On a Multivariate Strong Renewal Theorem

Abstract

This paper takes the so-called probabilistic approach to the strong renewal theorem (SRT) for multivariate distributions in the domain of attraction of a stable law. A version of the SRT is obtained that allows any kind of lattice–nonlattice composition of a distribution. A general bound is derived to control the so-called small-n contribution, which arises from random walk paths that have a relatively small number of steps but make large cumulative moves. The asymptotic negligibility of the small-n contribution is essential to the SRT. Applications of the SRT are given, including some that provide a unified treatment to known results but with substantially weaker assumptions.

Appendices

Appendix 1: Proofs for the Lattice–Nonlattice Composition

For a set E in a Euclidean space, denote by $\mathrm {span}(E)$ the linear subspace spanned by elements of E. If M is a matrix, denote by $\mathrm {csp}(M)$ the linear subspace spanned by the column vectors of M. If the rank of M is equal to its number of columns, then M is said to be of full column rank.

Proof of Proposition 2.1

Let $\Gamma = \Gamma _X$, where

$$\begin{aligned} \begin{aligned} \Gamma _X&= \{v\in \mathbb {R}^d: \text {there is}\,\, a\in \mathbb {R}\,\, \text {such that}\,\, \langle {v},{X}\rangle \in a + {\mathbb {Z}}\text { a.s.}\}\\&= \{v\in \mathbb {R}^d: |\varphi _X(2\pi v)|=1\}. \end{aligned} \end{aligned}$$

(8.1)

As in the proof of [31], T6.1, $\Gamma$ plays an important role. The first step is to show that it is a lattice. The first line in (8.1) implies that $\Gamma$ is an additive subgroup of $\mathbb {R}^d$, so it suffices to show that 0 is not a cluster point of $\Gamma$. Let $u_n\in \Gamma$ such that $u_n\rightarrow 0$. Let $V_n = \mathrm {span}(u_i,\,i\ge n)$. Since $\mathbb {R}^d\supset V_1\supset V_2 \supset \cdots$, there is k, such that $V_k = V_{k+1} = \cdots$. Let $X_*$ be i.i.d. $\sim X$ and $\xi = X - X_*$. Then almost surely, $\langle {u_n},{\xi }\rangle \in {\mathbb {Z}}$ for all n. Since $\langle {u_n},{\xi }\rangle \rightarrow 0$, this implies $\langle {u_n},{\xi }\rangle =0$ for $n\ge k$ large enough. But then $\xi \in V^\perp _n = V^\perp _k$. By assumption, $\xi$ is not concentrated in any linear subspace of dimension $d-1$. Then $V_k = \{0\}$, giving $u_k = u_{k+1} = \cdots = 0$, so 0 is not a cluster point of $\Gamma$.

Let $V=\mathrm {span}(\Gamma )$ and $r=\dim (V)$. Suppose $r\ge 1$. By a fundamental theorem on lattices (cf. [34], Lemma 3.4), $\Gamma = M{\mathbb {Z}}^r$ for some $M\in \mathbb {R}^{d\times r}$ of rank r. Let $v_1, \ldots , v_{r}$ be the column vectors of M and $a=(a_1, \ldots , a_{r})$ such that $\langle {v_i},{X}\rangle \in a_i + {\mathbb {Z}}$. Then $X\in \Lambda = \{x\in \mathbb {R}^d: M'x \in a + {\mathbb {Z}}^r\}$. By $\pi _V(x) = H M' x$, where $H = M(M'M)^{-1}$, $\pi _V(X) = H M' X \in H(a +{\mathbb {Z}}^r)$, so $\pi _V(X)$ is lattice. If $v \in \mathbb {R}^d\setminus V$, then $v\not \in \Gamma$, so $|\varphi _X(2\pi v)|<1$. Thus V has the property stated in (1). To continue, assume the following result is true for now.

Lemma 8.1

There is $K\in {\mathbb {Z}}^{r\times r}$ with $\det K=\pm 1$ such that $K a = (0, \ldots , 0, z_\nu , z_{\nu +1}, \ldots , z_r)$, where $z_\nu \in [0,\infty )\cap \mathbb {Q}$ and $z_{\nu +1}, \ldots , z_r\in (0,\infty )\setminus \mathbb {Q}$ are rationally independent.

