Convergence to a self-normalized G-Brownian motion

Abstract

G-Brownian motion has a very rich and interesting new structure that nontrivially generalizes the classical Brownian motion. Its quadratic variation process is also a continuous process with independent and stationary increments. We prove a self-normalized functional central limit theorem for independent and identically distributed random variables under the sub-linear expectation with the limit process being a G-Brownian motion self-normalized by its quadratic variation. To prove the self-normalized central limit theorem, we also establish a new Donsker’s invariance principle with the limit process being a generalized G-Brownian motion.

Introduction

Let {X _n ;n≥1} be a sequence of independent and identically distributed random variables on a probability space . Set $S_{n}=\sum _{j=1}^{n} X_{j}$. Suppose EX ₁=0 and $EX_{1}^{2}=\sigma ^{2}>0$. The well-known central limit theorem says that

$$ \frac{S_{n}}{\sqrt{n}}\overset{d}\rightarrow N\left(0,\sigma^{2}\right), $$

(1)

or, equivalently, for any bounded continuous function ψ(x),

$$ E\left[\psi\left(\frac{S_{n}}{\sqrt{n}}\right)\right]\rightarrow E\left[\psi(\xi)\right], $$

(2)

where ξ∼N(0,σ ²) is a normal random variable. If the normalization factor $\sqrt {n}$ is replaced by $\sqrt {V_{n}}$, where $V_{n}=\sum _{j=1}^{n} X_{j}^{2}$, then

$$ \frac{S_{n}}{\sqrt{V_{n}}}\overset{d}\rightarrow N(0,1). $$

(3)

Giné et al. (1997) proved that (3) holds if and only if EX ₁=0 and

$$ {\lim}_{x\rightarrow \infty} \frac{x^{2}P\left(|X_{1}|\ge x\right)}{EX_{1}^{2}I\{|X_{1}|\le x\}}=0. $$

(4)

The result (3) is refered to as the self-normalized central limit theorem. The purpose of this paper is to establish the self-normalized central limit theorem under the sub-linear expectation.

The sub-linear expectation, or also called G-expectation, is a nonlinear expectation generalizing the notions of backward stochastic differential equations, g-expectations, and provides a flexible framework to model non-additive probability problems and the volatility uncertainty in finance. Peng (2006, 2008a,b) introduced a general framework of the sub-linear expectation of random variables and the notions of the G-normal random variable, G-Brownian motion, independent and identically distributed random variables, etc., under the sub-linear expectation. The construction of sub-linear expectations on the space of continuous paths and discrete-time paths can also be founded in Yan et al. (2012) and Nutz and van Handel (2013). For basic properties of the sub-linear expectation, one can refer to Peng (2008b, 2009, 2010a etc.). For stochastic calculus and stochastic differential equations with respect to a G-Brownian motion, one can refer to Li and Peng (2011), Hu et al. (2014a, b), etc., and a book by Peng (2010a).

The central limit theorem under the sub-linear expectation was first established by Peng (2008b). It says that (2) remains true when the expectation E is replaced by a sub-linear expectation $\hat {\mathbb {E}}$ if {X _n ;n≥1} are independent and identically distributed under $\hat {\mathbb {E}}$, i.e.,

$$ \frac{S_{n}}{\sqrt{n}}\overset{d}\rightarrow \xi~\text{under}~\hat{\mathbb{E}}, $$

(5)

where ξ is a G-normal random variable.

In the classical case, when $\textsf {E}[X_{1}^{2}]$ is finite, (3) follows from the cental limit theorem (1) directly by Slutsky’s lemma and the fact that

$$ \frac{V_{n}}{n}\overset{P}\rightarrow \sigma^{2}. $$

The latter is due to the law of large numbers. Under the framework of the sub-linear expectation, $\frac {V_{n}}{n}$ no longer converges to a constant. The self-normalized central limit theorem cannot follow from the central limit theorem (5) directly. In this paper, we will prove that

$$ \frac{S_{n}}{\sqrt{V_{n}}}\overset{d}\rightarrow \frac{W_{1}}{\sqrt{\langle W\rangle_{1}}}~\text{under}~\hat{\mathbb{E}}, $$

(6)

where W _t is a G-Brownian motion and 〈W〉 _t is its quadratic variation process. A very interesting phenomenon of G-Brownian motion is that its quadratic variation process is also a continuous process with independent and stationary increments, and thus can still be regarded as a Brownian motion. When the sub-linear expectation $\hat {\mathbb {E}}$ reduces to a linear one, W _t is the classical Brownian motion with W ₁∼N(0,σ ²) and 〈W〉 _t =t σ ², and then (6) is just (3). Our main results on the self-normalized central limit theorem will be given in Section “Main results”, where the process of the self-normalized partial sums ${S_{[nt]}}/{\sqrt {V_{n}}}$ is proved to converge to a self-normalized G-Brownian motion ${W_{t}}/{\sqrt {\langle W\rangle _{1}}}$. We also consider the case in which the second moments of X _i ’s are infinite and obtain the self-normalized central limit theorem under a condition similar to (4). In the next section, we state basic settings in a sub-linear expectation space, including capacity, independence, identical distribution, G-Brownian motion, etc. One can skip this section if these concepts are familiar. To prove the self-normalized central limit theorem, we establish a new Donsker’s invariance principle in Section “Invariance principle” with the limit process being a generalized G-Brownian motion. The proof is given in the last section.

Basic settings

We use the framework and notations of Peng (2008b). Let $(\Omega,\mathcal F)$ be a given measurable space and let be a linear space of real functions defined on $(\Omega,\mathcal F)$ such that if , then for each $\varphi \in C_{b}(\mathbb {R}^{n})\bigcup C_{l,Lip}(\mathbb {R}^{n})$, where $C_{b}(\mathbb R^{n})$ denotes the space of all bounded continuous functions and $C_{l,Lip}(\mathbb {R}^{n})$ denotes the linear space of (local Lipschitz) functions φ satisfying

$$\begin{array}{@{}rcl@{}} & |\varphi(\boldsymbol{x}) - \varphi(\boldsymbol{y})| \le C(1 + |\boldsymbol{x}|^{m} + |\boldsymbol{y}|^{m})|\boldsymbol{x}- \boldsymbol{y}|, \;\; \forall \boldsymbol{x}, \boldsymbol{y} \in \mathbb R^{n},&\\ & \text {for some}~C > 0, m \in \mathbb N~\text{depending on}~\varphi. & \end{array} $$

is considered as a space of “random variables.” In this case, we denote . Further, we let $C_{b,Lip}(\mathbb R^{n})$ denote the space of all bounded and Lipschitz functions on $\mathbb R^{n}$.

Sub-linear expectation and capacity

Definition 1

A sub-linear expectation $\hat {\mathbb {E}}$ on is a function satisfying the following properties: for all , we have

• (a) Monotonicity: If X≥Y then $\hat {\mathbb {E}} [X]\ge \hat {\mathbb {E}} [Y]$;

• (b) Constant preserving:$\hat {\mathbb {E}} [c] = c$;

• (c) Sub-additivity:$\hat {\mathbb {E}}[X+Y]\le \hat {\mathbb {E}} [X] +\hat {\mathbb {E}} [Y ]$ whenever $\hat {\mathbb {E}} [X] +\hat {\mathbb {E}} [Y ]$ is not of the form +∞−∞ or −∞+∞;

• (d) Positive homogeneity:$\hat {\mathbb {E}} [\lambda X] = \lambda \hat {\mathbb {E}} [X]$, λ≥0.

Here $\overline {\mathbb R}=[-\infty, \infty ]$. The triple is called a sub-linear expectation space. Given a sub-linear expectation $\hat {\mathbb {E}}$, let us denote the conjugate expectation $\widehat {\mathcal {E}}$of $\hat {\mathbb {E}}$ by

Next, we introduce the capacities corresponding to the sub-linear expectations. Let $\mathcal G\subset \mathcal F$. A function $V:\mathcal G\rightarrow [0,1]$ is called a capacity if

$$ V(\emptyset)=0, \;V(\Omega)=1, \;~\text{and}~V(A)\le V(B)\;\; \forall\; A\subset B, \; A,B\in \mathcal G. $$

It is called sub-additive if $V(A\bigcup B)\le V(A)+V(B)$ for all $A,B\in \mathcal G$ with $A\bigcup B\in \mathcal G$.

Let be a sub-linear space and $\widehat {\mathcal {E}}$ be the conjugate expectation of $\hat {\mathbb {E}}$. We introduce the pair $(\mathbb {V},\mathcal {V})$ of capacities by setting

where A ^c is the complement set of A. Then, $\mathbb {V}$ is sub-additive and

Further, we define an extension of $\hat {\mathbb {E}}^{\ast }$ of $\hat {\mathbb {E}}$ by

where inf∅=+∞. Then,

Independence and distribution

Definition 2

(Peng (2006, 2008b))

• (i) (Identical distribution) Let X ₁ and X ₂ be two n-dimensional random vectors defined, respectively, in sub-linear expectation spaces and . They are called identically distributed, denoted by $\boldsymbol X_{1}\overset {d}= \boldsymbol X_{2}$ if

  $$ \hat{\mathbb{E}}_{1}[\varphi(\boldsymbol X_{1})]=\hat{\mathbb{E}}_{2}[\varphi(\boldsymbol X_{2})], \;\; \forall \varphi\in C_{l,Lip}(\mathbb R^{n}), $$

  whenever the sub-expectations are finite. A sequence {X _n ;n≥1} of random variables is said to be identically distributed if $X_{i}\overset {d}= X_{1}$ for each i≥1.

• (ii) (Independence) In a sub-linear expectation space , a random vector is said to be independent to another random vector under $\hat {\mathbb {E}}$ if for each test function $\varphi \in C_{l,Lip}(\mathbb R^{m} \times \mathbb R^{n})$ we have

  $$ \hat{\mathbb{E}} [\varphi(\boldsymbol{X}, \boldsymbol{Y})] = \hat{\mathbb{E}} \left[\hat{\mathbb{E}}[\varphi(\boldsymbol{x}, \boldsymbol{Y})]\big|_{\boldsymbol{x}=\boldsymbol{X}}\right], $$

  whenever $\overline {\varphi }(\boldsymbol {x}):=\hat {\mathbb {E}}\left [|\varphi (\boldsymbol {x}, \boldsymbol {Y})|\right ]<\infty$ for all x and $\hat {\mathbb {E}}\left [|\overline {\varphi }(\boldsymbol {X})|\right ]<\infty$.

• (iii) (IID random variables) A sequence of random variables {X _n ;n≥1} is said to be independent and identically distributed (IID), if $X_{i}\overset {d}=X_{1}$ and X _i+1 is independent to (X ₁,…,X _i ) for each i≥1.

G-normal distribution, G-Brownian motion and its quadratic variation

Let $0<\underline {\sigma }\le \overline {\sigma }<\infty$ and $G(\alpha)=\frac {1}{2}\left (\overline {\sigma }^{2} \alpha ^{+} - \underline {\sigma }^{2} \alpha ^{-}\right)$. X is called a normal $N\left (0, \left [\underline {\sigma }^{2}, \overline {\sigma }^{2}\right ]\right)$ distributed random variable (written as $X\sim N\left (0, \left [\underline {\sigma }^{2}, \overline {\sigma }^{2}\right ]\right)$) under $\hat {\mathbb {E}}$, if for any bounded Lipschitz function φ, the function $u(x,t)=\hat {\mathbb {E}}\left [\varphi \left (x+\sqrt {t} X\right)\right ]$ ($x\in \mathbb R, t\ge 0$) is the unique viscosity solution of the following heat equation:

$$ \partial_{t} u -G\left(\partial_{xx}^{2} u\right) =0, \;\; u(0,x)=\varphi(x). $$

Let C[0,1] be a function space of continuous functions on [0,1] equipped with the supremum norm $\|x\|=\sup \limits _{0\le t\le 1}|x(t)|$ and C _b (C[0,1]) is the set of bounded continuous functions $h(x):C[0,1]\rightarrow \mathbb R$. The modulus of the continuity of an element x∈C[0,1] is defined by

$$\omega_{\delta}(x)=\sup_{|t-s|<\delta}|x(t)-x(s)|. $$

It is showed that there is a sub-linear expectation space with $\widetilde {\Omega }= C[0,1]$ and such that is a Banach space, and the canonical process $W(t)(\omega) = \omega _{t} (\omega \in \widetilde {\Omega })$ is a G-Brownian motion with $W(1)\sim N\left (0, \left [\underline {\sigma }^{2}, \overline {\sigma }^{2}\right ]\right)$ under $\widetilde {\mathbb E}$, i.e., for all 0≤t ₁<…<t _n ≤1, $\varphi \in C_{l,lip}(\mathbb R^{n})$,

$$ \widetilde{\mathbb E}\left[\varphi\left(W(t_{1}),\ldots, W(t_{n-1}), W(t_{n})-W(t_{n-1})\right)\right] =\widetilde{\mathbb E}\left[\psi\left(W(t_{1}),\ldots, W(t_{n-1})\right)\right], $$

(8)

where $\psi \left (x_{1},\ldots, x_{n-1}\right)\big)=\widetilde {\mathbb {E}}\left [\varphi \left (x_{1},\ldots, x_{n-1}, \sqrt {t_{n}-t_{n-1}}W(1)\right)\right ]$ (cf. Peng (2006, 2008a, 2010a), Denis et al. (2011)).

