1 Introduction and Results

The p-adic field \({\mathbf {Q}}_p\) is identified with the set of formal series \(x=\sum _{i=-m}^\infty \alpha _i p^i\) with integers m and \(\alpha _i=0, 1, \dots , p-1\), equipped with the p-adic norm \(\left| \sum _{i=-m}^\infty \alpha _i p^i \right| _p=p^m\) if \(\alpha _{-m} \ne 0\), and \(|0|_p=0\). For a p-adic number a and an integer l, let \(B\left( a, p^l \right) :=\left\{ x \in {\mathbf {Q}}_p \ | \ |x-a|_p \le p^l \right\} \) be the ball of radius \(p^l\) centered at a. Let \(B_l:=B\left( 0, p^l \right) \) denote the ball centered at the origin, and \(H_l:=B_l \setminus B_{l-1}\) the sphere. By the ultra-metric property \(|x+y|_p \le \max \{|x|_p, |y|_p \}\), all balls and spheres are compact and open. For the p-adic field and related fundamental subjects, we can refer to [4].

The p-adic field is a separable and complete metric space where a lot of standard methods of stochastic analysis are available. On the other hand, the ultra-metric property brings some unique phenomena different from the case of Euclidean spaces. The behavior of sums of p-adic-valued random variables is one of the interesting topics which has been studied since 1990s. Analysis based on p-adic-valued measures has been developed with reference to mathematical physics, and sums of random variables are discussed in this framework [2, 3]. On the other hand, limit theorems on p-adic numbers in the context of real-valued measures have been established. A p-adic analog of the law of large numbers is covered in [11]. The p-adic central limit theorem is concerned in [5, 9], and [8] gives estimates of convergence. As refinements of the central limit theorem, [10] derives a p-adic analog of the law of iterated logarithms, and this article proceeds to estimates for large deviations. Besides, related works on abstract probabilities are also remarkable [7].

Let \(\xi _i (i=1, 2, \dots )\) be independent identically distributed (I.I.D.) random variables on the p-adic field. We suppose its law is invariant by rotations around the origin; namely, \(u\xi _i\) has the same law as \(\xi _i\) for any p-adic number u satisfying \(|u|_p=1\). We also suppose its tail probabilities satisfy

$$\begin{aligned} T_1(m):=P \left( |\xi _i|_p \ge p^m \right) =p^{-\alpha m} L(m), \end{aligned}$$

for a constant positive number \(\alpha \) and a function L on \({\mathbf {Z}}\) such that \(\lim _{m \rightarrow \infty }\frac{L(m+1)}{L(m)}=1\). Define a sequence \(N(n):=\frac{p^{2\alpha }(p-1)}{p^{\alpha +1}-1} T_1(n)^{-1}\) for \(n=1, 2, \dots \), and let [N(n)] be its integer part. Under these assumptions, the law of the scaled sum \(p^n \sum _{i=1}^{[N(n)]} \xi _i\) converges to a rotation-symmetric \(\alpha \)-semi-stable law as \(n \rightarrow \infty \) [9].

Let \(c_n (n=1, 2, \dots )\) be a sequence of nonnegative integers. Concerning the growth rate of the scaled sum, \(\limsup _{n \rightarrow \infty } \left| p^{n+c_n}\sum _{i=1}^{[N(n)]} \xi _i \right| _p =0\) if \(c_n\) diverges faster than \(\tilde{c}_n:=\left[ \frac{\log n}{\alpha \log p} \right] \), and \(=+\infty \) if \(c_n\) is slower than \(\tilde{c}_n\). At the critical order \(c_n=\tilde{c}_n\), the result differs by the rate of convergence of \(\frac{L(m+1)}{L(m)}\) to 1 [10].

In this article, we deal with general sequences \(c_n (n=1, 2, \dots )\) satisfying \(\lim _{n \rightarrow \infty }c_n=+\infty \) and give estimates for large deviations of the law of the scaled sum. Let \(P_n\) be the law of the scaled sum \(p^{n+c_n}\sum _{i=1}^{[N(n)]} \xi _i\) and put \(\theta _-:=\liminf _{n \rightarrow \infty } \frac{c_n}{n}\), \(\theta _+ :=\limsup _{n \rightarrow \infty } \frac{c_n}{n}\).

Proposition 1

For any compact open set K in \({\mathbf {Q}}_p\) including the origin,

$$\begin{aligned} \liminf _{n \rightarrow \infty } \frac{1}{n} \log P_n\left( K^c \right)= & {} -\alpha \theta _+ \log p, \\ \limsup _{n \rightarrow \infty } \frac{1}{n} \log P_n\left( K^c \right)= & {} -\alpha \theta _- \log p, \end{aligned}$$

where \(K^c\) is the complement of K.

Proposition 2

  1. i.

    For any closed set B in \({\mathbf {Q}}_p\) not including the origin,

    $$\begin{aligned} \liminf _{n \rightarrow \infty } \frac{1}{n} \log P_n(B)\le & {} -\alpha \theta _+ \log p, \\ \limsup _{n \rightarrow \infty } \frac{1}{n} \log P_n(B)\le & {} -\alpha \theta _- \log p. \end{aligned}$$
  2. ii.

