2.1. ‘Standard’ cat-and-mouse Markov chain on $\mathbb{Z}$ $(C\to M)$

Let $\xi = \pm 1$ w.p. $1/2$ . Let $\big\{\xi^{(1)}_n\big\}_{n=1}^\infty$ and $\big\{\xi^{(2)}_n\big\}_{n=1}^\infty$ be two mutually independent sequences of independent copies of $\xi$ . Given $C_0 = M_0 = 0$ , we define the dynamics of CM MC $(C_n, M_n)$ as follows:

\begin{equation*}C_{n} = C_{n-1} + \xi^{(1)}_{n},\end{equation*}

\begin{equation*}M_n = M_{n-1} + \begin{cases}0, & \text{if $C_{n-1} \neq M_{n-1},$}\\ \\[-7pt] \xi^{(2)}_n, & \text{if $C_{n-1} = M_{n-1},$}\end{cases}\end{equation*}

for $n \geq 1$ .

Let $D[[0,\infty), \mathbb{R}]$ denote the space of all right-continuous functions on $[0, \infty)$ having left limits (RCLL or càdlàg functions).

Let $M(nt) = M_{[nt]}$ , $t\geq0$ , be a continuous-time stochastic process taking values $M_k$ , $k\geq 0$ , for $t\in[k/n, (k+1)/n)$ . Clearly, it is piecewise constant and its trajectories belong to $D[[0,\infty), \mathbb{R}]$ .

[Reference Litvak and Robert26] have proved convergence of trajectories

\begin{equation*}\left\{\frac{1}{\sqrt[4]{n}}M(nt), t\geq 0\right\} \overset{\mathcal{D}}{\Rightarrow} \left\{B_1 (L_{B_2}(t)), t\geq 0\right\}, \ \text{as $n\to\infty$},\end{equation*}

where $B_1(t)$ and $B_2(t)$ are independent standard Brownian motions on $\mathbb{R}$ and $L_{B_2}(t)$ is the local time process of $B_2(t)$ at 0.

2.2. Cat-and-mouse model with a general jump distribution of the mouse $(C \to M)$

In this subsection we introduce our results for CM MCs with more general distributions of RVs $\xi^{(1)}_n$ and $\xi^{(2)}_n$ . We start with the same distribution of $\xi^{(1)}_n$ and generalise the distribution of $\xi^{(2)}_n$ . Thus, the cat is a simple random walk and the mouse is a general random walk. We then proceed to generalise the distribution of $\xi^{(1)}_n$ as well.

2.2.1.

We continue to assume that the dynamics of the cat is described by a simple random walk on $\mathbb{Z}$ . Let $\xi = \pm 1$ w.p. $1/2$ . Let $C_0 = 0, \ C_{n} = C_{n-1} + \xi^{(1)}_n$ , where $\xi, \xi^{(1)}_1, \xi^{(1)}_2, \ldots$ are i.i.d. RVs.

Let $M_0 = 0, \ M_n = M_{n-1} + \xi^{(2)}_n I[C_{n-1} = M_{n-1}]$ , where $\big\{\xi^{(2)}_n\big\}_{n=1}^\infty$ are i.i.d. RVs independent of $\big\{\xi^{(1)}_n\big\}_{n=1}^\infty$ . Assume that

(1) \begin{equation}\mu = \mathbb{E} \xi^{(2)}_1 \ \text{is finite}\end{equation}

and that the distribution of $\xi_1^{(2)}$ belongs to the domain of attraction of a strictly stable distribution, i.e. there exist a function $b(c) >0$ , $c\geq0$ , and an RV $A^{(2)}$ having a strictly stable distribution with index $\alpha \in [1,2]$ such that the weak convergence

(2) \begin{equation}\frac{\sum_{k=1}^{n} \Big(\xi^{(2)}_k - \mu\Big)}{b(n)} \Rightarrow A^{(2)}, \ \text{as $n\to\infty$}\end{equation}

holds. Define

\begin{align*}\tau(0) = 0 \ \text{and} \ \tau(n) = \inf\{m > \tau(n-1): \ C_m = M_m\}, \ \text{for $n\ge 1$}.\end{align*}

Given (1), we show that the tail-distribution of $\tau(1)$ is regularly varying with index $1/2$ . It follows then that there exists an RV $D^{(2)}$ having a strictly stable distribution with index $1/2$ such that

(3) \begin{equation}\frac{\tau(n)}{n^2} \Rightarrow D^{(2)}, \ \text{as $n\to\infty$.}\end{equation}

In the proof of Theorem 2 we show that in fact there is a weak convergence of two-dimensional random vectors

\begin{align*}\left(\frac{\sum_{k=1}^{n} \Big(\xi^{(2)}_k - \mu\Big)}{b(n)}, \ \frac{\tau(n)}{n^2}\right) \ \Rightarrow \big(A^{(2)}, D^{(2)}\big), \ \text{as $n\to\infty$,}\end{align*}

where the RVs on the right-hand side are independent. Further, let $\big\{\big(A^{(2)}(t), D^{(2)}(t)\big)\big\}_{t\geq 0}$ denote a stochastic process with independent increments such that $\big(A^{(2)}(1), D^{(2)}(1)\big)$ has the same distribution as $\big(A^{(2)}, D^{(2)}\big)$ , or, equivalently, the Lévy process generated by $\big(A^{(2)}, D^{(2)}\big)$ . Let $E^{(2)}(s) = \inf\big\{t\geq0 \,{:}\,D^{(2)}(t)> s\big\}$ , which is a multiple of $L_{B_2}(s)$ , the local time at zero of a standard Brownian motion. Thus, the following result is a natural extension of the result by [Reference Litvak and Robert26]; see remark below.

Theorem 1. Assume that (1) and (2) hold. Then

• if $\mu = 0$ , we have

  (4) \begin{equation} \left\lbrace \frac{M(nt + 1)}{b(\sqrt{n})}, \ t\geq 0\right\rbrace \overset{\mathcal{D}}{\Rightarrow} \big\{A^{(2)}\big(E^{(2)}(t)\big), t\geq 0\big\}, \ \text{as $n\to\infty$;}\end{equation}

• if $\mu \neq 0$ , we have

  (5) \begin{equation}\left\lbrace \frac{M(nt + 1)}{\sqrt{n}}, \ t\geq 0\right\rbrace \overset{\mathcal{D}}{\Rightarrow} \big\{\mu E^{(2)}(t), t\geq 0\big\}, \ \text{as $n\to\infty$.}\end{equation}

Remark. If the RVs $\xi_n^{(2)}$ are bounded, then the function b(n) is proportional to $\sqrt{n}$ , and it is easy to show that the processes

\begin{equation*}\frac{M(nt + 1)}{\sqrt[4]{n}}, \qquad \frac{M(nt)}{\sqrt[4]{n}}\end{equation*}

are equivalent (see Appendix B for details). Then the results of Theorem 1 may be reformulated for the scaled process M(nt) in place of $M(nt + 1)$ .

2.2.2.

Assume now that both $\xi^{(1)}_1$ and $\xi^{(2)}_1$ have general distributions on the integer lattice. The main difference for the mouse is that we now need to assume that the second moment of $\xi^{(2)}_1$ is finite. Our main result in this setting is that replacing the simple random walk with a general random walk does not change the scaling if we assume aperiodicity and finite second moments for the increments. By aperiodicity (strong aperiodicity in the sense of [Reference Spitzer32]) we mean here that

\begin{align*}G.C.D. \{ k: \ {\mathbb P} (\xi_1^{(1)}=k)>0\} =1,\end{align*}

where ‘G.C.D.’ stands for ‘greatest common divisor’.

Theorem 2. Assume that $\mathbb{E} \xi^{(1)} =0$ , $0 <\mathbf{Var} \xi^{(1)}_1 < \infty$ and $\xi^{(1)}_1$ has an aperiodic distribution. Assume $0< \mathbf{Var} \xi^{(2)}_1 < \infty$ and, therefore, (2) holds with $b(n) = \sqrt{n\mathbf{Var} \xi^{(2)}_1}$ and a standard normal RV $A^{(2)}$ . Then the statements (4) and (5) of Theorem 1 continue to hold, with $b(\sqrt{n}) = n^{1/4}\sqrt{\mathbf{Var} \xi^{(2)}_1}$ in (4).

