Stable Lévy Motion with Values in the Skorokhod Space: Construction and Approximation

Abstract

In this article, we introduce an infinite-dimensional analogue of the $\alpha$-stable Lévy motion, defined as a Lévy process $Z=\{Z(t)\}_{t \ge 0}$ with values in the space ${\mathbb {D}}$ of càdlàg functions on [0, 1], equipped with Skorokhod’s $J_1$ topology. For each $t \ge 0$, Z(t) is an $\alpha$-stable process with sample paths in ${\mathbb {D}}$, denoted by $\{Z(t,s)\}_{s\in [0,1]}$. Intuitively, Z(t, s) gives the value of the process Z at time t and location s in space. This process is closely related to the concept of regular variation for random elements in ${\mathbb {D}}$ introduced in de Haan and Lin (Ann Probab 29:467–483, 2001) and Hult and Lindskog (Stoch Proc Appl 115:249–274, 2005). We give a construction of Z based on a Poisson random measure, and we show that Z has a modification whose sample paths are càdlàg functions on $[0,\infty )$ with values in ${\mathbb {D}}$. Finally, we prove a functional limit theorem which identifies the distribution of this modification as the limit of the partial sum sequence $\{S_n(t)=\sum _{i=1}^{[nt]}X_i\}_{t\ge 0}$, suitably normalized and centered, associated with a sequence $(X_i)_{i\ge 1}$ of i.i.d. regularly varying elements in ${\mathbb {D}}$.

Appendices

Some Auxiliary Results

In this section, we include some auxiliary results which are used in this article.

The first result shows that the measure ${\overline{\nu }}$ which appears in the definition of regular variation for random elements in ${\mathbb {D}}$ must be of product form. This result is probably well known. We include its proof since we could not find it in the literature.

Lemma A.1

If $c={\overline{\nu }}((1,\infty )\times {\mathbb {S}}_{{\mathbb {D}}})>0$, then the measure ${\overline{\nu }}$ in Definition 1.4 must be of the product from (7), with probability measure $\Gamma _1$ given by (8).

Proof

Let ${{{\mathcal {P}}}}$ be the class of sets $A_{r,S}=(r,\infty ) \times S$ with $r>0$ and $S \in {\mathcal {B}}({\mathbb {S}}_{{\mathbb {D}}})$. Note that $A_{r,s}={\mathcal {T}}(V_{r,s})$ where $V_{r,S}=\{x \in {\mathbb {D}}; \Vert x\Vert >r, \frac{x}{\Vert x\Vert } \in S\}$. The sets $V_{r,S}$ have the scaling property $aV_{r,S}=V_{ar,S}$ for any $a>0$. To see that the sets $A_{r,S}$ have a similar property, we define $aA=\{(ar,z); (r,z) \in A\}$ for any $a>0$ and $A \in {\mathcal {B}}(\overline{{\mathbb {D}}}_0)$. Then

$$\begin{aligned} aA_{r,S}=\{(as,z); as>ar,z \in {\mathbb {S}}_{{\mathbb {D}}}\}=A_{ar,S}. \end{aligned}$$

In particular, $A_{r,S}=rA_{1,S}$. By the scaling property of ${\overline{\nu }}$ and Definition (8) of $\Gamma _1$,

$$\begin{aligned} {\overline{\nu }}(A_{r,S})=r^{-\alpha }{\overline{\nu }}(A_{1,S})=r^{-\alpha }c\Gamma _1(S)=(c \nu _{\alpha } \times \Gamma _1)(A_{r,S}). \end{aligned}$$

Hence, when restricted to $(0,\infty ) \times {\mathbb {S}}_{{\mathbb {D}}}$, the measures ${\overline{\nu }}$ and $c \nu _{\alpha } \times \Gamma _1$ coincide for sets in the class ${{{\mathcal {P}}}}$. Since ${{{\mathcal {P}}}}$ is a $\pi$-system which generates the Borel $\sigma$-field of $(0,\infty ) \times {\mathbb {S}}_{{\mathbb {D}}}$ (with respect to distance $d_{\overline{{\mathbb {D}}}_0}$), it follows that ${\overline{\nu }}=c \nu _{\alpha } \times \Gamma _1$ on $(0,\infty ) \times {\mathbb {S}}_{{\mathbb {D}}}$. Finally, these measures coincide on the entire space $\overline{{\mathbb {D}}}_0$ since they are zero on $\{\infty \} \times {\mathbb {S}}_{{\mathbb {D}}}$. $\square$

The next result is an extension of Lemma 5.2 of [21] to the case of functions with values in an arbitrary metric space.