Let $Q\in \mathbb {R}^{d\times (d-r)}$ be of full column rank such that $Q' M = O$. Define

$$\begin{aligned} T = \begin{pmatrix} K M' \\ Q'\end{pmatrix},\quad Y = K M'X, \quad Z = Q' X, \quad \beta _i = z_i - \left\lfloor z_i\right\rfloor , \ i=\nu , \ldots , r. \end{aligned}$$

(8.2)

Then $TX = (Y, Z)$, $\beta _\nu \in \mathbb {Q}\cap [0,1)$, and $\beta _{\nu +1}$, ..., $\beta _r\in (0,1) \setminus \mathbb {Q}$ are rationally independent. Put $\beta =(0,\ldots , 0, \beta _\nu , \beta _{\nu +1}, \ldots , \beta _r)$. Since $Y\in K (a + {\mathbb {Z}}^r) = \beta + {\mathbb {Z}}^r$, T has the property stated in (3).

To show T has the property stated in (2), if $u = (k, 0)\in {\mathbb {Z}}^r\times \{0\}$, then by $\langle {u},{T X}\rangle = \langle {k},{Y}\rangle \in \langle {k},{\beta }\rangle + {\mathbb {Z}}$, $|\varphi _{T X}(2\pi u)|=1$. Conversely, if $|\varphi _{T X}(2\pi u)|=1$, then $|\varphi _X(2\pi T' u)|=1$, so $T' u = M k\in \Gamma$ for some $k\in {\mathbb {Z}}^r$. Write $u = (w, v)$ with $w\in \mathbb {R}^r$. By (8.2), $H'T' u= H'(M K' w + Q v) = K' w$. As the LHS is also $H'M k=k$, $K' w= k$, giving $w = (K')^{-1} k\in {\mathbb {Z}}^r$. On the other hand, $(\mathrm {Id}_d -M H')T' u = (\mathrm {Id}_d - M H') (M K' w + Q v) = Qv$ and the LHS is also $(\mathrm {Id}_d - MH')M k = 0$. Thus $Q v=0$. Since Q is of full column rank, $v=0$ and hence $u = (w, 0)\in {\mathbb {Z}}^r\times \{0\}$.

So far it has been assumed that $r=\dim (V)>0$. If $r=0$, then $\Gamma = V = \{0\}$. Consequently, $|\varphi _X(2\pi v)|<1$ for $v\ne 0$ and $T = \mathrm {Id}_d$ has the property stated in (2)–(3).

To show that V is unique, let W be a linear subspace such that $\pi _W(X)$ is lattice and $|\varphi _X(2\pi v)|<1$ for $v\not \in W$. By definition, $\Gamma \subset W$, so $V = \mathrm {span}(\Gamma ) \subset W$. If $V\ne W$, then $W\cap V^\perp \ne \emptyset$ and $\pi _{W\cap V^\perp }(X) = \pi _{W\cap V^\perp } (\pi _W(X))$ is lattice. It follows that there is $0\ne u\in W\cap V^\perp$, such that $\langle {u},{X}\rangle = \langle {u},{\pi _{W\cap V^\perp }(X)}\rangle \in c + {\mathbb {Z}}$ for some c. But then $u\in \Gamma \subset V$. The contradiction shows $V=W$ and hence the uniqueness of V.

To show that $\nu$, r, and q are unique, suppose $0\le \mu \le s\le d$, $q_*\in \mathbb {N}$, and $B\in \mathbb {R}^{d\times d}$ is nonsingular, such that

$$\begin{aligned} |\varphi _{B X}(2\pi u)|=1\Longleftrightarrow u\in {\mathbb {Z}}^s \times \{0\}, \end{aligned}$$

(8.3)

and $B X = (Y_*, Z_*)$ with $Y_*\in \gamma + {\mathbb {Z}}^s$, $Z^* \in \mathbb {R}^{d-s}$, and $\gamma = (0,\ldots , 0, \gamma _\mu , \gamma _{\mu + 1}, \ldots , \gamma _s)$, where $\gamma _\mu = p_*/q_*$ with $0\le p_* < q_*$ being coprime, and $\gamma _{\nu +1}, \ldots , \gamma _s\in (0,1) \setminus \mathbb {Q}$ are rationally independent. By $B' u\in \Gamma \Longleftrightarrow |\varphi _{BX}(2\pi u)|=1$ and (8.3), $\Gamma = B'({\mathbb {Z}}^s \times \{0\})$. Since B is nonsingular, a comparison of dimensions yields $s=\dim (V)=r$.