The quadratic variation process of a G-Brownian motion W is defined by

$$\langle W \rangle_{t}={\lim}_{\|\Pi_{t}^{N}\|\rightarrow 0}\sum_{j=1}^{N-1} \left(W\left(t_{j}^{N}\right)-W\left(t_{j-1}^{N}\right)\right)^{2}=W^{2}(t)-2\int_{0}^{t} W(t) dW(t), $$

where $\Pi _{t}^{N}=\left \{t_{0}^{N},t_{1}^{N},\ldots, t_{N}^{n}\right \}$ is a partition of [0,t] and $\left \|\Pi _{t}^{N}\right \|=\max _{j}\left |t_{j}^{N}-t_{j-1}^{N}\right |$, and the limit is taken in L ₂, i.e.,

$$ {\lim}_{\left\|\Pi_{t}^{N}\right\|\rightarrow 0}\widetilde{\mathbb{E}}\left[\left(\sum_{j=1}^{N-1}\left(W\left(t_{j}^{N}\right)-W\left(t_{j-1}^{N}\right)\right)^{2}-\langle W \rangle_{t}\right)^{2}\right]=0. $$

The quadratic variation process 〈W〉 _t is also a continuous process with independent and stationary increments. For the properties and the distribution of the quadratic variation process, one can refer to a book by Peng (2010a).

Denis et al. (2011) showed the following representation of the G-Brownian motion (cf. Theorem 52).

Lemma 1

Let be a probability measure space and {B(t)} _t≥0 is a P-Brownian motion. Then, for all bounded continuous functions $\varphi : C_{b}[0,1]\rightarrow \ \mathbb R$,

$$\widetilde{\mathbb E}\left[\varphi\left(W(\cdot)\right)\right]=\sup_{\theta\in \Theta}\mathsf{E}_{P}\left[\varphi\left(W_{\theta}(\cdot)\right)\right],\;\; W_{\theta}(t) = \int_{0}^{t}\theta(s) dB(s), $$

where

For the reminder of this paper, the sequences {X _n ;n≥1}, {Y _n ;n≥1}, etc., of the random variables are considered in . Without specification, we suppose that {X _n ;n≥1} is a sequence of independent and identically distributed random variables in with $\hat {\mathbb {E}}[X_{1}]=\widehat {\mathcal {E}}[X_{1}]=0$, $\hat {\mathbb {E}}\left [X_{1}^{2}\right ]=\overline {\sigma }^{2}$, and $\widehat {\mathcal {E}}\left [X_{1}^{2}\right ]=\underline {\sigma }^{2}$. Denote $S_{0}^{X}=0$, $S_{n}^{X}=\sum _{k=1}^{n} X_{k}$, V ₀=0, $V_{n}=\sum _{k=1}^{n} X_{k}^{2}$. And suppose that is a sub-linear expectation space which is rich enough such that there is a G-Brownian motion W(t) with $W(1)\sim N\left (0,\left [\underline {\sigma }^{2},\overline {\sigma }^{2}\right ]\right)$. We denote a pair of capacities corresponding to the sub-linear expectation $\widetilde {\mathbb E}$ by $\left (\widetilde {\mathbb {V}},\widetilde {\mathcal {V}}\right)$, and the extension of $\widetilde {\mathbb E}$ by $\widetilde {\mathbb {E}}^{\ast }$.

Main results

We consider the convergence of the process $S_{[nt]}^{X}$. Because it is not in C[0,1], it needs to be modified. Define the C[0,1]-valued random variable $\widetilde {S}_{n}^{X}(\cdot)$ by setting

$$\widetilde{S}_{n}^{X}(t)= \left\{\begin{array}{cc} \sum_{j=1}^{k} X_{j}, \; \text{if}~t=k/n \; (k=0,1,\ldots, n);\\ \text{extended by linear interpolation in each interval }\\ \qquad \quad \left[[k-1]n^{-1}, kn^{-1}\right]. \end{array}\right.$$

Then, $\widetilde {S}_{n}^{X}(t)=S_{[nt]}^{X}+(nt-[nt])X_{[nt]+1}$. Here [nt] is the largest integer less than or equal to nt. Zhang (2015) obtained the functional central limit theorem as follows.

Theorem 1

Suppose $\hat {\mathbb {E}}\left [\left (X_{1}^{2}-b\right)^{+}\right ]\rightarrow 0$ as b→∞. Then, for all bounded continuous functions $\varphi :C[0,1]\rightarrow \mathbb {R}$,

$$ \hat{\mathbb{E}}\left[\varphi\left(\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{n}}\right)\right]\rightarrow \widetilde{\mathbb{E}}\left[{\phantom{2_{1}^{2}}}\!\!\!\!\!\varphi\left(W(\cdot) \right)\! \right]. $$

(9)

Replacing the normalization factor $\sqrt {n}$ by $\sqrt {V_{n}}$, we obtain the self-normalized process of partial sums:

$$W_{n}(t)=\frac{\widetilde{S}_{n}^{X}(t)}{\sqrt{V_{n}}}, $$

where $\frac {0}{0}$ is defined to be 0. Our main result is the following self-normalized functional central limit theorem (FCLT).

Theorem 2

Suppose $\hat {\mathbb {E}}\left [\left (X_{1}^{2}-b\right)^{+}\right ]\rightarrow 0$ as b→∞. Then, for all bounded continuous functions $\varphi :C[0,1]\rightarrow \mathbb {R}$,

$$ \hat{\mathbb{E}}^{\ast}\left[\varphi\left(W_{n}(\cdot)\right)\right]\rightarrow\widetilde{\mathbb{E}}\left[\varphi\left(\frac{W(\cdot)}{\sqrt{\langle W \rangle_{1}}}\right) \right]. $$

(10)

In particular, for all bounded continuous functions $\varphi :\mathbb {R}\rightarrow \mathbb {R}$,

$$ \begin{aligned} \hat{\mathbb{E}}^{\ast}\left[\varphi\left(\frac{S_{n}^{X}}{\sqrt{V_{n}}}\right)\right]\rightarrow & \widetilde{\mathbb{E}}\left[\varphi\left(\frac{W(1)}{\sqrt{\langle W \rangle_{1}}}\right) \right]\\ &=\sup_{\theta\in \Theta}\textsf{E}_{P}\left[\varphi\left(\frac{\int_{0}^{1}\theta(s) d B(s)}{\sqrt{\int_{0}^{1} \theta^{2}(s) ds }}\right) \right]. \end{aligned} $$

(11)

Remark 1

It is obvious that

$$\widetilde{\mathbb{E}}\left[\varphi\left(\frac{W(\cdot)}{\sqrt{\langle W \rangle_{1}}}\right) \right] \ge \textsf{E}_{P} \left[\varphi\left(B(\cdot)\right) \right]. $$

An interesting problem is how to estimate the upper bounds of the expectations on the right hand side of (10) and (11).

Further, $\frac {W(\cdot)}{\sqrt {\langle W\rangle _{1}}}\overset {d}=\frac {\overline {W}(\cdot)}{\sqrt {\langle \overline {W}\rangle _{1}}}$, where $\overline {W}(t)$ is a G-Brownian motion with $\overline {W}(1)\sim N(0,[r^{-2},1])$, $r^{2}=\overline {\sigma }^{2}/\underline {\sigma }^{2}$.

For the classical self-normalized central limit theorem, Giné et al. (1997) showed that the finiteness of the second moments can be relaxed to the condition (4). Csörgő et al. (2003) proved the self-normalized functional central limit theorem under (4). The next theorem gives a similar result under the sub-linear expectation and is an extension of Theorem 2.

Theorem 3

Let {X _n ;n≥1} be a sequence of independent and identically distributed random variables in the sub-linear expectation space with $\hat {\mathbb {E}}[X_{1}]=\widehat {\mathcal {E}}[X_{1}]=0$. Denote $l(x)=\hat {\mathbb {E}}\left [X_{1}^{2}\wedge x^{2}\right ]$. Suppose

• (I) $x^{2}\mathbb {V}(|X_{1}|\ge x)=o\left (l(x)\right)$ as x→∞;

• (II) ${\lim }_{x\rightarrow \infty } \frac {\hat {\mathbb {E}}\left [X_{1}^{2}\wedge x^{2}\right ]}{\widehat {\mathcal {E}}\left [X_{1}^{2}\wedge x^{2}\right ]}=r^{2}<\infty$;

• (III) $\hat {\mathbb {E}}[(|X_{1}|-c)^{+}]\rightarrow 0$ as c→∞.

Then, the conclusions of Theorem 2 remain true with W(t) being a G-Brownian motion such that W(1)∼N(0,[r ⁻²,1]).

Remark 2

Note for c>1, $l(cx)=\hat {\mathbb {E}}\left [X_{1}^{2}\wedge (cx)^{2}\right ]\le l(x)+(cx)^{2}\mathbb {V}(|X_{1}|\ge x)$. Condition (I) implies that l(cx)/l(x)→1 as x→∞, i.e., l(x) is a slowly varying function. Therefore, there is a constant C such that $\int _{x}^{\infty }y^{-2}l(y)dy \le C x^{-1} l(x)$ if x is large enough. So, $\int _{x}^{\infty }\mathbb {V}(|X_{1}|\ge y)dy=o(x^{-1}l(x))$. Also, by Lemma 3.9 (b) of Zhang (2016), condition (III) implies that $\hat {\mathbb {E}}\left [(|X_{1}|-x)^{+}\right ]\le \int _{x}^{\infty }\mathbb {V}(|X_{1}|\ge y)dy$. Hence, $\hat {\mathbb {E}}\left [\left (|X_{1}|-x\right)^{+}\right ]=o(x^{-1}l(x))$ if conditions (I) and (III) are satisfied. When $\hat {\mathbb {E}}$ is a continuous sub-linear expectation, then for any random variable Y we have $\hat {\mathbb {E}}[|Y|]\le \int _{0}^{\infty }\mathbb {V}(|Y|\ge y)dy$ by Lemma 3.9 (c) of Zhang (2016), and so the condition (III) can be removed. Here, $\hat {\mathbb {E}}$ is called continuous if, for any with $\hat {\mathbb {E}}[X_{n}],\hat {\mathbb {E}}[X]<\infty$, $\hat {\mathbb {E}}[X_{n}]\nearrow \hat {\mathbb {E}}[X]$ whenever 0≤X _n ↗X, and, $\hat {\mathbb {E}}[X_{n}]\searrow \hat {\mathbb {E}}[X]$ whenever X _n ↘X.

Invariance principle

To prove Theorems 2 and 3, we will prove a new Donsker’s invariance principle. Let {(X _i ,Y _i );i≥1} be a sequence of independent and identically distributed random vectors in the sub-linear expectation space with $\hat {\mathbb {E}}[X_{1}]=\hat {\mathbb {E}}[-X_{1}]=0$, $\hat {\mathbb {E}}[X_{1}^{2}]=\overline {\sigma }^{2}$, $\widehat {\mathcal {E}}[X_{1}^{2}]=\underline {\sigma }^{2}$, $\hat {\mathbb {E}}[Y_{1}]=\overline {\mu }$, $\widehat {\mathcal {E}}[Y_{1}]=\underline {\mu }$. Denote

$$ G(p,q)=\hat{\mathbb{E}}\left[\frac{1}{2} q X_{1}^{2}+pY_{1}\right], \;\; p,q\in \mathbb R. $$

(12)

Let ξ be a G-normal distributed random variable, η be a maximal distributed random variable such that the distribution of (ξ,η) is characterized by the following parabolic partial differential equation (PDE) defined on $[0,\infty)\times \mathbb {R}\times \mathbb {R}$:

$$ \partial_{t} u -G\left(\partial_{y} u, \partial_{xx}^{2} u\right) =0, $$

(13)

i.e., if for any bounded Lipschitz function $\varphi (x,y):\mathbb {R}^{2}\rightarrow \mathbb {R}$, the function $u(x,y,t)=\widetilde {\mathbb {E}}\left [\varphi \left (x+\sqrt {t} \xi, y+t\eta \right)\right ]$ ($x,y \in \mathbb {R}, t\ge 0$) is the unique viscosity solution of the PDE (13) with Cauchy condition u| _t=0=φ.

