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