Let $r\ge 1$, otherwise nothing remains to be shown. Then by $\Gamma = T'({\mathbb {Z}}^r\times \{0\}) = B'({\mathbb {Z}}^r\times \{0\})$, $T'_1{\mathbb {Z}}^r = B'_1{\mathbb {Z}}^r$, where $T_1, B_1\in \mathbb {R}^{r\times d}$ consist of the first r rows of T and B, respectively. Then there are $J, J_*\in {\mathbb {Z}}^{r\times r}$ such that $T_1 = J B_1$ and $J_* T_1 = B_1$, giving $J_* J B_1 = J_* T_1 = B_1$. Since the rows of $B_1$ are linearly independent, $J_* J=\mathrm {Id}_r$. Thus $J^{-1} = J_*$. On the other hand, $B_1 X = Y_*$. Then $T_1 X = J B_1 X = J Y_*\in J(\gamma + {\mathbb {Z}}^r) = J\gamma + J{\mathbb {Z}}^r = J\gamma + {\mathbb {Z}}^r$. Since $T_1 X = Y \in \beta + {\mathbb {Z}}^r$, then $\beta - J\gamma = (b_1, \ldots , b_{r}) \in {\mathbb {Z}}^r$. Let $J=(g_{ij})$. Each $\beta _i = c_i + g_{i,\mu +1} \gamma _{\mu +1} + \cdots + g_{ir}\gamma _r$ with $c_i = b_i + g_{i1} \gamma _1 + \cdots + g_{i\mu }\gamma _\mu = b_i + g_{i\mu }\gamma _\mu \in {\mathbb {Z}}$. Since $\beta _i$, $i>\nu$, are rationally independent, this leads to $\mu \le \nu$. Likewise, $\nu \le \mu$. Thus $\mu = \nu$. For $i\le \nu$, $g_{i,\nu +1} \gamma _{\nu +1} + \cdots + g_{ir} \gamma _r = \beta _i - c_i\in {\mathbb {Z}}$. By rational independence of $\gamma _{\nu +1}$, ..., $\gamma _r$, $\beta _i = c_i$. In particular, $\beta _\nu = k\gamma _\nu - l$ with $k = g_{\nu \nu }$ and $l = -b_\nu$. Likewise, $\gamma _\nu = k_* \beta _\nu - l_*$ with $k_*, l_*\in {\mathbb {Z}}$. As a result $(kk_*-1) \beta _\nu = k(\gamma _\nu + l_*) - \beta _\nu = l + k l_*\in {\mathbb {Z}}$. Since $\beta _\nu = p/q$ with $0\le p <q$ being coprime, $q\,|\,k k_* -1$, so $k_*$ and q are coprime. Then $\gamma _\nu =p_*/q$ with $p_* = k_* p - l_*q$ being coprime with q. Thus $q_* = q$, completing the proof. $\square$