Further, let B _t and b _t be two random processes such that the distribution of the process (B _·,b _·) is characterized by

• (i) B ₀=0, b ₀=0;

• (ii) for any 0≤t ₁≤…≤t _k ≤s≤t+s, (B _s+t −B _s ,b _s+t −b _s ) is independent to $(B_{t_{j}}, b_{t_{j}}), j=1,\ldots,k$, in sense that, for any $\varphi \in C_{l,Lip}(\mathbb {R}^{2(k+1)})$,

  $$ \begin{aligned} & \widetilde{\mathbb{E}}\left[\varphi\left((B_{t_{1}}, b_{t_{1}}),\ldots,(B_{t_{k}}, b_{t_{k}}), (B_{s+t}-B_{s}, b_{s+t}-b_{s})\right)\right]\\ &\qquad = \widetilde{\mathbb{E}}\left[\psi\left((B_{t_{1}}, b_{t_{1}}),\ldots,(B_{t_{k}}, b_{t_{k}})\right)\right], \end{aligned} $$

  (14)

  where

  $$ \begin{aligned} \psi\left((x_{1}, y_{1}),\ldots,(x_{k}, y_{k})\right)= \widetilde{\mathbb{E}}\left[\varphi\left((x_{1}, y_{1}),\ldots,(x_{k}, y_{k})\right.\right.,\\ \left.\left.(B_{s+t}-B_{s}, b_{s+t}-b_{s})\right)\right]; \end{aligned} $$

• (iii) for any t,s>0, $(B_{s+t}-B_{s},b_{s+t}-b_{s})\overset {d}\sim (B_{t},b_{t})$ under $\widetilde {\mathbb {E}}$;

• (iv) for any t>0, $(B_{t},b_{t})\overset {d}\sim \left (\sqrt {t}B_{1}, tb_{1}\right)$ under $\widetilde {\mathbb {E}}$;

• (v) the distribution of (B ₁,b ₁) is characterized by the PDE (13).

It is easily seen that B _t is a G-Brownian motion with $B_{1}\sim N\left (0,[\underline {\sigma }^{2},\overline {\sigma }^{2}]\right)$, and (B _t ,b _t ) is a generalized G-Brownian motion introduced by Peng (2010a). The existence of the generalized G-Brownian motion can be found in Peng (2010a).

Theorem 4

Suppose $\hat {\mathbb {E}}\left [(X_{1}^{2}-b)^{+}\right ]\rightarrow 0$ and $\hat {\mathbb {E}}\left [(|Y_{1}|-b)^{+}\right ]\rightarrow 0$ as b→∞. Let

$$\widetilde{\boldsymbol{W}}_{n}(t)=\left(\frac{\widetilde{S}_{n}^{X}(t)}{\sqrt{n}}, \frac{\widetilde{S}_{n}^{Y}(t)}{n}\right). $$

Then, for any bounded continuous function $\varphi :C[0,1]\times C[0,1]\rightarrow \mathbb R$,

$$ {\lim}_{n\rightarrow \infty}\hat{\mathbb{E}}\left[\varphi\left(\widetilde{\boldsymbol{W}}_{n}(\cdot) \right)\right]= \widetilde{\mathbb{E}}\left[\varphi\left(B_{\cdot},b_{\cdot}\right)\right]. $$

(15)

Further, let p≥2, q≥1, and assume $\hat {\mathbb {E}}[|X_{1}|^{p}]<\infty$, $\hat {\mathbb {E}}[|Y_{1}|^{q}]<\infty$. Then, for any continuous function $\varphi :C[0,1]\times C[0,1]\rightarrow \mathbb R$ with |φ(x,y)|≤C(1+∥x∥^p+∥y∥^q),

$$ {\lim}_{n\rightarrow \infty}\hat{\mathbb{E}}^{\ast}\left[\varphi\left(\widetilde{\boldsymbol{W}}_{n}(\cdot) \right)\right]= \widetilde{\mathbb{E}}\left[\varphi\left(B_{\cdot},b_{\cdot}\right)\right]. $$

(16)

Here ∥x∥= sup0≤t≤1|x(t)| for x∈C[0,1].

Remark 3

When X _k and Y _k are random vectors in $\mathbb R^{d}$ with $\hat {\mathbb {E}}[X_{k}]=\hat {\mathbb {E}}[-X_{k}]=0$, $\hat {\mathbb {E}}[(\|X_{1}\|^{2}-b)^{+}]\rightarrow 0$ and $\hat {\mathbb {E}}[(\|Y_{1}\|-b)^{+}]\rightarrow 0$ as b→∞. Then, the function G in (12) becomes

$$ G(p,A)=\hat{\mathbb{E}}\left[\frac{1}{2}\langle AX_{1},X_{1}\rangle+\langle p,Y_{1}\rangle\right],\;\; p\in \mathbb R^{d}, A\in\mathbb S(d), $$

where $\mathbb S(d)$ is the collection of all d×d symmetric matrices. The conclusion of Theorem 4 remains true with the distribution of (B ₁,b ₁) being characterized by the following parabolic partial differential equation defined on $[0,\infty)\times \mathbb {R}^{d}\times \mathbb {R}^{d}$:

$$ \partial_{t} u -G\left(D_{y} u, D_{xx}^{2} u\right) =0,\;\; u|_{t=0}=\varphi, $$

where $D_{y} =(\partial _{y_{i}})_{i=1}^{n}$ and $D_{xx}^{2}=(\partial _{x_{i}x_{j}}^{2})_{i,j=1}^{d}$.

Remark 4

As a conclusion of Theorem 4, we have

$$ \hat{\mathbb{E}}\left[\varphi\left(\frac{S_{n}^{X}}{\sqrt{n}},\frac{S_{n}^{Y}}{n}\right)\right]\rightarrow \widetilde{\mathbb{E}}\left[{\phantom{2_{0}^{0}}}\!\!\!\!\!\varphi(B_{1},b_{1})\right],\;\; \varphi\in C_{b}(\mathbb{R}^{2}). $$

This is proved by Peng (2010a) under the conditions $\hat {\mathbb {E}}\left [\left |X_{1}\right |^{2+\delta }\right ]<\infty$ and $\hat {\mathbb {E}}\left [|Y_{1}|^{1+\delta }\right ]<\infty$ (cf. Theorem 3.6 and Remark 3.8 therein).

When Y ₁≡0, (15) becomes

$${\lim}_{n\rightarrow \infty}\hat{\mathbb{E}}\left[\varphi\left(\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{n}}\right)\right]=\widetilde{\mathbb{E}} \left[\varphi\left(B_{\cdot}\right)\right],\;\; \varphi\in C_{b}(C[0,1]), $$

which is proved by Zhang (2015).

Before the proof, we need several lemmas. For random vectors X _n in and X in , we write $\boldsymbol X_{n}\overset {d}\rightarrow \boldsymbol {X}$ if

$$ \hat{\mathbb{E}}\left[\varphi(\boldsymbol{X}_{n})\right]\rightarrow \widetilde{\mathbb{E}}\left[\varphi(\boldsymbol{X})\right] $$

for any bounded continuous φ. Write $\boldsymbol X_{n} \overset {\mathbb {V}}\rightarrow \boldsymbol {x}$ if $\mathbb {V}(\|\boldsymbol {X}_{n}-\boldsymbol {x}\|\ge \epsilon)\rightarrow 0$ for any ε>0. {X _n } is called uniformly integrable if

$${\lim}_{b\rightarrow \infty}\limsup_{n\rightarrow \infty} \hat{\mathbb{E}}\left[(\|\boldsymbol{X}_{n}\|-b)^{+}\right]= 0. $$

The following three lemmas are obvious.

Lemma 2

If $\boldsymbol {X}_{n}\overset {d}\rightarrow \boldsymbol {X}$ and φ is a continuous function, then $\varphi (\boldsymbol {X}_{n})\overset {d}\rightarrow \varphi (\boldsymbol {X})$.

Lemma 3

(Slutsky’s Lemma) Suppose $\boldsymbol {X}_{n}\overset {d}\rightarrow \boldsymbol {X}$, $\boldsymbol {Y}_{n} \overset {\mathbb {V}}\rightarrow \boldsymbol {y}$, $\eta _{n}\overset {\mathbb {V}}\rightarrow a$, where a is a constant and y is a constant vector, and $\widetilde {\mathbb {V}}(\|\boldsymbol {X}\|>\lambda)\rightarrow 0$ as λ→∞. Then, $(\boldsymbol {X}_{n}, \boldsymbol {Y}_{n}, \eta _{n})\overset {d}\rightarrow (\boldsymbol {X},\boldsymbol {y}, a)$, and as a result, $\eta _{n}\boldsymbol {X}_{n}+\boldsymbol {Y}_{n}\overset {d}\rightarrow a\boldsymbol {X}+\boldsymbol {y}$.

Remark 5

Suppose $\boldsymbol {X}_{n}\overset {d}\rightarrow \boldsymbol {X}$. Then, $\widetilde {\mathbb {V}}(\|\boldsymbol {X}\|>\lambda)\rightarrow 0$ as λ→∞ is equivalent to the tightness of {X _n ;n≥1}, i.e.,

$$ {\lim}_{\lambda\rightarrow \infty} \limsup_{n\rightarrow \infty} \mathbb{V}\left(\|\boldsymbol{X}_{n}\|>\lambda\right)=0, $$

because for all ε>0, we can define a continuous function φ(x) such that I{x>λ+ε}≤φ(x)≤I{x>λ] and so

$$\begin{aligned} &\widetilde{\mathbb{V}}(\|\boldsymbol{X}\|>\lambda+\epsilon)\le \widetilde{\mathbb{E}}[\varphi(\|\boldsymbol{X}\|)]={\lim}_{n\rightarrow \infty} \hat{\mathbb{E}}[\varphi(\|\boldsymbol{X}_{n}\|)] \le \limsup_{n\rightarrow \infty} \mathbb{V}\left(\|\boldsymbol{X}_{n}\|> \lambda\right), \\ &\limsup_{n\rightarrow \infty} \mathbb{V}\left(\|\boldsymbol{X}_{n}\|> \lambda+\epsilon\right)\le {\lim}_{n\rightarrow \infty} \hat{\mathbb{E}}[\varphi(\|\boldsymbol{X}_{n}\|)]=\widetilde{\mathbb{E}}[\varphi(\|\boldsymbol{X}\|)] \le \widetilde{\mathbb{V}}(\|\boldsymbol{X}\|> \lambda). \end{aligned} $$

Lemma 4

Suppose $\boldsymbol {X}_{n}\overset {d}\rightarrow \boldsymbol {X}$.

• (a) If {X _n } is uniformly integrable and $\widetilde {\mathbb {E}}[((\|\boldsymbol {X}\|-b)^{+}]\rightarrow 0$ as b→∞, then,

  $$ \hat{\mathbb{E}}[\boldsymbol{X}_{n}]\rightarrow \widetilde{\mathbb{E}}[\boldsymbol{X}]. $$

  (17)

• (b) If $\sup _{n}\hat {\mathbb {E}}[|\boldsymbol X_{n}\|^{q}<\infty$ and $\widetilde {\mathbb {E}} [|\boldsymbol {X}\|^{q}<\infty$ for some q>1, then (17) holds.

The following lemma is proved by Zhang (2015).