2.3. Dog-and-cat-and-mouse model ( $D \to C \to M$ )

In this subsection we present a generalisation of the CM MC to the case of three dimensions. Let $\xi = \pm 1$ w.p. $1/2$ . Let $\big\{\xi^{(1)}_n\big\}_{n=1}^\infty$ , $\big\{\xi^{(2)}_n\big\}_{n=1}^\infty$ , and $\big\{\xi^{(3)}_n\big\}_{n=1}^\infty$ be mutually independent sequences of independent copies of $\xi$ . Given $D_0 = C_0 = M_0 = 0$ , we can define the dynamics of the DCM MC $\{(D_n, C_n, M_n)_n\}_{n=1}^\infty$ as follows: for $n\ge 1$ ,

\begin{equation*}D_{n} = D_{n-1} + \xi^{(1)}_{n},\end{equation*}

\begin{equation*}C_n = C_{n-1} + \begin{cases}0, & \text{if $D_{n-1} \neq C_{n-1},$}\\ \\[-7pt] \xi^{(2)}_n, & \text{if $D_{n-1} = C_{n-1},$}\end{cases}\end{equation*}

\begin{equation*}M_n = M_{n-1} + \begin{cases}0, & \text{if $C_{n-1} \neq M_{n-1},$}\\ \\[-7pt] \xi^{(3)}_n, & \text{if $C_{n-1} = M_{n-1}.$}\end{cases}\end{equation*}

Let $T^{(3)}(0) = 0$ and $T^{(3)}(k) = \min\{n >T^{(3)}(k-1): \ D_n = C_n = M_n\}$ , for $k\geq 1$ . We show that the tail-distribution of $T^{(3)}(1)$ is regularly varying with index $1/4$ . Further, we show that there exists a positive RV $D^{(3)}$ with a stable distribution and Laplace transform $\exp\big(\!-s^{1/4}\big)$ such that

(6) \begin{equation}\frac{T^{(3)}(k)}{2^7 k^4} \Rightarrow D^{(3)}, \ \text{as $k\to\infty$.}\end{equation}

Let $\big\{D^{(3)}(t)\big\}_{t\geq 0}$ be a Lévy process generated by $D^{(3)}$ and $E^{(3)}(s) = \inf\big\{t\geq0 : \ D^{(3)}(t)> s\big\}$ .

Theorem 3. We have $\mathbb{E} M\big(T^{(3)}(1)\big) = 0$ , $\sigma^2 = \mathbf{Var} M\big(T^{(3)}(1)\big) = 2$ , and

\begin{align*}\left\lbrace \frac{M(nt)}{2^{-7/8} n^{1/8}\sigma}, t\geq 0\right\rbrace \overset{\mathcal{D}}{\Rightarrow} \big\{B(E^{(3)}(t)), t\geq 0\big\},\ \text{as $n\to\infty$, }\end{align*}

where B(t) is a standard Brownian motion, independent of $E^{(3)}(t)$ .

2.4. Linear hierarchical chains $\big(\boldsymbol{X}^{(1)} \to \boldsymbol{X}^{(2)} \to \ldots \to \boldsymbol{X}^{(\boldsymbol{N})}\big)$ of length N

In this subsection we consider a generalisation of the CM MC to the case of N dimensions. Owing to the complexity of sample paths for $N>3$ , it does not seem to be possible to prove an analogue of (3) and (6). For this setting, we prove the convergence for every fixed $t>0$ .

Let $\xi = \pm 1$ w.p. $1/2$ . Let $\big\{\big\{\xi^{(j)}_n\big\}_{n=1}^\infty\big\}_{j=1}^N$ be mutually independent sequences of independent copies of $\xi$ . Assume $X^{(1)}_0 = \ldots = X^{(N)}_0 = 0$ . Then MC $\big(X^{(1)}_n, \ldots, X^{(N)}_n\big)$ is defined as follows:

\begin{equation*}X^{(1)}_{n} = X^{(1)}_{n-1} + \xi^{(1)}_{n},\end{equation*}

\begin{equation*}X^{(j)}_n = X^{(j)}_{n-1} + \begin{cases}0, & \text{if $X^{(j-1)}_{n-1} \neq X^{(j)}_{n-1},$}\\ \\[-6pt]\xi^{(j)}_n, & \text{if $X^{(j-1)}_{n-1} = X^{(j)}_{n-1},$}\end{cases}\end{equation*}

for $j \in \{2, \ldots, N\}$ and for $n \geq 1$ .

For the result below we need the following distribution. Let $G_\alpha$ be the distribution function of a one-sided strictly stable law satisfying the condition $x^\alpha (1- G_\alpha(x)) \to (2-\alpha) / \alpha$ , as $x\to\infty$ .

Theorem 4. Let $\{\zeta_i\}_{i=1}^\infty$ be i.i.d. RVs satisfying $\mathbb{P}\{\zeta_i \geq y\} = G_{1/2}(9/y^2)$ . Let $\psi$ be an RV with a standard normal distribution and independent of $\{\zeta_i\}_{i=1}^\infty$ . Then, for any fixed $t>0$ , we have

\begin{equation*}\frac{X^{(N)}_{[nt]}}{n^{1/2^N}} \Rightarrow t^{1/2^N} \psi \prod_{i=1}^{N} \sqrt{\sqrt{\frac{\pi}{2}}\zeta_i}, \ \text{as $n\to\infty$.}\end{equation*}

4.1. Distribution of the random variable $J^{(3)}_1$

We assume that $D_0 = C_0 = M_0 = 0$ . Let $V_n = (V_{n1}, V_{n2}) = (|D_n - C_n|, |C_n - M_n|)$ . Then the following recursion holds:

\begin{equation*}(D_{n+1}, C_{n+1}, M_{n+1}) = \Big(D_n + \xi^{(1)}_{n+1}, C_n + \xi^{(2)}_{n+1} I[V_{n1} = 0], M_n + \xi^{(3)}_{n+1} I[V_{n2} = 0]\Big).\end{equation*}

Note further that

(7) \begin{equation}V_{n+1} \overset{d}{=} \begin{cases}\Big(1 + \xi^{(1)}_{n+1}, 1+\xi^{(2)}_{n+1}\Big), \ \text{if $V_{n1} = V_{n2} = 0$,} \\ \\[-7pt] \Big(1 + \xi^{(1)}_{n+1}, V_{n2} + \xi^{(2)}_{n+1}\Big), \ \text{if $V_{n1} = 0$ and $V_{n2} \neq 0$,}\\ \\[-7pt] \Big(V_{n1} + \xi^{(1)}_{n+1}, 1\Big), \ \text{if $V_{n1} \neq 0$ and $V_{n2} = 0$,}\\ \\[-7pt]\Big(V_{n1} + \xi^{(1)}_{n+1}, V_{n2}\Big), \ \text{if $V_{n1} \neq 0$ and $V_{n2} \neq 0$.}\end{cases}\end{equation}

Thus, $\{V_n\}_{n=0}^\infty$ is a MC. Let $p_{ij}(k,l) = \mathbb{P}\{V_{n+1}=(k, l) | V_n = (i, j)\}$ , for $i,j, k,l\geq 0$ . Note that $p_{00}(k,l) = p_{01}(k,l)$ for any $k,l \ge 0$ .

Let

\begin{align*}U^{(3)}(0) = 0\ \text{and} \ U^{(3)}(k) = \min\big\{n> U^{(3)}(k-1):\ V_n \in \{(0,0), (0,1)\}\big\}.\end{align*}

Since $p_{00}(m,l) = p_{01}(m,l)$ for any m,l, we have that the RVs $\big\{U^{(3)}(k) - U^{(3)}(k-1)\big\}_{k=1}^\infty$ are i.i.d., and the RV $\big(U^{(3)}(k) - U^{(3)}(k-1)\big)$ does not depend on $V_{U^{(3)}(k-1)}$ , for $k\ge 1$ .

In other words, each time $n=U^{(3)}(k)$ , the DCM process visits either a state on the ‘main diagonal’ $D_n=C_n=M_n$ or an auxiliary state $D_n = C_n = M_n \pm 1$ . To find the tail asymptotics for $T^{(3)}_1$ , we find tail asymptotics for $U^{(3)}(1)$ and then use the natural link between time instants $T^{(3)}(1)$ and $\big\{U^{(3)}(k)\big\}_{k=1}^\infty$ .

Lemma 1. Let $V_0 \in \{(0,0), (0,1)\}$ . Then we have

\begin{equation*}\mathbb{P}\big\{U^{(3)}(1) > n\big\} \sim \frac{1}{2^{1/4}\Gamma(3/4) n^{1/4}}, \ \text{as $n\to\infty$.}\end{equation*}

Further, $U^{(3)}(1) = 1$ if and only if $V_{U^{(3)}(1)} = (0,0)$ .

Proof. Let $V_0 = (0,0)$ . It is apparent from the first line of Equation (7) that

\begin{align*}\mathbb{P}\{V_1 = (0,0)\} = \mathbb{P}\{V_1 = (2,0)\} = \mathbb{P}\{V_1 = (0,\ 2)\} = \mathbb{P}\{V_1 = (2,2)\} = \frac{1}{4}.\end{align*}

Since $p_{00}(k,l) = p_{01}(k,l)$ , the RV $V_1$ has the same distribution given $V_0 = (0, 1)$ .

Let $V_1 = (0, 2)$ (Figure 1c). From the second and the fourth lines of Equation (7) we know that $|V_{(k+1)2} - V_{k2}| \in \{0, 1\}$ , for $k\ge 1$ , given $V_{k2} \neq 0$ . Therefore $V_{k2}$ arrives at 1 before hitting 0, and $V_{U^{(3)}(1)} = (0, 1)$ . Let $\tau, \tau_1, \tau_2, \ldots$ be independent copies of $\tau^{(1)}_1$ . Then $U^{(3)}(1)$ has the same distribution as $\sum_{k=1}^\tau \tau_k$ , and we have that

\begin{equation*}\mathbb{P}\bigg\{\sum_{k=1}^\tau \tau_k > n\bigg\} \sim n^{-1/4} \frac{\Gamma^{1/2}(1/2) \Gamma(1/2)}{\Gamma(3/4)} \sqrt{\sqrt{\frac{2}{\pi}}} \sqrt{\frac{2}{\pi}} = \frac{2^{3/4}}{\Gamma(3/4) n^{1/4}},\end{equation*}

as $n\to\infty$ (see Appendix C).

