Assume $\varepsilon \in [0,1/2]$. Consider the discrete-time random walk $X\_0 = 0$, $X\_{t+1} - X\_t \sim f(X\_t) \delta\_0 + (1-f(X\_t))\operatorname{Rademacher}$, where $\delta\_0$ is the Dirac delta on zero and $\operatorname{Rademacher}$ is the Rademacher distribution, and
$$
f(x) = \begin{cases} 1-\varepsilon & x > 0 \\ \frac{1}{2} & x=0 \\ \varepsilon & x<0. \end{cases}
$$
I am interested in the probability $O\_T = \mathbb{P}[X\_T > 0]$, and would like to show that $O\_T > 1/2$ for sufficiently large $T$.


I tried to prove this via a continuum approximation to SDE and generalized arcsine laws and it seemed very hard for me.


I am wondering whether there is more general theory, as it seems to me that higher diffusion in some part of the space and appropriate boundary behavior should lead to the random walk spending more time in the "less diffusive" part.


I'm grateful for any pointers or proof ideas.



Occupation time of non-stationary random walk
Here is a back of envelope computation that you should be able to make rigorous without too much trouble.


Let's condition upon the last hit of $0$. The passing time from it to $T$ is large with probability close to $1$ (all you need to show for this is that the probability to be in any particular position tends to $0$). Let that time be $S$.


Let $p$ be the probability of moving when you are to the right of $0$ and let $q$ be the probability of moving when you are to the left of $0$. Then the probability to end up on the right divided by the probability to end up on the left under such conditioning is the same as the probability to stay above $0$ for $S$ steps in a lazy random walk with the probability $p$ of movement divided by the same probability with $q$ instead of $p$.


The first random walk is effectively the standard random walk with about $pS$ steps and the second one is the standard random walk with about $qS$ steps. The corresponding probabilities for the standard random walk are well-known to be approximately proportional to the inverse square root of the number of steps, so we get $\frac{P(X\_T>0)}{P(X\_T<0)}\to\sqrt{\frac qp}=\sqrt{\frac{1-\varepsilon}{\varepsilon}}$ as $T\to\infty$ in your setting.


It all should be well-known with quantitative estimates of the convergence speed but I leave it to better experts than I to provide a reference.