Lemma A.2

Let (S, d) be a complete metric space. We denote by ${\mathbb {D}}([0,\infty );S)$ the set of functions $x:[0,\infty ) \rightarrow S$ which are right-continuous and have left-limits (with respect to d). If $(x_n)_{n\ge 1}$ is a sequence in ${\mathbb {D}}([0,\infty );S)$ and the function $x:[0,\infty ) \rightarrow S$ is such that

$$\begin{aligned} \sup _{t\le T}d(x_n(t),x(t)) \rightarrow 0 \quad \text{ as } \quad n \rightarrow \infty , \end{aligned}$$

(74)

for any $T>0$, then $x \in {\mathbb {D}}([0,\infty );S)$.

Proof

We first prove that x is right-continuous. Let $t\ge 0$ be arbitrary and $(t_k)_{k \ge 1}$ be such that $t_k \rightarrow t$ and $t_k \ge t$ for all k. Let $T>0$ be such that $t_k \in [0,T]$ for all k. Let $\varepsilon >0$ be arbitrary. By (74), there exists $n_0$ such that $d(x_{n_0}(t),x(t))<\varepsilon$ for all $t \le T$. Since $x_{n_0}$ is right-continuous at t, there exists $K_{\varepsilon }$ such that $d(x_{n_0}(t_k),x_{n_0}(t))<\varepsilon$ for all $k\ge K_{\varepsilon }$. By the triangle inequality, for any $k\ge K_{\varepsilon }$,

$$\begin{aligned} d(x(t_k),x(t)) \le d(x(t_k),x_{n_0}(t_k)) +d(x_{n_0}(t_k),x_{n_0}(t))+d(x_{n_0}(t),x(t))<3\varepsilon . \end{aligned}$$

Next, we prove that x has left-limit at $t>0$. Let $(t_k)_{k \ge 1}$ be such that $t_k \rightarrow t$ and $t_k < t$ for all k. Let $T>0$ be such that $t_k \in [0,T]$ for all k. Let $\varepsilon >0$ be arbitrary. Choose $n_0$ as above. Since $x_{n_0}(t_{k}) \rightarrow x(t-)$, $\{x_{n_0}(t_k)\}_{k}$ is a Cauchy sequence. Then, there exists $L_{\varepsilon }$ such that $d(x_{n_0}(t_k),x_{n_0}(t_l))<\varepsilon$ for all $k,l\ge L_{\varepsilon }$. Then

$$\begin{aligned} d(x(t_k),x(t_l)) \le d(x(t_k),x_{n_0}(t_k)) +d(x_{n_0}(t_k),x_{n_0}(t_l))+d(x_{n_0}(t_l),x(t_l))<3\varepsilon , \end{aligned}$$

for any $k,l\ge L_{\varepsilon }$, and hence $\{x(t_k)\}_{k}$ is a Cauchy sequence. Since S is complete, there exists $l=\lim _{k \rightarrow \infty }x(t_k)$. We must show that l does not depend on $(t_k)_k$. Let $l'=\lim _{k \rightarrow \infty }x(t_k')$, where $(t_k')_k$ is another sequence such that $t_k' \rightarrow t$ and $t_k'< t$ for all k. Since both sequences $\{x_{n_0}(t_k)\}_{k}$ and $\{x_{n_0}(t_k')\}_{k}$ converge to $x_{n_0}(t-)$, there exists $M_{\varepsilon }$ such that $d(x_{n_0}(t_k),x_{n_0}(t_k'))<\varepsilon$ for all $k\ge M_{\varepsilon }$. Hence,

$$\begin{aligned} d(x(t_k),x(t_k')) \le d(x(t_k),x_{n_0}(t_k)) +d(x_{n_0}(t_k),x_{n_0}(t_k'))+d(x_{n_0}(t_k'),x(t_k'))<3\varepsilon , \end{aligned}$$

for any $k\ge M_{\varepsilon }$. This proves that $l=l'$. $\square$

The following result is probably well known. We include its proof since we could not find it in the literature.