Figure 1. The positioning after the first jump.

Let $V_1 = (2,2)$ (Figure 1d). From the fourth line of Equation (7), $V_{k2}$ remains at 2 (the cat and the mouse do not move) until $V_{k1}$ reaches 0. This happens after a time which has the same distribution as $\tau^{(1)}_2 = \min\big\{n > 0: \ \sum_{k=1}^n \xi^{(1)}_k = 2\big\}$ . Thus, we travel from (2,2) to (0, 2) while never hitting (0,0). We also know that the tail asymptotics of the travel time are $\mathbb{P}\big\{\tau^{(1)}_2 > n\big\} \sim \sqrt{8/\pi n}$ , as $n\to\infty$ . Therefore, we travel from (2, 2) to (0, 2) much faster than from (0, 2) to (0, 1), and given $V_1 = (2,2)$ , we again have $\mathbb{P}\big\{U^{(3)}(1) > n\big\} \sim \mathbb{P}\big\{\sum_{k=1}^\tau \tau_k > n\big\}$ , as $n\to\infty$ .

Finally, let $V_1 = (2, 0)$ (Figure 1b). From the third line of Equation (7) we have $V_{2} \overset{d}{=} \big(2+ \xi^{(1)}_{2}, 1\big)$ and $V_{U^{(3)}(1)} = (0, 1)$ , where, given $V_1 = (2, 0)$ , $\mathbb{P}\big\{U^{(3)}(1)>n\big\} \sim \sqrt{8/\pi n}$ , as $n\to\infty$ .

Thus,

\begin{align*}\mathbb{P}\big\{U^{(3)}(1) > n\big\} \sim \frac{1}{2}\mathbb{P}\bigg\{\sum_{k=1}^\tau \tau_k > n\bigg\} \sim \frac{1}{2^{1/4}\Gamma(3/4) n^{1/4}}, \ \text{as $n\to\infty$.}\\[-40pt]\end{align*}

We note the following relation between time instants $T^{(3)}(1)$ and $\big\{U^{(3)}(k)\big\}_{k=1}^\infty$ . At each time n when we are at the auxiliary state (when $V_n = (0, 1)$ ), we have a probability of $1/4$ of jumping into the state $D_{n+1} = C_{n+1} = M_{n+1}$ , independently of anything else. Using Lemma 1 and the results in Section XIII.6 of [Reference Feller13], we get the following result.

Proposition 1. Let

\begin{align*}\nu = \inf\big\{k \ge 1: \ U^{(3)}(k) - U^{(3)}(k-1) = 1\big\}.\end{align*}

Then $\nu$ has a geometric distribution with parameter $1/4$ , $T^{(3)}\equiv J^{(3)}=U^{(3)}(\nu)$ a.s., and

\begin{align*}\mathbb{P}\Big\{J^{(3)}_1 > n\Big\} = \mathbb{P}\big\{U^{(3)}(\nu) > n\big\} \sim 4\mathbb{P}\big\{U^{(3)}(1) > n\big\}, \ \text{as $n\to\infty$.}\end{align*}

Therefore, there exists a positive RV $D^{(3)}$ having a stable distribution with Laplace transform $\exp\big(\!-s^{1/4}\big)$ such that

\begin{align*}\frac{T^{(3)}(n)}{2^7 n^4} = \frac{\sum_{k=1}^n J^{(3)}_k}{2^7 n^4} \Rightarrow D^{(3)}, \ as\ \text{$n\to\infty$}.\end{align*}

4.2. Distribution of the random variable $Y^{(3)}_1$

In the previous subsection we analysed the time our process spends between auxiliary states. In this subsection we analyse the total displacement of the mouse by time $T_1^{(3)}$ . Note that the mouse may have zero, one, or two jumps between consecutive visits to auxiliary states; therefore, the total number of jumps of the mouse by time $T_1^{(3)}$ , say $\varkappa$ , admits the following simple upper bound:

\begin{align*}\varkappa \le 2\nu \quad \mbox{a.s.}\end{align*}

It follows that the sequence $Z_k = M_{U^{(3)}(k)} - C_{U^{(3)}(k)}$ , $k=0,1,\ldots$ , forms a time-homogeneous MC, and $Z_k \in \{-1, 0, 1\}$ a.s. Further, $Z_0 = Z_\nu = 0$ , and for any $k \in \{1, \nu - 1\}$ , we have $Z_k = \pm 1$ .

Let

\begin{align*}\gamma^{(3)}_k = M_{U^{(3)}(k)} - M_{U^{(3)}(k-1)} \ \text{for $k\geq 1$.}\end{align*}

Our analysis of trajectories of the DCM MC in Section 4.1 shows that the $\gamma^{(3)}_k$ are conditionally independent given values of the auxiliary MC $\{Z_k\}_{k = 0}^\infty$ . Clearly,

\begin{equation*}Y^{(3)}_1 = M_{T^{(3)}(1)} = \sum_{k=1}^\nu \gamma^{(3)}_k.\end{equation*}

We now find the first and second moments of $Y_1^{(3)}$ and show that it has a light-tailed distribution.

Since $D_0 = C_0 = M_0 = 0$ , we get $\nu=1$ if and only if $Z_1 = 0$ and $\gamma^{(3)}_1 = \pm 1$ . Additionally, we have

\begin{align*}\mathbb{P}\Big\{\gamma^{(3)}_1 = \pm 1, \ Z_1 = 0 \ | \ Z_0 = 0\Big\} = \mathbb{P}\{D_1 = C_1 = M_1 = \pm 1\} = \frac{1}{8}.\end{align*}

Another case of exactly one jump is when the cat and the mouse jump in different directions. Here we have

\begin{align*}\mathbb{P}\Big\{\gamma^{(3)}_1 = \pm 1, \ Z_1 = \pm 1 \ | \ Z_0 = 0\Big\} = \mathbb{P}\{C_1 = \mp 1, M_1 = \pm 1\} = \frac{1}{4}.\end{align*}

Further, the mouse can have two jumps in the same direction w.p.

\begin{align*}\mathbb{P}\Big\{\gamma^{(3)}_1 = \pm 2, \ Z_1 = \pm 1 \ | \ Z_0 = 0\Big\} = \mathbb{P}\{D_1 = \mp 1, C_1 = M_1 = \pm 1, M_2 = \pm 2\} = \frac{1}{16},\end{align*}

or two jumps in opposite directions w.p.

\begin{align*}\mathbb{P}\Big\{\gamma^{(3)}_1 = 0, \ Z_1 = \pm 1 \ | \ Z_0 = 0\Big\} = \mathbb{P}\{D_1 = \pm 1, C_1 = M_1 = \mp 1, M_2 = 0\} = \frac{1}{16},\end{align*}

Thus, given $Z_0 = 0$ we have

\begin{align*}\mathbb{P}\{Z_1 = 0\} = \frac{1}{4} \quad \text{and} \quad \mathbb{P}\{Z_1 = \pm 1\} = \frac{3}{8},\end{align*}

(8) \begin{align}\mathbb{P}\Big\{\gamma^{(3)}_1 = 0\Big\} = \frac{1}{8}, \qquad \mathbb{P}\Big\{\gamma^{(3)}_1 = \pm 1\Big\} = \frac{3}{8}, \quad \text{and} \quad \mathbb{P}\Big\{\gamma^{(3)}_1 = \pm 2\Big\} = \frac{1}{16}.\end{align}

From the above, we get that $\mathbb{E} \gamma^{(3)}_1 = 0$ and $\mathbf{Var} \gamma^{(3)}_1 = 5/4$ .

We now analyse the distribution of $\gamma^{(3)}_k$ , $k\ge 2$ . So as not to make the notation cumbersome, we assume that $D_0 = C_0 = 0$ and $M_0 = 1$ , so $Z_0 = 1$ . The case $Z_0 = -1$ is analogous by symmetry. Then the mouse can make either zero jumps or exactly one jump before the next visit of the process to an auxiliary state. In the case of zero jumps we have

\begin{align*}\mathbb{P}\Big\{\gamma^{(3)}_1 = \pm 0, \ Z_1 = 0 \ | \ Z_0 = 1\Big\} = \mathbb{P}\{D_1 = C_1 = 1\} = \frac{1}{4},\end{align*}

\begin{align*}\mathbb{P}\Big\{\gamma^{(3)}_1 = \pm 0, \ Z_1 = 1 \ | \ Z_0 = 1\Big\} = \mathbb{P}\{C_1 = -1\} = \frac{1}{2}.\end{align*}

In the case of exactly one jump we have

\begin{align*}\mathbb{P}\Big\{\gamma^{(3)}_1 = 1, \ Z_1 = 1 \ | \ Z_0 = 1\Big\} = \mathbb{P}\{D_1 =-1, C_1 = 1, M_2 =2 \} = \frac{1}{8},\end{align*}

\begin{align*}\mathbb{P}\Big\{\gamma^{(3)}_1 = -1, \ Z_1 = -1 \ | \ Z_0 = 1\Big\} = \mathbb{P}\{D_1 =-1, C_1 = 1, M_2 =0\} = \frac{1}{8}.\end{align*}

Thus, given $Z_0 = 1$ we have

\begin{align*}\mathbb{P}\{Z_1 = 0\} = \frac{1}{4}, \ \mathbb{P}\{Z_1 = 1\} = \frac{5}{8} \ \text{and} \ \mathbb{P}\{Z_1 = -1\} = \frac{1}{8},\end{align*}

(9) \begin{align}\mathbb{P}\Big\{\gamma^{(3)}_1 = 0\Big\} = \frac{3}{4} \ \text{and} \ \mathbb{P}\Big\{\gamma^{(3)}_1 = \pm 1\Big\} = \frac{1}{8}.\end{align}

Thus, $\mathbb{E} \left( \gamma^{(3)}_1 \big| Z_0 = 1\right) = 0$ and $\mathbf{Var} \left( \gamma^{(3)}_1\big| Z_0 =1\right) = 1/4$ .