Lemma 5

Suppose that $\boldsymbol {X}_{n}\overset {d}\rightarrow \boldsymbol {X}$, $\boldsymbol {Y}_{n}\overset {d}\rightarrow \boldsymbol {Y}$, Y _n is independent to X _n under $\hat {\mathbb {E}}$ and $\widetilde {\mathbb {V}}(\|\boldsymbol {X}\|>\lambda)\rightarrow 0$ and $\widetilde {\mathbb {V}}(\|\boldsymbol {Y}\|>\lambda)\rightarrow 0$ as λ→∞. Then $(\boldsymbol {X}_{n},\boldsymbol {Y}_{n})\overset {d}\rightarrow (\overline {\boldsymbol {X}},\overline {\boldsymbol {Y}}),$ where $\overline {\boldsymbol {X}}\overset {d}=\boldsymbol {X}$, $\overline {\boldsymbol {Y}}\overset {d}=\boldsymbol {Y}$ and $\overline {\boldsymbol {Y}}$ is independent to $\overline {\boldsymbol {X}}$ under $\widetilde {\mathbb {E}}$.

The next lemma is about the Rosenthal-type inequalities due to Zhang (2016).

Lemma 6

Let {X ₁,…,X _n } be a sequence of independent random variables in .

• (a) Suppose p≥2. Then,

  $$ \begin{aligned} \hat{\mathbb{E}}\left[\max_{k\le n} \left|S_{k}\right|^{p}\right]&\le C_{p}\left\{ \sum_{k=1}^{n} \hat{\mathbb{E}} \left[|X_{k}|^{p}\right]+\left(\sum_{k=1}^{n} \hat{\mathbb{E}} \left[|X_{k}|^{2}\right]\right)^{p/2} \right. \\ & \qquad \left. +\left(\sum_{k=1}^{n} \left[\left(\widehat{\mathcal{E}} [X_{k}]\right)^{-}+\left(\hat{\mathbb{E}} [X_{k}]\right)^{+}\right]\right)^{p}\right\}. \end{aligned} $$

  (18)

• (b) Suppose $\hat {\mathbb {E}}[X_{k}]\le 0$, k=1,…,n. Then,

  $$ \hat{\mathbb{E}}\left[\left|\max_{k\le n} (S_{n}-S_{k})\right|^{p}\right] \le 2^{2-p}\sum_{k=1}^{n} \hat{\mathbb{E}} [|X_{k}|^{p}], \;\; \text{for} 1\le p\le 2 $$

  (19)

  and

  $$ \begin{aligned} \hat{\mathbb{E}}\left[\left|\max_{k\le n}(S_{n}- S_{k})\right|^{p}\right] &\le C_{p}\left\{ \sum_{k=1}^{n} \hat{\mathbb{E}} \left[|X_{k}|^{p}\right]+\left(\sum_{k=1}^{n} \hat{\mathbb{E}} \left[|X_{k}|^{2}\right]\right)^{p/2}\right\} \\ &\le C_{p} n^{p/2-1} \sum_{k=1}^{n} \hat{\mathbb{E}} [|X_{k}|^{p}], \;\; \text{for}~p\ge 2. \end{aligned} $$

  (20)

Lemma 7

Suppose $\hat {\mathbb {E}}[X_{1}]=\hat {\mathbb {E}}[-X_{1}]=0$ and $\hat {\mathbb {E}}\left [X_{1}^{2}\right ]<\infty$. Let $\overline {X}_{n,k}=(-\sqrt {n})\vee X_{k}\wedge \sqrt {n}$, $\widehat {X}_{n,k}=X_{k}-\overline {X}_{n,k}$, $\overline {S}_{n,k}^{X}=\sum _{j=1}^{k} \overline {X}_{n,j}$ and $\widehat {S}_{n,k}^{X}=\sum _{j=1}^{k}\widehat {X}_{n,j}$, k=1,…,n. Then

$$\begin{aligned} \hat{\mathbb{E}}\left[\max_{k\le n} \left|\frac{\overline{S}_{n,k}^{X}}{\sqrt{n}}\right|^{q}\right]\le C_{q}, \;\; ~\text{for all} ~q\ge 2, \end{aligned} $$

and

$$ {\lim}_{n\rightarrow \infty} \hat{\mathbb{E}}\left[\max_{k\le n} \left|\frac{\widehat{S}_{n,k}^{X}}{\sqrt{n}}\right|^{p}\right]=0 $$

whenever $\hat {\mathbb {E}}[(|X_{1}|^{p}-b)^{+}]\rightarrow 0$ as b→∞ if p=2, and $\hat {\mathbb {E}}[|X_{1}|^{p}]<\infty$ if p>2.

Proof

Note $\hat {\mathbb {E}}[X_{1}]=\widehat {\mathcal {E}}[X_{1}]=0$. So, $|\widehat {\mathcal {E}}[\overline {X}_{n,1}]|=|\widehat {\mathcal {E}}[X_{1}]-\widehat {\mathcal {E}}[\overline {X}_{n,1}]|\le \hat {\mathbb {E}}|\widehat {X}_{n,1}|\le \hat {\mathbb {E}}[(|X_{1}|^{2}-n)^{+}]n^{-1/2}$ and $|\hat {\mathbb {E}}[\overline {X}_{n,1}]|=|\hat {\mathbb {E}}[X_{1}]-\hat {\mathbb {E}}[\overline {X}_{n,1}]|\le \hat {\mathbb {E}}|\widehat {X}_{n,1}|\le \hat {\mathbb {E}}[(|X_{1}|^{2}-n)^{+}]n^{-1/2}$. By Rosenthal’s inequality (cf. (18)),

$$\begin{aligned} & \hat{\mathbb{E}}\left[\max_{k\le n} \left|\overline{S}_{n,k}^{X}\right|^{q}\right] \le C_{p}\left\{ n \hat{\mathbb{E}} [|\overline{X}_{n,1}|^{q}+\left(n \hat{\mathbb{E}} \left[|\overline{X}_{n,1}|^{2}\right]\right)^{q/2}\right.\\ & \qquad\qquad \qquad\qquad \left.+\left(n\left[\left(\widehat{\mathcal{E}} [\overline{X}_{n,1}]\right)^{-}+\left(\hat{\mathbb{E}} [\overline{X}_{n,1}]\right)^{+}\right]\right)^{q}\right\}\\ & \;\; \le C_{q}\left\{ n n^{q/2-1}\hat{\mathbb{E}} \left[|X_{1}|^{2}\right]+n^{q/2}\left(\hat{\mathbb{E}} \left[X_{1}^{2}\right]\right)^{q/2}+\left(nn^{-1/2}\hat{\mathbb{E}}\left[\left(X_{1}^{2}-n\right)^{+}\right]\right)^{q}\right\} \\ & \;\; \le C_{q} n^{q/2}\left\{\hat{\mathbb{E}} \left[\left|X_{1}\right|^{2}\right]+\left(\hat{\mathbb{E}} \left[X_{1}^{2}\right]\right)^{q}\right\}, \;\; \text{for all~} q\ge 2 \end{aligned} $$

and

$$\begin{aligned} \hat{\mathbb{E}}\left[\max_{k\le n} \left|\widehat{S}_{n,k}^{X}\right|^{p}\right] \le & C_{p}\left\{ n \hat{\mathbb{E}} \left[|\widehat{X}_{n,1}|^{p}\right]+\left(n \hat{\mathbb{E}} \left[|\widehat{X}_{n,1}|^{2}\right]\right)^{p/2}\right. \\ & \left.\qquad +\left(n\left[\left(\widehat{\mathcal{E}} [\widehat{X}_{n,1}]\right)^{-}+\left(\hat{\mathbb{E}} [\widehat{X}_{n,1}]\right)^{+}\right]\right)^{p}\right\}\\ \le & C_{p}\left\{ n\hat{\mathbb{E}} \left[\big(|X_{1}|^{p}-n^{p/2}\big)^{+}\right]+n^{p/2}\left(\hat{\mathbb{E}} \left[(X_{1}^{2}-n)^{+}\right]\right)^{p/2}\right.\\ & \left.\qquad +n^{p/2}\left(\hat{\mathbb{E}} \left[(X_{1}^{2}-n)^{+}\right]\right)^{p}\right\},\; p\ge 2. \end{aligned} $$

The proof is completed. □

Lemma 8

(a) Suppose p≥2, $\hat {\mathbb {E}}[X_{1}]=\hat {\mathbb {E}}[-X_{1}]=0$, $\hat {\mathbb {E}}[(X_{1}^{2}-b)^{+}]\rightarrow 0$ as b→∞ and $\hat {\mathbb {E}}[|X_{1}|^{p}]<\infty$. Then,

$$ \left\{\max_{k\le n}\left|\frac{S_{k}^{X}}{\sqrt{n}}\right|^{p}\right\}_{n=1}^{\infty} \; \text{is uniformly integrable and therefore is tight}. $$

(b) Suppose p≥1, $\hat {\mathbb {E}}\left [(|Y_{1}|-b)^{+}\right ]\rightarrow 0$ as b→∞, and $\hat {\mathbb {E}}[|Y_{1}|^{p}]<\infty$. Then,

$$ \left\{\max_{k\le n}\left|\frac{S_{k}^{Y}}{n}\right|^{p}\right\}_{n=1}^{\infty} \; \text{is uniformly integrable and therefore is tight}. $$

Proof

(a) follows from Lemma 6. (b) is obvious by noting

$$\begin{aligned} & \hat{\mathbb{E}}\left[\left(\left(\frac{\max_{k\le n}|S_{k}^{Y}|}{n}-b\right)^{+}\right)^{p}\right] \le \hat{\mathbb{E}}\left[\left(\frac{\sum_{k=1}^{n} (|Y_{k}|-b)^{+}}{n}\right)^{p}\right]\\ \le& C_{p} \left(\frac{\sum_{k=1}^{n} \hat{\mathbb{E}}[(|Y_{k}|-b)^{+}]}{n}\right)^{p} \\ & \qquad +C_{p} \frac{\hat{\mathbb{E}}\Big[\Big|\left(\sum_{k=1}^{n}\{ (|Y_{k}|-b)^{+}-\hat{\mathbb{E}}[(|Y_{k}|-b)^{+}]\}\right)^{+}\Big|^{p}\Big]}{n^{p}}\\ \le & C_{p}\left(\hat{\mathbb{E}}\left[(|Y_{1}|-b)^{+}\right]\right)^{p}+C_{p}\big(n^{-p/2}+n^{1-p}\big)\hat{\mathbb{E}}\big[(|Y_{1}|^{p}-b^{p})^{+}\big] \end{aligned} $$

by the Rosenthal-type inequalities (19) and (20). □

Lemma 9

Suppose $\hat {\mathbb {E}}\left [(|Y_{1}|-b)^{+}\right ]\rightarrow 0$ as b→∞. Then, for any ε>0,

$$\mathbb{V}\left(\frac{S_{n}^{Y}}{n}>\hat{\mathbb{E}}[Y_{1}]+\epsilon\right)\rightarrow 0~\text{and}~\mathbb{V}\left(\frac{S_{n}^{Y}}{n}<\widehat{\mathcal{E}}[Y_{1}]-\epsilon\right)\rightarrow 0. $$

Proof

Let Y _k,b =(−b)∨Y _k ∧b, $S_{n,1}=\sum _{k=1}^{n} Y_{k,b}$ and $S_{n,2}=S_{n}^{Y}-S_{n,1}$. Note $\hat {\mathbb {E}}\big [Y_{1,b}]\rightarrow \hat {\mathbb {E}}[Y_{1}]$ as b→∞. Suppose $\left |\hat {\mathbb {E}}[Y_{1,b}] - \hat {\mathbb {E}}[Y_{1}]\right |<\epsilon /4$. Then, by Kolmogorov’s inequality (cf. (19)),

$$\begin{aligned} &\mathbb{V}\left(\frac{S_{n,1}}{n}>\hat{\mathbb{E}}[Y_{1}]+\epsilon/2\right)\le \mathbb{V}\left(\frac{S_{n,1}}{n}>\hat{\mathbb{E}}[Y_{1,b}]+\epsilon/4\right)\\ \le & \frac{16}{n^{2}\epsilon^{2}} \hat{\mathbb{E}}\left[\left(\Big(\sum_{k=1}^{n} \big(Y_{k,b}-\hat{\mathbb{E}}[Y_{k,b}]\big)\Big)^{+}\right)^{2}\right] \\ \le & \frac{32}{n^{2}\epsilon^{2}} \sum_{k=1}^{n} \hat{\mathbb{E}}\left[\big(Y_{k,b}-\hat{\mathbb{E}}[Y_{k,b}]\big)^{2}\right]\le \frac{32(2b)^{2}}{n\epsilon^{2}}\rightarrow 0. \end{aligned} $$

Also,

$$\begin{aligned} \mathbb{V}\left(\frac{S_{n,2}}{n}> \epsilon/2\right)\le & \frac{2}{n\epsilon} \sum_{k=1}^{n} \hat{\mathbb{E}}|Y_{k}-Y_{k,b}| \le \frac{2}{\epsilon}\hat{\mathbb{E}}\left[(|Y_{1}|-b)^{+}\right] \rightarrow 0~\text{as}~b \rightarrow \infty. \end{aligned} $$

It follows that

$$\mathbb{V}\left(\frac{S_{n}^{Y}}{n}>\hat{\mathbb{E}}[Y_{1}]+\epsilon\right)\rightarrow 0. $$

By considering {−Y _k } instead, we have

$$\mathbb{V}\left(\frac{S_{n}^{Y}}{n}<\widehat{\mathcal{E}}[Y_{1}]-\epsilon\right)=\mathbb{V}\left(\frac{-S_{n}^{Y}}{n}>\hat{\mathbb{E}}[-Y_{1}]+\epsilon\right)\rightarrow 0.$$

□

Proof of Theorem 4.

