Let $S^{k}_t$ be a subfractional Brownian motion (sfBm) with parameter $k\in (-\frac{1}{2}, \frac{1}{2})$ on $\mathbb{R}^d$. $S^{k}_t$ is a centered Gaussian process with representation

$\begin{equation} \begin{aligned} S_t^{k}=\frac{1}{c(k)}\int_{\mathbb{R}}[(t-s)^{k}_++(t+s)^{k}_--2(-s) ^{k}_+]dB_s, \end{aligned} \end{equation}$

(1.1)

where $c(k)=[2(\displaystyle\int_0^{\infty}((1+s)^{k}-s^{k})^2+\frac{1}{2k+1})]^{\frac{1}{2}}, a_+=\max\{a, 0\}, a_-=\max\{-a, 0\}$ and $B_s$ is a Brownian motion (Bm). It is well known that subfractional Brownian motion is an extension of Bm.

The object of study in this article is the local time of mixed Bm $B_t$ and sfBm $S^{k}_t$ (mBs), which is formally defined as

$\begin{equation} \begin{aligned} L_k(T, b)=\displaystyle\int_0^T\delta(\alpha B_t+\beta S^{k}_t-b)dt, \end{aligned} \end{equation}$

(1.2)

where b is some fixed point, T > 0, $\delta(x)$ is a Dirac delta function, $\alpha$ and $\beta$ are two real constants such that $(\alpha, \beta)\neq(0, 0)$. The local time $L_k(T, b)$ characterizes time of the mixed process spending on the interval [0, T].

In recent years, some authors focused on the research on fractional Brownian motion (fBm), due to its interesting properties and its applications. Moreover, some authors introduce another Gaussian process so called sfBm instead of fBm. In spite of sfBm has many properties analogous of fBm such as self-similar and long-rang dependent, sfBm has non-stationary increments and is weakly correlated in comparing with fBm. Indeed, sfBm is intermediate between Bm and fBm. At the same time, local times of sfBm were studied by many authors as well, see e.g. Liu et al. [1], Yan et al. [2] and Guo et al. [3].

In this article, motivated by [3-4], in order to deal with problem of complexity structure of sfBm, we use an alternative expression of sfBm in Guo et al. (2015), and study the existence of local time of mBs in white noise analysis framework. The outline of the paper is as follows. In Section 2, we give some background materials in white noise analysis. In Section 3, we present the main results and their demonstrations.

We briefly recall some notions and facts, and for detail see Refs. [4-6]. The first real Gelfand triple is $\mathcal{S}_{2d}(\mathbb{R})\subset L^2(\mathbb{R}, \mathbb{R}^{2d})\subset\mathcal{S}_{2d}^{\ast}(\mathbb{R}), $ where $\mathcal{S}_{2d}(\mathbb{R})$ (resp. $\mathcal{S}_{2d}^{\ast}(\mathbb{R})$) is the space of the vector valued Schwartz test functions (resp. tempered distributions). We consider $2d$-tuple of Gaussian white noise $\vec{\omega}= (\vec{\omega}_1, \vec{\omega}_2)=(\omega_{1, 1}, \cdots, \omega_{1, d}, \omega_{2, 1}, \cdots, \omega_{2, d}).$

Introduce the following notations

$\mathbf{n}=(n_1, \cdots, n_d), ~~~~n=\sum_{i=1}^{d}n_i, ~~~~\mathbf{n}!=\prod_{i=1}^{d}{n_i}!.$

Let $(L^2)\equiv L^2(\mathcal{S}_{2d}^{\ast}(\mathbb{R}), d \mu)$ be the Hilbert space of square integrable functionals on $\mathcal{S}_{2d}^{\ast}(\mathbb{R})$ with respect to measure $\mu$. By the Wiener-Itô-Segal isomorphism theorem, we have the chaos expansion $F(\vec{\omega}_1, \vec{\omega}_2)=\sum\limits_{\mathbf{m}\in\mathbb{N} ^d} \sum\limits_{\mathbf{k}\in\mathbb{N}^d}\langle :\vec{\omega}^ {\otimes \mathbf{m}}: \otimes:\vec{\omega}^{\otimes \mathbf{k}}:, \mathbf{f}_{\mathbf{m}, \mathbf{k}} \rangle$ for each $F\in (L^2)$.