Let us return to the case $D_0 = C_0 = M_0 = 0$ . Combining the results $(8)$ and $(9)$ we get

\begin{align*}\mathbb{E} Y^{(3)}_1 = \mathbb{E} \sum_{k=1}^\nu \gamma^{(3)}_k = \mathbb{E} \sum_{k = 1}^\infty \left( \gamma^{(3)}_k I[\nu \ge k]\right) = \mathbb{E} \gamma^{(3)}_1 + \sum_{k = 2}^\infty \mathbb{E} \left( \gamma^{(3)}_k I[\nu \ge k]\right)& \\ & \kern-164pt = \sum_{k = 2}^\infty \mathbb{E} \left( \gamma^{(3)}_k I[Z_{k-1} = 1] + \gamma^{(3)}_kI[Z_{k-1} = -1]\right) = 0.\end{align*}

In a similar manner we transform the second moment:

(10) \begin{align}\mathbb{E} \Big(Y^{(3)}_1\Big)^2 = \sum_{k=1}^\infty \mathbb{E}\left( \Big( \gamma^{(3)}_k\Big)^2 I[\nu\ge k]\right) + 2\sum_{k=1}^\infty \sum_{m=k+1}^\infty \mathbb{E} \left( \gamma^{(3)}_k \gamma^{(3)}_m I[\nu \ge m]\right).\end{align}

If we fix the values of $\{Z_k\}_{k=0}^\infty$ , the RVs $ \gamma^{(3)}_k$ and $ \gamma^{(3)}_m$ become independent. Combined with the fact that $\mathbb{E} \left( \gamma^{(3)}_m \big| Z_{m-1} = \pm 1 \right) = 0$ , we get that the second sum in $(10)$ equals zero. Now we can use the conditional second moments obtained above and the fact that the RV $\nu$ has a geometric distribution with parameter $1/4$ . We obtain

\begin{align*}\mathbb{E} \Big(Y^{(3)}_1\Big)^2 = \mathbb{E} \Big(\gamma^{(3)}_1\Big)^2 + \sum_{k=2}^\infty \mathbb{E}\left( \Big(\gamma^{(3)}_k\Big)^2\Bigg| \ Z_{k-1} = \pm 1\right) \mathbb{P}\{\nu\ge k\} = \frac{5}{4} + \frac{1}{4}(\mathbb{E} \nu -1) = 2.\end{align*}

Finally, we observe that $\big|\gamma^{(3)}_1\big| \le 2$ . Therefore, $\big|Y^{(3)}_1\big| = \Big|M_{J^{(3)}_1}\Big| \le 2\nu$ a.s. and $|Y^{(3)}|$ has a finite exponential moment. In particular, the following holds.

5.1. Proofs of Theorems 1 and 2

We start with the general idea of the proofs of Theorems 1 and 2. For $i=1,2$ , let $S^{(i)}_0 = 0$ and $S^{(i)}_n = \sum_{k=1}^n \xi^{(i)}_k$ , for $n\geq1$ . Let

\begin{align*}\tau(0) = 0 \ \text{and} \ \tau(n) = \inf\Big\{m > \tau(n-1): \ S^{(1)}_m = S^{(2)}_n\Big\}, \ \text{for $n\ge 1$}.\end{align*}

Since $\big\{\xi^{(1)}_k\big\}_{k=1}^\infty$ and $\big\{\xi^{(2)}_k\big\}_{k=1}^\infty$ are independent sequences of i.i.d. RVs, we have that $\tau(n) - \tau(n-1) \overset{d}{=} \tau(1)$ , for $n \ge 1$ .

Let $\eta(t) = \max\{k\geq 0 :\ \tau(k) \leq t\}$ for $t\geq 0$ . Define a continuous-time process M $^{\prime}$ (t) by

\begin{align*}M'(t) = 0 \ \text{for $t\in [0, 1)$, } \qquad M'(t) = S^{(2)}_{\eta(t-1) + 1} = \sum_{k=1}^{\eta(t-1)+1} \xi^{(2)}_k \ \text{for $t\geq 1$.}\end{align*}

It is straightforward to verify that $\{M'(n), \ n\ge 0\} \overset{d}{=} \{M(n), \ n\ge 0\}$ . In the rest of the section we will omit the prime symbol and simply write M(t). The process $\{\widehat{M}(t)\}_{t\geq 0} = \{M'(t+1)\}_{t\geq 0}$ is a so-called oracle continuous-time random walk (see, e.g., [Reference Jurlewicz, Kern, Meerschaert and Scheffler23].

First, consider the case $\mathbb{E}\xi^{(2)}_1 = 0$ . We want to show that

(11) \begin{equation}\left(\frac{S^{(2)}_n}{b(n)}, \frac{\tau(n)}{n^2} \right) \Rightarrow \big(A^{(2)}, D^{(2)}\big), \ \text{as $n\to\infty$.}\end{equation}

Given that, we will show that the first part of Theorem 1 follows from the next proposition.

Proposition 3. (Theorem 3.1, [Reference Jurlewicz, Kern, Meerschaert and Scheffler23].) Assume (11) holds. Then

\begin{align*}\left\lbrace \frac{\widehat{M}(nt)}{b(\sqrt{n})}, t\geq 0 \right\rbrace = \left\lbrace \frac{M(nt + 1)}{b(\sqrt{n})}, t\geq 0 \right\rbrace \overset{\mathcal{D}}{\Rightarrow} \left\lbrace A^{(2)}\big(E^{(2)}(t)\big), t\geq 0 \right\rbrace, \ \text{as $n\to\infty$. }\end{align*}

A similar result is proven in Theorem 3.6 of [Reference Henry and Straka21].

We will now show that the relation (11) holds and that the RVs $A^{(2)}$ and $D^{(2)}$ are independent. We show that for certain functions $f_1$ and $f_2$ ,

(12) \begin{align}\mathbb{E} \exp\left( i\left(\lambda_1\frac{S^{(2)}_n}{b(n)} + \lambda_2 \frac{\tau(n)}{n^2}\right) \right) &= \mathbb{E} \exp\left( i\left(\lambda_1\frac{\xi^{(2)}_1}{b(n)} + \lambda_2 \frac{\tau(1)}{n^2}\right)\right)^n\\ &\qquad\qquad\qquad = \left(1 + \frac{f_1(\lambda_1) + f_2(\lambda_2)}{n} + o\left(\frac{1}{n} \right) \right)^n,\end{align}

as $n\to\infty$ , for any $\lambda_1, \lambda_2 \in \mathbb{R}$ . Indeed, convergence of characteristic functions is equivalent to weak convergence of RVs, and for independence of RVs it is sufficient to verify that the characteristic function of the sum is equal to the product of the respective characteristic functions. Since the right-hand side of (12) converges to $\exp(f_1(\lambda_1))\exp(f_2(\lambda_2))$ , this will prove the convergence and the independence of the limits $A^{(2)}$ and $D^{(2)}$ .

5.1.1. Proof of Theorem 1

From the condition (2) we have the weak convergence of $S_n^{(2)} / b(n)$ to the RV $A^{(2)}$ . Again, this is equivalent to the convergence of the characteristic functions. Thus, (2) implies

\begin{align*}\mathbb{E}\exp\left(i\lambda_1\frac{\sum_{k=1}^n\xi^{(2)}_k}{b(n)} \right) = \left[ \mathbb{E}\exp\left(i\lambda_1\frac{\xi^{(2)}_1}{b(n)} \right)\right]^n \to \mathbb{E}\exp\left(i\lambda_1 A^{(2)}\right), \ \text{as $n\to\infty$.}\end{align*}

Additionally, if $B^n(n) \to z$ as $n\to\infty$ , then $n\log B(n) \to \log z$ , which leads to $\log B(n) \sim n^{-1}\log z $ and hence, finally, $B(n) \sim 1 + n^{-1}\log z$ , as $n\to\infty$ . Thus, we have the following:

(13) \begin{equation}\mathbb{E}\exp\left(i\lambda_1\frac{\xi^{(2)}_1}{b(n)} \right) \sim 1 + \frac{l_1(\lambda_1)}{n}, \ \text{as $n\to\infty$,}\end{equation}

where $l_1(\lambda) = \log \mathbb{E} \exp\big(i\lambda A^{(2)}\big)$ , the logarithmic characteristic function of $A^{(2)}$ .