We first show the tightness of $\widetilde {\boldsymbol W}_{n}$. It is easily seen that

$$w_{\delta}\left(\frac{\widetilde{S}_{n}^{Y}(\cdot)}{n}\right) \le 2\delta b+\frac{\sum_{k=1}^{n} (|Y_{k}|-b)^{+}}{n}. $$

It follows that for any ε>0, if δ<ε/(4b), then

$$\sup_{n}\mathbb{V}\left(w_{\delta}\left(\frac{\widetilde{S}_{n}^{Y}(\cdot)}{n}\right)\ge \epsilon\right) \le \sup_{n} \mathbb{V}\left(\sum_{k=1}^{n} (|Y_{k}|-b)^{+}\ge n\frac{\epsilon}{2}\right)\le \frac{2}{\epsilon}\hat{\mathbb{E}}\left[(|Y_{1}|-b)^{+}\right]. $$

Letting δ→0 and then b→∞ yields

$$\sup_{n}\mathbb{V}\left(w_{\delta}\left(\frac{\widetilde{S}_{n}^{Y}(\cdot)}{n}\right)\ge \epsilon\right)\rightarrow 0~\text{as}~\delta \rightarrow 0. $$

For any η>0, we choose δ _k ↓0 such that, if

$$A_{k}=\left\{x: \omega_{\delta_{k}}(x)<\frac{1}{k}\right\}, $$

then $\sup _{n}\mathbb {V}\left (\widetilde {S}_{n}^{Y}(\cdot)/n \in A_{k}^{c}\right)\le \eta /2^{k+1}$. Let A={x:|x(0)|≤a}, $K_{2}=A\bigcap _{k=1}^{\infty }A_{k}$. Then, by the Arzelá-Ascoli theorem, K ₂⊂C _b (C[0,1]) is compact. It is obvious that $\{ \widetilde {S}_{n}^{Y}(\cdot)/n\not \in A\}=\emptyset$, because $\widetilde {S}_{n}^{Y}(0)/n=0$. Next, we show that

$$\mathbb{V}\left(\widetilde{S}_{n}^{Y}(\cdot)/n\in K_{2}^{c}\right)\le \mathbb{V}\left(\widetilde{S}_{n}^{Y}(\cdot)/n\in A^{c}\right)+\sum_{k=1}^{\infty}\mathbb{V}\left(\widetilde{S}_{n}^{Y}(\cdot)/n\in A_{k}^{c}\right). $$

Note that when δ<1/(2n),

$$ \omega_{\delta}\left(\widetilde{S}_{n}^{Y}(\cdot)/n\right)\le 2n |t-s|\max_{i\le n} |Y_{i}|/n \le 2 \delta \max_{i\le n} |Y_{i}|. $$

Choose a k ₀ such that δ _k <1/(2Mk) for k≥k ₀. Then, on the event E={maxi≤n|Y _i |≤M}, $\{ \widetilde {S}_{n}^{Y}(\cdot)/n\in A_{k}^{c}\}=\emptyset$ for k≥k ₀. So, by the (finite) sub-additivity of $\mathbb {V}$,

$$\begin{aligned} &\mathbb{V}\left(E \bigcap \left\{ \widetilde{S}_{n}^{Y}(\cdot)/n\in K^{c}\right\}\right)\\ \le & \mathbb{V}\left(E \bigcap\left\{ \widetilde{S}_{n}^{Y}(\cdot)/n \in A^{c}\right\}\right)+\sum_{k=1}^{k_{0}}\mathbb{V}\left(E\bigcap \left\{ \widetilde{S}_{n}^{Y}(\cdot)/n \in A_{k}^{c}\right\}\right) \\ \le & \mathbb{V}\left(\widetilde{S}_{n}^{Y}(\cdot)/n \in A^{c}\right)+\sum_{k=1}^{\infty}\mathbb{V} \left(\widetilde{S}_{n}^{Y}(\cdot)/n \in A_{k}^{c}\right). \end{aligned} $$

On the other hand,

$$\mathbb{V}(E^{c})\le \frac{\hat{\mathbb{E}}[\max_{i\le n} |Y_{i}|]}{M}\le \frac{n\hat{\mathbb{E}}[|Y_{1}|]}{M}. $$

It follows that

$$\begin{aligned} \mathbb{V}\left(\widetilde{S}_{n}^{Y}(\cdot)/n\in K_{2}^{c} \right) \le \mathbb{V}\left(\widetilde{S}_{n}^{Y}(\cdot)/n \in A^{c}\right)+\sum_{k=1}^{\infty}\mathbb{V}\left(\widetilde{S}_{n}^{Y}(\cdot)/n \in A_{k}^{c}\right)+ \frac{n \hat{\mathbb{E}}[|Y_{1}|]}{M}. \end{aligned} $$

Letting M→∞ yields

$$\begin{aligned} \mathbb{V}\left(\widetilde{S}_{n}^{Y}(\cdot)/n\in K_{2}^{c} \right) & \le \mathbb{V}(\widetilde{S}_{n}^{Y}(\cdot)/n \in A^{c})+\sum_{k=1}^{\infty}\mathbb{V}\left(\widetilde{S}_{n}^{Y}(\cdot)/n \in A_{k}^{c}\right)\\ &< 0+\sum_{k=1}^{\infty} \frac{\eta}{2^{k+1}}<\frac{\eta}{2}. \end{aligned} $$

We conclude that for any η>0, there exists a compact K ₂⊂C _b (C[0,1]) such that

$$ \sup_{n} \hat{\mathbb{E}}^{\ast}\left[I\left\{\frac{\widetilde{S}_{n}^{Y}(\cdot)}{n}\not\in K_{2}\right\}\right]=\sup_{n} \mathbb{V}\left \{\frac{\widetilde{S}_{n}^{Y}(\cdot)}{n}\not\in K_{2}\right\}<\eta/2. $$

(21)

Next, we show that for any η>0, there exists a compact K ₁⊂C _b (C[0,1]) such that

$$ \sup_{n} \hat{\mathbb{E}}^{\ast}\left[I\left\{\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{n}}\not\in K_{1}\right\}\right]=\sup_{n} \mathbb{V}\left \{\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{n}}\not\in K_{1}\right\}<\eta/2. $$

(22)

Similar to (21), it is sufficient to show that

$$ \sup_{n}\mathbb{V}\left(w_{\delta}\left(\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{n}}\right)\ge \epsilon\right)\rightarrow 0 ~\text{as}~\delta \rightarrow 0. $$

(23)

With the same argument of Billingsley (1968, Pages 56–59, cf. (8.12)), for large n,

$$ \begin{aligned} &\mathbb{V}\left(w_{\delta}\left(\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{n}}\right)\ge 3\epsilon\right) \le \frac{2}{\delta} \mathbb{V}\left(\max_{i\le [n\delta]} \frac{|S_{i}^{X}|}{\sqrt{[n\delta]}}\ge \epsilon \frac{\sqrt{n}}{\sqrt{[n\delta]}} \right) \\ \le &\frac{2}{\delta} \mathbb{V}\left(\max_{i\le [n\delta]} \frac{\left|S_{i}^{X}\right|}{\sqrt{[n\delta]}}\ge \frac{\epsilon }{\sqrt{2 \delta }} \right) \le \frac{4}{\epsilon^{2}}\hat{\mathbb{E}}\left[\left(\max_{i\le [n\delta]} \Big|\frac{S_{i}^{X}}{\sqrt{[n\delta]}}\Big|^{2}-\frac{\epsilon^{2} }{ 2 \delta }\right)^{+}\right]. \end{aligned} $$

It follows that

$$ {\lim}_{\delta\rightarrow 0} \limsup_{n\rightarrow \infty} \mathbb{V}\left(w_{\delta}\left(\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{n}}\right)\ge 3\epsilon\right)=0 $$

by Lemma 8 (a), where p=2. On the other hand, for fixed n, if δ<1/(2n), then

$$ \omega_{\delta}(\widetilde{S}_{n}^{X}(\cdot)/\sqrt{n})\le 2n |t-s|\max_{i\le n} |X_{i}|/\sqrt{n} \le 2 \delta \sqrt{n} \max_{i\le n} |X_{i}|. $$

We have

$$ {\lim}_{\delta\rightarrow 0} \mathbb{V}\left(w_{\delta}\left(\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{n}}\right)\ge \epsilon\right)=0 $$

for each n. It follows that (23) holds.

Now, by combining (21) and (22) we obtain the tightness of $\widetilde {\boldsymbol W}_{n}$ as follows.

$$ \sup_{n} \hat{\mathbb{E}}^{\ast}\Big[I\Big\{\widetilde{\boldsymbol W}_{n}(\cdot)\not\in K_{1}\times K_{2}\Big\}\Big]<\eta. $$

(24)

Define $\hat {\mathbb {E}}_{n}$ by

$$ \hat{\mathbb{E}}_{n}[\varphi]=\hat{\mathbb{E}}\Big[\varphi\big(\widetilde{\boldsymbol W}_{n}(\cdot)\big)\Big],\;\; \varphi\in C_{b}\big(C[0,1]\times C[0,1]\big). $$

Then, the sequence of sub-linear expectations $\{\hat {\mathbb {E}}_{n}\}_{n=1}^{\infty }$ is tight by (24). By Theorem 9 of Peng (2010b), $\{\hat {\mathbb {E}}_{n}\}_{n=1}^{\infty }$ is weakly compact, namely, for each subsequence $\{\hat {\mathbb {E}}_{n_{k}}\}_{k=1}^{\infty }$, n _k →∞, there exists a further subsequence $\left \{\hat {\mathbb {E}}_{m_{j}}\right \}_{j=1}^{\infty } \subset \left \{\hat {\mathbb {E}}_{n_{k}}\right \}_{k=1}^{\infty }$, m _j →∞, such that, for each φ∈C _b (C[0,1]×C[0,1]), $\{\hat {\mathbb {E}}_{m_{j}}[\varphi ]\}$ is a Cauchy sequence. Define ${\mathbb F}[\cdot ]$ by

$$ {\mathbb F}[\varphi]={\lim}_{j\rightarrow \infty}\hat{\mathbb{E}}_{m_{j}}[\varphi], \; \varphi\in C_{b}\big(C[0,1]\times C[0,1]\big). $$

Let $\overline {\Omega }=C[0,1]\times C[0,1]$, and (ξ _t ,η _t ) be the canonical process $\xi _{t}(\omega) = \omega _{t}^{(1)}$, $\eta _{t}(\omega)=\omega _{t}^{(2)}\left (\omega =\left (\omega ^{(1)},\omega ^{(2)}\right)\in \overline {\Omega }\right)$. Then,

$$\hat{\mathbb{E}}\Big[\varphi\big(\widetilde{\boldsymbol W}_{m_{j}}(\cdot)\big)\Big]\rightarrow {\mathbb F}\left[\varphi(\xi_{\cdot},\eta_{\cdot})\right],\;\;\varphi\in C_{b}\big(C[0,1]\times C[0,1]\big). $$

(25)

The topological completion of $C_{b}(\overline {\Omega })$ under the Banach norm ${\mathbb F}[\|\cdot \|]$ is denoted by $L_{\mathbb F} (\overline {\Omega })$. ${\mathbb F}[\cdot ]$ can be extended uniquely to a sub-linear expectation on $L_{\mathbb F} (\overline {\Omega })$.