Let $\Gamma(A)$ be the second quantization of A, where A is defined by $(Ag)_i(t)=(-\frac{d^2}{dt^2}+t^2+1)g_i(t).$ For each integer p, let $(\mathcal{S}_p)$ be the completion of Dom$\Gamma(A)^p$ with respect to the Hilbert norm $\parallel \cdot\parallel_p=\parallel \Gamma(A)^p\cdot\parallel_0$, where we denote the norm of $(L^2)$ by $\parallel \cdot\parallel_0$. Let $(\mathcal{S})=\bigcap\limits_{p\geq 0}(\mathcal{S}_p)$ be the projective limit of $\{(\mathcal{S}_p)\mid p\geq 0\}$, and let $(\mathcal{S})^{\ast}=\bigcup\limits_{p\geq 0}(\mathcal{S}_{-p})$ be the inductive limit of $\{(\mathcal{S}_{-p})\mid p\geq 0\}$. Thus there is the second Gelfand triple $(\mathcal{S})\subset(L^2)\subset(\mathcal{S})^{\ast}.$ Elements of $(\mathcal{S})$ (resp. $(\mathcal{S})^{\ast})$ are called Hida testing (resp. generalized) functionals. For each test function $\mathbf{f}=(\mathbf{f}_1, \mathbf{f}_2)=(f_{1, 1}, \cdots, f_{1, d}, f_{2, 1}, \cdots, f_{2, d})\in \mathcal{S}_{2d}(\mathbb{R})$, S-transform of Hida generalized functionals $\Phi$ is defined by $S\Phi(\mathbf{f})=\ll \Phi, :\exp \langle \cdot, \mathbf{f}\rangle:\gg.$

Lemma 2.1[4]  Let $(\Omega, \mathfrak{F}, \mu)$ be a measure space, and let $\Phi$ be a mapping defined on $\Omega$ with values in $(\mathcal{S})^{\ast}$. We assume S-transform of $\Phi$:

(1) $\Phi$ is an $\mu$-measurable function of $\lambda$ for $\mathbf{f}\in\mathcal{S}_{2d}(\mathbb{R});$ and

(2) $\Phi$ obeys a U-functional estimate

$\begin{equation} \begin{aligned} \mid S\Phi(z\mathbf{f})\mid\leq C_1(\lambda)\exp \{C_2(\lambda)\mid z\mid^2\mid A^p\mathbf{f}\mid_0^2\} \end{aligned} \end{equation}$

(2.1)

for some fixed p and for $C_1\in L^1(\mu), ~C_2\in L^{\infty}(\mu).$ Then

$\displaystyle\int_{\Omega}\Phi d\mu(\lambda)\in(\mathcal{S})^{\ast}~~\mbox{and}~~~S(\displaystyle\int_{\Omega}\Phi d\mu(\lambda))(\mathbf{f})=\displaystyle\int_{\Omega}(S\Phi) (\mathbf{f})d\mu(\lambda).$

Let Bm and sfBm be independent processes. For any $\varepsilon>0$, we define

$\begin{equation} \begin{aligned} L_{k, \varepsilon}(T, b)=\displaystyle\int_0^Tp_{\varepsilon}(\alpha B_t+\beta S^k_t-b)dt, \end{aligned} \end{equation}$

(3.1)

where $p_{\varepsilon}(x)=\frac{1}{\sqrt{2\pi \varepsilon}} \exp{\{-\frac{x^2}{2\varepsilon}\}}.$

Lemma 3.1[3]  Let $k\in(-\frac{1}{2}, \frac{1}{2})$ be given. $S^k_t$ has a continuous version of $\langle \cdot, \frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)}\rangle, $ where $\mathbb{I}^o_{[0, t)}$ denotes the odd extension of $\mathbb{I}_{[0, t)}$ and $c(k)=[2(\displaystyle\int_0^{\infty}((1+s)^{k}-s^{k})^2ds+\frac{1}{2k+1})] ^{\frac{1}{2}}$.

The following lemma is very useful to prove our main results. Bender [7] gave the similar estimate in discussion the local time of fBm.

