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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 $\mathbb{P}(\tilde{S}_n>-b\;\;\forall n>0) \leq \mathbb{P}({S}_n>-b\;\;\forall n>0)$.

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    $\begingroup$ 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 $b \gg \lambda$). $\endgroup$ – Sam T Feb 20 at 22:30
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    $\begingroup$ 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 $\endgroup$ – Sam T Feb 20 at 22:31

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