As in Section 3, we reformulate the distribution of the RV $\tau(1)$ using the time needed for the simple random walk to hit a fixed point. Let $\big\{\tau^{(1)}_k\big\}_{k=1}^\infty$ be independent copies of $\tau$ , the time needed for the simple random walk to hit 0 if it starts from 1, independent of $\big\{\xi^{(2)}_n\big\}_{n=1}^\infty$ . Given $\xi^{(2)}_1 = m \neq 0$ , the RV $\tau(1)$ is the time needed for the random walk $S^{(1)}_n$ to reach m from zero, and thus, $\tau(1) \overset{d}{=} \sum_{k=1}^{|m| } \tau^{(1)}_k$ . Given $\xi^{(2)}_1 = 0$ , we have $\tau(1) \overset{d}{=} 1 + \tau^{(1)}_1$ . Then we have the following relation for $\tau(1)$ :

\begin{align*}\tau(1) \overset{d}{=} I\Big[\xi^{(2)}_1 \neq 0\Big] \sum_{k=1}^{\big|\xi^{(2)}_1\big| } \tau^{(1)}_k + I\Big[\xi^{(2)}_1 = 0\Big]\Big(1 + \tau^{(1)}_1\Big).\end{align*}

Since $\mathbb{P}\big\{\tau^{(1)}_1 > n\big\} \sim \sqrt{2/(\pi n)}$ , as $n\to\infty$ , we conclude from Proposition 8 (see appendix) that

\begin{align*}\mathbb{P}\{\tau(1) > n\} \sim \Big(\mathbb{E}\Big|\xi^{(2)}_1\Big| + \mathbb{P}\Big\{\xi^{(2)}_1 = 0\Big\}\Big) \mathbb{P}\{\tau > n\}.\end{align*}

Thus, there exists an RV $D^{(2)}$ having a stable distribution with index $1/2$ such that

\begin{align*}\frac{\tau(n)}{n^2} \Rightarrow D^{(2)}, \ \text{as $n\to\infty$}.\end{align*}

Using the same argument as for (13), we get

(14) \begin{equation}\mathbb{E} \exp\left(i\lambda_2\frac{\tau}{n^2} \right) \sim 1 + \frac{l_2(\lambda_2)}{n}, \ \text{as $n\to\infty$,}\end{equation}

where $\lambda_2(\lambda) = \log \mathbb{E} \exp\big(i\lambda D^{(2)}/ \big(\mathbb{E}|\xi^{(2)}_1 \big| + \mathbb{P}\big\{\xi^{(2)}_1 = 0\big\}\big)\big)$ , the logarithmic characteristic function of $D^{(2)}/ \big(\mathbb{E}|\xi^{(2)}_1\big| + \mathbb{P}\big\{\xi^{(2)}_1 = 0\big\}\big)$ . We proceed as follows:

\begin{align*}\mathbb{E} \exp &\left( i\left[\lambda_1\frac{\xi^{(2)}_1}{b(n)} + \lambda_2\frac{ \tau(1)}{n^2}\right]\right)\\& =\sum_{-\infty}^{\infty} \exp\left(i\lambda_1\frac{k}{b(n)} \right)\mathbb{P}\Big\{\xi^{(2)}_1 = k\Big\}\mathbb{E} \exp\left( \left. i\lambda_2\frac{\tau(1)}{n^2} \right| \xi^{(2)}_1 = k \right) \\& = \mathbb{P}\Big\{\xi^{(2)}_1 = 0\Big\}\mathbb{E} \exp\left( i\lambda_2 \frac{1 + \tau}{n^2}\right) \\& + \sum_{k\neq 0} \exp\left(i\lambda_1\frac{k}{b(n)} \right)\mathbb{P}\Big\{\xi^{(2)}_1 = k\Big\}\left(\mathbb{E} \exp\left( i\lambda_2\frac{\tau}{n^2} \right)\right)^{|k|}.\end{align*}

In order to transform the last sum in the last equation, we use (14) and get that, uniformly in $m>0$ ,

\begin{align*}\left(\mathbb{E} \exp\left( i\lambda_2\frac{\tau}{n^2} \right)\right)^m & = \left( 1 + \frac{l_2(\lambda_2)}{n} + o\left(\frac{1}{n}\right)\right)^m \notag \\[3pt]& = \exp\left( m\ln\bigg(1 + \frac{l_2(\lambda_2)+o(1)}{n}\bigg)\right) \notag \\[3pt] & = \exp\left( \frac{m l_2(\lambda_2)}{n}(1 + o(1)) \right) \notag \\[3pt] & = 1 + \frac{m l_2(\lambda_2)(1 + o(1))}{n} + \frac{1}{n^2}\sum_{j=2}^\infty \frac{(m l_2(\lambda_2)(1 + o(1)))^j}{n^{j-2} j!},\end{align*}

as $n\to\infty$ . Since the latter equation is uniform in $m>0$ , we get

\begin{align*}\sum_{k\neq 0} & \exp\left(i\lambda_1\frac{k}{b(n)} \right)\mathbb{P}\Big\{\xi^{(2)}_1 = k\Big\}\left(\mathbb{E} \exp\left( i\lambda_2\frac{\tau}{n^2} \right)\right)^{|k|}\\[3pt] & = \left(\mathbb{E}\exp\left(i\lambda_1\frac{\xi^{(2)}_1}{b(n)} \right) -\mathbb{P}\Big\{\xi^{(2)}_1 = 0\Big\}\right)\\ & + \frac{l_2(\lambda_2)}{n} \sum_{k\neq 0} |k| \exp\left(i\lambda_1\frac{k}{b(n)} \right)\mathbb{P}\Big\{\xi^{(2)}_1 = k\Big\} + \frac{1}{n^2}\sum_{k\neq 0}\sum_{j=2}^\infty \frac{(k l_2(\lambda_2)(1 + o(1)))^j}{n^{j-2} j!}+o\left(\frac{1}{n}\right),\end{align*}

as $n\to\infty$ . Now we use the fact that if $\sum_{-\infty}^{\infty} A_n = \sum_{-\infty}^{\infty} B_n + \sum_{-\infty}^{\infty} C_n$ and if the series $\sum_{-\infty}^{\infty}A_n$ and $\sum_{-\infty}^{\infty}B_n$ converge, then $\sum_{-\infty}^{\infty} C_n$ converges too. Thus,

\begin{equation*}\frac{1}{n^2}\sum_{k\neq 0}\sum_{j=2}^\infty \frac{(k l_2(\lambda_2)(1 + o(1)))^j}{n^{j-2} j!} = o\left(\frac{1}{n}\right)\end{equation*}

and

\begin{align*}\sum_{k\neq 0} & \exp\left(i\lambda_1\frac{k}{b(n)} \right)\mathbb{P}\Big\{\xi^{(2)}_1 = k\Big\}\left(\mathbb{E} \exp\left( i\lambda_2\frac{\tau}{n^2} \right)\right)^{|k|}\\& =\left(\mathbb{E}\exp\left(i\lambda_1\frac{\xi^{(2)}_1}{b(n)} \right) -\mathbb{P}\Big\{\xi^{(2)}_1 = 0\Big\}\right) +\mathbb{E}\Big|\xi^{(2)}_1\Big| \frac{l_2(\lambda_2)}{n} + o\left(\frac{1}{n}\right),\end{align*}

as $n\to\infty$ . Using (13) and (14), we have

\begin{align*}\mathbb{E} \exp\left( i\left[\lambda_1\frac{\xi^{(2)}_1}{b(n)} + \lambda_2\frac{ \tau(1)}{n^2}\right]\right)& \\ &\kern-10pt = 1 + \frac{l_1(\lambda_1)}{n} + \Big(\mathbb{E}\Big|\xi^{(2)}_1\Big| + \mathbb{P}\Big\{\xi^{(2)}_1 = 0\Big\}\Big) \frac{l_2(\lambda_2)}{n} + o\left(\frac{1}{n}\right),\end{align*}

as $n\to\infty$ . We have proved that Equation (12) holds with $f_1(\lambda_1) = l_1(\lambda_1)$ and $f_2(\lambda_2) = \big(\mathbb{E}\big|\xi^{(2)}_1\big| + \mathbb{P}\big\{\xi^{(2)}_1 = 0\big\}\big) l_2(\lambda_2)$ . Therefore, Equation (11) holds and we can use Proposition 3 to prove the first part of Theorem 1.

