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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.

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  • $\begingroup$ When you are writing $\delta_0$, you just mean that you stay in place, right? $\endgroup$
    – fedja
    Commented Aug 23, 2021 at 19:34
  • $\begingroup$ Yes, exactly. I wanted to stay in $\mathbb{Z}$, but say that $\operatorname{Var}[X_{t+1} | X_t = x]$ is decreasing in $x$. $\endgroup$
    – Andreas Haupt
    Commented Aug 23, 2021 at 22:21

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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.

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  • $\begingroup$ Thank you @fedja, this has been very helpful! Could you expand a bit why the probability of being in a particular position tending to zero implies that $S$ is large? Also, do you have a reference on the square-root statement? $\endgroup$
    – Andreas Haupt
    Commented Aug 24, 2021 at 14:28
  • $\begingroup$ @AndreasHaupt Sure: $P(S<B)\le\sum_{s=0^B}P(X_{T-s}=0)$, for example. As to the square root, the standard application of the reflection principle for the simple random walk yields that $P(S_1,S_2,\dots,S_{2n}>0)=2^{-2n}{2n-1\choose n-1}$ and Stirling finishes the story. $\endgroup$
    – fedja
    Commented Aug 24, 2021 at 15:44
  • $\begingroup$ One more request for a pointer: I am seeking to generalize the above argument to general random walks (i.e. non-Rademacher increments). What is a good text on reflection principle and probability of staying positive for a general random walk? $\endgroup$
    – Andreas Haupt
    Commented Jul 17, 2023 at 2:50

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