Next, it is sufficient to show that (ξ _t ,η _t ) defined on the sub-linear space $(\overline {\Omega }, L_{\mathbb F} (\overline {\Omega }), {\mathbb F})$ satisfies (i)-(v) and so $(\xi _{\cdot },\eta _{\cdot })\overset {d}=(B_{\cdot },b_{\cdot })$, which means that the limit distribution of any subsequence of $\widetilde {\boldsymbol W}_{n}(\cdot)$ is uniquely determined.

The conclusion in (i) is obvious. For (ii) and (iii), we let 0≤t ₁≤…≤t _k ≤s≤t+s. By (25), for any bounded continuous function $\varphi :\mathbb R^{2(k+1)}\rightarrow \mathbb R$ we have

$$\begin{aligned} & \hat{\mathbb{E}}\left[\varphi\big(\widetilde{W}_{m_{j}}(t_{1}),\ldots, \widetilde{W}_{m_{j}}(t_{k}), \widetilde{W}_{m_{j}}(s+t)-\widetilde{W}_{m_{j}}(s)\big)\right] \\ \rightarrow &{\mathbb F} \left[\varphi\big((\xi_{t_{1}},\eta_{t_{1}}), \ldots, (\xi_{t_{k}},\eta_{t_{k}}),(\xi_{s+t}-\xi_{s},\eta_{s+t}-\eta_{s})\big)\right]. \end{aligned} $$

Note

$$\begin{aligned} & \sup_{0\le t\le 1}\frac{\left|\widetilde{S}_{n}^{X}(t)-S_{[nt]}^{X}\right|}{\sqrt{n}}\le \frac{\max_{k\le n}|X_{k}|}{\sqrt{n}}\overset{\mathbb{V}}\rightarrow 0,\\ &\sup_{0\le t\le 1}\frac{\left|\widetilde{S}_{n}^{Y}(t)-S_{[nt]}^{Y}\right|}{n}\le \frac{\max_{k\le n}|Y_{k}|}{n}\overset{\mathbb{V}}\rightarrow 0. \end{aligned} $$

It follows that by Lemmas 3 and 8,

$$ \begin{aligned} & \hat{\mathbb{E}}\left[\varphi\left(\left(\frac{S_{[m_{j}t_{1}]}^{X}}{\sqrt{m_{j}}},\frac{S_{[m_{j}t_{1}]}^{Y}}{m_{j}}\right),\ldots, \left(\frac{S_{[m_{j}t_{k}]}^{X}}{\sqrt{m_{j}}},\frac{S_{[m_{j}t_{k}]}^{Y}}{m_{j}}\right),\right.\right.\\ &\ \ \ \ \ \ \ \ \ \ \ \left.\left. \left(\frac{S_{[m_{j}(s+t)]}^{X}-S_{[m_{j}s]}^{X}}{\sqrt{m_{j}}},\frac{S_{[m_{j}(s+t)]}^{Y}-S_{[m_{j}s]}^{Y}}{m_{j}}\right)\right)\right] \\ & \qquad \rightarrow {\mathbb F} \left[\varphi\big((\xi_{t_{1}},\eta_{t_{1}}), \ldots, (\xi_{t_{k}},\eta_{t_{k}}),(\xi_{s+t}-\xi_{s},\eta_{s+t}-\eta_{s})\big)\right]. \end{aligned} $$

(26)

In particular,

$$\begin{aligned} \left(\frac{S_{[m_{j}(s+t)]-[m_{j}s]}^{X}}{\sqrt{m_{j}}},\frac{S_{[m_{j}(s+t)]-[m_{j}s]}^{Y}}{m_{j}}\right) \overset{d}=& \left(\frac{S_{[m_{j}(s+t)]}^{X}-S_{[m_{j}s]}^{X}}{\sqrt{m_{j}}}, \frac{S_{[m_{j}(s+t)]}^{Y}-S_{[m_{j}s]}^{Y}}{m_{j}}\right)\\ &\overset{d}\rightarrow \big(\xi_{s+t}-\xi_{s}, \eta_{s+t}-\eta_{s}\big). \end{aligned} $$

It follows that

$$ \left(\frac{S_{[m_{j}t]}^{X}}{\sqrt{m_{j}}},\frac{S_{[m_{j}t]}^{Y}}{m_{j}}\right) \overset{d}\rightarrow \big(\xi_{s+t}-\xi_{s}, \eta_{s+t}-\eta_{s}\big). $$

(27)

On the other hand,

$$\left(\frac{S_{[m_{j}t]}^{X}}{\sqrt{m_{j}}},\frac{S_{[m_{j}t]}^{Y}}{m_{j}}\right) \overset{d}\rightarrow \big(\xi_{t}, \eta_{t}\big), $$

by (26). Hence,

$$ {\mathbb F}\left[\phi(\xi_{s+t}-\xi_{s},\eta_{s+t}-\eta_{s})\right]={\mathbb F}\left[\phi(\xi_{t},\eta_{t})\right]\;\; \text{for all}~\phi\in C_{b}\left(\mathbb R^{2}\right). $$

(28)

Next, we show that

$$ {\mathbb F}[|\xi_{s+t}-\xi_{s}|^{p}]\le C_{p} t^{p/2}\; \text{and }\; {\mathbb F}[|\eta_{s+t}-\eta_{s}|^{p}]\le C_{p} t^{p},\;\;\text{for all}~p\ge 2~\text{and}~t,s \ge 0. $$

(29)

By Lemma 9,

$$ \widetilde{\mathcal{V}}\big(t\underline{\mu}-\epsilon\le \eta_{s+t}-\eta_{s}\le t\overline{\mu}+\epsilon\big)=1\;\; \text{for all} \; \epsilon>0. $$

(30)

It follows that

$${\mathbb F}[|\eta_{s+t}-\eta_{s}|^{p}]\le t^{p}\big|\hat{\mathbb{E}}[|Y_{1}|]\big|^{p}. $$

For considering ξ _s+t −ξ _s , we let $\overline {S}_{n,k}^{X}$ and $\widehat {S}_{n,k}^{X}$ be defined as in Lemma 7. Then, $S_{k}^{X}=\overline {S}_{n,k}^{X}+ \widehat {S}_{n,k}^{X}$. By (27) and Lemmas 7 and 3,

$$\frac{\overline{S}_{[m_{j}t],[m_{j}t]}^{X}}{\sqrt{m_{j}}} \overset{d}\rightarrow \xi_{s+t}-\xi_{s}\; \text{and }\; \hat{\mathbb{E}}\left[\left|\frac{\overline{S}_{[m_{j}t],[m_{j}t]}^{X}}{\sqrt{m_{j}}}\right|^{p}\right]\le C_{p} t^{p/2},\; p\ge 2. $$

It follows that

$$ {\mathbb F}\left[| \xi_{s+t}-\xi_{s}|^{p}\wedge b\right]={\lim}_{n\rightarrow\infty}\hat{\mathbb{E}}\left[\left|\frac{\overline{S}_{[m_{j}t],[m_{j}t]}^{X}}{\sqrt{m_{j}}}\right|^{p}\wedge b\right]\le C_{p} t^{p/2}, \;\text{ for any}~b>0. $$

Hence,

$$ {\mathbb F}\left[| \xi_{s+t}-\xi_{s}|^{p} \right]={\lim}_{b\rightarrow \infty} {\mathbb F}\left[| \xi_{s+t}-\xi_{s}|^{p}\wedge b\right]\le C_{p} t^{p/2} $$

by the completeness of $(\overline {\Omega }, L_{\mathbb F} (\overline {\Omega }), {\mathbb F})$. (29) is proved.

Now, note that (X _i ,Y _i ),i=1,2,…, are independent and identically distributed. By (26) and Lemma 5, it is easily seen that (ξ _·,η _·) satisfies (14) for $\varphi \in C_{b}(\mathbb R^{2(k+1)})$. Note that, by (29), the random variables concerned in (14) and (28) have finite moments of each order. The function space $C_{b}(\mathbb R^{2(k+1)})$ and $C_{b}(\mathbb R^{2})$ can be extended to $C_{l,Lip}(\mathbb R^{2(k+1)})$ and $C_{l,Lip}(\mathbb R^{2})$, respectively, by elemental arguments. So, (ii) and (iii) are proved.

For (iv) and (v), we let $\varphi :\mathbb R^{2}\rightarrow \mathbb R$ be a bounded Lipschitz function and consider

$$ u(x,y,t)={\mathbb F}\left[\varphi(x+\xi_{t},y+\eta_{t})\right]. $$

It is sufficient to show that u is a viscosity solution of the PDE (13). In fact, due to the uniqueness of the viscosity solution, we will have

$${\mathbb F}\left[\varphi(x+\xi_{t},y+\eta_{t})\right]=\widetilde{\mathbb E}\left[\varphi(x+\sqrt{t} \xi,y+t\eta)\right], \;\; \varphi\in C_{b,Lip}(\mathbb R^{2}). $$

Letting x=0 and y=0 yields (iv) and (v).

To verify PDE (13), first it is easily seen that

$$ \hat{\mathbb{E}}\left[\frac{q}{2} \left(\frac{S_{[nt]}^{X}}{\sqrt{n}}\right)^{2}+p \frac{S_{[nt]}^{Y}}{n}\right]=\frac{[nt]}{n}\hat{\mathbb{E}}\left[\frac{q}{2} \left(\frac{S_{[nt]}^{X}}{\sqrt{[nt]}}\right)^{2}+p \frac{S_{[nt]}^{Y}}{[nt]}\right]=\frac{[nt]}{n}G(p,q). $$

Note that $\left \{\frac {q}{2} \left (\frac {S_{[nt]}^{X}}{\sqrt {n}}\right)^{2}+p \frac {S_{[nt]}^{Y}}{n}\right \}$ is uniformly integrable by Lemma 8. By Lemma 4, we conclude that

$${\mathbb F}\left[\frac{q}{2}\xi_{t}^{2}+p\eta_{t}\right]={\lim}_{m_{j}\rightarrow \infty}\hat{\mathbb{E}}\left[\frac{q}{2} \left(\frac{S_{[m_{j}t]}^{X}}{\sqrt{m_{j}}}\right)^{2}+p \frac{S_{[m_{j}t]}^{Y}}{m_{j}}\right]=t G(p,q). $$

It is obvious that if q ₁≤q ₂, then G(p,q ₁)−G(p,q ₂)≤G(0,q ₁−q ₂)≤0. Also, it is easy to verify that $|u(x,y,t)-u(\overline {x},\overline {y},t)|\le C (|x-\overline {x}|+|y-\overline {y}|)$, $|u(x,y,t)-u(x,y,s)|\le C\sqrt {|t-s|}$ by the Lipschitz continuity of φ, and

$$\begin{aligned} u(x,y,t)=&{\mathbb F}\left[\varphi(x+\xi_{s}+\xi_{t}-\xi_{s},y+\eta_{s}+\eta_{t}-\eta_{s})\right]\\ =& {\mathbb F}\left[ {\mathbb F} \left[\varphi(x+\overline{x}+\xi_{t}-\xi_{s},y+\overline{y}+\eta_{t}-\eta_{s})\right]\big|_{(\overline{x},\overline{y})=(\xi_{s},\eta_{s})}\right]\\ = & {\mathbb F}\left[u(x+\xi_{s},y+\eta_{s}, t-s)\right],\; 0\le s\le t. \end{aligned} $$

Let $\psi (\cdot,\cdot,\cdot)\in C_{b}^{3,3,2}(\mathbb R,\mathbb R,[0,1])$ be a smooth function with ψ≥u and ψ(x,y,t)=u(x,y,t). Then,

$${{\begin{aligned} 0=& {\mathbb F}\left[u(x+\xi_{s},y+\eta_{s}, t-s)-u(x,y,t)\right]\le {\mathbb F}\left[\psi(x+\xi_{s},y+\eta_{s}, t-s)-\psi(x,y,t)\right]\\ = & {\mathbb F}\left[\partial_{x}\psi(x,y,t)\xi_{s}+\frac{1}{2} \partial_{xx}^{2}\psi(x,y,t)\xi_{s}^{2}+\partial_{y}\psi(x,y,t)\eta_{s}-\partial_{t} \psi(x,y,t) s+I_{s}\right]\\ \le & {\mathbb F}\left[\partial_{x}\psi(x,y,t)\xi_{s}+\frac{1}{2} \partial_{xx}^{2}\psi(x,y,t)\xi_{s}^{2}+\partial_{y}\psi(x,y,t)\eta_{s}-\partial_{t} \psi(x,y,t) s\right]+ {\mathbb F}[|I_{s}|]\\ =& {\mathbb F}\left[\frac{1}{2} \partial_{xx}^{2}\psi(x,y,t)\xi_{s}^{2}+\partial_{y}\psi(x,y,t)\eta_{s}\right]-\partial_{t} \psi(x,y,t) s+ {\mathbb F}[|I_{s}|]\\ =& sG(\partial_{y}\psi(x,y,t),\partial_{xx}^{2}\psi(x,y,t)) -s\partial_{t} \psi(x,y,t)+ {\mathbb F}[|I_{s}|], \end{aligned}}} $$

where

$$|I_{s}|\le C\left(|\xi_{s}|^{3}+|\eta_{s}|^{2}+s^{2}\right). $$

By (29), we have ${\mathbb F}[|I_{s}|]\le C\big (s^{3/2}+s^{2}+s^{2}\big)=o(s).$ It follows that $[\partial _{t} \psi - G(\partial _{y}\psi,\partial _{xx}^{2})](x,y,t)\le 0$. Thus, u is a viscosity subsolution of (13). Similarly, we can prove that u is a viscosity supersolution of (13). Hence, (15) is proved.