Proof of Proposition 2.2

(1) Let $K\xi = (\zeta _1, \ldots , \zeta _{\nu -1}, p + q \zeta _\nu )$, where $K\in {\mathbb {Z}}^{\nu \times \nu }$ with $\det K=\pm 1$, $0\le p < q$ are coprime, and $\zeta = (\zeta _1, \ldots , \zeta _{\nu })\in {\mathbb {Z}}^\nu$ is strongly aperiodic. By $\xi \in {\mathbb {Z}}^\nu$, $\langle {t},{\xi }\rangle \in {\mathbb {Z}}$ for $t\in {\mathbb {Z}}^\nu$. Conversely, if $\langle {t},{\xi }\rangle \in {\mathbb {Z}}$, then letting $s= (K')^{-1} t$, $\langle {s},{K\xi }\rangle = s_1\zeta _1 + \cdots + s_{\nu -1} \zeta _{\nu -1} + q s_\nu \zeta _\nu + p s_\nu \in {\mathbb {Z}}$. The strong aperiodicity of $\zeta$ implies $s_1, \ldots , s_{\nu -1}, q s_\nu , p s_\nu \in {\mathbb {Z}}$. Since p and q are coprime, then $s_\nu \in {\mathbb {Z}}$. Thus $s\in {\mathbb {Z}}^\nu$ and $t = K' s\in {\mathbb {Z}}^\nu$. This shows $\xi$ is aperiodic.

Conversely, let $\xi$ be aperiodic. Define $\Gamma = \Gamma _\xi$ as in (8.1). Then $\Gamma$ is a lattice. Since ${\mathbb {Z}}^\nu \subset \Gamma$, by Smith normal form ( [34], Th. 3.7), there are linearly independent $u_1, \ldots , u_{\nu }\in \Gamma$ and integers $1\le n_1 \le \cdots \le n_\nu$ with $n_i\,|\,n_{i+1}$, such that, letting $M = (u_1, \ldots , u_{\nu })$ and $D = \mathrm {diag}(n_1, \ldots , n_{\nu })$,

$$\begin{aligned} \Gamma = M{\mathbb {Z}}^\nu , \quad {\mathbb {Z}}^\nu = M D {\mathbb {Z}}^\nu . \end{aligned}$$

(8.4)

By $u_i\in \Gamma$, $\langle {u_i},{\xi }\rangle \in s_i + {\mathbb {Z}}$ for some $s_i$. In matrix form, $M'\xi \in s + {\mathbb {Z}}^\nu$, where $s=(s_1, \ldots , s_{\nu })$. Define $K = D M'$, $Z = K\xi$, and $b = D s$. From the second identity in (8.4), ${\mathbb {Z}}^\nu = K'{\mathbb {Z}}^\nu$, giving $K, K^{-1}\in {\mathbb {Z}}^{\nu \times \nu }$. Then $Z\in {\mathbb {Z}}^\nu$ and is aperiodic. Meanwhile, $Z = D M'\xi \in D(s + {\mathbb {Z}}^\nu ) = b + D {\mathbb {Z}}^\nu$. Then from $Z\in (b + D{\mathbb {Z}}^\nu )\cap {\mathbb {Z}}^\nu$ and $D {\mathbb {Z}}^\nu \subset {\mathbb {Z}}^\nu$, $b\in {\mathbb {Z}}^\nu$. Let $Z_1, Z_2, \ldots$, $Z'_1, Z'_2, \ldots$ be i.i.d. $\sim Z$. For $m,n\ge 0$, $S_m(Z) - S_n(Z')\in {\mathbb {Z}}b + D{\mathbb {Z}}^\nu \subset {\mathbb {Z}}^\nu$. By aperiodicity of Z, for every standard base vector $e_i$ of $\mathbb {R}^\nu$, there are m and n such that $\mathbb {P}\{S_m(Z) - S_n(Z') = e_i\}>0$ ([31], p. 20). This yields $e_i \in {\mathbb {Z}}b + D {\mathbb {Z}}^\nu$. As a result, ${\mathbb {Z}}b+ D{\mathbb {Z}}^\nu = {\mathbb {Z}}^\nu$. Let $s_i\in {\mathbb {Z}}$ and $v_i = (v_{i1}, \ldots , v_{i\nu })\in {\mathbb {Z}}^\nu$, such that $b s_i + D v_i = e_i$. Write $b = (b_1, \ldots , b_{\nu })$. By comparing the coordinates,

$$\begin{aligned} b_i s_i + n_i v_{ii} = 1, \quad b_j s_i + n_j v_{ij}=0,\ j\ne i. \end{aligned}$$

