_{i} {\mathop {\longrightarrow }\limits ^{\mathcal {S} - \mathcal {L}}}M_{t},$ where $E_{i-1}[\cdot ]$ := $E[\cdot |\mathscr {F}_{(i-1)\Delta _{n}}].$

For $S_{1}$, due to the martingale property, we can easily know that,

$$\begin{aligned} E_{i-1}[q_{i}] \equiv 0. \end{aligned}$$

Hence,

$$\begin{aligned} S_{1}=\sup _{1\le i\le n}|\sum ^{n}_{i=1}E_{i-1}[q_{i}]|{\mathop {\rightarrow }\limits ^{P}}0 \end{aligned}$$

For $S_{2}$,

$$\begin{aligned} S_{2}=&\sum ^{n}_{i=1}E_{i-1}[q_{i}^{2}]\\ =&\sum ^{n}_{i=1}E_{i-1}\left[ 4nR^{2}(h_{n})\left( K_{G(x/h_{n}+1,h_{n})}^{2}(x_{i-1})-2K_{G(x/h_{n}+1,h_{n})}^{2}(x_{i-1})(x_{i-1}-x)\frac{1}{x}\right. \right. \\&\left. \left. +K_{G(x/h_{n}+1,h_{n})}^{2}(x_{i-1})(x_{i-1}-x)^{2}\frac{1}{x^{2}}\right) \left( \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}\right) ^{2}\right] \\ =&\sum ^{n}_{i=1}E_{i-1}[q_{i1}^{2}+q_{i2}^{2}+q_{i3}^{2}] \end{aligned}$$

One can show that $q_{i1}^{2}$ is larger than the others according to Lemma 4 and the equation (A.8), so we only deal with the dominant one $q_{i1}^{2},$

$$\begin{aligned}&\sum ^{n}_{i=1}E_{i-1}[q_{i1}^{2}]\nonumber \\&\quad =4nR^{2}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{2}(x_{i-1})E_{i-1} \left[ \left( \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}\right) ^{2}\right] \\&\quad =4nR^{2}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{2}(x_{i-1})E_{i-1} \left[ \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})^{2}\sigma ^{2}(x_{s})ds\right] . \end{aligned}$$

Using the Itô formula on $(x_{s}-x_{i-1})^{2}$, we have

$$\begin{aligned}&\sum ^{n}_{i=1}E_{i-1}[q_{i1}^{2}]\\&=4nR^{2}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{2}(x_{i-1})\bigg \{E_{i-1} \left[ \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}\int ^{s\Delta _{n}}_{(i-1)\Delta _{n}}\sigma ^{2}(x_{u})du\sigma ^{2}(x_{s})ds\right] \\&\quad +E_{i-1}\left[ \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}\int ^{s\Delta _{n}}_{(i-1)\Delta _{n}}(x_{u}-x_{i-1})\mu (x_{u})du\sigma ^{2}(x_{s})ds\right] \\&\quad +E_{i-1}\left[ \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}\int ^{s\Delta _{n}}_{(i-1)\Delta _{n}}(x_{u}-x_{i-1})\sigma (x_{u})dW_{u}\sigma ^{2}(x_{s})ds\right] \bigg \}\\&=B_{1}+B_{2}+B_{3}. \end{aligned}$$

We can deduce that $B_{3}\equiv 0$ by Fubini theorem, and $B_{2}=o_{a.s.}(B_{1})$ by UBI property and Assumption 1. Applying the mean value theorem and UBI property for $B_{1},$ we can reach

$$\begin{aligned} B_{1}=&4nR^{2}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{2}(x_{i-1})E_{i-1}\left[ \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}} \int ^{s\Delta _{n}}_{(i-1)\Delta _{n}}\sigma ^{2}(x_{u})du\sigma ^{2}(x_{s})ds \right] \\ \quad \rightarrow&4nR^{2}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{2}(x_{i-1})\frac{\sigma ^{4}(x_{i-1})}{2}\Delta _{n}^{2}\\&=2R^{2}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{2}(x_{i-1})\sigma ^{4}(x_{i-1})\Delta _{n}\\ \quad \rightarrow&2R^{2}(h_{n})B_{n}(x)\sum ^{n}_{i=1}K_{G(2x/h_{n}+1,h_{n}/2)(x_{i-1})}\sigma ^{4}(x_{i-1})\Delta _{n}, \end{aligned}$$

where

$$\begin{aligned} B_{n}(x):=\left\{ \begin{aligned}&\frac{1}{2\sqrt{\pi }}h_{n}^{-\frac{1}{2}}x^{-\frac{1}{2}},\qquad interior~x,\\&\frac{\Gamma (2k+1)}{2^{2k+1}\Gamma ^{2}(k+1)}h_{n}^{-1}, \qquad boundary~x. \end{aligned} \right. \end{aligned}$$