Lemma 3.2[3]  Let $k\in(-\frac{1}{2}, \frac{1}{2})$ and $f\in\mathcal{S}_1(\mathbb{R})$ be given. Then there exists a non-negative constant $C_{k}$ such that

$\begin{equation} \begin{aligned} \mid\displaystyle\int_\mathbb{R}f(x)\frac{1}{c(k)}(I_{-}^k\mathbb{I}_{[0, t)}^o)(x)dx\mid \leq C_k\mid t\mid\parallel f\parallel, \end{aligned} \end{equation}$

(3.2)

where $c(k)=[2(\displaystyle\int_0^{\infty}((1+s)^k-s^k)^2ds+\frac{1}{2k+1})]^{\frac{1}{2}}, C_k$ is some constant independent of f and $\parallel f\parallel\equiv\sup\limits_{x\in\mathbb{R}}\mid f(x)\mid+\sup\limits_{x\in\mathbb{R}}\mid f'(x)\mid.$

Theorem 3.1  For each $k\in(-\frac{1}{2}, \frac{1}{2})$ and given $b\in\mathbb{R}$, the local time of mBs given by

$\begin{equation} \begin{aligned} L_{k, \varepsilon}(T, b)=&\displaystyle\int^T_0p_\varepsilon(\alpha B_t+\beta S^k_t -b)dt\\ =&\frac{1}{(2\pi\varepsilon)^{\frac{d}{2}}}\displaystyle\int^T_0 \exp\{-\frac{(\alpha B_t+\beta S^k_t-b)^2}{2\varepsilon}\}dt \end{aligned} \end{equation}$

(3.3)

is a Hida distribution. Moreover, S-transform of $L_{k, \varepsilon}(T, b)$ is given by

$\begin{equation} \begin{aligned} S(L_{k, \varepsilon}(T, b))(\mathbf{f}) =&\displaystyle\int^T_0\frac{1}{(2\pi(\varepsilon+\alpha^2t+\beta^2(2-2^{2k})t^{2k+1})) ^{\frac{d}{2}}}\\ \cdot &\exp\{-\frac{(\alpha\displaystyle\int_0^t\mathbf{f}(s)ds+\beta \displaystyle\int_{\mathbb{R}}\mathbf{f}(s)(\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)}) (s)ds-b)^2} {2(\varepsilon+\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))}\}dt \end{aligned} \end{equation}$

(3.4)

for all $\mathbf{f}=(\mathbf{f}_1, \mathbf{f}_2)=(f_{1, 1}, \cdots, f_{1, d}, f_{2, 1}, \cdots, f_{2, d})\in\mathcal{S}_{2d}(\mathbb{R})$.

Proof  Set

$\begin{equation*} \begin{aligned} \Phi_{k, \varepsilon}(\vec{\omega})\equiv(\frac{1}{2\pi\varepsilon})^{\frac{d}{2}} \exp\{-\frac{(\alpha B_t+\beta S^k_t-b)^2}{2\varepsilon}\}, \end{aligned} \end{equation*}$

where $\vec{\omega}=(\vec{\omega}_1, \vec{\omega}_2)= (\omega_{1, 1}, \cdots, \omega_{1, d}, \omega_{2, 1}, \cdots, \omega_{2, d})$. For each $\mathbf{f}\in\mathcal{S}_{2d}(\mathbb{R})$, calculate S-transform of $L_{k, \varepsilon}(T, b)$ as follows

$\begin{equation*} \begin{aligned} S(\Phi_{k, \varepsilon}(\vec{\omega}))(\mathbf{f})=&\frac{1}{(2\pi(\varepsilon+\alpha^2 t+\beta^2(2-2^{2k})t^{2k+1}))^{\frac{d}{2}}}\\ &\cdot \exp\{-\frac{(\alpha\displaystyle\int_0^t\mathbf{f}(s)ds+\beta\displaystyle\int_{\mathbb{R}} \mathbf{f}(s)(\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)}) (s)ds-b)^2} {2(\varepsilon+\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))}\}. \end{aligned} \end{equation*}$