As for (16), let $\varphi :C[0,1]\times C[0,1]\rightarrow \mathbb R$ be a continuous function with |φ(x,y)|≤C ₀(1+∥x∥^p+∥y∥^q). For λ>4C ₀, let φ _λ (x,y)=(−λ)∨(φ(x,y)∧λ)∈C _b (C[0,1]). It is easily seen that φ(x,y)=φ _λ (x,y) if |φ(x,y)|≤λ. If |φ(x,y)|>λ, then

$$\begin{aligned} |\varphi(x,y)- & \varphi_{\lambda}(x,y)|=|\varphi(x,y)|-\lambda\le C_{0}(1+\|x\|^{p}+\|y\|^{q})-\lambda\\ \le & C_{0}\Big\{\Big(\|x\|^{p}-\lambda/(4C_{0})\Big)^{+} +\Big(\|y\|^{q}-\lambda/(4C_{0})\Big)^{+}\Big\}. \end{aligned} $$

Hence,

$$|\varphi(x,y)-\varphi_{\lambda}(x,y)| \le C_{0}\Big\{\Big(\|x\|^{p}-\lambda/(4C_{0})\Big)^{+} +\Big(\|y\|^{q}-\lambda/(4C_{0})\Big)^{+}\Big\}.$$

It follows that

$$\begin{aligned} &{\lim}_{\lambda\rightarrow \infty} \limsup_{n\rightarrow \infty}\left|\hat{\mathbb{E}}^{\ast}\Big[\varphi\Big(\widetilde{\boldsymbol W}_{n}(\cdot) \Big)\Big]- \hat{\mathbb{E}}\Big[\varphi_{\lambda}\left(\widetilde{\boldsymbol W}_{n}(\cdot) \right)\Big]\right| \\ \le & {\lim}_{\lambda\rightarrow \infty} \limsup_{n\rightarrow \infty} C_{0}\left\{ \hat{\!\mathbb{E}}\left[\!\left(\max_{k\le n}\left|\frac{S_{k}^{X}}{\sqrt{n}}\right|^{p}-\frac{\lambda}{4C_{0}}\right)^{+}\right]+\hat{\mathbb{E}}\left[\left(\max_{k\le n}\left|\frac{S_{k}^{Y}}{n}\right|^{q}-\frac{\lambda}{4C_{0}}\right)^{+}\right]\!\right\}\\ =&0, \end{aligned} $$

by Lemma 8. Further, by (15),

$${\lim}_{n\rightarrow \infty}\hat{\mathbb{E}}\left[\varphi_{\lambda} \left(\widetilde{\boldsymbol W}_{n}(\cdot) \right)\right]= \widetilde{\mathbb E}\left[\varphi_{\lambda} \left(B_{\cdot},b_{\cdot}\right)\right]\rightarrow \widetilde{\mathbb E}\left[\varphi \left(B_{\cdot},b_{\cdot}\right)\right]\;\; \text{as}~\lambda\rightarrow \infty. $$

(16) is proved, and the proof of Theorem 4 is now completed. □

Proof of Theorem 4.

When X _k and Y _k are d-dimensional random vectors, the tightness (24) of $\widetilde {\boldsymbol W_{n}}(\cdot)$ also follows, because each sequence of the components of vector $\widetilde {\boldsymbol W_{n}}(\cdot)$ is tight. Also, (29) remains true, because each component has this property. Moreover, it follows that

$$ \begin{aligned} {\mathbb F}\left[\frac{1}{2}\left\langle A\xi_{t},\xi_{t}\right\rangle+\left\langle p,\eta_{t}\right\rangle\right] = & {\lim}_{m_{j}\rightarrow \infty} \hat{\mathbb{E}}\left[\frac{1}{2}\left\langle A\frac{S_{[m_{j}t]}^{X}}{\sqrt{m_{j}}},\frac{S_{[m_{j}t]}^{X}}{\sqrt{m_{j}}}\right\rangle+\left\langle p,\frac{S_{[m_{j}t]}^{Y}}{m_{j}}\right\rangle\right]\\ = & {\lim}_{m_{j}\rightarrow \infty} \frac{[m_{j}t]}{m_{j}} G(p,A)=t G(p,A). \end{aligned} $$

The remaining proof is the same as that of Theorem 4. □

Proof of the self-normalized FCLTs

Let $Y_{k}=X_{k}^{2}$. The function G(p,q) in (12) becomes

$$G(p,q)=\hat{\mathbb{E}}\left[\left(\frac{q}{2} +p\right) X_{1}^{2}\right]=\left(\frac{q}{2} +p\right)^{+}\overline{\sigma}^{2}-\left(\frac{q}{2} +p\right)^{-}\underline{\sigma}^{2}, \;\; p,q\in \mathbb R. $$

Then, the process (B _t ,b _t ) in (15) and the process (W(t),〈W〉 _t ) are identically distributed.

In fact, note

$$ \langle W\rangle_{t+s}-\langle W\rangle_{t}=(W(t+s)-W(t))^{2}-2\int_{0}^{s} (W(t+x)-W(t))d(W(t+x)-W(t)). $$

It is easy to verify that (W(t),〈W〉 _t ) satisfies (i)-(iv) for (B _·,b _·). It remains to show that $(B_{1}, b_{1})\overset {d}= (W(1), \langle W\rangle _{1})$. Let {X _n ;n≥1} be a sequence of independent and identically distributed random variables with $X_{1}\overset {d}= W(1)$. Then, by Theorem 4,

$$ \left(\frac{\sum_{k=1}^{n}X_{k}}{\sqrt{n}},\frac{\sum_{k=1}^{n} X_{k}^{2}}{n}\right)\overset{d}\rightarrow (B_{1}, b_{1}). $$

Further, let $t_{k}=\frac {k}{n}$. Then,

$$\left(\frac{\sum_{k=1}^{n}X_{k}}{\sqrt{n}},\frac{\sum_{k=1}^{n} X_{k}^{2}}{n}\right)\overset{d}= \left(W(1), \sum_{k=1}^{n} (W(t_{k})-W(t_{k-1}))^{2}\right)\overset{L_{2}}\rightarrow (W(1), \langle W\rangle_{1}). $$

Hence, $(B_{\cdot },b_{\cdot })\overset {d}=(W(\cdot), \langle W\rangle _{\cdot })$. We conclude the following proposition from Theorem 4.

Proposition 1

Suppose $\hat {\mathbb {E}}[(X_{1}^{2}-b)^{+}]\rightarrow 0$ as b→∞. Then, for any bounded continuous function $\psi :C[0,1]\times C[0,1]\rightarrow \mathbb R$,

$$\hat{\mathbb{E}}\left[\psi\left(\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{n}},\frac{\widetilde{V}_{n} (\cdot)}{n}\right)\right]\rightarrow \widetilde{\mathbb E}\left[\psi\Big(W(\cdot), \langle W \rangle_{\cdot} \Big) \right], $$

where $\widetilde {V}_{n}(t)=V_{[nt]}+(nt-[nt])X^{2}_{[nt]+1}$, and, in particular, for any bounded continuous function $\psi :C[0,1]\times \mathbb R\rightarrow \mathbb R$,

$$ \hat{\mathbb{E}}\left[\psi\left(\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{n}},\frac{V_{n}}{n}\right)\right]\rightarrow \widetilde{\mathbb E}\Big[\psi\Big(W(\cdot), \langle W \rangle_{1}\Big) \Big]. $$

(31)

Now, we begin the proof of Theorem 2. Let $a=\underline {\sigma }^{2}/2$ and $b=2\overline {\sigma }^{2}$. According to (30), we have $\widetilde {\mathcal {V}}\big (\underline {\sigma }^{2}-\epsilon < \langle W \rangle _{1}<\overline {\sigma }^{2}+\epsilon \big)=1$ for all ε>0. Let $\varphi :C[0,1]\rightarrow \mathbb R$ be a bounded continuous function. Define

$$ \psi\big(x(\cdot),y\big)=\varphi\left(\frac{x(\cdot)}{\sqrt{a\vee y\wedge b}}\right), \;\; x(\cdot)\in C[0,1],\;y\in \mathbb R. $$

Then, $\psi :C[0,1]\times \mathbb R\rightarrow \mathbb R$ is a bounded continuous function. Hence, by Proposition 1,

$$ \hat{\mathbb{E}}\left[\varphi\left(\frac{\widetilde{S}_{n}^{X}(\cdot)/ \sqrt{n}}{\sqrt{a\vee(V_{n}/n)\wedge b}}\right)\right]\rightarrow \widetilde{\mathbb E}\left[\varphi\left(\frac{W(\cdot)}{\sqrt{a\vee (\langle W \rangle_{1})\wedge b}} \right) \right]=\widetilde{\mathbb E}\left[\varphi\left(\frac{W(\cdot)}{\sqrt{ \langle W \rangle_{1})}} \right) \right]. $$

Also,

$$\begin{aligned} \limsup_{n\rightarrow \infty} & \left|\hat{\mathbb{E}}^{\ast}\left[\varphi\left(\frac{\widetilde{S}_{n}^{X}(\cdot)/ \sqrt{n}}{\sqrt{ V_{n}/n }}\right)\right]-\hat{\mathbb{E}}\left[\varphi\left(\frac{\widetilde{S}_{n}^{X}(\cdot)/ \sqrt{n}}{\sqrt{a\vee(V_{n}/n)\wedge b}}\right)\right]\right|\\ &\le C \limsup_{n\rightarrow \infty} \mathbb{V}\left(V_{n}/n\not\in (a,b)\right)\\ &\le C\widetilde{\mathbb{V}}\left(\langle W \rangle_{1}\ge 3\overline{\sigma}^{2}/2\right) + C\widetilde{\mathbb{V}}\left(\langle W \rangle_{1}\le 2\underline{\sigma}^{2}/3\right)=0. \end{aligned} $$

It follows that

$$ \hat{\mathbb{E}}^{\ast}\left[\varphi\left(\frac{\widetilde{S}_{n}^{X}(\cdot)}{\sqrt{V_{n} }}\right)\right]\rightarrow \widetilde{\mathbb E}\left[\varphi\left(\frac{W(\cdot)}{\sqrt{ \langle W \rangle_{1})}} \right) \right]. $$

The proof is now completed. □

Proof of Theorem 3.