Hence,

$$\begin{aligned} B_{1}{\mathop {\rightarrow }\limits ^{P}} \left\{ \begin{aligned}&\frac{\sigma ^{4}(x)}{\sqrt{\pi }\sqrt{x}}\bar{L}_{X}(T,x), \qquad interior~x, \\&\frac{\sigma ^{4}(x)\Gamma (2k+1)}{2^{2k}\Gamma ^{2}(k+1)}\bar{L}_{X}(T,x), \qquad boundary~x. \end{aligned} \right. \end{aligned}$$

For $S_{3},$ by using BDG and Hölder inequalities, we have

$$\begin{aligned} S_{3}=&16n^{2}R^{4}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{4}(x_{i-1})E_{i-1} \left[ \left( \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}\right) ^{4}\right] \\ \le&16n^{2}R^{4}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{4}(x_{i-1})E_{i-1} \left[ \sup _{s\in [(i-1)\Delta _{n},i\Delta _{n}]}\left( \int ^{s}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}\right) ^{4}\right] \\ \le&16n^{2}R^{4}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{4}(x_{i-1})E_{i-1} \left[ \left( \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})^{2}\sigma ^{2}(x_{s})ds \right) ^{2}\right] \\ \le&16n^{2}R^{4}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{4}(x_{i-1})E_{i-1} \left[ \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})^{4}\sigma ^{4}(x_{s})ds\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}1^{2}ds\right] \\ =&C nR^{4}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{4}(x_{i-1})E_{i-1}\left[ \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})^{4}\sigma ^{4}(x_{s})ds \right] \\ =&CnR^{4}(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{4}(x_{i-1})\left( \Delta _{n} \log \frac{1}{\Delta _{n}} \right) ^{2} \Delta _{n}\\ =&C R^{4}(h_{n})\Delta _{n} \left( \log \frac{1}{\Delta _{n}} \right) ^{2} \sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}^{4}(x_{i-1})\Delta _{n}\\ =&C R^{4}(h_{n})\Delta _{n} \left( \log \frac{1}{\Delta _{n}} \right) ^{2} B(4,h_{n},x) \sum ^{n}_{i=1}K_{G(4x/h+1,h/4)}\Delta _{n}\\ =&C\frac{R^{4}(h)}{n}B(4,h,x)\bar{L}_{X}(T,x) \left( \log \frac{1}{\Delta _{n}} \right) ^{2}, \end{aligned}$$

where

$$\begin{aligned} B(4,h,x)= \left\{ \begin{aligned}&h_{n}^{-\frac{3}{2}},\qquad interior~x,\\&h_{n}^{-3}, \qquad boundary~x,\\ \end{aligned} \right. \end{aligned}$$

which one can refer to Rosa and Nogueira (2016) for more details.

Hence

$$\begin{aligned} S_{3} \le C \bar{L}_{X}(T,x) \left( \log \frac{1}{\Delta _{n}} \right) ^{2} \left\{ \begin{aligned}&\frac{1}{nh^{\frac{1}{2}}}{\mathop {\rightarrow }\limits ^{P}} 0, \qquad interior~x,\\&\frac{1}{nh}{\mathop {\rightarrow }\limits ^{P}} 0, \qquad boundary~x.\\ \end{aligned} \right. \end{aligned}$$

