## Local Functional Law of the Iterated Logarithm for Increments of Two-Parameter Brownian Motion

Liu Yonghong, Tang Yiheng, Zhang Qingqing

 基金资助: 广西自然科学基金.  2020GXNSFAA159118广西自然科学基金.  2018GXNSFBA281076桂林电子科技大学研究生教育创新计划项目.  2020YCXS083

 Fund supported: the NSF of Guangxi.  2020GXNSFAA159118the NSF of Guangxi.  2018GXNSFBA281076the Innovation Project of GUET Graduate Education.  2020YCXS083

Abstract

In this paper, using large deviations for two-parameter Brownian motion and increments of two-parameter Brownian motion, we obtain local functional law of the iterated logarithm for increments of two-parameter Brownian motion.

Keywords： Two-parameter Brownian motion ; Increments ; Local functional law of the iterated logarithm

Liu Yonghong, Tang Yiheng, Zhang Qingqing. Local Functional Law of the Iterated Logarithm for Increments of Two-Parameter Brownian Motion. Acta Mathematica Scientia[J], 2021, 41(3): 874-881 doi:

## 1 引言与结果

$\{w(s, t); s\ge 0, t\ge 0\}$是定义在概率空间$(\Omega, {\cal F}, P)$上的两参数Brown运动. 对这一过程轨道性质的研究已有许多深刻结果, 参见文献[1-3, 5]. 本文中, 我们也对两参数Brown运动Csörg\H{o}-Révész型增量的泛函极限问题进行了研究, 得到了一种局部泛函重对数律. 在文中, 我们设${\cal C} = \{f; f:[0, 1]^2\to {{\Bbb R}}, f\;\rm{连续}\}$, 赋予一致范数$\|f\|: = \mathop{\sup}\limits_{(x, y)\in[0, 1]^2}| f(x, y)|$.${\cal C}_0: = \{f \in {\cal C}; f(0, t) = f(s, 0) = 0\}$, ${\cal H} = \{f \in {\cal C}_0; f\ \rm{绝对连续, }\ \|f\|_{\mathcal H}^2 = \int _0^1\int_0^1 |\frac{\partial^2f}{\partial y\partial x}|^2{\rm d}x{\rm d}y < +\infty\}$, 设函数$I: {\cal C}_0\to [0, \infty]$, 定义如下

(i) $a_u\leq u, \rm{对任何} u\in(0, 1)$, $\rm{并且}\lim\limits_{u\to0}a_u = 0$;

(ii) $\frac{u}{a_u}$非增;

(iii) $\lim\limits_{u\to 0}\frac{\log(u/a_u)}{\log\log u^{-1}} = \frac{r}{2}$, $r$为非负常数. \\ 对$0\le s\le u-a_u, 0\leq t\leq u-a_u$, $(x, y)\in [0, 1]^2$, $\Delta(s, t, a_ux, a_uy)$记下面轨道

$\ell_u = \log\frac{u^2\log u^{-1}}{{a_u^2}}$, $\beta_u = (2{a_u^2}{\ell_u})^{-\frac{1}{2}}$. 定义$Y_{s, t, u}(x, y) = \beta_u\Delta(s, t, a_ux, a_uy)$.

$$$\liminf\limits_{u\to0}\sup\limits_{s, t\in [0, u-a_u]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u}(\cdot, \cdot)-f(\cdot, \cdot)\| = 0, \; \; a.s.,$$$

$$$\lim\limits_{u\to0}\inf\limits_{s, t\in [0, u-a_u]^2}\|Y_{s, t, u}(\cdot, \cdot)-f(\cdot, \cdot)\| = 0, \; \; a.s..$$$

