Abstract
We introduce a new type of random walk where the definition of edge repellence/reinforcement is very different from the one in the “traditional” reinforced random walk models and investigate its basic properties, such as null versus positive recurrence, transience, as well as the speed. The two basic cases will be dubbed “impatient” and “ageing” random walks.
Keywords
Random walk Impatient random walk Ageing random walk Passage generating function Transience RecurrenceMathematics Subject Classification (2010)
60J101 Introduction
1.1 Model
Consider an infinite connected graph G. The set of edges will be denoted by E(G). With some slight abuse of notation, \(v\in G\) will mean that v is a vertex of G. Consider a random walk \(X=\{X_n\}_{n\ge 0}\) on the vertex set of G, with the jumps restricted to E(G).
Assumption 1
(Nondegeneracy) The jumps have strictly positive probabilities for each edge in both directions. In particular, X is an irreducible Markov chain on G.
Fix a vertex \(v_0\in G\) which we call “origin”, and assume that the walk starts at this point, \(X_0=v_0\); for \(G={\mathbb {Z}}^d\), the default will be \(v_0:={\mathbf {0}}\).
Definition 1
(Passage times) A sequence \(s_0,s_1,s_2,\ldots \) of nonnegative real numbers will be called a sequence of passage times if \(s_0=1\).
Definition 2
(Walk modified by passage times) We will modify the walk in such a way that if it has crossed an edge e exactly k times before, then it takes \(s_k\) units of time (as opposed to 1) to cross this edge again, in either direction; in particular, it takes one unit of time to cross the edge for the first time.
The two basic cases are as follows:
Definition 3
 (i)
impatient^{1} when \(s_k\downarrow 0\);
 (ii)
ageing when \(s_k\uparrow \infty \).
Remark 1
If \(s_k\downarrow s_\infty >0\) or \(s_k\uparrow s_\infty <\infty \), then the questions (about recurrence, speed, etc.) we investigate in this paper will be equivalent to the corresponding ones related to the original random walk; hence, these cases are not interesting and are not considered in our paper.
To have a more formal definition, note the main feature of the process we are studying: the system’s “actual” time depends on the local time of the walk.
Definition 4
Let the random function \(U:[0,\infty )\rightarrow [0,\infty )\) denote the right continuous generalized inverse of T, that is, \(U(t):=\sup \{s: T(s)\le t\}.\) With the exception of one case, we will work with \(s_k>0\) for all \(k\ge 0\), and then T is strictly increasing in t and \(U=T^{1}\).
Definition 5
1.2 Motivation
Imagine that at every edge, one has to perform a certain task. For example, the edge represents a piece of road, where driving through is not trivial for some reason. Or that piece of connection between the vertices is itself a small maze, one has to learn to solve. Then, the more one solved it in the past, the quicker it goes, and so the impatient walk models a learning process.^{3}
Similarly, one can think of a model where the roads, or paths which are often used deteriorate with time, and therefore passing them becomes harder and harder and thus takes more and more time. It seems that ageing random walk can provide a good model for this situation.
While the model we introduce is somewhat reminiscent of the famous edgereinforced random walk (see [7] for a survey) as well as the “cookie” walk (introduced in [8]), the behaviour in our model differs significantly from these latter ones, since in our case the transition probabilities remain intact, while “reinforcement” affects only passage times.
1.3 Notation
As usual, \({\mathbb {Z}}_+\) will denote the set of nonnegative integers and \({\mathbb {Z}}^d\) will denote the ddimensional integer lattice. We will write \(a_n\asymp b_n\) if \(\lim _{n\rightarrow \infty } a_n/b_n=1\) and \(a_n\sim b_n\) if \(a_n/b_n=O(1)\) and \(b_n/a_n=O(1)\). The letters \({\mathbb {P}}\) and \({{\mathbb {E}}}\) will denote the probability and expectation corresponding to the impatient/ageing random walk, respectively; when the starting point is emphasized, we will write \({\mathbb {P}}^{v_{0}}\) and \({{\mathbb {E}}}^{v_{0}}\).
1.4 Questions We Investigate
One object we would like to study is \({{\widetilde{\tau }}}\), the return time to the origin for the impatient walk, which is defined precisely as follows.