    For any open set A in \({\mathbf {Q}}_p\) including the origin,

    $$\begin{aligned} \lim _{n \rightarrow \infty } \log P_n(A)=0. \end{aligned}$$

In order to determine the asymptotics of \(P_n(B)\) for sets B off the origin, we require an additional assumption to the rate of convergence \(\frac{L(m+1)}{L(m)} \rightarrow 1\). Define \(\delta _n:=\sup _{m \ge n} \left| 1-\frac{L(m+1)}{L(m)} \right| \) for \(n \ge 1\).

Theorem 1

Assume that either of the following two conditions is satisfied :

  1. i.

    \(\theta _-=\theta _+=+\infty \),

  2. ii.

    \(\limsup _{n \rightarrow \infty }c_n \log \frac{1+\delta _n}{1-\delta _n}<\alpha \log p\).

Then for any \(a \in {\mathbf {Q}}_p\), \(a \ne 0\), and any integer l such that \(0 \notin B\left( a, p^l \right) \),

$$\begin{aligned} \liminf _{n \rightarrow \infty } \frac{1}{n} \log P_n\left( B \left( a,p^l \right) \right)= & {} -\alpha \theta _+ \log p, \\ \limsup _{n \rightarrow \infty } \frac{1}{n} \log P_n\left( B \left( a,p^l \right) \right)= & {} -\alpha \theta _- \log p. \end{aligned}$$

By this result, we can discuss the large deviation principle of the sequence of probability measures \(P_n\). For a general theory of the large deviation principle for a sequence of random variables or probability measures on a metric space, we can refer to [1].

Corollary 1

Under the assumption of Theorem 1, the distributions \(P_n (n=1, 2, \dots )\) satisfy the large deviation principle if and only if the limit \(\theta :=\lim _{n \rightarrow \infty } \frac{c_n}{n} \in [0, +\infty ]\) exists. If that is the case, the rate function is given by

$$\begin{aligned} I(x)=\left\{ \begin{array}{ll} 0, &{} x=0, \\ \alpha \theta \log p, &{} x \ne 0, \end{array} \right. \end{aligned}$$

and I is good only for the case \(\theta =+\infty \).

2 Tail Probabilities of the Sum of I.I.D.

For \(n \ge 1\) and integers m, let \(T_n(m):=P\left( \left| \sum _{i=1}^n \xi _i \right| _p \ge p^m \right) \) be the tail probabilities of the sum of \(\xi _i\). As a preparation for a formula for \(T_n(m)\), we shall derive an inductive relation of \(U_n(m):=P\left( \left| \sum _{i=1}^n \xi _i \right| _p = p^m \right) =T_n(m)-T_n(m+1)\).

Lemma 1

$$\begin{aligned} U_n(m)= & {} U_1(m)(1-T_{n-1}(m))+U_{n-1}(m)(1-T_1(m)) \\&+\frac{p-2}{p-1}U_{n-1}(m)U_1(m)+\sum _{k=m+1}^\infty p^{m-k}U_{n-1}(k)U_1(k). \end{aligned}$$

Proof

By the ultra-metric property, p-adic numbers x and y satisfy \(|x+y|_p=p^m\) if and only if one of the following exclusive events happens :

  1. i.

    \(|x|_p<p^m ,\ |y|_p=p^m\),

  2. ii.

    \(|x|_p=p^m, \ |y|_p<p^m\),

  3. iii.

    \(|x|_p=|y|_p \ge p^m, \ |x+y|_p=p^m\).

Putting \(x=\sum _{i=1}^{n-1} \xi _i\) and \(y=\xi _n\), by the independence of \(\xi _i\) we have

$$\begin{aligned} U_n(m)= & {} P\left( \left| \sum _{i=1}^{n-1} \xi _i \right| _p< p^m, \ |\xi _n|_p=p^m \right) +P\left( \left| \sum _{i=1}^{n-1} \xi _i \right| _p = p^m, \ |\xi _n|_p<p^m \right) \nonumber \\&+P\left( \left| \sum _{i=1}^{n-1} \xi _i \right| _p=|\xi _n|_p \ge p^m, \ \left| \sum _{i=1}^n \xi _i \right| _p = p^m \right) \nonumber \\= & {} U_1(m)\sum _{k=-\infty }^{m-1}U_{n-1}(k)+U_{n-1}(m)\sum _{k=-\infty }^{m-1}U_1(k) \nonumber \\&+\sum _{l=0}^\infty P\left( \left| \sum _{i=1}^{n-1} \xi _i \right| _p=|\xi _n|_p = p^{m+l}, \ \left| \sum _{i=1}^n \xi _i \right| _p = p^m \right) . \end{aligned}$$
(1)

For the case \(l=0\) in the last sum, the sphere \(H_m\) consists of \(p-1\) disjoint balls \(B\left( \alpha _{-m}p^{-m}, p^{m-1} \right) \) (\(\alpha _{-m}=1, 2, \dots , p-1\)) of radius \(p^{m-1}\). We can see that the event \(|x|_p=|y|_p=|x+y|_p=p^m\) happens if and only if \(x \in B\left( \alpha _{-m}p^{-m}, p^{m-1} \right) \) and \(-y \in B\left( \alpha ^\prime _{-m}p^{-m}, p^{m-1} \right) \) for some \(\alpha _{-m} \ne \alpha ^\prime _{-m}\). Since the balls \(B\left( \alpha _{-m}p^{-m}, p^{m-1} \right) \) are mapped to each other by rotations around the origin, and the law of \(\xi _i\) is invariant by the rotations,