For the bounded condition we observe that

$\begin{equation*} \begin{aligned} &\mid S(\Phi_{k, \varepsilon}(\vec{\omega}))(z\mathbf{f})\mid =(\frac{1}{2\pi (\varepsilon+\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))})^\frac{d}{2}\\ &\cdot \mid\exp\{-\frac{(z\alpha\displaystyle\int_0^t\mathbf{f}(s)ds+z\beta\displaystyle\int_{\mathbb{R}}\mathbf{f}(s) (\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(s)ds -b)^2}{2(\varepsilon+\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))}\}\mid\\ \leq & (\frac{1}{2\pi(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1})})^{\frac{d}{2}}\exp\{\frac{b^2+2\mid z\mid \mid b\mid \mid\alpha\mid\mid\displaystyle\int_0^t\mathbf{f}(s)ds\mid}{2(\varepsilon+ \alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))}\}\\ \cdot&\exp\{\frac{2\mid z\mid\mid b\mid\mid\beta\mid\mid\displaystyle\int_{\mathbb{R}}\mathbf{f}(s)(\frac{1}{c(k)}I^{k} _{-}\mathbb{I}^o_{[0, t)})(s) ds\mid}{2(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}\\ \cdot&\exp\{\frac{2\mid z\mid^2\mid\alpha\mid\mid\beta\mid\mid \displaystyle\int_0^t\mathbf{f}(s)ds\mid\mid\displaystyle\int_{\mathbb{R}}\mathbf{f}(s) (\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(s) ds\mid}{2(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}\\ \cdot&\exp\{\frac{\mid z\mid^2\alpha^2\mid\displaystyle\int_0^t\mathbf{f}(s)ds\mid^2+\mid z\mid^2 \beta^2\mid\displaystyle\int_{\mathbb{R}}\mathbf{f}(s) (\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(s)ds\mid^2}{2(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\} \end{aligned} \end{equation*}$

for any complex number z and each $\mathbf{f}\in\mathcal{S}_{2d}(\mathbb{R})$. By Lemma 3.1 and Lemma 3.2, we can obtain the following inequalities

$\begin{eqnarray*} &&\exp\{\frac{2\mid z\mid\mid a\mid\mid\beta\mid\mid\displaystyle\int_{\mathbb{R}}\mathbf{f}(s)(\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(s) ds\mid}{2(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\} \leq\exp\{\frac{2\mid z\mid\mid a\mid\mid\beta\mid C_{k, 1}t\parallel \mathbf{f} \parallel}{2(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}, \\ && \exp\{\frac{\mid z\mid^2\alpha^2\mid\displaystyle\int_0^t\mathbf{f}(s)ds\mid^2} {2(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\} \leq\exp\{\frac{\mid z\mid^2\alpha^2\mid \mathbf{f}\mid^2}{2(\varepsilon+\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))}\}, \\ &&\exp\{\frac{\mid z\mid^2 \beta^2\mid\displaystyle\int_{\mathbb{R}}\mathbf{f}(s) (\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(s)ds\mid^2}{2(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\} \leq \exp\{\frac{C_{k, 1}^2t^2\mid z\mid^2 \beta^2\parallel\mathbf{ f}\parallel^2}{2(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}, \\ && \exp\{\frac{2\mid z\mid \mid a\mid \mid\alpha\mid\mid\displaystyle\int_0^t\mathbf{f}(s)ds\mid}{2(\varepsilon+ \alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))}\}\leq\exp\{\frac{2\mid z\mid \mid a\mid \mid\alpha\mid\mid \mathbf{f}\mid}{2(\varepsilon+ \alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))}\}, \end{eqnarray*}$

$\begin{eqnarray*} &&\exp\{\frac{2\mid z\mid^2\mid\alpha\mid\mid\beta\mid\mid \displaystyle\int_0^t\mathbf{f}(s)ds\mid\mid\displaystyle\int_{\mathbb{R}}\mathbf{f}(s) (\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(s) ds\mid}{2(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}\\ &\leq &\exp\{\frac{2C_{k, 1}t\mid z\mid^2\mid\alpha\mid\mid\beta\mid\mid \mathbf{f}\mid\parallel\mathbf{ f}\parallel}{2(\varepsilon+\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}, \end{eqnarray*}$

where $\mathbf{f}=(\mathbf{f}_1, \mathbf{f}_2)\in\mathcal{S}_{2d} (\mathbb{R}), ~~~\mid \mathbf{f}\mid\equiv\sum\limits_{i=1}^{2d}\mid f_i\mid, $

$\begin{equation*} \begin{aligned} \parallel \mathbf{f}\parallel\equiv(\sum_{i=1}^{2d}(\sup_{x\in\mathbb{R}}\mid f_i(x)\mid+\sup_{x\in\mathbb{R}}\mid f_i'(x)\mid)^2)^{\frac{1}{2}}, \end{aligned} \end{equation*}$

and $C_{k, 1}$ is some constant, which dependents on parameter k. Hence we obtain