We now turn to the second part and assume $\mathbb{E}\xi^{(2)} = \mu\neq 0$ . Then the above arguments are applicable to $\sum_{k=1}^{\eta(t) + 1} \big(\xi^{(2)}_k - \mu\big)$ . Thus, we have shown that the process $\left( \left(\sum_{k=1}^{\eta(nt) + 1} \big(\xi^{(2)}_k - \mu\big) \right)/ b(\sqrt{n}), \ t\ge 0 \right) $ weakly converges to the limiting one (see Appendix A for the corresponding definitions). Since $\mu < \infty$ , we have $b(n) = o(n)$ , as $n\to\infty$ . Indeed, by the strong law of large numbers,

\begin{equation*}\frac{\sum_{k=1}^{n} \Big(\xi^{(2)}_k - \mu\Big)}{n} \overset{a.s.}{\to} 0, \ \text{as $n\to\infty$.}\end{equation*}

Since $A^{(2)}$ in (2) is not deterministic, the denominator in the left-hand side of (2) must be an o(n) function. Therefore the process

\begin{align*}\left( \frac{\sum_{k=1}^{\eta(nt) + 1} \Big(\xi^{(2)}_k - \mu\Big)}{\sqrt{n}}, \ t\ge 0 \right) =\left( \frac{\sum_{k=1}^{\eta(nt) + 1} \Big(\xi^{(2)}_k - \mu\Big)}{b(\sqrt{n})} \frac{b(\sqrt{n})}{\sqrt{n}} , \ t\ge 0\right)\end{align*}

converges to the zero-valued process. Thus, it follows from the representation

\begin{align*}\frac{\widehat{M}(nt)}{\sqrt{n}} = \frac{\sum_{k=1}^{\eta(nt)+1} \Big(\xi^{(2)}_k - \mu\Big)}{\sqrt{n}} + \frac{\mu (\eta(nt) +1)}{\sqrt{n}}\end{align*}

and from the corollary to Theorem 3.2 of [Reference Meerschaert and Scheffler29] that

\begin{align*}\left\{ \frac{\widehat{M}(nt)}{\sqrt{n}}, \ t\geq 0 \right\} \overset{\mathcal{D}}{\Rightarrow} \big\{\mu E^{(2)}(t), \ t\geq 0\big\}, \ \text{as $n\to\infty$}.\end{align*}

5.1.2. Proof of Theorem 2

Under the assumption of the finiteness of second moments, we can extend our result to the case where both $\xi^{(1)}$ and $\xi^{(2)}$ have general distributions. Assume now that $\{S^{(1)}_n\}_{n=0}^\infty = \{\sum_{k=1}^n \xi^{(1)}_k\}_{n=0}^\infty$ is an aperiodic random walk with zero-mean and finite-variance- $\sigma_1^2$ increments. The theory of general random walks and their hitting times is well developed. Nevertheless, results that are uniform in terms of the hitting point are rather scarce. It follows from Section 3.3 of [Reference Uchiyama33] that, uniformly in x,

(15) \begin{equation}\mathbb{E}\left[ \exp\left( it \tau(1)\right)\Big | \ \xi^{(2)}_1= x\right] = 1 - (a^*(x) + e_x(t))\Big(\sigma_1\sqrt{-2it} + o\Big(\sqrt{|t|}\Big)\Big), \ \text{as $t\to 0$,}\end{equation}

where

(16) \begin{align}a^*(x) = 1 + \sum_{n=1}^\infty \left(\mathbb{P}\Big\{S^{(1)}_n = 0\Big\} - \mathbb{P}\Big\{S^{(1)}_n = -x\Big\} \right),\end{align}

(17) \begin{align}e_x(t)= c_x(t) + is_x(t),\end{align}

(18) \begin{align}|c_x(t)| = O\left(x^2\sqrt{|t|} \right), \ \text{as $t\to 0$, uniformly in $x$,}\end{align}

(19) \begin{align}s_0(t) = 0 \ \text{and} \ \frac{s_x(t)}{x} = o(1), \ \text{as $t\to 0$, uniformly in $x\neq 0$.}\end{align}

Following steps similar to those used in the previous part, we take $t = \lambda_2 / n^2$ and, eventually, let n become large. A very important relation here is (18). When we take the characteristic function

\begin{equation*}\mathbb{E} \exp\left( i\left[ \lambda_1\frac{\xi^{(2)}_1}{\sigma_2\sqrt{n}} + \lambda_2 \frac{\tau(1)}{n^2}\right]\right)\end{equation*}

and start to separate it into different summands, the relation (18) leads to the summand (see details below)

\begin{align*}\sum_{x \in \mathbb{Z}} O\left(\frac{x^2}{n^2}\right) \mathbb{P}\Big\{\xi^{(2)}_1 = x\Big\}, \ \text{as $n\to\infty$,}\end{align*}

and this is the main reason we need to assume that $\xi^{(2)}_1$ has a finite second moment.

Assume now that $\mathbb{E} \xi^{(2)}_1 = 0$ and $\sigma_2 = \mathbf{Var} \xi^{(2)}_1 < \infty$ . We have (see, e.g., Proposition 7.2 of [Reference Uchiyama34])

(20) \begin{align}\sigma^2_1(a^*(x) - I(x=0)) \sim |x|, \ \text{ as $|x|\to\infty$.}\end{align}

As a consequence, we get $\mathbb{E} a^*\big(\xi^{(2)}_1\big) < \infty$ . Let $p^{(2)}(x) = \mathbb{P}\big\{\xi^{(2)}_1 = x\big\}$ . Then the total probability formula gives us

\begin{align*}\mathbb{E} \exp\left( i\left[ \lambda_1\frac{\xi^{(2)}_1}{\sigma_2\sqrt{n}} + \lambda_2 \frac{\tau(1)}{n^2}\right]\right)& \\ &\kern-50pt = \sum_{x\in\mathbb{Z}}\exp\left( i\lambda_1 \left[\frac{x}{\sigma_2\sqrt{n}} \right] \right)\mathbb{E}\left[ \exp\left.\left( i\left[\lambda_2 \frac{\tau(1)}{n^2}\right]\right) \right| \ \xi^{(2)}_1= x\right] p^{(2)}(x).\end{align*}

Now we use (15)–(19) to get

\begin{align*}\mathbb{E} \exp\left( i\left[ \lambda_1\frac{\xi^{(2)}_1}{\sigma_2\sqrt{n}} + \lambda_2 \frac{\tau(1)}{n^2}\right]\right) & = \mathbb{E} \left[\exp\left( i\lambda_1 \left[\frac{\xi^{(2)}_1}{\sigma_2\sqrt{n}} \right]\right)\right] \\[3pt] & \quad - \frac{\sigma_1 \sqrt{-2i\lambda_2}}{n} \mathbb{E} \left[a^*\Big(\xi^{(2)}_1\Big) \exp\left( i\lambda_1 \left[\frac{\xi^{(2)}_1}{\sigma_2\sqrt{n}} \right]\right)\right] \\[3pt] &\quad + O\left( \frac{1}{n^2} \mathbb{E} \left[\left( \xi^{(2)}_1\right)^2 \exp\left( i\lambda_1 \left[\frac{\xi^{(2)}_1}{\sigma_2\sqrt{n}} \right]\right)\right]\right) \\[3pt] &\quad + o\left( \frac{1}{n} \mathbb{E} \left[\xi^{(2)}_1 \exp\left( i\lambda_1 \left[\frac{\xi^{(2)}_1}{\sigma_2\sqrt{n}} \right]\right)\right]\right) + o\left(\frac{1}{n} \right),\end{align*}

as $n\to\infty$ . Next, we use the relation (20) and the Taylor expansion for the exponent to get

\begin{align*} \mathbb{E} \left[a^*\Big(\xi^{(2)}_1\Big) \exp\left( i\lambda_1 \left[\frac{\xi^{(2)}_1}{\sigma_2\sqrt{n}} \right]\right)\right] = \mathbb{E} \left[a^*\Big(\xi^{(2)}_1\Big) \right] + o(1), \ \text{as $n\to\infty$.}\end{align*}

Since $\mathbb{E} \xi^{(2)}_1 = 0$ and $\mathbf{Var} \xi^{(2)}_1 < \infty$ , the central limit theorem holds. Thus, we have the analogue of (14) with $l_1$ being a logarithmic characteristic function of an RV with a standard normal distribution. Finally, we get

\begin{multline*}\mathbb{E} \exp\left( i\left[ \lambda_1\frac{\xi^{(2)}_1}{\sigma_2\sqrt{n}} + \lambda_2 \frac{\tau(1)}{n^2}\right]\right) = 1 + \frac{l_1(\lambda_1)}{n} - \frac{\sigma_1 \sqrt{-2i\lambda_2}}{n} \mathbb{E} \left[a^*\Big(\xi^{(2)}_1\Big) \right] + o\left(\frac{1}{n}\right).\end{multline*}

Thus, we have proved Equation (12) for this case. The rest of the proof follows the same argument as in the previous case.