Lemma A.3

Let (S, d) be a separable metric space. Let $X_n^{(1)}, \ldots , X_n^{(k)}$ and $X^{(1)},\ldots , X^{(k)}$ be random elements in S defined on a probability space $(\Omega ,{\mathcal {F}},P)$, such that $d(X_n^{(i)},X^{(i)}) \rightarrow 0$ a.s. for any $i=1,\ldots ,k$. If $X_n^{(1)},\ldots ,X_n^{(k)}$ are independent for any $n \ge 1$, then $X^{(1)},\ldots ,X^{(k)}$ are independent.

Proof

We assume for simplicity that $k=2$, the general case being similar. To simplify the notation, we let $X_n=X_n^{(1)}$ and $Y_n=X_n^{(2)}$. Clearly, $d(X_n,X) {\mathop {\rightarrow }\limits ^{P}} 0$ and $d(Y_n,Y) {\mathop {\rightarrow }\limits ^{P}} 0$. Note that the space $S \times S$ equipped with the product metric is separable and $(X_n,X)$ is a random element in $S \times S$ (see p.225 of [4]). By Corollary to Theorem 3.1 of [5], $X_n {\mathop {\rightarrow }\limits ^{d}} X$ and $Y_n {\mathop {\rightarrow }\limits ^{d}} Y$. By Theorem 3.2 of [4],

$$\begin{aligned} (P\circ X_n^{-1}) \times (P \circ Y_n^{-1}) {\mathop {\rightarrow }\limits ^{w}}(P \circ X^{-1}) \times (P \circ Y^{-1}) \quad \text{ on } \quad S \times S. \end{aligned}$$

(75)

On the other hand, $(X_n,Y_n) \rightarrow (X,Y)$ a.s. with respect to the product distance in $S \times S$. Hence, again by Corollary to Theorem 3.1 of [5], $(X_n,Y_n) {\mathop {\rightarrow }\limits ^{d}} (X,Y)$ in $S \times S$, i.e.,

$$\begin{aligned} P\circ (X_n,Y_n)^{-1} {\mathop {\rightarrow }\limits ^{w}}P \circ (X,Y)^{-1} \quad \text{ on } \quad S \times S. \end{aligned}$$

(76)

Finally, $P\circ (X_n,Y_n)^{-1}=(P\circ X_n^{-1}) \times (P \circ Y_n^{-1})$ for any $n\ge 1$, since $X_n$ and $Y_n$ are independent for any $n \ge 1$. The fact that $P \circ (X,Y)^{-1}=(P \circ X^{-1}) \times (P \circ Y^{-1})$ follows from (75) and (76), by the uniqueness of the limit. $\square$

The $\alpha$-Stable Lévy Sheet

In this section, we show that the $\alpha$-stable Lévy sheet can be viewed as an example of a ${\mathbb {D}}$-valued $\alpha$-stable Lévy motion restricted to the time interval [0, 1].

First, we recall briefly the construction of the $\alpha$-stable Lévy sheet, as described in Section 4.8 of [19]. Let $M=\sum _{i\ge 1}\delta _{(T_i,S_i,J_i)}$ be a Poisson random measure on $[0,\infty ) \times [0,\infty ) \times {\overline{{\mathbb {R}}}}_0$ of intensity $\mathrm{Leb} \times \mathrm{Leb} \times \nu _{\alpha ,p}$, where $\nu _{\alpha ,p}$ is given by (69), for some $\alpha \in (0,2),\alpha \not =1$ and $p \in [0,1]$, with $q=1-p$. Let $(\varepsilon _j)_{j\ge 0}$ be a sequence of real numbers such that $\varepsilon _j \downarrow 0$ and $\varepsilon _0=1$. Let $I_j=(\varepsilon _j,\varepsilon _{j-1}]$ for $j\ge 1$ and $I_0=(1,\infty )$. For any $t,s\in [0,1]$ and $j\ge 0$, let

$$\begin{aligned} L_j(t,s)=\int _{[0,t] \times [0,s] \times \Gamma _j}zM(\mathrm{d}u_1,\mathrm{d}u_2,\mathrm{d}z)=\sum _{i\ge 1}J_i 1_{\{J_i \in \Gamma _j\}}1_{\{T_i \le t, S_i \le s\}}. \end{aligned}$$