$$\begin{aligned} P\left( \sum _{i=1}^{n-1} \xi _i \in B \left( \alpha _{-m}p^{-m}, p^{m-1} \right) \right) =\frac{U_{n-1}(m)}{p-1} \end{aligned}$$

and

$$\begin{aligned} P\left( -\xi _n \in B \left( \alpha ^\prime _{-m}p^{-m}, p^{m-1} \right) \right) =\frac{U_1(m)}{p-1} \end{aligned}$$

hold for all \(\alpha _{-m}\) and \(\alpha ^\prime _{-m}\). Therefore,

$$\begin{aligned}&P\left( \left| \sum _{i=1}^{n-1} \xi _i \right| _p=|\xi _n|_p= \left| \sum _{i=1}^n \xi _i \right| _p = p^m \right) \\&\quad =\sum _{\alpha _{-m}} \sum _{\alpha ^\prime _{-m} \ne \alpha _{-m}} P\left( \sum _{i=1}^{n-1} \xi _i \in B\left( \alpha _{-m}p^{-m}, p^{m-1} \right) \right) \\&\qquad \times P\left( -\xi _n \in B\left( \alpha ^\prime _{-m}p^{-m}, p^{m-1} \right) \right) \\&\quad =(p-1)(p-2) \frac{U_{n-1}(m)}{p-1} \frac{U_1(m)}{p-1} \\&\quad =\frac{p-2}{p-1} U_{n-1}(m)U_1(m). \end{aligned}$$

As for \(l \ge 1\), the sphere \(H_{m+l}\) consists of \(p^l (p-1)\) disjoint balls \(B\left( \sum _{i=-m-l}^{-m}\alpha _i p^i, p^{m-1} \right) \) (\(\alpha _{-m-l}=1, 2, \dots , p-1\), and \(\alpha _{-m-l+1}, \dots , \alpha _{-m}=0, 1, \dots , p-1\)) of radius \(p^{m-1}\). The event \(|x|_p=|y|_p=p^{m+l}\), \(|x+y|_p=p^m\) happens if and only if \(x \in B\left( \sum _{i=-m-l}^{-m}\alpha _i p^i, p^{m-1} \right) \) and \(-y \in B\left( \sum _{i=-m-l}^{-m-1}\alpha _i p^i+\alpha ^\prime _{-m}p^{-m}, p^{m-1} \right) \) for some \(\alpha _{-m-l}, \dots , \alpha _{-m}\), and \(\alpha ^\prime _{-m} \ne \alpha _{-m}\). Hence, we have

$$\begin{aligned}&P\left( \left| \sum _{i=1}^{n-1} \xi _i \right| =|\xi _n|_p = p^{m+l}, \ \left| \sum _{i=1}^n \xi _i \right| = p^m \right) \\&\quad =\sum _{\alpha _{-m-l}, \dots , \alpha _{-m}} \sum _{\alpha ^\prime _{-m} \ne \alpha _{-m}} P\left( \sum _{i=1}^{n-1} \xi _i \in B\left( \sum _{i=-m-l}^{-m}\alpha _i p^i, p^{m-1} \right) \right) \\&\qquad \times P\left( \xi _n \in B\left( \sum _{i=-m-l}^{-m-1}\alpha _i p^i+\alpha ^\prime _{-m}p^{-m}, p^{m-1} \right) \right) \\&\quad =p^l(p-1) \cdot (p-1) \frac{U_{n-1}(m+l)}{p^l (p-1)} \frac{U_1(m+l)}{p^l(p-1)} \\&\quad =p^{-l}U_{n-1}(m+l)U_1(m+l). \end{aligned}$$

Consequently, (1) leads to

$$\begin{aligned} U_n(m)= & {} U_1(m)\sum _{k=-\infty }^{m-1}U_{n-1}(k)+U_{n-1}(m)\sum _{k=-\infty }^{m-1}U_1(k) \\&+\frac{p-2}{p-1}U_{n-1}(m)U_1(m)+\sum _{l=1}^\infty p^{-l}U_{n-1}(m+l)U_1(m+l) \\= & {} U_1(m)(1-T_{n-1}(m))+U_{n-1}(m)(1-T_1(m)) \\&+\frac{p-2}{p-1}U_{n-1}(m)U_1(m)+\sum _{k=m+1}^\infty p^{m-k}U_{n-1}(k)U_1(k). \end{aligned}$$

\(\square \)

Proposition 3

$$\begin{aligned} T_n(m)=1-\left( 1-p^{-1} \right) \sum _{k=0}^\infty p^{-k} \left( 1- \frac{T_1(m+k)-p^{-1}T_1(m+k+1)}{1-p^{-1}} \right) ^n. \end{aligned}$$

Proof

Let us put \(V_n(m):=T_n(m)-p^{-1}T_n(m+1)\). Since \(T_n(m)=\sum _{l=m}^\infty U_n(l)\), Lemma 1 gives