## 2 若干引理

设$f(s, t) = \int^s_0\int^t_0g(u, v){\rm d}_u{\rm d}_v, 0\le s, t\le 1 $$\|f\|_{{\cal H}}^2 = \int^1_0\int^1_0g^2(u, v){\rm d}_u{\rm d}_v . 我们有 \begin{eqnarray} \left\|f(\cdot, \cdot)-f^*(\cdot, \cdot)\right\| & = &\sup\limits_{s, t\in[0, 1]^2}|f(s, t)-f(\lambda s, \lambda't)|{}\\ & = &\sup\limits_{s, t\in[0, 1]^2}\bigg|\int^s_0\int^t_0g(u, v){\rm d}_u{\rm d}_v-\int^{\lambda s}_0\int^{\lambda't}_0g(u, v){\rm d}_u{\rm d}_v\bigg|{}\\ & = &\sup\limits_{s, t\in[0, 1]^2}\bigg|\int^s_{\lambda s}\int^t_{\lambda't}g(u, v){\rm d}_u{\rm d}_v+\int^{s}_{\lambda s}\int^{\lambda't}_0g(u, v){\rm d}_u{\rm d}_v {}\\ && +\int^{\lambda s}_{0}\int^{t}_{\lambda' t}g(u, v){\rm d}_u{\rm d}_v\bigg|{}\\ &\le&\{((1-\lambda)(1-\lambda'))^{1/2}+(1-\lambda)^{1/2}+(1-\lambda')^{1/2}\}\|f\|_{{\cal H}}. \end{eqnarray} 引理2.3证毕. 引理 2.4[4, 引理4] 设 \{\xi_n\}_{n\ge 1} 为随机变量序列. 若 \lim\limits_{n\to \infty}P\{\xi_n\ge \xi_0\} = 0 , 则存在子列 \{\xi_{n_k}\} , 使得 { }\limsup_{k\to \infty}\xi_{n_k}\le \xi_0, \; a.s. , 因此 { }\liminf_{n\to \infty}\xi_n\le \xi_0, \; a.s. . ### 3 定理1.1的证明 ### 3.1 (1.1)式的证明 (1.1)式的证明分三种情形完成: (Ⅰ) 0<r<\infty , (Ⅱ) r = 0 , (Ⅲ) r = \infty . 情形(Ⅰ) 0<r<\infty . 引理 3.1 若条件 \rm(i)–(iii) 成立, 则存在 u_n^{-1} = e^{e^n} , 使得 $$\lim\limits_{n\to\infty}P\Big(\sup\limits_{(s, t)\in [0, u_n-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u_n}(\cdot, \cdot)-f(\cdot, \cdot)\|\ge \varepsilon\Big) = 0.$$ 设 A = \{g: \inf\limits_{f\in K_b}\|g-f\|\ge \varepsilon\} , 则 A 是闭集, 存在任意小的 \delta>0 , 使得 \inf\limits_{g\in A}I(g)>\frac{1}{2}b^2(r)+\delta . 为了简单, 我们设 \frac{a_{u_n}}{u_n-a_{u_{n+1}}+a_{u_n}} = p , 由引理2.2, 对 n 足够大, 我们有 \begin{eqnarray} &&P\left(\sup\limits_{(s, t)\in [0, {u_n}-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u_n}-f(\cdot, \cdot)\|\geq\varepsilon\right){}\\ & = &P\left(\sup\limits_{(s, t)\in [0, {u_n}-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\Big\|\frac{1}{\sqrt{2a_{u_n}^2\ell_{u_n}}}\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot) -f(\cdot, \cdot)\Big\|\geq\varepsilon\right){}\\ & = &P\left(\sup\limits_{(s, t)\in [0, 1-p]^2}\inf\limits_{f\in K_b}\Big\|\sqrt{\frac{1}{p^2}\frac{1}{2a_{u_n}^2\ell_{u_n}}}\Delta(s, t, p\cdot, p\cdot) -f(\cdot, \cdot)\Big\|\geq\varepsilon\right){}\\ & = &P\left(\bigcup\limits_{s\in [0, 1-p]}\bigcup\limits_{t\in [0, 1-p]} \left\{\sqrt{\frac{1}{p^2}\frac{1}{2a_{u_n}^2\ell_{u_n}}}\Delta(s, t, p\cdot, p\cdot)\in A\right\}\right){}\\ &\leq&\left(\frac{{u_n}-a_{u_{n+1}}+a_{u_n}}{a_{u_n}}\right)^2\left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{b^2(r)+\delta}{}\\ &\leq& 4\left(\frac{u_n}{a_{u_n}}\right)^{2(1-b^2(r)-\delta)} \left(\frac{1}{\log u_n^{-1}}\right)^{b^2(r)+\delta}. \end{eqnarray} 在(3.2)式中, 若 1-b^2(r)-\delta\leq 0 , 则我们有 \begin{eqnarray} &&P\left(\sup\limits_{(s, t)\in [0, u_n-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u_n}-f(\cdot, \cdot)\|\geq\varepsilon\right){}\\ &\le &4\left(\log u_n^{-1}\right)^{-(b^2(r)+\delta)} = 4e^{-(b^2(r)+\delta)n}\to 0, (n\to\infty). \end{eqnarray} 否则, 若 1-b^2(r)-\delta> 0 , 由条件(iii), 对任意小的 \mu>0 , 我们有 \begin{eqnarray} &&P\left(\sup\limits_{s, t\in [0, {u_n}-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u_n}-f(\cdot, \cdot)\|\geq\varepsilon\right){}\\ &\le &4\left(\frac{1}{\log u_n^{-1}}\right)^{-(r+\mu)(1-b^2(r)-\delta)+b^2(r)+\delta} = 4\left(\frac{1}{\log u_n^{-1}}\right)^{(\delta-\frac{1}{r+1})\mu+(r+1)\delta}, \end{eqnarray} \delta<\frac{1}{r+1} , \mu\le\frac{1}{2}\frac{(r+1)^2\delta}{1-\delta(r+1)} , 使得 (\delta-\frac{1}{r+1})\mu+(r+1)\delta>0 , 我们有 \begin{eqnarray} &&P\left(\sup\limits_{s, t\in [0, {u_n}-a_{u_{n+1}}]^2}\inf\limits_{f\in K_b}\|Y_{s, t, u_n}-f(\cdot, \cdot)\|\geq\varepsilon\right){}\\ &\le &4\left(\frac{1}{\log u_n^{-1}}\right)^{(\delta-\frac{1}{r+1})\mu+(r+1)\delta}\to 0, (n\to\infty). \end{eqnarray} 从而(3.1)式获证. 由引理3.1和引理2.4, 得到(1.1)式. 情形(Ⅱ) r = 0 . 类似情形(Ⅰ)的证明, 结论显然成立. 情形(Ⅲ) r = \infty . r = \infty 时, b = 1 , 此时 K_b = \{f\in{\cal C}_0;2I(f)\leq 1\} . 为证(1.1)式, 只需证明下面的引理. 引理3.2 若条件(i)–(iii)成立, 设 u_n = \frac{1}{\theta^n}, \theta>1 , 我们有 $$\limsup\limits_{n\rightarrow \infty}\sup\limits_ {(s, t)\in [0, u_n-a_{u_{n+1}}]^2}\left\|\frac{\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)}{\sqrt{2a_{u_n}^2\ell_{u_n}}}-K_b\right\| = 0, \; \; \; \; a.s..$$ 设 A_1 = \{g: \|g-K_b\|\ge \varepsilon\} , 则 A_1 是闭集, 存在 \delta'>0 , 使得 \inf\limits_{g\in A_1}I(g)> \frac{1}{2}+\delta' .为了简单, 我们设 \frac{a_{u_n}}{u_n-a_{u_{n+1}}+a_{u_n}} = p , \begin{eqnarray} &&P\left(\sup\limits_ {(s, t)\in [0, u_n-a_{u_{n+1}}]^2}\left\|\frac{1}{(2a_{u_n}^2\ell_{u_n})^{1/2}}\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)-K_b\right\|\ge \varepsilon\right){}\\ & = &P\left(\sup\limits_{(s, t)\in [0, 1-\frac{a_{u_n}}{u_n-a_{u_{n+1}}+a_{u_n}}]^2}\left\|\sqrt{\frac{1}{2a_{u_n}^2\ell_{u_n}p^2}}\Delta\left(s, t, pa_{u_n}\cdot, pa_{u_n}\cdot\right)-K_b\right\|\ge \varepsilon\right){}\\ & = &P\left(\mathop{\bigcup}\limits_{\mathop{0\le s\le 1-p}}\mathop{\bigcup}\limits_{\mathop{0\le t\le 1-p}} \left\{\sqrt{\frac{1}{p}\frac{1}{p}\frac{1}{2\log\frac{u_n^2\log u_n^{-1}}{a_{u_n}^2}}}\Delta\left(s, t, p\cdot, p\cdot\right)\in A_1\right\}\right){}\\ &\le&4\left(\frac{u_n}{a_{u_n}}\right)^2\left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{1+\delta'}\le 4\left(\frac{1}{\log u_n^{-1}}\right)^{1+\delta'}, \end{eqnarray} 由Borel-Cantelli引理, 结束引理3.2的证明. 由引理3.2和引理2.4, 得到(1.1)式. ### 3.2 (1.2)式的证明 (1.2)式的证明分三种情形完成： (Ⅰ') 0<r<\infty , (Ⅱ') r = 0 , (Ⅲ ' ) r = \infty . 情形(Ⅰ') 0<r<\infty (1.2)式的证明由下面引理完成. 引理 3.3 如果条件(i)–(iii)成立, 那么对任何 f\in K_b , 存在子列 \{a_{u_n} = \theta^{-n}; \theta>1,$$ n\ge 1\}$, 使得