5.2. Proof of Theorem 3

Recall that the random vectors $\big\{Y^{(3)}_n, J^{(3)}_n\big\}_{n=1}^\infty$ are i.i.d., where $Y^{(3)}_1 = \sum_{k=1}^\nu \gamma^{(3)}_k$ and $J^{(3)}_1 = T^{(3)}(1) = U^{(3)}(\nu)$ . We have

\begin{align*}N(t) = \max\!\left\{n> 0: \ \sum_{k=1}^n J^{(3)}_k \leq t\right\} \text{and} \ \widetilde{M}(t) = \sum_{k = 1}^{N(t)} Y^{(3)}_k.\end{align*}

From Propositions 1 and 2 we have

\begin{align*}\mathbb{E} Y^{(3)}_1 = 0, \ \mathbf{Var} Y^{(3)}_1 = 2, \ \mathbb{E} \Big(Y^{(3)}_1\Big)^m < \infty, \ \text{for $m\geq 2$,} \ \text{and} \ \mathbb{P}\Big\{J^{(3)}_1 > n\Big\} \sim \frac{2^{7/4}}{\Gamma(3/4) n^{1/4}},\end{align*}

as $n\to\infty$ . From Theorem 5.1 from [Reference Kasahara24] we have

(21) \begin{equation}\left\lbrace \frac{\widetilde{M}(nt)}{2^{-7/8}n^{1/8}\sqrt{\mathbf{Var} Y^{(3)}_1}}, t\geq 0\right\rbrace \overset{\mathcal{D}}{\Rightarrow} \{B(E^{(3)}(t)), t\geq 0\}, \ \text{as $n\to\infty$,}\end{equation}

where B(t) is a standard Brownian motion, independent of $E^{(3)}(t)$ .

We show now that (21) holds with M(nt) in the place of $\widetilde{M}(nt)$ . It is sufficient to prove that, for any fixed $T>0$ ,

\begin{align*}\frac{\max_{1 \leq k \leq [nT]}\left\lbrace \widetilde{M}_{k} - M_{k}\right\rbrace}{n^{1/8}} \overset{a.s.}{\to} 0, \ \text{as $n\to\infty$. }\end{align*}

For $k=1,2,\ldots$ , let $\nu_k$ be the number of visits to the set $\{(0,0),(0,1)\}$ by the MC $V_n$ within the time interval $\big(T_{k-1}^{(3)}, T_k^{(3)}\big]$ , and let $\varkappa_k$ be the total number of jumps of the mouse within this time interval. Further, let

\begin{equation*}R_k = \max_{T^{(3)}(k-1) \leq l \leq T^{(3)}(k)} |\widetilde{M}_l - M_l|.\end{equation*}

The triples $(\nu_k,\varkappa_k,R_k)$ are i.i.d., the pairs $(\nu_k,\varkappa_k)$ are i.i.d. copies of the pair $(\nu,\varkappa)$ introduced earlier, and

(22) \begin{align}R_k \le \varkappa_k \le 2\nu_k \ \ \mbox{a.s.}\end{align}

Further, all RVs in (22) have a finite exponential moment, ${\mathbb E} \exp (cR_1)<\infty$ for some $c>0$ .

For $K>0$ , let $\widehat{J}_j = \min \big(J^{(3)}_j, K\big)$ . Since ${\mathbb E} J^{(3)}_1=\infty$ , one can choose K such that $A:={\mathbb E} \widehat{J}_1> T$ . Let $\widehat{S}_n=\sum_{i=1}^n \widehat{J}_i$ . Then, for any $C>0$ ,

\begin{align*}{\mathbb P} (\eta (nT)+1 >n) &\le {\mathbb P}(\widehat{S}_n\le nT) \le \left(e^{CT}{\mathbb E} e^{-C\widehat{J}_1} \right)^n,\end{align*}

where the term in the parentheses on the right-hand side may be made less than 1 for $C>0$ sufficiently small. Next,

\begin{align*}{\mathbb P}\bigg( \max_{1\le k \le n} R_k \ge n^{1/8} \varepsilon \bigg)&\le n{\mathbb P} \big(R_1 \ge n^{1/8} \varepsilon \big)\le n {\mathbb E} e^{cR_1} \cdot e^{-c\varepsilon n^{1/8}}.\end{align*}

Then, for any $\varepsilon >0$ ,

\begin{align*}{\mathbb P} \left(\frac{\max_{1 \leq k \leq [nT]}\left\lbrace \widetilde{M}_{k} - M_{k}\right\rbrace}{n^{1/8}}\ge \varepsilon\right)\le{\mathbb P}\bigg( \max_{1\le k \le n} R_k \ge n^{1/8}\varepsilon \bigg) +{\mathbb P} (\eta (nT)\ge n),\end{align*}

where both terms on the right-hand side are summable in n. Therefore, by the 0–1 law and by the arbitrariness of $\varepsilon >0$ , (22) follows. This completes the proof of Theorem 3.

5.3. Proof of Theorem 4

The random process $X^{(1)}$ is a simple random walk on $\mathbb{Z}$ , and for $j \in [2, N]$ we have

\begin{align*}\mathbb{P}\big\{X^{(j)}(n) - X^{(j)}(n-1) &= 1\big| \ X^{(j)}(n-1) = X^{(j-1)}(n-1)\big\}\\[2pt] &= \mathbb{P}\big\{X^{(j)}(n) - X^{(j)}(n-1) = - 1\big| \ X^{(j)}(n-1) = X^{(j-1)}(n-1)\big\} = \frac{1}{2}.\end{align*}

Let us give a new representation of such a process. Let $X^{(1)}(0) = X^{(2)}(0) = \ldots = X^{(N)}(0) = 0$ , and let the RV $T_j(n)$ denote the time when $X^{(j)}$ takes the nth step. Let $T_j(0) = 0$ . Note the difference between $T_j$ and $T^{(3)}$ . Thus, $\big\{X^{(j)}\big(T_j(k)\big)\big\}_{k=0}^\infty$ is a simple random walk on $\mathbb{Z}$ , and if $X^{(j)}(n) \neq X^{(j)}(n-1)$ , then $n\in \{T_j(k)\}_{k=1}^\infty$ . Let

\begin{equation*}\xi^{(j)}_k = X^{(j)}(T_j(k)) - X^{(j)}(T_j(k-1)) =X^{(j)}(T_j(k)) - X^{(j)}(T_j(k)-1)\end{equation*}

for $j\geq 1$ and $k\geq 1$ . By definition, $\big\{\big\{\xi^{(j)}_k\big\}_{k=0}^\infty\big\}_{j=1}^N$ are mutually independent and equal $\pm 1$ w.p. $1/2$ .

Since $X^{(1)}$ jumps every time, $T_1(k) = k$ for $k\geq 0$ . Let $\tau$ be the time at which the simple random walk goes from the point 1 to 0. From Section 3 it is easy to deduce that the time between the meeting-time instants of the cat and the mouse has the same distribution as $\tau$ . Thus, if we look at the system only at the times $\{T_j(k)\}_{k=0}^\infty$ the time between meeting-time instants of $X^{(j)}$ and $X^{(j+1)}$ has the same distribution as $\tau$ .

Let us define $\nu_j(n) = \max\{k\ge 0: \ T_j(k) \le n\}$ , the number of time instants up to time n at which $X^{(j)}$ changes its value. Then we can rewrite the dynamics of the jth coordinate as

\begin{equation*}X^{(j)}(n) = \sum_{k=1}^{\nu_j(n)} \xi^{(j)}_{T_j(k)}.\end{equation*}

Our assumptions on the distribution of the increments $\xi^{(j)}_k$ , for $k\ge 1$ , provide us with the next important property of our process.

This property comes from the space-symmetry of the model. Indeed, for $j=1$ the result is trivial, since $\nu_1(n) = n$ . We show the result for $j=2$ and then extend it onto $j>2$ . Define

\begin{align*}{}^1\tau(0) = 0 \ \text{and} \ {}^1\tau(k) = \inf\big\{n > {}^1\tau(k-1): \ X^{(1)}(n) = X^{(2)}(n)\big\}, \ \text{for $k\ge 1$}.\end{align*}

One can see that in our model $T_2(k) = 1 + {}^1\tau(k-1)$ , for $k\ge 1$ . In the time interval $[1, {}^1\tau(1)]$ the second coordinate changes its value only at the time $T_2(1) = 1$ . Thus, the time ${}^1\tau(1)$ does not depend on $\xi^{(2)}_k$ , for $k\ge 2$ . Additionally, the trajectory $\{X^{(1)}(n)\}_{n=0}^\infty$ has the same distribution as $\{-X^{(1)}(n)\}_{n=0}^\infty$ . Thus,

\begin{align*}\mathbb{P}\Big\{{}^1\tau(1) = n, \ \xi^{(2)}_1 = 1\Big\} = \mathbb{P}\Big\{{}^1\tau(1) = n, \ \xi^{(2)}_1 = -1\Big\}.\end{align*}

As a corollary of the last equation, we get that ${}^1\tau(1)$ has the same distribution as the time that is needed for the simple random walk to hit 0 if it starts from 1. This implies that

(23) \begin{align}\mathbb{P}\big\{{}^1\tau(1) > n\big\} \sim \sqrt{\frac{2}{\pi n}}, \ \text{as $n\to\infty$.}\end{align}

From the symmetry of our model, it further follows that the sequence $\{{}^1\tau(k)\}_{k=0}^\infty$ , and subsequently the sequences $\{T_2(k)\}_{k=0}^\infty$ and $\{\nu_2(n)\}_{n=1}^\infty$ , do not depend on $\big\{\xi^{(2)}_k\big\}_{k=1}^\infty$ (or, therefore, on $\big\{\xi^{(j)}_k\big\}_{k\ge 1, j\ge2}$ ).