Thus, each pair of $b_i$ and $n_i$ are coprime. For $j>i$, as $n_j\ne 0$, $n_j\,|\,b_j s_i$, so $n_j \,|\,s_i$. Then by $b_i (s_i/n_j) n_j + n_i v_{ii}=1$, $n_j$ and $n_i$ are coprime. By $n_i\,|\,n_j$, this gives $n_i=1$. As a result, $n_1 = \cdots = n_{\nu -1}=1$. Put $q=n_\nu$ and let $0\le p<q$ such that $q\,|\,(b_\nu - p)$. Let $\zeta = D^{-1}(Z - p e_\nu )$. By $Z\in b + D{\mathbb {Z}}^\nu$ and $b- p e_\nu \in D{\mathbb {Z}}^\nu$, $\zeta \in {\mathbb {Z}}^\nu$. If $\langle {t},{\zeta }\rangle \in s+{\mathbb {Z}}$, where $t\in \mathbb {R}^\nu$ and $s\in \mathbb {R}$, then $\langle {M t},{\xi }\rangle = \langle {t},{M'\xi }\rangle = \langle {t},{D^{-1} Z}\rangle \in c + {\mathbb {Z}}$ with $c = s + p\langle {t},{D^{-1} e_\nu }\rangle$. Then $M t\in \Gamma = M {\mathbb {Z}}^\nu$, so $t\in {\mathbb {Z}}^\nu$. Thus $\zeta$ is strongly aperiodic. By $K\xi = Z = p e_\nu + D\zeta$, the proof is complete.

(2) Let $\zeta = L - \beta _\nu e_\nu$. Then $\zeta \in {\mathbb {Z}}^\nu$. Since for $u\in \mathbb {R}^d$, $|\varphi _{TX}(2\pi u)|=1\Longleftrightarrow u\in {\mathbb {Z}}^\nu \times \{0\}$, then for $v\in \mathbb {R}^\nu$, $|\varphi _\zeta (2\pi v)| = |\varphi _L(2\pi v)|=1\Longleftrightarrow v\in {\mathbb {Z}}^\nu$, so $\zeta$ is strongly aperiodic. Then by (1), $D L= (\zeta _1, \ldots , \zeta _{\nu -1}, p + q \zeta _\nu )\in {\mathbb {Z}}^\nu$ is aperiodic. $\square$

Proof of Lemma 8.1

Recall $\mathbb {Q}$, $\mathbb {R}$, and their quotient $\mathbb {R}/\mathbb {Q}$ are vector spaces over the field $\mathbb {Q}$, Let ${\bar{a}} = ({\bar{a}}_1, \ldots , {\bar{a}}_{r})$ with ${\bar{a}}_i = a_i + \mathbb {Q}\in \mathbb {R}/\mathbb {Q}$. First, if $\bar{a}\ne 0$, then there are linearly independent ${\bar{u}}_1, \ldots , {\bar{u}}_{s}\in \mathbb {R}/\mathbb {Q}$, $1\le s\le r$, such that ${\bar{a}}= A {\bar{u}}$, where $A \in \mathbb {Q}^{r\times s}$ is of full column rank and ${\bar{u}}=(\bar{u}_1, \ldots , \bar{u}_{s})$. Equivalently, $a - A u\in \mathbb {Q}^r$. Note that $u_i$ are rationally independent. By multiplying A by a large $m\in \mathbb {N}$ and dividing u by m, A can be assumed to be in ${\mathbb {Z}}^{r\times s}$. It is known that there are $P\in {\mathbb {Z}}^{r\times r}$ and $R\in {\mathbb {Z}}^{s\times s}$ with $|\det P| = |\det R|= 1$, such that $P A = \left( {\begin{array}{c}D\\ O\end{array}}\right) R$, where $D=\mathrm {diag}(d_1, \ldots , d_{s})$ with $d_i\in \mathbb {N}$ and $d_i\,|\,d_{i+1}$ (cf. [23], Th. III.5). Let $D R u = v$. Then $P (a -A u) = P a - (v, 0)\in \mathbb {Q}^r$, so $P a = ({\tilde{v}}, w)$, where ${\tilde{v}} = v + y$ for some $y\in \mathbb {Q}^s$ and $w \in \mathbb {Q}^{r-s}$. The coordinates of ${\tilde{v}}$ are rationally independent. On the other hand, similar to A, there is $M\in {\mathbb {Z}}^{(r-s)\times (r-s)}$ with $\det M = \pm 1$, such that $M w = (q, 0, \ldots , 0)\in \mathbb {Q}^{r-s}$ with $q\ge 0$. Then gives $K_0 a = (\tilde{v}, q, 0, \ldots , 0)$. By permuting the coordinates, the lemma follows. Finally, if ${\bar{a}}=0$, then $a\in \mathbb {Q}^r$. Following the treatment of the above w, the result follows. $\square$