Note that $L_j(t,s)$ is a compound Poisson random variable with characteristic function

$$\begin{aligned} E[\mathrm{e}^{iu L_j(t,s)}]=\exp \left\{ ts \int _{\Gamma _j} (\mathrm{e}^{iuz} -1) \nu _{\alpha ,p}(\mathrm{d}z) \right\} , \quad u \in {\mathbb {R}}. \end{aligned}$$

By Kolmogorov’s criterion, the series $\sum _{j\ge 1}\big (L_j(t,s)-E(L_{j}(t,s))\big )$ converges a.s., since $\mathrm{Var}\big (L_j(t,s) \big )=ts \int _{\Gamma _j}z^2 \nu _{\alpha ,p}(\mathrm{d}z)$ for any $j\ge 1$ and $\int _{|z| \le 1}z^2 \nu _{\alpha ,p}(\mathrm{d}z)<\infty$.

We define ${\overline{L}}(t,s)=\sum _{j\ge 0}L_j(t,s)$ if $\alpha <1$ and ${\overline{L}}(t,s)=\sum _{j\ge 0}\big ( L_j(t,s)-E(L_j(t,s))\big )$ if $\alpha >1$. It can be proved that there exists a process $\{L(t,s)\}_{(t,s) \in [0,1]^2}$ with sample paths in ${\mathbb {D}}([0,1]^2)$ such that $L(t,s)={\overline{L}}(t,s)$ a.s. for any $t,s \in [0,1]$, and

$$\begin{aligned}&\sup _{(t,s)\in [0,1]^2}|L^{(\varepsilon _k)}(t,s)-L(t,s)| \rightarrow 0 \quad \text{ a.s. } \quad \text{ if } \quad \alpha <1, \end{aligned}$$

(77)

$$\begin{aligned}&\sup _{(t,s)\in [0,1]^2}|{\overline{L}}^{(\varepsilon _k)}(t,s)-L(t,s)| \rightarrow 0 \quad \text{ a.s. } \quad \text{ if } \quad \alpha >1, \end{aligned}$$

(78)

where $L^{(\varepsilon _k)}(t,s)=\sum _{j=0}^k L_j(t,s)$ and ${\overline{L}}^{(\varepsilon _k)}(t,s)=L^{(\varepsilon _k)}(t,s)-E(L^{(\varepsilon _k)}(t,s))$ (if $\alpha >1$). Here ${\mathbb {D}}([0,1]^2)$ is the space of functions $x:[0,1]^2 \rightarrow {\mathbb {R}}$ which are continuous at any point (t, s) when this point is approached from the upper right quadrant, and have limits when the point is approached from the other three quadrants. Moreover,

$$\begin{aligned} E[\mathrm{e}^{iu L(t,s)}]= & {} \exp \left\{ ts \int _{{\mathbb {R}}}(\mathrm{e}^{iuz}-1)\nu _{\alpha ,p}(\mathrm{d}z) \right\} \quad \text{ if } \quad \alpha <1, \\ E[\mathrm{e}^{iu L(t,s)}]= & {} \exp \left\{ ts \int _{{\mathbb {R}}}(\mathrm{e}^{iuz}-1-iuz)\nu _{\alpha ,p}(\mathrm{d}z) \right\} \quad \text{ if } \quad \alpha >1. \end{aligned}$$

Consequently, L(t, s) has a $S_{\alpha }\big ((ts)^{1/\alpha }C_{\alpha }^{-1},\beta ,0\big )$-distribution with $\beta =p-q$ and $C_{\alpha }$ given by (30). The process $\{L(t,s)\}_{(t,s) \in [0,1]^2}$ is called a $\alpha$-stable Lévy sheet. Note that both processes $\{L(t,s)\}_{t \in [0,1]}$ and $\{L(t,s)\}_{s \in [0,1]}$ are $\alpha$-stable Lévy motions with paths in ${\mathbb {D}}$.

Theorem B.1

Let $L(t)=\{L(t,s)\}_{s \in [0,1]}$ for any $t \in [0,1]$. The process $\{L(t)\}_{t \in [0,1]}$ is an ${\mathbb {D}}$-valued $\alpha$-stable Lévy motion (according to Definition 1.1).

Proof

We show that $\{L(t)\}_{t \in [0,1]}$ satisfies conditions (i)–(iv) of Definition 1.1. We assume that $\alpha <1$, the case $\alpha >1$ being similar. Clearly, $L(0)=0$, so property (i) holds.

