# Escaping probability for a poisson random walk

Poisson random walk: Let independent random variables $$Z_i \sim Pois(\lambda)$$. Consider random walk $$S_n = \sum_{i=1}^{n}X_i,$$ where $$X_i = \begin{cases} Z_i &\text{w.p}\; p\\ -Z_i &\text{w.p}\; 1-p \end{cases}$$

Question: For $$b>0, p>1/2$$, what is the probability that $$S_n > -b,\;\;\forall n>0$$? Mathematically, $$\mathbb{P}\big(S_n > -b\; \forall n>0 \big)?$$

Binary random: Consider random walk: $$\tilde{S}_n = \sum_{i=1}^{n}\tilde{X}_i,$$ where $$\tilde{X}_i = \begin{cases} 1 &\text{w.p}\; p\\ -1 &\text{w.p}\; 1-p \end{cases}.$$

My Conjecture: The poisson random walk is a "stretched" version of the binary random walk and thus $$\mathbb{P}(\tilde{S}_n>-b\;\;\forall n>0) \leq \mathbb{P}({S}_n>-b\;\;\forall n>0)$$.

• It seems unlikely that your conjecture should be correct (although the intuition may not be bad). For example, take $\lambda \to \infty$. Then the $Z_i$ concentrate around $\lambda$, and $\lambda \gg b$, so basically this just says $\Pr(\text{standard biased SRW is always positive})$, which is just the usual escape probability. Now take $\lambda \to 0$. In order to reach a distance $b \gg \lambda$ from $0$, some large number of steps will be taken. By the LLN, may as well replace $\text{Poisson}(\lambda)$ with just $\lambda$. So this probability tends to $1$ (since biased and $b \gg \lambda$). – Sam T Feb 20 at 22:30
• On the other hand, your $\mathbb{P}(\tilde{S}_n>-b \: \forall n>0)$ lies between these two (providing $b > 2$). \\ For proceeding with your question, I would note that the sum of independent Poissons is a Poisson. So write $\xi_i = \pm1$ for whether you go up or down. Then all the "ups" sum to get a Poisson distribution, as do the "downs", conditional on how many up/down – Sam T Feb 20 at 22:31