$$\begin{aligned}&V_n(m) \nonumber \\&\quad =\sum _{l=m}^\infty \left\{ U_1(l)(1-T_{n-1}(l)) +U_{n-1}(l)(1-T_1(l)) \right. \nonumber \\&\qquad \left. +\,\frac{p-2}{p-1}U_{n-1}(l) U_1(l)+\sum _{k=l+1}^\infty p^{l-k}U_{n-1}(k)U_1(k) \right\} \nonumber \\&\qquad -p^{-1} \sum _{l=m+1}^\infty \left\{ U_1(l)(1-T_{n-1}(l))+U_{n-1}(l)(1-T_1(l)) \right. \nonumber \\&\qquad \left. +\,\frac{p-2}{p-1}U_{n-1}(l) U_1(l) +\sum _{k=l+1}^\infty p^{l-k}U_{n-1}(k)U_1(k) \right\} \nonumber \\&\quad =\,U_1(m)(1-T_{n-1}(m))+U_{n-1}(m)(1-T_1(m))+\frac{p-2}{p-1}U_{n-1}(m)U_1(m) \nonumber \\&\qquad +\,\left( 1-p^{-1} \right) \sum _{l=m+1}^\infty \left\{ U_1(l)(1-T_{n-1}(l))+U_{n-1}(l)(1-T_1(l))\right. \nonumber \\&\qquad \left. +\,U_{n-1}(l) U_1(l) \right\} \nonumber \\&\quad =(T_1(m)-T_1(m+1))(1-T_{n-1}(m))\nonumber \\&\qquad +\,(T_{n-1}(m)-T_{n-1}(m+1))(1-T_1(m)) \nonumber \\&\qquad +\,\frac{p-2}{p-1}(T_{n-1}(m)-T_{n-1}(m+1))(T_1(m)-T_1(m+1)) \nonumber \\&\qquad +\,\left( 1-p^{-1}\right) \sum _{l=m+1}^\infty \{(T_1(l)-T_1(l+1))(1-T_{n-1}(l)) \nonumber \\&\qquad +\,(T_{n-1}(l)-T_{n-1}(l+1))(1-T_1(l)) \nonumber \\&\qquad +\,(T_{n-1}(l)-T_{n-1}(l+1))(T_1(l)-T_1(l+1)\} \nonumber \\&\quad =T_1(m)-p^{-1}T_1(m+1)+T_{n-1}(m)-p^{-1}T_{n-1}(m+1) \nonumber \\&\qquad -\frac{1}{1-p^{-1}}\left( T_1(m)-p^{-1}T_1(m+1) \right) \left( T_{n-1}(m)-p^{-1}T_{n-1}(m+1) \right) \nonumber \\&\quad =V_1(m)+V_{n-1}(m)-\frac{1}{1-p^{-1}}V_1(m)V_{n-1}(m). \end{aligned}$$
(2)

By this equation, we can derive inductively that

$$\begin{aligned} V_n(m)=\left( 1-p^{-1} \right) \left( 1-\left( 1-\frac{V_1(m)}{1-p^{-1}}\right) ^n \right) . \end{aligned}$$
(3)

Indeed, this is trivial for \(n=1\). Provided it is true for \(n=n_0-1\), then (2) yields

$$\begin{aligned} V_n(m)= & {} V_1(m)+\left( 1-p^{-1}\right) \left( 1-\left( 1-\frac{V_1(m)}{1-p^{-1}}\right) ^{n_0-1} \right) \left( 1-\frac{V_1(m)}{1-p^{-1}}\right) \\= & {} \left( 1-p^{-1} \right) \left( 1-\left( 1-\frac{V_1(m)}{1-p^{-1}}\right) ^{n_0} \right) . \end{aligned}$$

By (3), taking the sum of \(p^{-k}V_n(m+k)=p^{-k}T_n(m+k)-p^{-k-1}T_n(m+k+1)\) for \(k=0, 1, \dots \), we obtain

$$\begin{aligned} T_n(m)= & {} \sum _{k=0}^\infty p^{-k}\left( 1-p^{-1} \right) \left( 1-\left( 1-\frac{V_1(m+k)}{1-p^{-1}}\right) ^n \right) \\= & {} 1-\left( 1-p^{-1}\right) \sum _{k=0}^\infty p^{-k}\left( 1-\frac{T_1(m+k)-p^{-1}T_1(m+k+1)}{1-p^{-1}} \right) ^n. \end{aligned}$$

\(\square \)

Remark 1

If we import a result of [9], Proposition 3 can be derived more concisely by using Fourier transform. Let \(\varphi \) be the character on the p-adic field defined by

$$\begin{aligned} \varphi \left( \sum _{i=-m}^\infty \alpha _i p^i \right) :=\left\{ \begin{array}{ll} \exp \left( 2 \pi \sqrt{-1} \sum \nolimits _{i=-m}^{-1} \alpha _i p^i \right) , &{} \text{ if }~ m \ge 1, \\ 1, &{} \text{ if }~ m \le 0, \end{array}\right. \end{aligned}$$

then the characteristic function of a probability measure \(\mu \) on \({\mathbf {Q}}_p\) is defined by

$$\begin{aligned} \hat{\mu }(y):=\int _{{\mathbf {Q}}_p} \varphi (xy) \mu (\hbox {d}x), \quad y \in {\mathbf {Q}}_p. \end{aligned}$$

We can see the Fourier transform of the indicator function \({\mathbf {1}}_{B_l}\) of the ball \(B_l\) is given by

$$\begin{aligned} {\mathcal {F}}{\mathbf {1}}_{B_l}(x):=\int _{{\mathbf {Q}}_p} {\mathbf {1}}_{B_l}(y) \varphi (xy) \hbox {d}y =p^l {\mathbf {1}}_{B_{-l}}(x), \end{aligned}$$