Definition 6
An interesting phenomenon which arises in our model is that, depending on the passage times, \(X^{{\mathrm {imp}}}\) can be positive recurrent even if the original walk was null recurrent.
Remark 2
The distribution of \({{\widetilde{\tau }}}_n\) will in general depend on the history of the process \({\mathcal {F}}_n:=\sigma \{X_0,X_1,\ldots ,X_{\tau _{n1}}\}\).
Another aspect of interest is the spatial speed (spread) of the process. We now need a definition.
Definition 7
(Infinitely impatient walk) Consider the walk on \(G:={\mathbb {Z}}_+\), and an extreme case, when \(s_k=0\) for all \(k\ge 1\). That is, old edges are passed instantaneously. We call this walk the “infinitely impatient walk” and denote it by \(X^{\mathrm {inf.imp}}\).
Clearly, \(X^{\mathrm {inf.imp}}\) just steps to the right every time unit, because excursions to the left happen in “infinitesimally small” times. Hence, it spreads with constant speed. On the other hand, when \(s_k=1\), for \(k=1,2,\ldots ,\) we get the classical random walk for which the range up to n scales with \(\sqrt{n}\). So, this indicates that the scaling is always between \(\sqrt{n}\) and n, and it depends on the passage times in some way. See also Remark 3 and Theorem 8. This latter theorem will also shed some light on how the classical ArcSine Law is modified in our setting.
1.5 Basic Notions and a Useful Lemma
When the passage times are summable, \(X^{{\mathrm {imp}}}\) can only spend a finite amount of time (uniformly bounded by S) on any given edge. Accordingly, we make the following definition.
Definition 8
 (A1)
\(1\le S:=\sum _{k=0}^{\infty } s_k <\infty ,\) and weakly impatient if
 (A2)
\(S:=\sum _{k=0}^{\infty } s_k =\infty .\)
Remark 3
(Strong/weak impatience and speed) Clearly, when the walk is strongly impatient, it spends at most S time on each edge, and so by time t (we are talking about “actual time” here), it has visited at least \(\lfloor t/S\rfloor \) different edges. This means that the spread (measured by the number of distinct edges crossed by the process up to time \(t>0\)) is linear, with the constant being between 1 / S and 1 (because it has spent at least one unit of time on each edge visited). It would be desirable to figure out how the constant depends on the passage times. A reasonable conjecture is that it is 1 / S. See Sect. 5.
In order to actually change linearity and get closer to order \(\sqrt{t}\), one needs the walk to be weakly impatient. And so the question, in this case, is how the passage times will determine the order between linear and square root.
Next, regarding recurrence, we make the following definitions.
Definition 9
 (B1)

recurrent if \({{\widetilde{\tau }}}_n(v_0)<\infty \), \({\mathbb {P}}^{v_{0}}\)a.s. for all \(n\ge 0\) and all \(v_0\in G\);
 (B2)

transient if it is not recurrent;
 (C1)

positive recurrent if \({{\mathbb {E}}}^{v_{0}} {{\widetilde{\tau }}}_n(v_0)<\infty \) for all \(n\ge 0\) and all \(v_0\in G\);
 (C2)

null recurrent if it is recurrent and \({{\mathbb {E}}}^{v_{0}} {{\widetilde{\tau }}}_n(v_0)=\infty \) for all \(n\ge 0\) and all \(v_0\in G\).
Clearly, if \({{\widetilde{\tau }}}_n(v_0)<\infty \) a.s. for some \(n\ge 0,v_0\in G\), then X must be recurrent, in which case \({{\widetilde{\tau }}}_n(v_0)<\infty \) a.s. for all \(n\ge 0\) and \(v_0\in G\).
A similar statement holds for positive recurrence.
Theorem 1
(Process property) Recall Assumption 1 and assume also that X on G is recurrent. Then, \(X^{\mathrm {imp}}\) is either positive recurrent or null recurrent. In other words, the properties in (C1–C2) do not depend on the choice of n or \(v_0\).