$\begin{equation*} \begin{aligned} &\mid S(\Phi_{\varepsilon}(z\mathbf{f})\mid\\ \leq &(\frac{1}{2\pi (\alpha^2 t+\beta^2 (2-2^{2k})t^{2k+1})})^{\frac{d}{2}} \exp\{\frac{b^2+\mid z\mid^2\alpha^2\mid f\mid^2 +2\mid z\mid\mid b\mid\mid \alpha\mid\mid f\mid} {2\pi (\alpha^2 t+\beta^2 (2-2^{2k})t^{2k+1})}\}\\ &\cdot\exp\{\frac{2(\mid z\mid\mid b\mid\mid\beta\mid \parallel f\parallel+C_{k, 1}\mid z\mid^2\mid \alpha\mid\mid\beta\mid\mid f\mid\parallel f\parallel)t +C_{k, 1}^2\mid z\mid^2\beta^2 \parallel f\parallel^2t^2} {2\pi (\alpha^2 t+\beta^2 (2-2^{2k})t^{2k+1})}\}. \end{aligned} \end{equation*}$

We see that the first part in above inequality is integrable as a function on [0, T], and the third part is bound on [0, T]. Then by an application of Lemma 2.1, the proof is completed.

Similar to [4, 5], we can prove that $L_{k, \varepsilon}(T, b)$ convergence to $L_k(T, b)$ in $(L^2).$ The following theorem provides the chaos expansion of the local time of mixed independent Bm and sfBm. For simplicity, we only consider the expansion of $L_k(T, 0)$ and case $a\neq0$ is similar.

Theorem 3.2  For each $k\in(-\frac{1}{2}, \frac{1}{2})$, the local time of mBs

$\begin{equation*} \begin{aligned} L_k(T, b)\equiv =\frac{1}{2\pi}\displaystyle\int_0^T\displaystyle\int _{\mathbb{R}}\exp\{i\lambda(\alpha B_t+\beta S^k_t -b)\}d\lambda dt\\ \end{aligned} \end{equation*}$

is a Hida distribution for $b\in\mathbb{R}$. Moreover, kernel functions of chaos expansion of $L_k(T, 0)$ are given by

$\begin{equation*} \begin{aligned} &\textbf{G}_{\textbf{m}, \textbf{k}}(u_1, \cdots, u_m, v_1, \cdots, v_k)\\ =&(\frac{1}{2\pi})^{\frac{d}{2}}(\frac{1}{2})^{\frac{m+k}{2}} \frac{1}{(\frac{\textbf{m}+\textbf{k}}{2})!} \frac{(\textbf{m}+\textbf{k})!}{\textbf{m}!\textbf{k}!}\\ &\cdot\displaystyle\int _{0}^T(\frac{1}{\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}})^{\frac{d+m+k}{2}} \alpha^m\prod _{i=1}^m\mathbb{I}_{[0, t]}(u_i)\beta^k\prod _{j=1}^k(\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(v_j)dt \end{aligned} \end{equation*}$

for each $\textbf{n}\in \mathbb{N}^d$ such that $n\geq N$. All other kernel functions $\textbf{G}_{\textbf{m}, \textbf{k}}$ are identically equal to zero.

Proof  To show this result, we need apply Lemma 2.1 to the S-transform of the integral with respect to Lebesgue measure dt on [0, T]. For $\mathbf{f}\in \mathcal{S}_{2d}(\mathbb{R})$ and any complex number z, by the definition of S-transform, these is