(see, e.g., Chapter XIV of [6]), where \(\int \cdot \hbox {d}y\) denotes the integration with respect to Haar measure of \({\mathbf {Q}}_p\) normalized so that \(\int {\mathbf {1}}_{B_l}(y)\hbox {d}y=p^l\). Let \(\mu \) be the law of \(\xi _i\), then we have

$$\begin{aligned} 1-T_n(m)= & {} \int _{{\mathbf {Q}}_p} {\mathbf {1}}_{B_{m-1}}(x) \mu ^{*n}(\hbox {d}x) \\= & {} \int _{{\mathbf {Q}}_p} p^{m-1} {\mathcal {F}}{\mathbf {1}}_{B_{-(m-1)}}(x) \mu ^{*n}(\hbox {d}x) \\= & {} p^{m-1} \int _{{\mathbf {Q}}_p} {\mathbf {1}}_{B_{-(m-1)}} (y) \left( \int _{{\mathbf {Q}}_p} \varphi (xy) \mu ^{*n}(\hbox {d}x) \right) \hbox {d}y \\= & {} p^{m-1} \int _{{\mathbf {Q}}_p} {\mathbf {1}}_{B_{-(m-1)}} (y) \hat{\mu }(y)^n \hbox {d}y \\= & {} p^{m-1} \sum _{k=m-1}^\infty \int _{|y|_p=p^{-k}} \hat{\mu }(y)^n \hbox {d}y. \end{aligned}$$

The characteristic function of \(\mu \) is calculated in Lemma 3 of [9] as

$$\begin{aligned} \hat{\mu }(y)=1-(p-1)^{-1}(pT_1(k+1)-T_1(k+2)), \quad \text{ if }~ |y|_p=p^{-k}, \end{aligned}$$

and then Proposition 3 follows immediately.

3 Proofs

For proofs of Propositions 1, 2, and Theorem 1, the following estimate is crucial.

Lemma 2

There exist positive constants \(C_1\) and \(C_2\) independent of \(n \ge 1\) and \(l \in {\mathbf {Z}}\) such that, for every fixed integer l,

$$\begin{aligned} C_1\left( p^{-\alpha }(1-\delta _n) \right) ^{c_n+l} \le P_n\left( B_l^c \right) \le C_2 \left( p^{-\alpha }(1+\delta _n) \right) ^{c_n+l} \end{aligned}$$

holds for sufficiently large n. Furthermore, the both constants \(C_1\) and \(C_2\) can be taken arbitrarily close to \(C:=C(\alpha ):=\frac{p^\alpha (p-1)}{p^{\alpha +1}-1}\), and accordingly,

$$\begin{aligned} \limsup _{n \rightarrow \infty } \frac{P_n\left( B_l^c \right) }{\left( p^{-\alpha }(1+\delta _n) \right) ^{c_n+l}} \le C \le \liminf _{n \rightarrow \infty } \frac{P_n\left( B_l^c \right) }{\left( p^{-\alpha }(1-\delta _n) \right) ^{c_n+l}} \end{aligned}$$

holds for any integer l.

Proof

Fix an integer l and take any \(\varepsilon >0\). By Proposition 3, we have

$$\begin{aligned} P_n\left( B_l^c \right)= & {} P\left( \left| p^{n+c_n} \sum _{i=1}^{[N(n)]} \xi _i \right| _p \ge p^{l+1} \right) \\= & {} T_{[N(n)]} (n+c_n+l+1) \\= & {} 1-\left( 1-p^{-1} \right) \sum _{k=0}^\infty p^{-k} (1-v(n+c_n+l+k))^{[N(n)]}, \end{aligned}$$

where

$$\begin{aligned} v(m):= & {} \left( 1-p^{-1}\right) ^{-1}(T_1(m+1)-p^{-1}T_1(m+2)) \\= & {} \left( 1-p^{-1}\right) ^{-1} \left( 1-p^{-\alpha -1}\frac{L(m+2)}{L(m+1)} \right) T_1(m+1) . \end{aligned}$$

Since v(m) tends to 0 as \(m \rightarrow \infty \),

$$\begin{aligned} e^{-(1+\varepsilon )}<\left( 1-v(n+c_n+l+k)\right) ^{\frac{1}{v(n+c_n+l+k)}}<e^{-(1-\varepsilon )} \end{aligned}$$

holds for sufficiently large n and any \(k \ge 0\), and then

$$\begin{aligned}&1-\left( 1-p^{-1} \right) \sum _{k=0}^\infty p^{-k}\left( e^{-(1-\varepsilon )} \right) ^{v(n+c_n+l+k)[N(n)]} \\&\quad<P_n\left( B_l^c \right) \\&\quad <1-\left( 1-p^{-1} \right) \sum _{k=0}^\infty p^{-k}\left( e^{-(1+\varepsilon )} \right) ^{v(n+c_n+l+k)[N(n)]}, \end{aligned}$$

namely,

$$\begin{aligned}&\left( 1-p^{-1} \right) \sum _{k=0}^\infty p^{-k}\left( 1-e^{-(1-\varepsilon )v(n+c_n+l+k)[N(n)]}\right) \nonumber \\&\quad<P_n\left( B_l^c \right) \nonumber \\&\quad <\left( 1-p^{-1} \right) \sum _{k=0}^\infty p^{-k}\left( 1-e^{-(1+\varepsilon )v(n+c_n+l+k)[N(n)]}\right) . \end{aligned}$$
(4)