Proof
 (i)
for given \(v_0\in G\), the property does not depend on \(n\ge 0\);
 (ii)
for \(n=0\), the property does not depend on choice of \(v_0\in G\).
 (i)
Assume first that \({{\mathbb {E}}}^{v_{0}} {{\widetilde{\tau }}}_0(v_0)<\infty \). Then, because of the monotonicity of s, \({{\mathbb {E}}}^{v_{0}} {{\widetilde{\tau }}}_n(v_0)<\infty \) for \(n>0\), as well.
Now assume that \({{\mathbb {E}}}^{v_{0}} {{\widetilde{\tau }}}_0(v_0)=\infty \). Let \(e=(v_0,v_1)\) be an outgoing edge from \(v_0\). If \(n\ge 1\), then (because of the nondegeneracy of X on G) the event(i.e. the first n excursions consist only of traversing e back and forth, so that this edge has been crossed 2n times) has a positive probability.$$\begin{aligned} A_n:=\{X_0=v_0,\quad X_1=v_1,\quad X_2=v_0,\quad X_3=v_1,\ldots , X_{2n1}=v_1,\quad X_{2n}=v_0\} \end{aligned}$$On \(A_n\), the next passage time on e has been set to \(s_{2n}\), while no other edge has been crossed. It is enough to show thatNow, (1) would certainly be true without the impatience mechanism of the model, as we assumed that \({{\mathbb {E}}}^{v_{0}} {{\widetilde{\tau }}}_0=\infty \), so our task is to prove that this mechanism does not change the validity of (1).$$\begin{aligned} {{\mathbb {E}}}({{\widetilde{\tau }}}_n(v_0)\mid A_n)=\infty . \end{aligned}$$(1)To this end, let \(\eta \) denote the number of crossings of e in the \(n+1\)st excursion starting at \(v_0\). Then, either \(\eta =0\) (when both the initial and the last edges are different from e), or \(\eta =1\) (when the final edge is e and the initial edge is not, or vice versa), or \(\eta =2\) (when both the initial and the last edges are e). Recall the notion of “actual time” from Definition 4. If \(p_j:={\mathbb {P}}(\eta =j)\) for \(j=0,1,2\), then the expected actual time spent on e in the excursion with \(s_0=1\) initial passage time is \(p_1 +p_2 (1+s_1)\), whereas with \(s_{2n}\) initial passage time it is \(p_1 s_{2n}+p_2(s_{2n}+s_{2n+1}).\) These are finite quantities, and therefore resetting the initial time to \(s_{2n}\) from \(s_0\) does not change the finiteness of the expected actual return time.
 (ii)
Assume now that \(a:={{\mathbb {E}}}^{v_0} {{\widetilde{\tau }}}_0(v_0)<\infty \). We will show that for any \(v_1\ne v_0\) we also have \({{\mathbb {E}}}^{v_1} {{\widetilde{\tau }}}_0(v_1)<\infty \). Since G is connected, without the loss of generality, we may (and will) assume that they \(v_1\) and \(v_0\) are neighbours on G.
Let \(p_{0\rightarrow 1}:={\mathbb {P}}^{v_{0}}(X_1=v_1)>0\). Note that \({{\mathbb {E}}}(\kappa \Vert X_0=v_0,X_1=v_1)<\infty ,\) where$$\begin{aligned} \kappa&:=T(\min \{k\ge 1:\ X_k=v_0\}),\quad \text {since }\\ a&\ge p_{0\rightarrow 1}\cdot {{\mathbb {E}}}(\kappa \Vert X_0=v_0,X_1=v_1). \end{aligned}$$
Given that the original walk is recurrent, so are the impatient and ageing random walks, since \(T(n)<\infty \), whenever \(n<\infty \). The question of positive versus null recurrence is not so trivial though. The statement below follows from the obvious inequality \({{\widetilde{\tau }}}\le \tau \) for \(X^{{\mathrm {imp}}}\).
Remark 4
If the original walk X is positive recurrent, then \(X^{{\mathrm {imp}}}\) is also positive recurrent.
Finally, define \(M:=\mathsf{card}\{(X_{i1},X_i),\ i=1,\ldots ,\tau _1(v_0)\}\), that is, M is the number of distinct edges crossed by X between consecutive visits to the origin. The following statement involving M will be useful later.