$$$\limsup\limits_{n\to \infty}\inf\limits_{s, t\in [0, u_{n+1}-a_{u_n}]^2}\left\|Y_{s, t, u_n}(\cdot, \cdot)-f\right\| = 0, \; \; a.s..$$$

设$s_i = ia_{u_{n}}, t_j = ja_{u_{n}}, i, j = 0, 1, 2, \cdot\cdot\cdot, k_n = [\frac{b_{u_{n+1}}}{a_{u_{n}}}]-1$. 我们有

$\begin{eqnarray} & & P\left(\inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|Y_{s, t, u_n}(\cdot, \cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ &\le & P\left(\min\limits_{0\le i, j\le k_n}\Big\|\frac{1}{\sqrt{2a_{u_n}^2\ell_{u_n}}}\Delta(s_i, t_j, a_{u_n}\cdot, a_{u_n}\cdot)-f\Big\|\ge \varepsilon\right){}\\ & = &\left\{1-P\left(\frac{1}{\sqrt{2\ell_{u_n}}}w(\cdot, \cdot)\in B\right)\right\}^{(k_n+1)^2}, \end{eqnarray}$

$\begin{eqnarray} & &P\left(\inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|Y_{s, t, u_n}(\cdot, \cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ &\le &\exp\left\{-\left(\Big[\frac{u_{n+1}-a_{u_n}}{a_{u_n}}\Big]+1\right)^2\left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{b^2(r)-\delta_0}\right\}{}\\ &\le &\exp\left\{-\frac{1}{4\theta^2}\left(\frac{u_n}{a_{u_n}}\right)^{2(1-b^2(r)+\delta_0)}\left(\frac{1}{\log u_n^{-1}}\right)^{b^2(r)-\delta_0}\right\}, \end{eqnarray}$

$\begin{eqnarray} & &P\left(\inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|Y_{s, t, u_n}(\cdot, \cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ &\le& \exp\left\{-\frac{1}{4\theta^2}\Big(\log u_n^{-1}\Big)^{(r-\mu')(1-b^2(r)+\delta_0)-b^2(r)+\delta_0}\right\}{}\\ & = &\exp\left\{-\frac{1}{4\theta^2}\Big(\log u_n^{-1}\Big)^{(r+1)\delta_0-(\delta_0+\frac{1}{r+1})\mu'}\right\}, \end{eqnarray}$