We have

$$\begin{aligned}&\left( p^{-\alpha }(1-\delta _n) \right) ^{c_n+l+k+1} \\&\quad \le \frac{T_1(n+c_n+l+k+1)}{T_1(n)}=p^{-\alpha (c_n+l+k+1)} \prod _{i=0}^{c_n+l+k} \frac{L(n+i+1)}{L(n+i)} \\&\quad \le \left( p^{-\alpha }(1+\delta _n) \right) ^{c_n+l+k+1}, \end{aligned}$$

and

$$\begin{aligned} 1-\varepsilon<1-\delta _{n+c_n+l} \le \frac{L(n+c_n+l+k+2)}{L(n+c_n+l+k+1)} \le 1+\delta _{n+c_n+l}<1+\varepsilon , \end{aligned}$$

for all \(k \ge 0\), if n is large enough so that \(\delta _{n+c_n+l}<\varepsilon \). We also have

$$\begin{aligned} (1-\varepsilon ) N(n) \le [N(n)] \le N(n) \end{aligned}$$
(5)

for large n, since \(N(n)=\frac{p^{2\alpha }(p-1)}{p^{\alpha +1}-1} T_1(n)^{-1} \rightarrow \infty \) as \(n \rightarrow \infty \). Applying these inequalities to \(v(n+c_n+l+k)[N(n)] =\left( 1-p^{-1}\right) ^{-1} \left( 1-p^{-\alpha -1} \frac{L(n+c_n+l+k+2)}{L(n+c_n+l+k+1)}\right) T_1(n+c_n+l+k+1) \left[ \frac{p^{2\alpha }(p-1)}{p^{\alpha +1}-1}T_1(n)^{-1} \right] \), we see that

$$\begin{aligned}&(1-\varepsilon )\left( 1-p^{-1}\right) ^{-1}\left( 1-p^{-\alpha -1}(1+\varepsilon ) \right) \frac{p^{2\alpha }(p-1)}{p^{\alpha +1}-1}\left( p^{-\alpha }(1-\delta _n)\right) ^{c_n+l+k+1} \nonumber \\&\quad<v(n+c_n+l+k+1)[N(n)] \nonumber \\&\quad < \left( 1-p^{-1}\right) ^{-1} \left( 1-p^{-\alpha -1}(1-\varepsilon )\right) \frac{p^{2\alpha }(p-1)}{p^{\alpha +1}-1} \left( p^{-\alpha }(1+\delta _n) \right) ^{c_n+l+k+1}. \end{aligned}$$
(6)

In particular, take n large enough so that \(\delta _n<\varepsilon \wedge (p^{\alpha }-1)\), then the assumption \(c_n \rightarrow +\infty \) implies that the right-hand side goes to 0 as \(n \rightarrow \infty \) uniformly for \(k \ge 0\). Since \(\lim _{t \rightarrow 0} \frac{1-e^{-t}}{t}=1\), we have

$$\begin{aligned} 1-e^{-(1-\varepsilon )v(n+c_n+l+k+1)[N(n)]}\ge & {} (1-\varepsilon )^2v(n+c_n+l+k+1)[N(n)], \\ 1-e^{-(1+\varepsilon )v(n+c_n+l+k+1)[N(n)]}\le & {} (1+\varepsilon )^2v(n+c_n+l+k+1)[N(n)] \end{aligned}$$

for large n, and therefore, inequalities (4) lead to

$$\begin{aligned}&\left( 1-p^{-1}\right) \sum _{k=0}^\infty p^{-k}(1-\varepsilon )^2v(n+c_n+l+k+1)[N(n)] \\&\quad<P_n\left( B_l^c \right) \\&\quad <\left( 1-p^{-1}\right) \sum _{k=0}^\infty p^{-k} (1+\varepsilon )^2v(n+c_n+l+k+1)[N(n)]. \end{aligned}$$

Applying (6) to the above, we obtain

$$\begin{aligned}&(1-\varepsilon )^3\left( 1-p^{-\alpha -1}(1+\varepsilon ) \right) \frac{p^{2\alpha }(p-1)}{p^{\alpha +1}-1} \left( p^{-\alpha }(1-\delta _n)\right) ^{c_n+l+1}\\&\qquad \times \frac{1}{1-p^{-\alpha -1}(1-\delta _n)} \\&\quad<P_n\left( B_l^c \right) \\&\quad <(1+\varepsilon )^2\left( 1-p^{-\alpha -1}(1-\varepsilon ) \right) \frac{p^{2\alpha }(p-1)}{p^{\alpha +1}-1} \left( p^{-\alpha }(1+\delta _n)\right) ^{c_n+l+1}\\&\qquad \times \frac{1}{1-p^{-\alpha -1}(1+\delta _n)}. \end{aligned}$$

Put \(C_1=(1-\varepsilon )^4\frac{1-p^{-\alpha -1}(1+\varepsilon )}{1-p^{-\alpha -1}(1-\varepsilon )}\frac{p^\alpha (p-1)}{p^{\alpha +1}-1}\) and \(C_2=(1+\varepsilon )^3 \frac{1-p^{-\alpha -1}(1-\varepsilon )}{1-p^{-\alpha -1}(1+\varepsilon )} \frac{p^\alpha (p-1)}{p^{\alpha +1}-1}\), then since \(\delta _n <\varepsilon \) we obtain

$$\begin{aligned} C_1\left( p^{-\alpha }(1-\delta _n) \right) ^{c_n+l} \le P_n\left( B_l^c \right) \le C_2 \left( p^{-\alpha }(1+\delta _n) \right) ^{c_n+l} \end{aligned}$$

for sufficiently large n.