Lemma 1
 (i)
positive recurrent, provided that the walk is strongly impatient and \({{\mathbb {E}}}M<\infty \);
 (ii)
null recurrent, provided that \({{\mathbb {E}}}M=\infty \).
Proof
Corollary 1
(Switching from null to positive, recurrence) If the original walk X is null recurrent but \({{\mathbb {E}}}M<\infty \) holds, then strong impatience turns null recurrence into positive recurrence (for \(X^{{\mathrm {imp}}}\)).
This is the case, for example, for the random walk mentioned in the second part of Theorem 4 later.
2 Impatient Simple Random Walk on \({\mathbb {Z}}^d\), \(d=1,2\), and Generalization
It turns out that for \(d=1,2\), impatience cannot change the nullrecurrent character of the simple random walk.
Theorem 2
The impatient simple random walks on \({\mathbb {Z}}^1\) and on \({\mathbb {Z}}^2\) are null recurrent for any sequence of passage times.
Proof

\(\underline{d=1:}\) We may assume that \(v_0={\mathbf {0}}\) and that the first step of the walk is to the right, \(X_1=1\). The probability to reach vertex \(m\ge 1\) before returning to the origin is 1 / m, hence \({\mathbb {P}}(M\ge m)=1/m\), \(m=1,2,\ldots \). (Note that the number of distinct edges visited by the onedimensional walk coincides with its maximum during the excursion). Consequently, \({{\mathbb {E}}}M=\infty \) and by Lemma 1 the impatient random walk is null recurrent.

\(\underline{d=2:}\) one can easily show (e.g. by coupling) that the quantity M for \(d=2\) is stochastically larger than for \(d=1\), so \({{\mathbb {E}}}M=\infty \) here and we again can use Lemma 1. \(\square \)
Remark 5
As the following example shows, one can easily construct a null recurrent random walk on \({\mathbb {Z}}^2\) (and similarly on \({\mathbb {Z}}^d\), \(d\ge 3\)) such that the strongly impatient random walk with the same transition probabilities has \({{\mathbb {E}}}M<\infty \) and hence is positive recurrent.
Indeed, consider the following random walk on \({\mathbb {Z}}^2\). Let \((x,y):= \max (x,y)\) be the “norm” of a vertex, let orbit \(O_k=\{(x,y): (x,y)=k\}\), \(k=1,2,3,\ldots \) that is, the squares with sides of length 2k parallel to the axes and the centre in the origin (please see the picture, each orbit has a distinctive colour).
Assume that the transition probabilities are the following: once the walker is on \(O_k\), it goes to the previous orbit \(O_{k1}\) with probability \(\frac{2}{3} \cdot 2^{k}\), or to the next orbit \(O_{k+1}\) with probability \(\frac{1}{3}\cdot 2^{k}\), and with the remaining probability \(12^{k}\), it goes clockwise staying on \(O_k\). (This needs to be adjusted somewhat at the four corners of \(O_k\).)
If we consider the embedded process, defined as the norm of the point whenever it changes, then it is a positive recurrent random walk with probability 2 / 3 going towards the origin, and 1 / 3 going away from the origin. Also, when the walker reaches orbit \(O_k\), it stays there a geometric number of steps, on average \(2^k\) steps.
Theorem 3
 (a)
If either \(I^r(b)=\infty \) or \(I^l(b)=\infty \), then \(X^{\mathrm {imp}}\) on \({\mathbb {Z}}^1\) is null recurrent as well, for any sequence of passage times.
 (b)
If \(I^r(b),I^l(b)<\infty \), then \(X^{\mathrm {imp}}\) on \({\mathbb {Z}}^1\) is positive recurrent whenever the walk is strongly impatient.
Proof
3 Lamperti and LampertiType Walks
In Sect. 2, we have seen that for the symmetric random walk in dimensions one and two, impatience cannot turn null recurrence into positive recurrence. We are, therefore, going to consider certain random walks which are “just barely null recurrent”, and show that in those cases impatience can actually make them positive recurrent. Our models will be related to the “Lamperti walk” (see [6]).
