Crossings States and Sets of States in Random Walks

Abstract

We consider a random walk in $\mathbb {Z}^2$ and establish a number of results for the distributions and expectations of the number of usual (undirected) and specifically defined in the paper directed state-crossings and different sets of states crossings. As well, we extend the results to d-dimensional random walks, $d\ge 2$, in bounded areas.

Appendix: Derivation of the Formula for the Transition Probability from \(\mathcal {N}(n)\) to \(\mathcal {N}(n+1)\) for the Random Walk in \(\mathbb {Z}^d\)

The results of the appendix are taken from Abramov (2018), pp. 1906–1908.

1.1 Description of the Model

The random walk in $\mathbb {Z}^d$ is modelled by the d independent queueing processes as follows. Assume that arrivals in the ith queueing system are Poisson with rate $\lambda$, and service times are exponentially distributed with the parameter $\lambda$. If a system becomes free, it is switched for a special service, which is exponentially distributed with the same parameter $\lambda$. This service is negative, and it results in a new customer in the queue. If during a negative service a new arrival occurs, the negative service remains unfinished and not resumed. The negative service models the reflection at zero and in fact implies the state-dependent arrival rate, which becomes equal to $2\lambda$ at the moment when the system is empty. It is associated with the situation, when an original one-dimensional random walk reaches zero at some time moment s, and at the next time moment $s + 1$ it must take one of the values $\pm 1$ that corresponds to value $+1$ for an one-dimensional random walk reflected at zero.

1.2 Queueing Systems with Finite Capacity and Characterization of the Level n Probability

Assume that the number of waiting places in each of the queueing system is N. The assumption on limited number of waiting places means that an arriving customer, who meets N customers in the system, is lost. For a vector $\mathbf {n}=\big (n^{(1)}, n^{(2)},\ldots , n^{(d)}\big )$ satisfying $\Vert \mathbf {n}\Vert <N$ let $P_N(\mathbf {n})$ denote the stationary probability to be in state $\mathbf {n}$ immediately before arrival of a customer in one of the d queueing systems. Application of the PASTA property Wolff (1982) enables us to first obtain the stationary probability for each single system to obtain then the required stationary probability $P_N(\mathbf {n})$. Let $P_N^{(i)}(n)$ denote the stationary probability to be in state $n<N$ for the ith queueing system. We have

$$\begin{aligned} P_N^{(i)}(n)={\left\{ \begin{array}{ll}\frac{2}{2N+1}, &{}\text {for} \quad 1\le n\le N,\\ \frac{1}{2N+1}, &{}\text {for} \quad n=0. \end{array}\right. } \end{aligned}$$

Hence the queueing systems are independent, the product form solution for $P_N(\mathbf {n})$ is

$$\begin{aligned} P_N(\mathbf {n})=2^{d-d_0(\mathbf {n})}\frac{1}{(2N+1)^d}, \end{aligned}$$

(69)

where $d_0(\mathbf {n})$ denotes the number of zero components in the presentation of the vector $\mathbf {n}$.

The total number of vectors having norm n in $\mathbb {Z}^d_+$ is

$$\begin{aligned} \sum _{i=1}^{d}\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) . \end{aligned}$$

(70)

(The formula sums over i being the number of nonzero components of the vector.) Hence, denoting the stationary state probability to belong to the set $\mathcal {N}^+(n)$ by $P_N[\mathcal {N}^+(n)]$, we have

$$\begin{aligned} P_N[\mathcal {N}^+(n)]=\sum _{\mathbf {n}\in \mathcal {N}^+(n)}=\frac{1}{(2N+1)^d}\sum _{i=1}^{d}2^i\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) , \end{aligned}$$

(71)

where the term

$$\begin{aligned} \sum _{i=1}^{d}2^i\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) \end{aligned}$$

on the right-hand side of (71) characterizes the total number of vectors in $\mathbb {Z}^d$ having norm n.