We can see \(C_1\) and \(C_2\) both approach C as \(\varepsilon \rightarrow 0\); then, the second assertion is clear. \(\square \)

Now let us give proofs to Propositions 1, 2, and Theorem 1, using the estimates of Lemma 2.

Proof

(Proposition 1) Since K is assumed to be compact open and \(0 \in K\), there exist integers \(l_1 \ge l_2\) such that \(B_{l_2} \subset K \subset B_{l_1}\). Then, Lemma 2 implies

$$\begin{aligned} C_1 \left( p^{-\alpha }(1-\delta _n)\right) ^{c_n+l_1} \le P_n \left( K^c \right) \le C_2 \left( p^{-\alpha }(1+\delta _n)\right) ^{c_n+l_2}. \end{aligned}$$

Take their logarithm and divide by n, then we have

$$\begin{aligned}&\frac{c_n+l_1}{n}(-\alpha \log p+\log (1-\delta _n))+\frac{\log C_1}{n} \\&\quad \le \frac{1}{n} \log P_n \left( K^c \right) \\&\quad \le \frac{c_n+l_2}{n}(-\alpha \log p+\log (1+\delta _n))+\frac{\log C_2}{n}. \end{aligned}$$

Since \(\delta _n \rightarrow 0\) as \(n \rightarrow \infty \), taking \(\limsup \) and \(\liminf \) of each side, the assertion is proved. \(\square \)

Proof

(Proposition 2) (i) Since the complement \(B^c\) is an open set including the origin, we can take an integer l such that \(B_l \subset B^c\). Apply Proposition 1 to \(K=B_l\), then

$$\begin{aligned} \liminf _{n \rightarrow \infty }\frac{1}{n} \log P_n(B) \le \liminf _{n \rightarrow \infty }\frac{1}{n} \log P_n \left( B_l^c \right) =-\alpha \theta _+ \log p, \end{aligned}$$

and

$$\begin{aligned} \limsup _{n \rightarrow \infty }\frac{1}{n} \log P_n(B) \le \limsup _{n \rightarrow \infty }\frac{1}{n} \log P_n \left( B_l^c \right) =-\alpha \theta _- \log p. \end{aligned}$$

(ii) We can take an integer l such that \(B_l \subset A\), and Lemma 2 implies

$$\begin{aligned} \log \left( 1-C_2\left( p^{-\alpha }(1+\delta _n) \right) ^{c_n+l} \right) \le \log P_n(B_l) \le \log P_n(A) \le 0. \end{aligned}$$

By the assumption \(\lim _{n \rightarrow \infty }c_n=+\infty \), we have \(C_2 \left( p^{-\alpha }(1+\delta _n) \right) ^{c_n+l} \rightarrow 0\) as \(n \rightarrow \infty \), and hence, the left-hand side tends to 0. \(\square \)

Proof

(Theorem 1) For the case (i), our claim is trivial by Proposition 2 (i). Let us assume (ii) and put \(|a|_p=p^k\). The sphere \(H_k\) is the disjoint union of \(p^{k-l-1}(p-1)\) balls of radius \(p^l\). Since each of these balls is mapped to the ball \(B\left( a, p^l \right) \) by a rotation around the origin, and the law of \(\xi _i\) is invariant by the rotation, it follows that

$$\begin{aligned} P_n\left( B \left( a, p^l \right) \right) =\frac{P_n(H_k)}{p^{k-l-1}(p-1)} =\frac{P_n\left( B_{k-1}^c \right) -P_n \left( B_k^c \right) }{p^{k-l-1}(p-1)}. \end{aligned}$$
(7)

Applying Lemma 2 to \(l=k-1\) and \(l=k\), we obtain

$$\begin{aligned}&C_1\left( p^{-\alpha }(1-\delta _n) \right) ^{c_n+k-1}-C_2\left( p^{-\alpha }(1+\delta _n) \right) ^{c_n+k} \nonumber \\&\quad \le P_n\left( B_{k-1}^c \right) -P_n\left( B_k^c \right) \nonumber \\&\quad \le C_2 \left( p^{-\alpha }(1+\delta _n)\right) ^{c_n+k-1}-C_1 \left( p^{-\alpha }(1-\delta _n)\right) ^{c_n+k}. \end{aligned}$$
(8)

Here let us verify the left-hand side of this inequality is positive for large n. By the assumption \(\limsup _{n \rightarrow \infty } c_n \log \frac{1+\delta _n}{1-\delta _n}<\alpha \log p\), we have \(p^\alpha \left( \frac{1-\delta _n}{1+\delta _n}\right) ^{c_n}>1\) for sufficiently large n. Since \(\frac{C_1}{C_2}\) can be arbitrarily close to 1 and \(\delta _n \rightarrow 0\) as \(n \rightarrow \infty \), the ratio

$$\begin{aligned} \frac{C_1\left( p^{-\alpha }(1-\delta _n) \right) ^{c_n+k-1}}{C_2\left( p^{-\alpha }(1+\delta _n) \right) ^{c_n+k}} =\frac{C_1}{C_2}\frac{(1-\delta _n)^{k-1}}{(1+\delta _n)^k} \cdot p^\alpha \left( \frac{1-\delta _n}{1+\delta _n} \right) ^{c_n} \end{aligned}$$

is greater than 1 for sufficiently large n. Therefore, the left-hand side of (8) is positive, and then we can estimate the logarithm of (7) as

$$\begin{aligned}&\log \left( C_1\left( p^{-\alpha }(1-\delta _n) \right) ^{c_n+k-1} -C_2 \left( p^{-\alpha }(1+\delta _n) \right) ^{c_n+k} \right) \\&\qquad -(k-l-1) \log p-\log (p-1) \\&\quad \le \log P_n \left( B \left( a, p^l \right) \right) \\&\quad \le \log \left( C_2 \left( p^{-\alpha }(1+\delta _n)\right) ^{c_n+k-1}-C_1 \left( p^{-\alpha }(1-\delta _n)\right) ^{c_n+k} \right) \\&\qquad -(k-l-1) \log p-\log (p-1). \end{aligned}$$