First, note that

$$\begin{aligned} \hat{\mathbb{E}}\left[X_{1}^{2}\wedge x^{2}\right]& \le \hat{\mathbb{E}}\left[X_{1}^{2}\wedge(kx)^{2}\right]\le \hat{\mathbb{E}}\left[X_{1}^{2}\wedge x^{2}\right]+k^{2}x^{2}\mathbb{V}(|X_{1}|>x),\;\; k\ge 1, \\ \hat{\mathbb{E}}\left[|X_{1}|^{r}\wedge x^{r}\right]& \le \hat{\mathbb{E}}\left[|X_{1}|^{r}\wedge (\delta x)^{r}\right]+\hat{\mathbb{E}}\left[(\delta x)^{r}\vee |X_{1}|^{r}\wedge x^{r}\right]\\ & \le \delta^{r-2} x^{r-2}l(\delta x)+x^{r} \mathbb{V}(|X_{1}|\ge \delta x), \;\;0<\delta<1,\; r>2. \end{aligned} $$

The condition (I) implies that l(x) is slowly varying as x→∞ and

$$ \hat{\mathbb{E}}[|X_{1}|^{r}\wedge x^{r}]=o(x^{r-2}l(x)), \; r>2. $$

Further,

$$ \frac{\hat{\mathbb{E}}^{\ast}[X_{1}^{2}I\{|X_{1}|\le x\}]}{l(x)}\rightarrow 1, $$

$$ C_{\mathbb{V}}\big(|X_{1}|^{r}I\{|X_{1}|\ge x\}\big)=\int_{x^{r}}^{\infty} \mathbb{V}(|X_{1}|^{r}\ge y)dy =o(x^{2-r} l(x)),\;\; 0<r<2. $$

If conditions (I) and (III) are satisfied, then

$$ \hat{\mathbb{E}}[(|X_{1}|-x)^{+}]\le \hat{\mathbb{E}}^{\ast}[|X_{1}|I\{|X|\ge x\}] \le C_{\mathbb{V}}\big(|X_{1}|I\{|X_{1}|\ge x\}\big)=o(x^{-1} l(x)). $$

Now, let d _t = inf{x:x ⁻² l(x)=t ⁻¹}. Then, $nl(d_{n})=d_{n}^{2}$. Similar to Theorem 2, it is sufficient to show that for any bounded continuous function $\psi :C[0,1]\times C[0,1]\rightarrow \mathbb R$,

$$ \hat{\mathbb{E}}\left[\psi\left(\frac{\widetilde{S}_{n}^{X}(\cdot)}{d_{n}},\frac{\widetilde{V}_{n}(\cdot)}{d_{n}^{2}}\right)\right]\rightarrow \widetilde{\mathbb E}\left[\psi(W(\cdot), \langle W\rangle_{\cdot})\right]\;\; \text{with}~W(1)\sim N(0,[r^{-2}, 1]). $$

Let $\overline {X}_{k}=\overline {X}_{k,n}=(-d_{n})\vee X_{k}\wedge d_{n}$, $\overline {S}_{k} =\sum _{i=1}^{k} \overline {X}_{i}$, $\overline {V}_{k}=\sum _{i=1}^{k} \overline {X}_{i}^{2}$. Denote $\overline {S}_{n}(t)=\overline {S}_{[nt]}+(nt-[nt])\overline {X}_{[nt]+1}$ and $\overline {V}_{n}(t)=\overline {V}_{[nt]}+(nt-[nt])\overline {X}^{2}_{[nt]+1}$. Note

$$\mathbb{V}\left(X_{k}\ne \overline{X}_{k}~\text{for some}~k\le n\right)\le n \mathbb{V}\left(|X_{1}|\ge d_{n}\right)=n\cdot o\left(\frac{l(d_{n})}{d_{n}^{2}}\right)=o(1). $$

It is sufficient to show that for any bounded continuous function $\psi :C[0,1]\times C[0,1]\rightarrow \mathbb R$,

$$ \hat{\mathbb{E}}\left[\psi\left(\frac{\overline{S}_{n}(\cdot)}{d_{n}},\frac{\overline{V}_{n}(\cdot)}{d_{n}^{2}}\right)\right]\rightarrow \widetilde{\mathbb E}\left[\psi(W(\cdot), \langle W\rangle_{\cdot})\right]. $$

Following the line of the proof of Theorem 4, we need only to show that

• (a) for any 0<t≤1,

  $${} \limsup_{n\rightarrow \infty} \hat{\mathbb{E}}\left[\max_{k\le [nt]}\left|\frac{\overline{S}_{k}}{d_{n}}\right|^{p}\right]\le C_{p} t^{p/2},\;\; \limsup_{n\rightarrow \infty}\hat{\mathbb{E}}\left[\max_{k\le [nt]}\left|\frac{\overline{V}_{k}}{d_{n}^{2}}\right|^{p}\right]\le C_{p} t^{p},\;\; \forall p\ge 2; $$

• (b) for any 0<t≤1,

  $$ {\lim}_{n\rightarrow \infty} \hat{\mathbb{E}}\left[ \frac{q}{2} \left(\frac{\overline{S}_{[nt]}}{d_{n}}\right)^{2}+p \frac{\overline{V}_{[nt]}}{d_{n}^{2}}\right]=tG(p,q), $$

  where

  $$G(p,q)=\left(\frac{q}{2}+p\right)^{+} - r^{-2}\left(\frac{q}{2}+p\right)^{-}; $$
• (c)
  $${\kern-3.8cm} \max_{k\le n} \frac{|X_{k}|}{d_{n}}\overset{\mathbb{V}}\rightarrow 0. $$

In fact, (a) implies the tightness of $\left (\frac {\widetilde {S}_{n}^{X}(\cdot)}{d_{n}},\frac {\widetilde {V}_{n}(\cdot)}{d_{n}^{2}}\right)$ and (29), and (b) implies the distribution of the limit process is uniquely determined.

First, (c) is obvious, because

$$\mathbb{V}\Big(\max_{k\le n}|X_{k}|\ge \epsilon d_{n}\Big)\le n \mathbb{V}\Big(|X_{1}|\ge \epsilon d_{n}\Big)=o(1) n \frac{l(\epsilon d_{n})}{\epsilon^{2} d_{n}^{2}} =o(1) n \frac{l(d_{n})}{d_{n}^{2}}=o(1). $$

As for (a), by the Rosenthal-type inequality (18),

$$ {{\begin{aligned} &\hat{\mathbb{E}} \left[\max_{k\le [nt]}\left|\frac{\overline{S}_{k}}{d_{n}}\right|^{p}\right] \le C_{p}d_{n}^{-p}\left\{ [nt] \hat{\mathbb{E}}\left[|X_{1}|^{p}\wedge d_{n}^{p}\right]+\left([nt] \hat{\mathbb{E}}\left[|X_{1}|^{2}\wedge d_{n}^{2}\right]\right)^{p/2}\right.\\ & \quad + \left. \Big([nt] (\widehat{\mathcal{E}}[(-d_{n})\vee X_{1}\wedge d_{n}])^{+}+[nt] (\hat{\mathbb{E}}[(-d_{n})\vee X_{1}\wedge d_{n}])^{+}\Big)^{p}\right\}\\ & \quad \le C_{p}d_{n}^{-p}\left\{ [nt] \hat{\mathbb{E}}\left[|X_{1}|^{p}\wedge d_{n}^{p}\right]+\left([nt] \hat{\mathbb{E}}\left[|X_{1}|^{2}\wedge d_{n}^{2}\right]\right)^{p/2} + \left([nt] \hat{\mathbb{E}}\left[(|X_{1}|-d_{n})^{+}\right] \right)^{p}\right\}\\ & \quad \le C_{p}d_{n}^{-p}\left\{ [nt] o\left(d_{n}^{p-2}l(d_{n})\right) +\left([nt] l(d_{n})\right)^{p/2} + \left([nt] o\left(\frac{l(d_{n})}{d_{n}}\right) \right)^{p}\right\}\\ & \quad = o(1)[nt]\frac{l(d_{n})}{d_{n}^{2}}+\left(\frac{[nt]}{n} \right)^{p/2} \left(\frac{nl(d_{n})}{d_{n}^{2}}\right)^{p/2}+o(1)\left([nt] \frac{l(d_{n})}{d_{n}^{2}}\right)^{p}\le C_{p}t^{p/2}+o(1), \end{aligned}}} $$

and similarly,

$$\begin{aligned} \hat{\mathbb{E}} \left[\max_{k\le [nt]}\left|\frac{\overline{V}_{k}}{d_{n}^{2}}\right|^{p}\right] & \le C_{p}d_{n}^{-2p}\left\{ [nt] \hat{\mathbb{E}}\left[|X_{1}|^{2p}\wedge d_{n}^{2p}\right]+\left([nt] \hat{\mathbb{E}}\left[|X_{1}|^{4}\wedge d_{n}^{4}\right]\right)^{p/2}\right.\\ &\quad + \left. \left([nt] \widehat{\mathcal{E}}\left[ X_{1}^{2}\wedge d_{n}^{2}\right] \right)+[nt] \left(\hat{\mathbb{E}}\left[ X_{1}^{2}\wedge d_{n}^{2}\right] \right)^{p}\right\}\\ & = o(1)+ C_{p} \left([nt] \frac{l(d_{n})}{d_{n}^{2}}\right)^{p}\le C_{p}t^{p}+o(1). \end{aligned} $$

Thus (a) follows.

As for (b), note

$$\begin{aligned} \frac{q}{2} \left(\frac{\overline{S}_{[nt]}}{d_{n}}\right)^{2}+p \frac{\overline{V}_{[nt]}}{d_{n}^{2}} =\left(\frac{q}{2}+p\right)\frac{\overline{V}_{[nt]}}{d_{n}^{2}}+q\frac{\sum_{k=1}^{[nt]-1}\overline{S}_{k-1}\overline{X}_{k}}{d_{n}^{2}}. \end{aligned} $$

By (32),

$$\begin{aligned} \hat{\mathbb{E}}\left[ \sum_{k=1}^{[nt]-1}\overline{S}_{k-1}\overline{X}_{k} \right]\le& \sum_{k=1}^{[nt]-1}\hat{\mathbb{E}}\left[\overline{S}_{k-1}\overline{X}_{k} \right]\\ \le & \sum_{k=1}^{[nt]-1}\left\{\hat{\mathbb{E}}\left[\left(\overline{S}_{k-1}\right)^{+}\right]\hat{\mathbb{E}}\left[\overline{X}_{k}\right]-\hat{\mathbb{E}}\left[(\overline{S}_{k-1})^{-}\right]\widehat{\mathcal{E}}\left[\overline{X}_{k}\right]\right\}\\ \le & \sum_{k=1}^{[nt]-1} \left(\hat{\mathbb{E}}\left[| \overline{S}_{k-1}|^{2}\right]\right)^{1/2}\hat{\mathbb{E}}\left[(|X_{1}|-d_{n})^{+}\right]\\ =& O\left(\left(d_{n}^{2}\right)^{1/2}\right)\cdot n\hat{\mathbb{E}}\left[\left(|X_{1}|-d_{n}\right)^{+}\right]\\ =& O(d_{n})\cdot n\cdot o\left(\frac{l(d_{n})}{d_{n}}\right)=o\left(d_{n}^{2}\right), \end{aligned} $$

and similarly,

$$ \hat{\mathbb{E}}\left[- \sum_{k=1}^{[nt]-1}\overline{S}_{k-1}\overline{X}_{k} \right]=o\left(d_{n}^{2}\right). $$

Further,

$$ \frac{\hat{\mathbb{E}}\left[V_{[nt]}\right]}{d_{n}^{2}}=\frac{[nt]\hat{\mathbb{E}}\left[X_{1}^{2}\wedge d_{n}^{2}\right]}{d_{n}^{2}}=\frac{[nt]}{n}\frac{nl(d_{n})}{d_{n}^{2}}=\frac{[nt]}{n}\rightarrow t $$

and

$$ \frac{\widehat{\mathcal{E}}\left[V_{[nt]}\right]}{d_{n}^{2}}= \frac{[nt]\widehat{\mathcal{E}}\left[X_{1}^{2}\wedge d_{n}^{2}\right]}{d_{n}^{2}}=\frac{[nt]}{n}\frac{\widehat{\mathcal{E}}\left[X_{1}^{2}\wedge d_{n}^{2}\right] }{\hat{\mathbb{E}}\left[X_{1}^{2}\wedge d_{n}^{2}\right]}\rightarrow t r^{-2}. $$

Hence, we conclude that

$$ \begin{aligned} \hat{\mathbb{E}}& \left[\frac{q}{2} \left(\frac{\overline{S}_{[nt]}}{d_{n}}\right)^{2}+p \frac{\overline{V}_{[nt]}}{d_{n}^{2}}\right] = \hat{\mathbb{E}}\left[\left(\frac{q}{2}+p\right)\frac{\overline{V}_{[nt]}}{d_{n}^{2}}\right]+o(1)\\ &\quad = t \left[ \left(\frac{q}{2}+p\right)^{+} - r^{-2}\left(\frac{q}{2}+p\right)^{-}\right] +o(1). \end{aligned} $$

(32)

Thus, (b) is statisfied, and the proof is completed. □