For property (ii), note that by (77), $L^{(\varepsilon _k)}(t_i) \rightarrow L(t_i)$ a.s. in $({\mathbb {D}},\Vert \cdot \Vert )$ as $k \rightarrow \infty$ for $i=1,\ldots ,K$, and hence $L^{(\varepsilon _k)}(t_i)-L^{(\varepsilon _k)}(t_{i-1}) \rightarrow L(t_i)-L(t_{i-1})$ a.s. in $({\mathbb {D}},\Vert \cdot \Vert )$ as $k \rightarrow \infty$, for any $i=2,\ldots ,K$. By Lemma A.3, $L(t_i)-L(t_{i-1}),i=2,\ldots ,K$ are independent, since $L^{(\varepsilon _k)}(t_i)-L^{(\varepsilon _k)}(t_{i-1}),i=2,\ldots ,K$ are independent for any k.

To verify property (iii), we observe that for any $t_1<t_2$ and $s \in [0,1]$,

$$\begin{aligned} L(t_2,s)-L(t_1,s)={\overline{L}}(t_1,s)-{\overline{L}}(t_2,s)=\sum _{j\ge 0} \int _{(t_1,t_2] \times [0,s] \times \Gamma _j} z M(\mathrm{d}u_1,\mathrm{d}u_2,\mathrm{d}z) \quad \text{ a.s. } \end{aligned}$$

From this, it can be proved that $L(t_2)-L(t_1)=\{L(t_2,s)-L(t_1,s)\}_{s \in [0,1]}$ is an $\alpha$-stable Lévy motion with characteristic function

$$\begin{aligned} E[\mathrm{e}^{iu (L(t_2,s)-L(t_1,s))}]=\exp \left\{ (t_2-t_1)s \int _{{\mathbb {R}}} (\mathrm{e}^{iuz}-1)\nu _{\alpha ,p}(\mathrm{d}z)\right\} , \quad \text{ for } \text{ all } \ u \in {\mathbb {R}}. \end{aligned}$$

On the other hand, $L(t_2-t_1)=\{L(t_2-t_1,s)\}_{s \in [0,1]}$ is also an $\alpha$-stable Lévy motion with the same characteristic function. Hence, $L(t_2)-L(t_1) {\mathop {=}\limits ^{d}}L(t_2-t_1)$.

To verify property (iv), we assume first that $t=1$. The process $L(1)=\{L(1,s)\}_{s \in [0,1]}$ is an $\alpha$-stable Lévy motion, so it is an $\alpha$-stable process. It follows that for any $s_1,\ldots ,s_m \in [0,1]$, $(L(1,s_1),\ldots ,L(1,s_m))$ has an $\alpha$-stable distribution in ${\mathbb {R}}^m$ with Lévy measure $\mu _{s_1,\ldots ,s_m}$:

$$\begin{aligned} E\big (\mathrm{e}^{iu_1 L(1,s_1)+\ldots +iu_m L(1,s_m)} \big )=\exp \left\{ \int _{{\mathbb {R}}^m}(\mathrm{e}^{iu \cdot y}-1)\mu _{s_1 \ldots ,s_m}(\mathrm{d}y)\right\} , \quad u=(u_1,\ldots ,u_m)\in {\mathbb {R}}^m. \end{aligned}$$

In particular, $(L(1,s_1),\ldots ,L(1,s_m))$ is regularly varying with limiting measure $\mu _{s_1,\ldots ,s_m}$.