Divide the each side by n, then we proceed to

$$\begin{aligned}&\frac{c_n+k-1}{n}(-\alpha \log p +\log (1-\delta _n)) \\&\qquad +\frac{1}{n} \Biggm (\log \left( C_1-C_2 p^{-\alpha }(1-\delta _n)\left( \frac{1+\delta _n}{1-\delta _n}\right) ^{c_n+k}\right) \\&\qquad -(k-l-1)\log p-\log (p-1)\Biggm ) \\&\quad \le \frac{1}{n} \log P_n \left( B\left( a, p^l \right) \right) \\&\quad \le \frac{c_n+k-1}{n}(-\alpha \log p +\log (1+\delta _n)) \\&\qquad +\frac{1}{n} \Biggm (\log \left( C_2-C_1 p^{-\alpha }(1+\delta _n)\left( \frac{1-\delta _n}{1+\delta _n}\right) ^{c_n+k}\right) \\&\qquad -(k-l-1)\log p-\log (p-1)\Biggm ). \end{aligned}$$

Since \(\delta _n \rightarrow 0\) and \(\limsup _{n \rightarrow \infty } \left( \frac{1+\delta _n}{1-\delta _n}\right) ^{c_n}<p^\alpha \), the second terms of the left- and the right-hand side go to 0 as \(n \rightarrow \infty \). Hence, taking \(\liminf \) and \(\limsup \) of the both sides, our assertion follows. \(\square \)

Proof

(Corollary 1) If we suppose \(P_n\)\((n=1, 2, \dots \)) satisfy the large deviation principle with some rate function I, then Theorem 1 requires

$$\begin{aligned} \alpha \theta _+ \log p \le \inf _{x \in B\left( a, p^l \right) } I(x) \le \alpha \theta _- \log p; \end{aligned}$$

therefore, \(\theta _+=\theta _-\). Thus for the large deviation principle, it is necessary that \(\lim _{n \rightarrow \infty } \frac{c_n}{n}\) exists.

Conversely, suppose that \(\theta :=\lim _{n \rightarrow \infty } \frac{c_n}{n}\) exists, and let \(I(x)=\alpha \theta \log p\) for \(x \ne 0\) and \(I(0)=0\). For the large deviation principle, what we have to show are

$$\begin{aligned} \liminf _{n \rightarrow \infty } \frac{1}{n} \log P_n(A) \ge -\inf _{x \in A} I(x) \end{aligned}$$
(9)

for any open set A in \({\mathbf {Q}}_p\), and

$$\begin{aligned} \limsup _{n \rightarrow \infty } \frac{1}{n} \log P_n(B) \le -\inf _{x \in B} I(x) \end{aligned}$$
(10)

for any closed set B.

For the empty set \(\phi \), (9) and (10) are trivial. If an open set A includes the origin, then by Proposition 2 (ii),

$$\begin{aligned} \liminf _{n \rightarrow \infty } \frac{1}{n} \log P_n(A) =0=-I(0)=-\inf _{x \in A} I(x). \end{aligned}$$

Suppose next \(A \ne \phi \) is an open set not including the origin. Take \(a \in {\mathbf {Q}}_p\) and an integer l such that \(B\left( a, p^l \right) \subset A\), then by Theorem 1,

$$\begin{aligned} \liminf _{n \rightarrow \infty } \frac{1}{n} \log P_n(A) \ge \liminf _{n \rightarrow \infty } \frac{1}{n} \log P_n \left( B \left( a, p^l \right) \right) =-\alpha \theta \log p =-\inf _{x \in A}I(x). \end{aligned}$$

Let B be a closed set including the origin, then trivially we have

$$\begin{aligned} \limsup _{n \rightarrow \infty } \frac{1}{n} \log P_n(B) \le 0=-I(0)=-\inf _{x \in B} I(x). \end{aligned}$$

In case a closed set \(B \ne \phi \) does not include the origin, we can take an integer l such that \(B_l \subset B^c\), and then Proposition 1 implies

$$\begin{aligned} \limsup _{n \rightarrow \infty } \frac{1}{n} \log P_n(B) \le \limsup _{n \rightarrow \infty } \frac{1}{n} \log P_n \left( B_l^c \right) =-\alpha \theta \log p =-\inf _{x \in B} I(x). \end{aligned}$$

Therefore, the large deviation principle holds with the rate function I.

In case \(\theta =+\infty \), it holds that \(\{x \in {\mathbf {Q}}_p \ | \ I(x) \le c\}=\{0\}\) for any \(c \ge 0\), and in case \(\theta <+\infty \), we have \(\{x \in {\mathbf {Q}}_p \ | \ I(x) \le c\}={\mathbf {Q}}_p\) for \(c \ge \alpha \theta \log p\). Therefore, I is good if and only if \(\theta =+\infty \). \(\square \)