We first need a general result, presented in the next subsection.
3.1 A General Formula for NearestNeighbour Walk on \({\mathbb {Z}}_+\)
Lemma 2
Proof
3.2 Lamperti Walk: \(b(x)\asymp c/x\) on \({\mathbb {Z}}_+\)
As an application to Lemma 2, we obtain a theorem concerning a case when short enough passage times can turn null recurrence into positive recurrence. But first we need the following statement (see e.g. Proposition 7.1 (i)–(iii) in [2]).
Proposition 1
Let X be a random walk on \({\mathbb {Z}}_+\), with drift \(b(x)\asymp c/x\). Then, X is positive recurrent if \(c<1/2\), null recurrent for \(c\in [1/2,1/2]\), and transient for \(c>1/2\).
Theorem 4
Let X be a recurrent random walk on \({\mathbb {Z}}_+\), with drift \(b(x)\asymp c/x\) (recurrence means \(c\le 1/2\)). Furthermore, let the passage times satisfy \(s_j\asymp j^{\alpha }\), \(\alpha >0\). Then, \(X^{{\mathrm {imp}}}\) is positive recurrent if and only if \(c<\min \left\{ 0,\frac{\alpha 1}{2}\right\} \).
In particular, when \(c\in [1/2,0)\), X is null recurrent but \(X^{{\mathrm {imp}}}\) is positive recurrent whenever \(\alpha >1+2c\) (for example, when \(\alpha >1\), i.e. the impatience is strong).
Remark 6
In fact, \(X^{{\mathrm {imp}}}\) is positive recurrent for any strongly impatient walk when \(c<0\) (the proof is similar).
Proof of Theorem 4

if \(\alpha >1\) then \(\mathsf {SUM}(m)\le \sum _{j=1}^{\infty } j^{\alpha }<\infty \);
 if \(\alpha =1\) then using Riemannsum approximation for the function \(f(x)=1/x\),$$\begin{aligned} \mathsf {SUM}(m)&\le \sum _{j=1}^{m} \frac{1}{j} + \sum _{j=m+1}^{\infty } \left( 1\frac{c_1}{m}\right) ^{j+1} \frac{1}{j} \le \sum _{j=1}^{m} \frac{1}{j} + \sum _{k=1}^{\infty } \gamma ^{k} \sum _{j=km+1}^{(k+1)m} \frac{1}{j}\\&\le \log m+\sum _{k=1}^{\infty } \gamma ^{k} \int _{km}^{(k+1)m} \frac{1}{j}= \log m \\&+ \sum _{k=1}^{\infty } \log (1+1/k)\gamma ^{k} \sim \log m = h_1(m); \end{aligned}$$
 if \(\alpha <1\) then$$\begin{aligned} \mathsf {SUM}(m)&\le \sum _{j=1}^{\infty } \left( 1\frac{c_1}{m}\right) ^{j+1} \frac{1}{j^{\alpha }} \le \sum _{k=0}^{\infty } \gamma ^{k} \sum _{j=km+1}^{(k+1)m} \frac{1}{j^{\alpha }} \le \sum _{k=0}^{\infty } \gamma ^k \frac{m}{(km+1)^{\alpha }}\\&\le m^{1\alpha } \sum _{k=0}^{\infty } \frac{\gamma ^k}{k^\alpha } \sim m^{1\alpha }=h_{\alpha }(m), \end{aligned}$$
It remains to consider the case \(c=\frac{1}{2}\). Now, we have \(p_m\sim \frac{1}{\log m}\) and \(\mathsf {SUM}(m)\ge O(1)\), so \({{\mathbb {E}}}{{\tilde{\tau }}} \sim \sum _{m} p_m \mathsf {SUM}(m)\ge \sum _m \frac{1}{\log m}=\infty \) for any value of \(\alpha \ge 0\), ruling out positive recurrence for \(X^{{\mathrm {imp}}}\). Since X is recurrent, so is \(X^{{\mathrm {imp}}}\), and thus \(X^{{\mathrm {imp}}}\) is in fact null recurrent in this case. \(\square \)
3.3 LampertiType Walk on \({\mathbb {Z}}_{+}\) with Very Small Inward Drift
Theorem 5
 (i)
null recurrent for any \(D\ge 1/2\) and any sequence of passage times;
 (ii)
positive recurrent for any \(D<1/2\), provided it is strongly impatient.