$\begin{eqnarray} &&P\left(\inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|Y_{s, t, u_n}(\cdot, \cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ & \le& \exp\left\{-\frac{1}{4\theta^2}\Big(\log u_n^{-1}\Big)^{(r+1)\delta_0-(\delta_0+\frac{1}{r+1})\mu'}\right\}{}\\ &\le&\exp\left\{-\frac{1}{4\theta^2}\Big(\log a_{u_n}^{-1}-\frac{1}{2}(r+\eta')\log\log a_{u_n}^{-1}\Big)^{(r+1)\delta_0-(\delta_0+\frac{1}{r+1})\mu'}\right\}, \end{eqnarray}$

$$$\sum\limits_{n = 1}^{\infty}P\left(\inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|Y_{s, t, u_n}(\cdot, \cdot)-f\Big\|\ge \varepsilon\Big)\right)<\infty.$$$

$$$\limsup\limits_{u\to 0}\inf\limits_{(s, t)\in[0, u-a_u]^2}\Big\|\frac{\Delta(s, t, a_u\cdot, a_u\cdot)}{\sqrt{2a_u^2\ell_u}}-f\Big\| = 0, \; \; a.s..$$$

设$u_n$如引理3.3中定义. 对$u\in(u_{n+1}, u_n]$, 注意到

$\begin{eqnarray} & &\inf\limits_{(s, t)\in[0, u-a_u]^2}\|Y_{s, t, u}(\cdot, \cdot)-f(\cdot, \cdot)\|{}\\ & = &\inf\limits_{(s, t)\in[0, u-a_u]^2}\Big\|\frac{\beta_u}{\beta_{u_n}}Y_{s, t, u_n}\Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)-f(\cdot, \cdot)\Big\|{}\\ &\le & \inf\limits_{(s, t)\in [0, u_{n+1}-a_{u_n}]^2}\frac{\beta_u}{\beta_{u_n}}\Big\|Y_{s, t, u_n} \Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)-f \Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)\Big\|{}\\ &&+\left|\frac{\beta_u}{\beta_{u_n}}-1\right|\left\|f\Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)\right\| +\Big\|f\Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)-f(\cdot, \cdot)\Big\|. \end{eqnarray}$

$\begin{eqnarray} \frac{\beta_u}{\beta_{u_n}}&\le &\frac{\beta_{u_{n+1}}}{\beta_{u_n}} = \left(\frac{a_{u_n}^2\log\frac{u_n^2\log u_n^{-1}}{a_{u_n}^2}}{a^2_{u_{n+1}}\log\frac{u_{n+1}^2\log u_{n+1}^{-1}}{a_{u_{n+1}}^2}}\right)^{1/2} \le \left(\frac{a^2_{{u_n}}}{a^2_{{u_{n+1}}}}\right)^{1/2} = \theta, \end{eqnarray}$

$\begin{eqnarray} \left\|f\Big(\frac{a_u}{a_{u_n}}\cdot, \frac{a_u}{a_{u_n}}\cdot\Big)-f(\cdot, \cdot)\right\|&\le &\left|1-\frac{a_{u_{n+1}}}{a_{u_n}}\right|+2\left|1-\frac{a_{u_{n+1}}}{a_{u_n}}\right|^{1/2}{}\\ &\le &\left|1-\frac{1}{\theta}\right|+2\left|1-\frac{1}{\theta}\right|^{1/2}. \end{eqnarray}$

$\theta\to 1$, 由(3.15)–(3.17)式和引理3.3, 我们得到(3.14)式. 从而(1.2)式获证.