$\begin{equation*} \begin{aligned} &S(\delta(\alpha B_t+\beta S^k_t -b))(\mathbf{f}) \\ =&(\frac{1}{2\pi (\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))})^\frac{d}{2}\\ &\cdot\mid\exp\{-\frac{(z\alpha\displaystyle\int_0^t\mathbf{f}(s)ds+z\beta \displaystyle\int_{\mathbb{R}}\mathbf{f}(s) (\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(s)ds -b)^2}{2(\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))}\}\mid\\ \leq& (\frac{1}{2\pi(\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))})^{\frac{d}{2}} \exp\{\frac{b^2+2\mid z\mid \mid b\mid \mid\alpha\mid\mid\displaystyle\int_0^t\mathbf{f}(s)ds\mid}{2( \alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))}\}\\ \cdot&\exp\{\frac{2\mid z\mid\mid b\mid\mid\beta\mid\mid\displaystyle\int_{\mathbb{R}}\mathbf{f}(s) (\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(s) ds\mid}{2(\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}\\ \cdot&\exp\{\frac{2\mid z\mid^2\mid\alpha\mid\mid\beta\mid\mid \displaystyle\int_0^t\mathbf{f}(s)ds\mid\mid\displaystyle\int_{\mathbb{R}}\mathbf{f}(s) (\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(s) ds\mid}{2(\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}\\ \cdot&\exp\{\frac{\mid z\mid^2\alpha^2\mid\displaystyle\int_0^t\mathbf{f}(s)ds\mid^2+\mid z\mid^2 \beta^2\mid\displaystyle\int_{\mathbb{R}}\mathbf{f}(s) (\beta^2(2-2^{2k})t^{2k+1})(s)ds\mid^2}{2(\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}\\ \leq&(\frac{1}{2\pi(\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))})^{\frac{d}{2}}\exp\{\frac{a^2+\mid z\mid^2\alpha^2\mid \mathbf{f}\mid^2+2\mid z\mid\mid b\mid\mid\beta\mid C_{k, 1}\parallel \mathbf{f} \parallel}{2(\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}\\ \cdot&\exp\{\frac{C_{k, 1}\mid z\mid^2 \beta^2\parallel\mathbf{ f}\parallel^2+2\mid z\mid \mid b\mid \mid\alpha\mid\mid \mathbf{f}\mid+2C_{k, 1}\mid z\mid^2\mid\alpha\mid\mid\beta\mid\mid \mathbf{f}\mid\parallel\mathbf{ f}\parallel}{2(\alpha^2t+ \beta^2(2-2^{2k})t^{2k+1}))}\}, \\ \end{aligned} \end{equation*}$

where the first part is integrable as a function on [0, T], and the third part is bound.

Given a $\textbf{f}=(f_{1, 1}, \cdots, f_{1, d}, f_{2, 1}, \cdots, f_{2, d}) \in\mathcal{S}_{2d}(\mathbb{R})$, there is

$\begin{equation*} \begin{aligned} &S(L(T, 0))(\textbf{f})\\ =&(\frac{1}{\pi})^\frac{d}{2}\displaystyle\int_{0}^T\sum_{n\geq N}\{(\frac{1}{2(\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))})^{n+\frac{d}{2}}\} \sum_{n_1, \cdots, n_d}\frac{1}{{n_1}!\cdots{n_d}!}\prod_{i=1}^d\sum_{m_i+k_i=2n_i}\\ &\cdot\frac{(m_i+k_i)!}{m_i!k_i!}\alpha^{m_i}(\displaystyle\int_{0}^tf_i(s)ds)^{m_i} \beta^{k_i}(\displaystyle\int_{\mathbb{R}}f_i(s)(\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)}) (s)ds)^{k_i}dt\\ =&(\frac{1}{\pi})^\frac{d}{2}\displaystyle\int_{0}^T\sum_{n\geq N}\sum_{n_1, \cdots, n_d}\sum_{m_i+k_i=2n_i} \{\prod_{i=1}^d(\frac{1}{2(\alpha^2t+\beta^2(2-2^{2k})t^{2k+1}))}) ^{\frac{1}{2}+\frac{m_i+k_i}{2}}\frac{1}{{\frac{m_i+k_i}{2}}!}\} \\ &\cdot\{\prod_{i=1}^d \frac{(m_i+k_i)!}{m_i!k_i!}\alpha^{m_i}({\displaystyle\int_0^tf_{i}(s)ds})^{m_i} \beta^{k_i}{(\displaystyle\int_{\mathbb{R}}f_{i}(s)(\frac{1}{c(k)}I^{k}_{-}\mathbb{I}^o_{[0, t)})(s)ds})^{k_i}\}. \end{aligned} \end{equation*}$

Comparing with the general form of the chaos expansion,

$L_k(T, 0)=\sum_\textbf{m}\sum_\textbf{k} \langle:\omega_1^{\otimes \textbf{m}}:\otimes:\omega_2^{\otimes\textbf{k}}:, {\rm G}_{\textbf{m}, \textbf{k}}\rangle, $

the kernel functions of chaos expansion are obtained.

Next we discuss the regularized condition of local time of mBs.