Proof
4 Expectation Calculations for Hitting Times and Positive Recurrence in OneDimension: The Passage Generating Function
In this section, we perform some calculations concerning onedimensional hitting times and analyse onesided positive/null recurrence for the walk.
 1.For twosided hitting times, definefor \(n\le x,m\le n\).$$\begin{aligned} \rho _m^{(x)}:={\mathbb {P}}_{x}(X\ \text {reaches}\ m\ \text {before}\ \pm n)={\mathbb {P}}_{x}(T_n>\tau _{m}), \end{aligned}$$
 2.For onesided hitting times definefor all \(x,m\in {\mathbb {Z}}^1\) and note that \(r_m^{(x)}=0\) for \(m\ge h\) and \(r_m^m=1\).$$\begin{aligned} r_m^{(x)}:={\mathbb {P}}_{u+x}(X\ \text {reaches}\ u+m\ \text {before}\ v)={\mathbb {P}}_{u+x}(\tau _v>\tau _{u+m}), \end{aligned}$$
Remark 7
In a concrete situation, the quantities \(\rho _m^{(x)},r_m^{(x)}\) are of course, computable as ratios involving resistors.
4.1 Passage Generating Function and Passage Radius
The following notion will be useful for impatient as well as ageing walks.
Definition 10
Remark 8
 (i)
It is clear that strong impatience implies that \(R^{\mathrm {pass}}\ge 1\), while weak impatience implies that \(R^{\mathrm {pass}}= 1.\)
 (ii)
In the strongly impatient case, we can normalize the passage times so that \(\sum s_j^*=\sum s_j=1\) (at the expense of speeding up time by a constant factor), and then \(\phi \) is actually a probability generating function. This has the practical advantage that we can use wellknown formulae for \(\phi \) in the strongly impatient case.
 (iii)
In the ageing case, it is possible that \(R^{\mathrm {pass}}=0\) (“superageing”).
4.2 “Positive Recurrence to the Right” (PRR) for Impatient Walk
Let X be a recurrent walk on \({\mathbb {Z}}^1\).
Definition 11
(Onesided positive recurrence) We say that \(X^{\mathrm {imp}}\) is positive recurrent to the right if \({{\mathbb {E}}}_u \sigma _v<\infty \) for \(u<v\) and null recurrent to the right if \({{\mathbb {E}}}_u \sigma _v=\infty \) for \(u<v\). We will show below (see Remark 9) that this definition does not actually depend on choice of u or v.
The definition for the positive/null recurrence to the left (when \(u>v\)) is analogous.
Our fundamental result about onesided positive recurrence is as follows.
Theorem 6
Proof

\((I):={{\mathbb {E}}}_u\)(actual time spent between u and v until hitting v);

\(({ II}):={{\mathbb {E}}}_u\)(actual time spent between \(\infty \) and u before hitting v).
Remark 9
(Consistency of the definition) It is easy to see that the condition (7) does not depend on the choice of u or v (for nondegenerate X). For example, for a fixed u, if we increase h, as this is shown in the above proof: only \(({ II})\) will change; (I) is always finite.
A similar calculation shows that the condition is invariant under fixing v and changing u.
We can refine Theorem 6 as follows.
Corollary 2
 (a)
Then (8) is a necessary condition for the PRR property for \(X^{\mathrm {imp}}\), no matter what the passage times are, as long as we rule out that \(\lim _{m\rightarrow \infty }r_{m1}^{(m)}=0\).
 (b)
If either the impatience is strong, or \(r:=\sup _{m\in {\mathbb {Z}}}r_{m1}^{(m)}<1\) then (8) is a sufficient condition for PRR for \(X^{\mathrm {imp}}\).
Proof
 (a)
This follows from Theorem 6 and the fact that \(\phi (t)\) is bounded away from zero if \(t>\epsilon >0\).
