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Let $\mathcal{B}_n^{(c)}$ be the set of $n$ step lattice bridges (starts at $(0,0)$ ends at $(n,0)$), taking steps $\{(1,1), (-1,1)\}$ bounded between $y = c$ and $y = -c$ for a constant $c \geq 0$. For any lattice bridge $b$ of $n$ steps and $0 \leq \alpha \leq 1/2$, let $b_{\alpha}$ be the lattice bridge obtained from $b$ by simultaneously choosing $\alpha n$ $(1,1)$ entries at random and changing them to $(-1,1)$ and choosing $\alpha n$ $(-1,1)$ entries at random and changing them to $(1,1)$. I would like to find:

$$\Pr\left[b_{\alpha} \in \mathcal{B}_n^{(c)}| b \in \mathcal{B}_n^{(c)}\right].$$ Searching the literature, there seems to two potential approaches (1) consider the problem in the continuous setting (two correlated, bounded Browanian bridges which are coupled $2\alpha$ of the time) and then show that the discrete case is well approximated by the continuous case or (2) use combinatorial methods a. la. Banderier and Faljolet to find the generating function for the number of bridges. Unfortunately there are obstacles to each. For the first method, there is a scale factor of $O(\frac{1}{\sqrt{n}})$ needed to go from the discrete to the continuous setting (similar to the scaling in the CLT) which causes the probability for my bounded bridges to exist to vanish too quickly (since $c$ is a constant it would be like requiring the supremum of the Brownian bridge to be tending to zero). For the second method, counting the number of bounded lattice paths $b_{\alpha} \in \mathcal{B}_n^{(c)}$ given a particular path $b \in \mathcal{B}_n^{(c)}$ depends on the particular path $b$ (whether it is in $\mathcal{B}_n^{(c-1)}$ or touches the lines $y = c$ or $y = -c$ many times for example) so there are many cases to consider.

For people who are familiar with such problems, what is the more standard approach? Are there any references that might circumvent either of the two obstacles?

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  • $\begingroup$ Since it is unlikely to compute this probability exactly, can you indicate what bound you would conjecture or hope to establish? $\endgroup$ Commented Jun 7, 2021 at 15:27
  • $\begingroup$ Yes sure, the hope is to get something similar to the result for a pair of random lattice walk. For such a pair of $n$ steps walk with $\alpha n$ positions taking the same steps, the probability that one walk is $c$-bounded given that the other is is $$\frac{c}{\sqrt{\pi(1-\alpha)\alpha n}}.$$ $\endgroup$ Commented Jun 8, 2021 at 14:49
  • $\begingroup$ The calculations are simpler for the walks since the steps are independent and we can use local limit theorem. It is possible that the dependence of the steps in the pair of bridges might make the probability at bit smaller. $\endgroup$ Commented Jun 8, 2021 at 14:54
  • $\begingroup$ The conditional probability will decay exponentially in $n$. $\endgroup$ Commented Jun 12, 2021 at 19:08
  • $\begingroup$ Is there a easy way to calculate the asymptotics? $\endgroup$ Commented Jun 13, 2021 at 20:08

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