It follows from (69) that for any two vectors $\mathbf {n}_1$ and $\mathbf {n}_2$ satisfying $\Vert \mathbf {n}_1\Vert <N$ and $\Vert \mathbf {n}_2\Vert <N$ we have

$$\begin{aligned} \frac{P_N(\mathbf {n}_1)}{P_N(\mathbf {n}_2)}=2^{d_0(\mathbf {n}_2)-d_0(\mathbf {n}_1)} \end{aligned}$$

(72)

independently on N. Hence,

$$\begin{aligned} \lim _{N\rightarrow \infty }\frac{P_N(\mathbf {n}_1)}{P_N(\mathbf {n}_2)}=2^{d_0(\mathbf {n}_2)-d_0(\mathbf {n}_1)}. \end{aligned}$$

(73)

For $N=\infty$, let $P_\infty (\mathbf {n},\tau )$ be the probability that at time $\tau$ the system with infinite capacity is in state $\mathbf {n}$. Then, for the limiting ratio of the final probabilities we also have

$$\begin{aligned} \lim _{\tau \rightarrow \infty }\frac{P_\infty (\mathbf {n}_1,\tau )}{P_\infty (\mathbf {n}_2,\tau )}=2^{d_0(\mathbf {n}_2)-d_0(\mathbf {n}_1)}. \end{aligned}$$

(74)

The result in (74) is true, since the limit in (73) due to Relation (72) is uniform, and interchanging the order of limits $\tau$ vs N is correct. (The last can also be established directly Abramov (2018)).

Let $p_n(d)$ denote the transition probability from the set of states $\mathcal {N}^+ (n)$ (level n) to the set of states $\mathcal {N}^+ (n+1)$ (level $n + 1$), and let $q_n(d)$ = $1-p_n(d)$ denote the transition probability from the level n to the level $n-1$. Derive the formula for $p_n(d)$.

The total number of vectors in the set $\mathcal {N}^+ (n)$ is given by (70). Each vector contains d components. Hence, the total number of components in the set of vectors in $\mathcal {N}^+ (n)$ is

$$\begin{aligned} d\sum _{i=1}^{d}\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) . \end{aligned}$$

(75)

Among them, the total number of zero components is

$$\begin{aligned} \sum _{i=1}^{d-1}(d-i)\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) , \end{aligned}$$

(76)

and the total number of nonzero components is

$$\begin{aligned} \sum _{i=1}^{d}i\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) . \end{aligned}$$

(77)

To derive the formula for $p_n(d)$ let us build the sample space. The components of all vectors in $\mathcal {N}^+ (n)$, the total number of which is given by (75), are not equally likely. According to (69), a nonzero component appears with two times higher probability than a zero component. To make the components equally likely, we are to extend the number of nonzero components by factor 2. Then the total number of equally likely components is to be equal to

$$\begin{aligned} d\sum _{i=1}^{d}2^i\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) . \end{aligned}$$

(78)

Following (78), the sample space for level n contains

$$\begin{aligned} 2d\sum _{i=1}^{d}2^i\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) , \end{aligned}$$

that is two times more than that given by (78). This is because it include possible transitions from each state (component) in the two directions. Specifically, let $C_0(n, d)$ denote the number of possible transitions associated with reflections at zero (the number of zero components given in (76) multiplied by two), and let 2C(n, d) be the rest of transitions, the half of which are associated with the transitions from level n to $n+1$ and another half with the transitions from level n to $n-1$. So, C(n, d) is the number of nonzero components given in (77)

That is, the expression in (77) is presented as

$$\begin{aligned} 2d\sum _{i=1}^{d}2^i\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) =C_0(n, d)+2C(n, d). \end{aligned}$$

(79)

Then, the total number of transitions from level n to $n+1$ is $C_0(n, d)+C(n, d)$ and the total number of transitions from level n to $n-1$ is C(n, d). Then, from (76), (77) and (79) we obtain

$$\begin{aligned} \begin{aligned} p_n(d)&=\frac{C_0(n,d)+C(n,d)}{C_0(n,d)+2C(n,d)}\\&=\frac{2\sum _{i=1}^{d-1}(d-i)2^i\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) +\sum _{i=1}^{d}i2^i\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) }{2\sum _{i=1}^{d-1}(d-i)2^i\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) +2\sum _{i=1}^{d}i2^i\left( {\begin{array}{c}d\\ i\end{array}}\right) \left( {\begin{array}{c}n-1\\ i-1\end{array}}\right) }. \end{aligned} \end{aligned}$$