 (b)
For strong impatience, \(\phi \left( r_{m1}^{(m)}\right) \le \phi (1)<\infty \) for all m, giving the assertion. If \(r<1\), then \(\phi \left( r_{m1}^{(m)}\right) \le \phi (r)<\infty \) for all m, and we are done again.
Remark 10
Example 1
More generally, we have
Corollary 3
4.3 Expected Exit Times: TwoSided
In this section, we consider ageing random walks.^{5} The first piece of information about the speed we are aiming to obtain is the expected actual time to reach \(\pm \, n\) starting from the origin, that is, \({{\mathbb {E}}}_0 T_n\). Let \(n\ge 2\) and \(0\le m\le n2\). Each edge \((m,m+1)\) can be crossed \(0,1,2,\ldots \) times before the walk reaches \(\pm \, n\), unlike the cases for recurrence to the right, where the parity of those times is fixed. Note also that the actual time spent on the edge \((n1,n)\) is always either 0 or \(s_0\) and hence finite, as the walk can traverse it at most once before exiting \((\,n,n)\). Similar statement holds for the edge \((\,n, n+1)\).
Recall Assumption 1.
Theorem 7
As a consequence, we can see that if ageing is either very slow or very fast, then the behaviour of the original walk becomes irrelevant.
Corollary 4
 1.
If \(R^{\mathrm {pass}}= 1\) (“slow ageing”), then \({{\mathbb {E}}}_0 {\widetilde{T}}_n<\infty \) holds whatever the original walk is.
 2.
If \(R^{\mathrm {pass}}=0\) (“superageing”), then \({{\mathbb {E}}}_0 {\widetilde{T}}_n=\infty \) holds whatever the original walk is.
5 The Spatial Spread of the Process
Let \(R_t\) denote the number of distinct edges crossed by \(X^{\mathrm {imp}}\) up to actual time \(t>0\). If \(G={\mathbb {Z}}\), then \(R_t=\max _{0\le s\le t}X^{\mathrm {imp}}_s\min _{0\le s\le t}X^{\mathrm {imp}}_s\).
Problem 1
Problem 2
Problem 3
(Weakly impatient walk) Assume that \(X^{\mathrm {imp}}\) is weakly impatient. What is the asymptotic behaviour of \(R_t\) as \(t\rightarrow \infty \)?
6 Comparison with Classical ArcSine Law
Now, in our case, the lefthand side depends on the passage times. Intuitively, if the random walk with the given passage times is impatient, then the limiting distribution of the proportions is more balanced than in the classical case.
In fact, if the excursion time (between returns to the origin) has finite expectation (e.g. when M has finite expectation and passage times are strongly impatient, or the cases computed in Sect. 3.1), then by the Renewal Theorem, the limit is completely balanced (\(=1/2\))!
Problem 4
(Modified ArcSine Law) What happens when X is simple random walk and the excursion time has infinite expectation? How will the passage times modify the ArcSine Law, making the limit more balanced?
The following theorem may be considered as an initial step in this direction; it indicates that for strong enough impatience, a behaviour much more balanced than for the classical ArcSine Law is exhibited.
Theorem 8
Proof
Therefore, identifying right and left with “heads” and “tails”, \(R_n\) can be identified with the number of heads in the following experiment: We first toss a fair coin. Then we turn it over with probability 1 / 3, and with probability 2 / 3 we do nothing. Next we turn it over with probability 1 / 4, etc. Finally, in the nth step we turn it over with probability \(1/(n1)\).
Using this equivalent formulation, the claim follows from Theorem 1 in [4], where a more general “coin turning” model is investigated. \(\square \)
Remark 11
Breiman [1] proved a generalization of the ArcSine Law in the 1960s, and this was picked up by Mason et al. [5] recently. The point is that one can have a nice limit even if the law of the excursion is different from the classical one for simple random walk (in [5], take \(X_i\) to be a random sign and \(Y_i\) the excursion time for the ith excursion).