For $S_{4},$ if $H=W,$ then

$$\begin{aligned} S_{4}&=2\sqrt{n}R(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}(x_{i-1})\left[ 1-(x_{i-1}-x)\frac{1}{x}\right] \\&\quad \times E_{i-1}\left[ \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}\Delta ^{n}_{i}H\right] \\&\le 2\sqrt{n}R(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}(x_{i-1})\left[ 1-(x_{i-1}-x)\frac{1}{x}\right] \\&\quad \times \sqrt{E_{i-1}\left[ \int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})^{2}\sigma ^{2}(x_{s})ds\right] }\sqrt{E_{i-1}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}ds}\\&=2\sqrt{n}R(h_{n})\sum ^{n}_{i=1}K_{G(x/h_{n}+1,h_{n})}(x_{i-1})\left[ 1-(x_{i-1}-x)\frac{1}{x}\right] \Delta _{n}^{\frac{3}{2}}\log ^{\frac{1}{2}}\frac{1}{\Delta _{n}}\\&\quad \rightarrow 2R(h_{n}) \bar{L}_{x}(T,x)\left[ 1-\frac{\alpha _{1}(x)}{x}\right] \log ^{\frac{1}{2}}\frac{1}{\Delta _{n}}\rightarrow 0, \end{aligned}$$

due to $R(h_{n}) \rightarrow 0.$

If H is orthogonal to W, $\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}$ is orthogonal to H, then we can get

$$\begin{aligned} S_{4}=\sum ^{n}_{i=1}E_{i-1}[q_{i}\Delta _{i}^{n}H] \equiv 0. \end{aligned}$$

Furthermore, we can observe

$$\begin{aligned} \bar{L}_{X}(T, x)&\cdot \frac{2\sqrt{n}R(h_{n})\Delta _{n}\sum \limits _{i=1}^{n}\omega _{i-1}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}}{\Delta _{n}^{2} \sum _{i=1}^{n}\omega _{i-1}}\\&\quad - 2\sqrt{n}R(h_{n})\sum \limits _{i=1}^{n}\omega _{i-1}^{+}\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}\\&= \left[ \bar{L}_{X}(T, x) \cdot \frac{\Delta _{n}\sum _{j=1}^{n}K_{G(x/h_{n}+1, h_{n})}(x_{j-1})(x_{j-1}-x)^{2}}{\Delta _{n}^{2} \sum _{i=1}^{n}\omega _{i-1}} - 1\right] \cdot \sum \limits _{i=1}^{n} q_{2, i}\\&\quad - \left[ \bar{L}_{X}(T, x) \cdot \frac{\Delta _{n}\sum _{j=1}^{n}K_{G(x/h_{n}+1, h_{n})}(x_{j-1})(x_{j-1}-x)}{\Delta _{n}^{2} \sum _{i=1}^{n}\omega _{i-1}} - \frac{1}{x}\right] \cdot \sum \limits _{i=1}^{n} q_{3, i}, \end{aligned}$$

where $q_{2, i} := 2\sqrt{n}R(h_{n})K_{G(x/h_{n}+1, h_{n})}(x_{i-1})\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}$ and $q_{3, i} := 2\sqrt{n}R(h_{n})K_{G(x/h_{n}+1, h_{n})}(x_{i-1})(x_{i-1} - x)\int ^{i\Delta _{n}}_{(i-1)\Delta _{n}}(x_{s}-x_{i-1})\sigma (x_{s})dW_{s}.$

Under the same proof procedure as $\sum \limits _{i=1}^{n} q_{i},$ we can obtain $\sum ^{n}_{i=1}q_{2, i} {\mathop {\longrightarrow }\limits ^{\mathcal {S} - \mathcal {L}}}M_{2, t}$ and $\sum ^{n}_{i=1}q_{3, i} {\mathop {\longrightarrow }\limits ^{\mathcal {S} - \mathcal {L}}}M_{3, t},$ which implies that $\sum ^{n}_{i=1}q_{2, i} = O_{p}(1)$ and $\sum ^{n}_{i=1}q_{3, i} = O_{p}(1).$ Moreover, based on Lemma 4 and its remark, we have $\bar{L}_{X}(T, x) \cdot \frac{\Delta _{n}\sum _{j=1}^{n}K_{G(x/h_{n}+1, h_{n})}(x_{j-1})(x_{j-1}-x)^{2}}{\Delta _{n}^{2} \sum _{i=1}^{n}\omega _{i-1}} - 1 {\mathop {\rightarrow }\limits ^{P}} 0$ and $\bar{L}_{X}(T, x) \cdot \frac{\Delta _{n}\sum _{j=1}^{n}K_{G(x/h_{n}+1, h_{n})}(x_{j-1})(x_{j-1}-x)}{\Delta _{n}^{2} \sum _{i=1}^{n}\omega _{i-1}} - \frac{1}{x} {\mathop {\rightarrow }\limits ^{P}} 0.$ Hence, we can get the main results in Theorem 2.3 due to the Remark A.1. $\square$