Theorem 4.1  For $k\in(-\frac{1}{2}, \frac{1}{2})$, local time of mBs is a H$\ddot{o}$lder continuous function with fractional order

$1-\frac{1}{2}\max\limits_{u\in(s, t)}\{\alpha^2 u, \beta^2 (2-2^{2k})u^{2k+1}\}, $

i.e., there exists some constant $C_{k, 2}$ such that

$E[\mid L_t-L_s\mid]\leq C_{k, 2}\mid t-s\mid^{1-\frac{1}{2}\max\limits_{u\in(s, t)} \{\alpha^2 u, \beta^2 (2-2^{2k})u^{2k+1}\}}.$

Proof  Set

$\begin{eqnarray*} L_t=\displaystyle\int_0^t\delta(\alpha B_u+\beta S_u^k)du, ~~ L_{t, \varepsilon}=\displaystyle\int_0^tp_{\varepsilon}(\alpha B_u+\beta S_u^k)du.\end{eqnarray*}$

For s < t, consider

$\begin{equation*} \begin{aligned} &E[\mid L_{t, \varepsilon}-L_{s, \varepsilon}\mid]\\ =&\frac{1}{2\pi}\displaystyle\int_s^t\displaystyle\int_{\mathbb{R}}E[\exp\{i\xi(\alpha B_u+\beta S_u^k)\}]\exp\{-\varepsilon\xi^2\}d\xi du\\ =&\frac{1}{2\pi}\displaystyle\int_s^t\displaystyle\int_{\mathbb{R}}E[\exp\{-\frac{1}{2}\xi^2(\alpha^2 Var(B_u)+\beta^2 Var(S_u^k))\}]\exp\{-\varepsilon \xi^2 \}d\xi du\\ \leq& \frac{1}{2\pi}\displaystyle\int_s^t\displaystyle\int_{\mathbb{R}}\exp\{-\frac{1}{2}\xi^2(\alpha^2 u+\beta^2 (2-2^{2k})u^{2k+1})\}d\xi du\\ =&\frac{1}{2\pi}\displaystyle\int_s^t\displaystyle\int_{\mathbb{R}}\sqrt{\frac{2}{\alpha^2 u+\beta^2 (2-2^{2k})u^{2k+1}}}\exp\{-x^2\}dx du\\ =&\frac{1}{\sqrt{2\pi}}\displaystyle\int_s^t\sqrt{\frac{1}{\alpha^2 u+\beta^2 (2-2^{2k})u^{2k+1}}}du\\ \leq&\frac{1}{\sqrt{2\pi}}\displaystyle\int_s^tu^{-\frac{1}{2}\max\limits_{u\in(s, t)}\{\alpha^2 u, \beta^2 (2-2^{2k})u^{2k+1}\}}du\\ \leq&\frac{1}{\sqrt{2\pi}}\frac{t^{1-\frac{1}{2}\max\limits_{u\in(s, t)}\{\alpha^2 u, \beta^2 (2-2^{2k})u^{2k+1}\}}-s^{1-\frac{1}{2}\max\limits_{u\in(s, t)}\{\alpha^2 u, \beta^2 (2-2^{2k})u^{2k+1}\}}}{1-\frac{1}{2}\max\limits_{u\in(s, t)}\{\alpha^2 u, \beta^2 (2-2^{2k})u^{2k+1}\}}\\ \leq&C_{k, 2}\mid t-s\mid^{1-\frac{1}{2}\max\limits_{u\in(s, t)}\{\alpha^2 u, \beta^2 (2-2^{2k})u^{2k+1}\}}, \end{aligned} \end{equation*}$

where $C_{k, 2}=\frac{1}{\sqrt{2\pi}( 1-\frac{1}{2}\max\limits_{u\in(s, t)}\{\alpha^2 u, \beta^2 (2-2^{2k})t^{2k+1}\})}.$

By Fatou's lemma, we get

$\begin{equation*} \begin{aligned} E[\mid L_{t}-L_{s}\mid]=&E[\lim_{\varepsilon\rightarrow 0}\mid L_{t, \varepsilon}-L_{s, \varepsilon}\mid]\\ \leq&\liminf_{{\varepsilon\rightarrow 0}} E[\mid L_{t, \varepsilon}-L_{s, \varepsilon}\mid]\\ \leq&C_{k, 2}\mid t-s\mid^{1-\frac{1}{2}\max\limits_{u\in(s, t)}\{\alpha^2 u, \beta^2 (2-2^{2k})u^{2k+1}\}}. \end{aligned} \end{equation*}$

Therefore $L_t$ is a Hôlder continuous function.