Appendix 2: Proofs Regarding Distributions in the Domain of Attraction

Proof of (3.1) and (3.2)

Let $X\sim F\in \mathcal {D}(\alpha )$. If $\alpha =2$, then (3.2) is part of [28], Th. 4.1. For any $c>1$, $V_X(s) \le V_X(c s) \le V_X(s) + c^2 s^2 q_X(s)$, which by (3.2) gives $V_X(c s)/V_X(s)\rightarrow 1$ as $s\rightarrow \infty$. Then (3.1) follows. If $\alpha \in (0,2)$, then Th. 4.2 of [28] states that $q_X\in \mathcal {R}_{-\alpha }$, which leads to both (3.1) and (3.2) (cf. [1], Th. 1.6.4). $\square$

Proof of (3.4)

For the univariate case, see [1], p. 347. For the multivariate case, first, let $\alpha \in (0,2)$. The proof of Th. 4.2 of [28] shows that a choice of $a_n$ is the infimum of all s such that

$$\begin{aligned} \mathbb {P}\{|X|>s, X/|X|\in E\} \le \gamma (E)/n \le \mathbb {P}\{|X|\ge s, X/|X|\in E\}, \end{aligned}$$

where $\gamma$ is a nonzero measure on $S^{d-1}$ and E is any fixed subset of $S^{d-1}$ with $\gamma (E)>0$. By Th. 14.10 of [29], $\gamma$ is finite. Letting $E = S^{d-1}$ and $c = \gamma (S^{d-1})$, it follows that $a_n$ can be any s satisfying $q_X(s) \le c/n \le q_X(s-)$. Then by (3.2) and (3.3), $a_n$ can (also) be taken to be any sequence such that $A(a_n)\sim cn$.

Let $\alpha =2$ and $b(u) = \langle {u},{\Sigma u}\rangle$, where $\Sigma$ is the covariance matrix of the limiting normal distribution. If $\mathbb {E}|X|^2<\infty$, then (3.4) follows from the central limit theorem. Suppose $\mathbb {E}|X|^2 = \infty$. By Th. 2.4 of [28], $a_n$ can be any sequence such that for any $\epsilon >0$, (i) $n q_X(\epsilon a_n)\rightarrow 0$ and (ii) $(n/a^2_n) [m_V(\epsilon a_n,u) - \langle {c_V(\epsilon a_n)},{u}\rangle ^2]\rightarrow b(u)$ for any $u\in S^{d-1}$. Since $|c_V(s)|^2 = o(V_X(s))$ as $s\rightarrow \infty$ ([28], (4.5)), by (3.1) and (4.6), (ii) is equivalent to $n/A(a_n) \rightarrow \sum _i b(e_i)$. Once (ii) is satisfied, by (3.2), (i) is satisfied. Then the claim on $a_n$ follows. $\square$

Proof of (3.5)

For the univariate case, see [1], p. 347. For the multivariate case, according to the last comment on p. 190 in [28], $b_n$ can be taken to be $(n/a_n) c_X(t a_n) + \gamma$, where $\gamma$ is any constant vector, and $t>0$ is any fixed number such that $\{|x|=t\}$ has measure 0 under the Lévy measure of the limiting stable law. From the characterization of the Lévy measure (cf. [28], (3.4)–(3.5)), t can be any positive number. It follows that any $b_n$ satisfying (3.4) must be of the form $(n/a_n) c_X(a_n) + \gamma + \epsilon _n$ for some constant vector $\gamma$, where $\epsilon _n\rightarrow 0$ as $n\rightarrow \infty$. This implies (3.5). $\square$