$r = \infty$时, $b = 1$, 此时$K_b = \{f\in{\cal C}_0;2I(f)\leq 1\}$. 为证(2)式, 只需证明下面的引理.

$$$\limsup\limits_{n\to \infty}\inf\limits_{(s, t)\in[0, u_{n+1}-a_{u_n}]^2}\left\|\frac{\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)}{\sqrt{2a_{u_n}^2\ell_{u_n}}}-f\right\| = 0, \; \; a.s..$$$

因为$\lim\limits_{u\to 0}\frac{\log\frac{u}{a_u}}{\log\log u^{-1}} = \infty$, 故存在递减子列$\{u_n, n\ge 1\}$使得$\frac{u_n}{a_{u_n}} = n^{d}$.$s_i = ia_{u_{n}}, $$t_j = ja_{u_{n}}, i, j = 0, 1, 2, \cdot\cdot\cdot , k_n = [\frac{u_{n+1}}{a_{u_{n}}}]-1 , h(n) = \frac{\log\frac{u_n}{a_{u_n}}}{\log\log u_n^{-1}} = \frac{\log n^{d}}{\log\log u_n^{-1}} . 我们有 u_n^{-1} = \exp(n^{\frac{d}{h(n)}})$$ h(n)\to \infty$, $n\to \infty$. 而且, 对任意小的$\alpha>0$, $\frac{n^\alpha}{\log u_n^{-1}}\to \infty$, $1\le \frac{u_n}{u_{n+1}} = \exp\{(n+1)^{\frac{d}{h(n+1)}}- n^{\frac{d}{h(n)}}\}\le\exp(n^{\frac{d}{h(n)}-1})\to 1, (n\to \infty)$.$n$足够大, 我们有

$\begin{eqnarray} & &P\left(\inf\limits_{(s, t)\in[0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|\frac{1}{\sqrt{2a_{u_n}^2\ell_{u_n}}}\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ &\le & P\left(\min\limits_{0\le i, j\le k_n}\Big\|\frac{1}{\sqrt{2a_{u_n}^2\ell_{u_n}}}\Delta(s_i, t_j, a_{u_n}\cdot, a_{u_n}\cdot)-f\Big\|\ge \varepsilon\right){}\\ & = &\left\{1-P\left(\frac{1}{\sqrt{2\ell_{u_n}}}w(\cdot, \cdot)\in B_1\right)\right\}^{(k_n+1)^2}, \end{eqnarray}$

$\begin{eqnarray} & & P\left(\inf\limits_{(s, t)\in[0, u_{n+1}-a_{u_n}]^2}\Big(\Big\|\frac{1}{\sqrt{2a_{u_n}^2\ell_{u_n}}}\Delta(s, t, a_{u_n}\cdot, a_{u_n}\cdot)-f\Big\|\ge \varepsilon\Big)\right){}\\ &\le &\exp\left\{-\left(\frac{a_{u_n}^2}{u_n^2\log u_n^{-1}}\right)^{\mu_0}\left(\left[\frac{u_{n+1}}{a_{u_n}}\right]\right)^2\right\}, \end{eqnarray}$

## 参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子

Chen B. On Strassen's version of the law of the iterated logarithm for the two-parameter Wiener process//Szyskowicz B. Asymptotic Methods in Probability and Statistics. North-Holland: Elsevier Science Publishers, 1998: 343-358

Gao F .

The functional Lévy's modulus for a multiparameter Wiener process

J Hubei Univ Nat Sci, 1998, 20 (4): 321- 324

Wang W .

On Strassen-type theorem for the increments of two-parameter Wiener processes

Chinese Ann Math, 2001, 22A (1): 27- 34

Wei Q .

Functional limit theorems for C-R increments of k-dimensional Brownian motion in Hölder norm

Acta Mathematica Sinica, 2000, 16 (4): 637- 654

Xu J .

Quasi sure functional modulus of continuity for a two-parameter Wiener process in Hölder norm

J Math Anal Appl, 2016, 434, 501- 515

/

 〈 〉