For the analysis of $\{T_j(k)\}_{k=0}^\infty$ , $j > 2$ , we need to define an ‘embedded version’ of ${}^1\tau(k)$ . Let

\begin{align*}{}^j\tau(0) = 0 \ \text{and} \ {}^j\tau(k) = \inf\Big\{m > {}^j\tau(k-1): \ X^{(j)}(T_j(m)) = X^{(j+1)}(T_j(m))\Big\}, \ \text{for $k\ge 1$}.\end{align*}

The process $\big\{{}^j\tau(k)\big\}_{k=0}^\infty$ counts the number of times that the process $X^{(j)}$ changes its value between the time instants when $X^{(j)}$ and $X^{(j+1)}$ have the same value. Using the same argument as before, we get that the sequence $\big\{{}^j\tau(k)\big\}_{k=0}^\infty$ does not depend on $\big\{\xi^{(j+1)}_k\big\}_{k=1}^\infty$ .

The jth coordinate $X^{(j)}$ changes its value for the kth time at a time instant n if and only if, up to time $n-1$ , the processes $X^{(j-1)}$ and $X^{(j)}$ have had the same value exactly $k-1$ times (not including $X^{(j-1)}(0) = X^{(j)}(0) = 0$ ), with the last time being at the time instant $n-1$ (which also means that at the time instant $n-1$ the process $X^{(j-1)}$ changes its value). This can be rewritten as

\begin{align*}T_j(k) = n \ \Leftrightarrow \ n-1 = T_{j-1}\big({}^{j-1}\tau(k-1)\big), \ \text{for $j\ge 2$, $k\ge 1$,}\end{align*}

and thus $T_j(k) = 1 + T_{j-1}({}^{j-1}\tau(k-1))$ . Thus, since the sequences $\{T_2(k)\}_{k=0}^\infty$ and $\big\{{}^2\tau(k)\big\}_{k=0}^\infty$ do not depend on $\big\{\xi^{(j)}_k\big\}_{k\ge 1, j\ge 3}$ , the same holds for $\{T_3(k)\}_{k=0}^\infty$ . Therefore, using induction, we get that the sequences $\{T_j(k)\}_{k=0}^\infty$ and $\big\{\xi^{(j)}_k\big\}_{k=1}^\infty$ are independent for any $j\ge 1$ .

As a corollary of this result we get

\begin{align*}X^{(j)}(n) = \sum_{k=1}^{\nu_j(n)} \xi^{(j)}_{T_j(k)} \overset{d}{=} \sum_{k=1}^{\nu_j(n)} \xi^{(j)}_k.\end{align*}

Let ${}^j\eta(n) = \max\{k\geq 0 :\ {}^j\tau(k) \leq n\}$ for $n\geq 0$ and $j \in [1, \ldots, N]$ . Since the sequence $\{{}^j\tau(k)\}_{k=0}^\infty$ depends only on the sequence $\big\{\xi^{(j)}_k\big\}_{k=1}^\infty$ , we have that $\{{}^j\eta(n)\}_{j=1}^{N-1}$ are i.i.d. RVs. For $n\geq 1$ and $j \in \{1, \ldots, N\}$ we have

\begin{align*}\nu_j(n) = \max\{k\geq 0:\ T_j(k) \leq n\} & = \max\Big\{k\geq 1:\ 1 + T_{j-1}\big({}^{j-1}\tau(k-1)\big) \leq n\Big\}\notag\\[2pt] & = 1 + \max\Big\{k\geq 0:\ T_{j-1}\big({}^{j-1}\tau(k)\big) \leq n - 1\Big\}\notag\\[2pt] & = 1 + \max\Big\{k\geq 0:\ {}^{j-1}\tau(k) \leq \nu_{j-1}(n - 1)\Big\}\notag\\[2pt] & = 1 + {}^{j-1}\eta\big(\nu_{j-1}(n-1)\big).\end{align*}

For $n < N-1$ we can iterate the process and get

\begin{align*}\nu_N(n) & = 1 + {}^{N-1}\eta\Big(1 + {}^{N-2}\eta\Big(\ldots \Big(1 + {}^{N-n}\eta(0)\Big)\ldots\Big)\Big)\notag\\[2pt] & = 1 + {}^{N-1}\eta\Big(1 + {}^{N-2}\eta\Big(\ldots \Big(1 + {}^{N-n+1}\eta(1)\Big)\ldots\Big)\Big)\notag\\[2pt] & \overset{d}{=} 1 + {}^{n-1}\eta\Big(1 + {}^{n-2}\eta(\ldots \Big(1 + {}^{1}\eta(1)\Big)\ldots\Big)\Big)\notag\\[2pt] & = \nu_n(n).\end{align*}

For $n \ge N-1$ we have

\begin{align*}\nu_N(n) = 1 + {}^{N-1}\eta\Big(1 + {}^{N-2}\eta\Big(\ldots + {}^1\eta(n-N+1)\Big)\Big).\end{align*}

We want to construct a process with the same distribution as $\{\nu_N(n)\}_{n=0}^\infty$ in a form of $\nu_{N-1}(\varphi(n))$ , where the process $\{\varphi(n)\}_{n=0}^\infty$ is independent of everything else. Define the process $\{\eta(n)\}_{n=0}^\infty \overset{d}{=} \big\{{}^{N-1}\eta(n)\big\}_{n=0}^\infty$ , which is independent of everything else. Then, for $n \ge N-1$ , we have

\begin{align*}\nu_N(n) \overset{d}{=} 1 + {}^{N-2}\eta\Big(1 + {}^{N-3}\eta\Big(\ldots + \eta(n-N+1)\Big)\Big).\end{align*}

Using the same formula for $\nu_{N-1}(m)$ with m such that $m- (N\,{-}\,1) + 1 = 1 + {}^{N-1}\eta(n-N\,{+}\,1)$ , we get

\begin{align*}\nu_N(n) \overset{d}{=} \nu_{N-1}(N-1 + \eta(n-N+1)), \ \text{for $n \ge N-1$.}\end{align*}

Then, for $n\geq N$ , we have $X^{(N)}(n) \overset{d}{=} X^{(N-1)}(N -1 + \eta(n-N+1))$ . There exists a nondegenerate RV $\zeta$ (see Section XI.5 in [Reference Feller13]) such that $\mathbb{P}\{\zeta_i \geq y\} = G_{1/2}\big(9/y^2\big)$ and

\begin{align*}\eta(n)\mathbb{P}\big\{{}^{N-1}\tau(1) > n\big\} \Rightarrow \zeta, \quad \text{as $n\to\infty$.}\end{align*}

Therefore, using (23) we get

(24) \begin{align}\frac{j -1 + \eta(n-j+1)}{\sqrt{n}} = \frac{j -1 + \eta(n-j+1)}{\sqrt{n-j+1}}\frac{\sqrt{n-j+1}}{\sqrt{n}} \Rightarrow \sqrt{\frac{\pi}{2}}\zeta,\end{align}

as $n\to\infty$ , for $j\ge 1$ . We now present a known result that we utilise to prove Theorem 4.

Proposition 5. ([Reference Dobrushin11], (v).) Let Y(t) and $\tau_n$ be independent sequences of RVs such that

(25) \begin{align}\frac{Y(t)}{bt^\beta} \Rightarrow Y, \ as\ \text{$t\to\infty$,}\ and \ \ \frac{\tau_n}{d n^\delta} \Rightarrow \tau, \ as\ \text{$n\to\infty$.}\end{align}

Then for independent Y and $\tau$ we have

\begin{align*}\frac{Y(\tau_n)}{b d^\beta n^{\beta\delta}} \Rightarrow Y \tau^\beta , \ as\ \text{$n\to\infty$.}\end{align*}

Indeed, by the central limit theorem, $X^{(1)}(n) / \sqrt{n}$ weakly converges to a normally distributed RV $\psi$ (we assume that $\psi$ and $\zeta$ are independent). Together with (24) and the independence of $X^{(1)}(n)$ and $\eta(n)$ , this ensures that the condition (25) holds with $Y(t) = X^{(1)}([t])$ , $\tau_n = 1 + \eta(n-1)$ , and $\beta = \delta = 1/2$ . By Proposition 5,

\begin{align*}\frac{X^{(2)}(n)}{n^{1/4}} \overset{d}{=} \frac{X^{(1)}(1 + \eta(n-1))}{n^{1/4}} \Rightarrow \psi \sqrt{\sqrt{\frac{\pi}{2}}\zeta}, \ \text{as $n\to\infty$.}\end{align*}

Let $\{\zeta_j\}_{j=2}^{N}$ be independent copies of $\zeta$ which are independent of $\psi$ . Next, we use the induction argument. For some $j\ge 1$ , the condition (25) holds with $Y(t) = X^{(j)}([t])$ , $\tau_n = j-1 + \eta(n-j+1)$ , $\beta = 2^{-j}$ , and $\delta = 2^{-1}$ . By Proposition 5, we get

\begin{align*}\frac{X^{(j+1)}(n)}{n^{2^{-(j+1)}}}\overset{d}{=} \frac{X^{(j)}(j-1 + \eta(n-j+1))}{n^{2^{-(j+1)}}} \Rightarrow \psi \prod_{i=2}^{j+1}\sqrt{\sqrt{\frac{\pi}{2}}\zeta_i}, \quad \text{as $n\to\infty$.}\end{align*}

This concludes the proof of Theorem 4.