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Let be $\{X_n\}_{n\in \mathbb{N}}$ random walk on $\mathbb{Z}$. Let be

$$P(X_{n+1} = k + a| X_n = k)= p_a$$ for $a\in \mathcal{A} \subset \mathbb{Z}$. Let say that $X_0 = 0$. I am interested in probabilities $$\tilde{p}_k = P(\exists n\in \mathbb{N}_0. X_n =k\;|\; X_0=0)\text{.}$$

If $\mathcal{A} = \{-1,1\}$, then this problem can be solved if one writes down recurrence relation

$$\tilde{p}_k = p_{-1}\tilde{p}_{k+1} +p_1\tilde{p}_{k-1}$$ with $\tilde{p}_0=1$ and solves it by finding $\lambda$'s satisfying

$$\lambda^k = p_{-1}\lambda^{k+1} +p_1\lambda^{k-1}\text{.}$$

If we devide the upper equation by $\lambda^{k-1}$, we get a quadratic equation, which can be easily solved. We should consider $3$ different cases:

  1. $p_1 = p_{-1} \; (\Leftrightarrow \lambda_{1,2} = 1)$

  2. $p_1 > p_{-1}$

  3. $p_{1} < p_{-1}$

In the case 1. ($\lambda_1 = \lambda_2 = 1$) we should (in general) find such constants $A^+$, $B^+$, $A^-$ and $B^-$ that

$$\tilde{p_k} = A^+ + B^+k,\;k\geq 0 $$ $$\tilde{p_k} = A^- + B^-k,\;k\leq 0$$

In the other two cases $\lambda_1\neq \lambda_2$ and it holds

$$\tilde{p}_k = A^+\lambda_1^k + B^+\lambda_2^k,\; k\geq 0$$ $$\tilde{p}_k = A^-\lambda_1^k + B^-\lambda_2^k,\; k\leq 0$$

In the case one, we can conclude from simetry that $A^+ = A^- = A$ and $B^+ = B^- = B$. Since $1\geq \tilde{p}\geq 0$, it follows $B=0$ and $A =\tilde{p}_0 = 1$. It turns out that it is enough to solve case 2 (in the third, we just change the roles of positive and negative numbers). By help of $\tilde{p}_0=1$ and strong law of large numbers, we can determine $A^{+,-}$ and $B^{+,-}$.

Since $\sum p_a = 1$, one $\lambda_i$ is $1$ for every set $\mathcal{A}$. So the problem occurs when we get an equation of a degree $4$ or more (Cardano's and Ferrari's formulas are quite complicated). This happens when

$$M:= \max \{|a_i -a_j|\; \mid a_i,a_j \in \mathcal{A}\}>3\text{.}$$

My questions were

$1$. How to compute $\tilde{p}_k$ when $M>3$? Are there any other approaches?

$2$. I assume that $|\lambda_i|\leq1$ should hold for every $\lambda_i$. How to prove that?

I can prove $2$) if all $\lambda_i\in \mathbb{R}$ knowing $0\leq \tilde{p}_k\leq 1$ for all $k\in \mathbb{Z}$ (To be more precise: there are nonzero constants only in front of the $|\lambda_i |\leq1$). Thanks in advance.

The second was answered. And the assumpzion is wrong. There is still the first.

The case $\mathcal{A} =\{-3,3,6\}$ with $M=9$ can be reduced to $\tilde{\mathcal{A}} = \{-1,1,2\}$, since $\text{GCD}(\mathcal{A}) = 3$. So from now on, we can assume that $\text{GCD}(\mathcal{A}) = 1$.

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Antoine
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  • How did you get the recurrence relation? The random walk might visit point $k-1$ many times; each time it visits $k-1$, there is a chance that it will go to point $k$. Also is $k$ a positive integer? If not, your formula might give values greater than 1 for $\tilde p_k$ (which doesn't make any sense). – Yury Feb 01 '14 at 05:30
  • @Yury Walk starts at $0$. For $k>0$, $\tilde{p}k$ is a probability of coming to $k$ at least once. Let be $a$ the first step. Then $\tilde{p}_k = P(a = -1)P(\text{walk visits k}| a = -1) + P(a = 1)P(\text{walk visits k}| a = 1)$. The steps are independent, so if the walk starts at $0$, then $P(\text{walk visits k}| a = 1) = \tilde{p}{k-1}$. Similarly, $P(\text{walk visits k}| a = -1) = \tilde{p}_{k+1}$. For $k<0$, the same. I will edit the definition of $\tilde{p}_k$. These are ment to be conditional probabilities. Why do you think that $\tilde{p}_k$ can be greater than $1$ for negative $k$? – Antoine Feb 01 '14 at 11:26
  • (1) I don't understand your formula. What is the event $a=-1$? When does it happen? (2) The expression $A \lambda_1^k + B \lambda_2^k$ is unbounded (either when $k > 0$, or $k < 0$, or both) unless $A \lambda_1^k + B \lambda_2^k \equiv 1$. – Yury Feb 01 '14 at 18:45
  • (1) The event $[a = -1]$ is the event $[\text{we go left on the first step}]$ and $[a = 1]$ is $[\text{we go right on the first step}]$. (2) Sorry, my mistake. There should be four constants: $A_+, B_+$ for $k>0$ and $A_-, B_-$ for $k<0$. I am sorry. I should edit the question again. – Antoine Feb 01 '14 at 22:41
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    Sounds good. Regarding question 2, even if $A={-1,1}$ and $p_{-1} > p_1$, we have $\lambda_1=1$ and $\lambda_2 > 1$. Consider another example: $A={-3,-1,2}$ and $p_{-3} = 1/4$, $p_{-1} = 1/4$, and $p_2 = 1/2$. Then values of $\lambda$ are given by $\lambda^{-2}/2 + \lambda/4 + \lambda^3/4 = 1$. The roots of this equation have absolute values approximately 1, 0.6388969200, and 1.769292354 (some are equal to 1, some are greater than 1, and some are less than 1). – Yury Feb 01 '14 at 23:35
  • @Yuri So there is just one question left:) – Antoine Feb 01 '14 at 23:52
  • @Antoine The recurrence relation is not true, at least not with the definition of $\tilde{p}_k$. The events ${\text{at some point, we reach }k-1\text{ and the next step is }+1}$ and ${\text{at some point, we reach }k+1\text{ and the next step is }-1}$ are not disjoint. – D Poole Feb 06 '14 at 16:14
  • I do not understand your reason. The events ${\text{the first step is }+1}$ and ${\text{the first step is }-1}$ are disjoint and I believe that $P(\exists n. X_n = k\mid X_0 = 0, X_1 = 1) = P(\exists n. X_n = k\mid X_1 = 1) = P(\exists n. X_n = k-1\mid X_0 = 0) $. Analogically for $-1$. – Antoine Feb 07 '14 at 22:08

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