7 SpaceDependent Impatience
Here, we modify the definition of impatience given in Sect. 1. Suppose now that the passage times for an edge e do not depend on the number of times the edge has been crossed, but rather on the location of this edge in the graph G; thus \(s_0(e)=s_1(e)=\cdots =s(e)\), following our definition of the passage times. As before, fix some specific vertex \(v_0\) of the graph and call it the origin. For a vertex v in G, let the distance from \(v_0\) to v, denoted by \(\Vert v\Vert \in \{0,1,2,\ldots \}\), be the number of edges on the shortest path connecting \(v_0\) and v, and for an edge \(e=(v_1,v_2)\) let \(\Vert e \Vert =\min \{\Vert v_1\Vert ,\Vert v_2\Vert \}\).
Let X be a unit time random walk on G, i.e. a Markov chain whose transitions are restricted to the edges of G. It is then conceivable that while \(X_n\) is null recurrent, \(X^{{\mathrm {imp}}}\) may still be positive recurrent, provided \(s(e)\rightarrow 0\) sufficiently quickly as \(\Vert e\Vert \rightarrow \infty \).
Assumption 2

(Uniform ellipticity) There is a universal constant \(\varepsilon >0\) such that for any edge \(e=(v_1,v_2)\) in the graph \({\mathbb {P}}(X_{n+1}=v_2\Vert X_n=v_1)\ge \varepsilon \), \({\mathbb {P}}(X_{n+1}=v_1\Vert X_n=v_2)\ge \varepsilon \).
 (Return symmetry) There is a universal constant \(\rho \ge 1\) such that for any v in the graph$$\begin{aligned} \rho ^{1} \, p(v,v_0)\le p(v_0,v)\le \rho \, p(v,v_0). \end{aligned}$$
Remark 12
 (a)
uniform ellipticity implies that the graph is of uniformly bounded degree, i.e. there is \(D\ge 1\) such that each vertex of the graph has at most D edges coming out of it and that there are no oriented edges on G;
 (b)
return symmetry implies that the underlying random walk cannot be positive recurrent.
Note that part (b) of the above remark follows from the following observation. For each \(v\ne v_0\), the walk starting at \(v_0\) has the probability \(p(v_0,v)\) of hitting v before returning to the origin \(v_0\). However, after reaching v, the walk makes a geometric number of returns to v itself before coming back to \(v_0\). The expected number of such returns, including the very first visit, is \(\frac{1}{p(v,v_0)}\). Thus, the total expected number of vertices visited by the walk (with multiplicity) equals \(\sum _{v\in G} \frac{p(v_0,v)}{p(v,v_0)}\) which is infinite since each term is at least \(\rho ^{1}\). Hence, the expected number of steps the walk makes before returning to \(v_0\) is also infinite.
Later we will show that SRW both on \({\mathbb {Z}}\) and \({\mathbb {Z}}^2\) satisfy the above assumptions.
Theorem 9
Proof
Next, note that the number of vertices at distance at most k from \(v_0\) is bounded above by \(\sum _{i=1}^k d(d1)^{i1}<\infty \), so we can safely assume from now on that \(p(v_0,v)\) is small.
Recall that the graph G is called transitive if, viewed from any vertex v in G, it is isomorphic to G viewed from \(v_0\).
Proposition 2
Let X be a recurrent simple^{6} random walk on the transitive graph G. Then, Assumption 2 is satisfied.
Proof
The uniform ellipticity assumption is trivially satisfied since X is a symmetric random walk, and each vertex is incident to the same number of edges, because of transitivity of G. Now, we have \(p(v_0,v)=p(v,v_0)\) by transitivity again, and thus one can set \(\rho =1\). \(\square \)
Corollary 5

\(\alpha >1\), in case of \({\mathbb {Z}}^1\);

\(\alpha >2\), in case of \({\mathbb {Z}}^2\).
Proof
Footnotes
 1.
The intuitive meaning is clear: the more the walker crosses the same edge, the faster it happens.
 2.
But one can also imagine that the walker is actually crossing the edge continuously with a speed depending on the passage time sequence.
 3.
Our original model was more mundane: a person window shopping who gets bored quickly by the stores of any street she has already visited.
 4.
We have already verified this with another method—see Theorem 2.
 5.
It is easy to see that in the impatient case, the expectation we study is always finite for any irreducible walk.
 6.
I.e. X jumps to each neighbour with the same probability.
Notes
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