On the other hand, by Lemma 2.1 of [14], L(1) is regularly varying in ${\mathbb {D}}$ (in the sense of Definition 1.4), i.e., $L(1) \in RV(\{a_n\},{\overline{\nu }},{\overline{{\mathbb {D}}}}_0)$ for a boundedly finite measure ${\overline{\nu }}$ on ${\overline{{\mathbb {D}}}}_0$ with ${\overline{\nu }} ({\overline{{\mathbb {D}}}}_0 \backslash T({\mathbb {D}}_0))=0$. Moreover, ${\overline{\nu }}=c\nu _{\alpha } \times \Gamma _1$ for some $c>0$ and a probability measure $\Gamma _1$ on ${\mathbb {S}}_{{\mathbb {D}}}$. Let $\nu ={\overline{\nu }} \circ S^{-1}$, where $S:(0,\infty ) \times {\mathbb {S}}_{{\mathbb {D}}} \rightarrow {\mathbb {D}}_0$ is the inverse of the map T, i.e., $S(r,z)=rz$. By Theorem 8 of [13], $(L(1,s_1),\ldots ,L(1,s_m))$ is regularly varying with limiting measure $\nu _{s_1,\ldots ,s_m}=\nu \circ \pi _{s_1,\ldots ,s_m}^{-1}$. By the unicity of the limit, $\mu _{s_1,\ldots ,s_m}=\nu _{s_1,\ldots ,s_m}$. Finally, property (iv) for general t follows using the scaling property of $\mu _{s_1,\ldots ,s_m}$ and the fact that $\{L(t,s)\}_{s \in [0,1]} {\mathop {=}\limits ^{d}}\{t^{1/\alpha }L(1,s)\}_{s \in [0,1]}$. $\square$

In relation to the simulation procedure described in Example 5.1, we include the following result, which can be proved using the same argument as in Section 48 of [19].

Theorem B.2

Let $\xi ,(\xi _{ij})_{i,j \ge 1}$ be i.i.d. regularly varying random variables, i.e.,

$$\begin{aligned} n P\left( \frac{\xi }{a_n} \in \cdot \right) {\mathop {\rightarrow }\limits ^{v}} \nu _{\alpha ,p} \quad \text{ in } \quad {\overline{{\mathbb {R}}}}_0, \end{aligned}$$

for some $a_n \uparrow \infty$, $\alpha \in (0,2),\alpha \not =1$ and $p \in [0,1]$, where $\nu _{\alpha ,p}$ is given by (69). For any $t,s \in [0,1]$, let $T_{n,m}(t,s)=a_n^{-1}a_m^{-1} \sum _{i=1}^{[nt]} \sum _{j=1}^{[ms]} (\xi _{ij}-\mu )$, where $\mu =0$ if $\alpha <1$ and $\mu =E(\xi )$ if $\alpha >1$. Let $L=\{L(t,s)\}_{(t,s) \in [0,1]^2}$. Then

$$\begin{aligned} T_{n,m} {\mathop {\rightarrow }\limits ^{d}}L \quad in \quad {\mathbb {D}}([0,1]^2), \quad as \quad n,m\rightarrow \infty . \end{aligned}$$

A Result About Brownian Motion

In this section, we include a result about the Brownian motion which is used in Example 5.2. This result is probably well known. We include its proof since we could not find it in the literature.

Lemma C.1

Let $W=\{W(s)\}_{s \in [0,1]}$ be the Brownian motion. Then,

$$\begin{aligned} E\Big (\sup _{s \in [0,1]}|W(s)|^{\alpha }\Big )<\infty \quad for \ all \ \alpha >0. \end{aligned}$$

Proof

Let $W^+(t)=\max (W(t),0)$ and $W^{-}(t)=\max (-W(t),0)$. For any $x>0$,

$$\begin{aligned} \{\sup _{t \in [0,1]} |W(t)|>&\,x\} \subset \{\sup _{t \in [0,1]} W^+(t)>x/2\} \cup \{\sup _{t \in [0,1]} W^{-}(t)>x/2\}\\&=\{\sup _{t \in [0,1]} W(t)>x/2\} \cup \{\sup _{t \in [0,1]} (-W(t))>x/2\}. \end{aligned}$$

Note that $\{-W(t)\}_{t \in [0,1]} {\mathop {=}\limits ^{d}} \{W(t)\}_{t \in [0,1]}$. By the reflection principle for the Brownian motion,

$$\begin{aligned} P(\sup _{t \in [0,1]} |W(t)|>x)&\le 2 P(\sup _{t \in [0,1]} W(t)>x/2)\\&=4P(W(1)>x/2) \le 4P(|W(1)|>x/2). \end{aligned}$$

Hence,

$$\begin{aligned} E\Big (\sup _{t \in [0,1]} |W(t)|^{\alpha }\Big )&=\int _0^{\infty }P(\sup _{t \in [0,1]} |W(t)|>x^{1/\alpha })\mathrm{d}x \le 4 \int _0^{\infty }P(|W(1)|>(x/2)^{1/\alpha })\mathrm{d}x\\&=8E|W(1)|^{\alpha }<\infty . \end{aligned}$$

$\square$