Pattern-avoiding permutations and Brownian excursion, part II: fixed points

Abstract

Permutations that avoid given patterns are among the most classical objects in combinatorics and have strong connections to many fields of mathematics, computer science and biology. In this paper we study fixed points of both 123- and 231-avoiding permutations. We find an exact description for a scaling limit of the empirical distribution of fixed points in terms of Brownian excursion. This builds on the connections between pattern-avoiding permutations and Brownian excursion developed in Hoffman et al. (Pattern-avoiding permutations and Brownian excursion, Part 1: Shapes and fluctuations. to appear Random Structures and Algorithms. arXiv:1406.5156, 2016) and strengthens the recent results of Elizalde (Electron J Comb 18(2):17, 2011) and Miner and Pak (Adv Appl Math 55:86–130, 2014) on fixed points of pattern-avoiding permutations.

Appendix A: Technical lemmas

We begin with a useful Lemma that will help count non-negative lattice paths between points. Let ${\mathcal {A}}_n$ denote the set of points $(i,m) \in {\mathbb {Z}} ^2$ such that $0<n^{0.6}<i<2n^{0.9}$ and $|m| < i^{0.55}.$

Lemma 6.1

For $(i,m) \in {\mathcal {A}}_n.$

$$\begin{aligned} { 2i-m \atopwithdelims ()i } = \frac{(1+ \Delta (i,m))4^{i}}{2^m\sqrt{\pi i}}e^{-\frac{m^2}{4i }} \end{aligned}$$

where $\Delta (i,m) = o(n^{-0.1})$ uniformly in i and m in ${\mathcal {A}}_n$.

For $i> n^{0.6}$ and $|m| > i^{0.55}$

$$\begin{aligned} {2i -m \atopwithdelims ()i} \le \frac{4^i}{2^m}e^{-n^{0.05}}. \end{aligned}$$

Proof

This first equality follows from $\triangleright$ IX.1 on page 615 of Flajolet and Sedgewick [10]. For the second equality we let $m = i^{0.55} + r$ or $m = -i^{0.55} -r$ for some $r>0.$ Using the first equality,

$$\begin{aligned} {2i - m \atopwithdelims ()i }= & {} {2i - i^{0.55} \atopwithdelims ()i } \prod _{k = 0}^{r-1} \frac{ i - i^{0.55} -k }{2i - i^{0.55} - k}\\\le & {} \frac{4^i}{2^{i^{0.55} +r}}e^{-i^{0.1}/4} \prod _{k=0}^{r-1}\frac{ 2i - 2i^{0.55} - 2k }{2i - i^{0.55} - k}.\\\le & {} \frac{4^i}{2^m}e^{-n^{0.05}}. \end{aligned}$$

A similar computation holds for $m = -i^{0.55} -r.$ $\square$

Consider a lattice path starting at $(v_0,h_0).$ Recall Definition 4.2. We may extend those definitions to general lattice paths with a slight modification. The definitions $v_i$ and $h_i$ remain the same, the position and the height after the ith up-step from the start of the path. For $l_i$ we do not necessarily have an excursion. If the path never returns below $h_i$ at some time later than $v_i$ then we say that $l_i = \infty .$

Lemma 6.2

Suppose $(i,m)\in {\mathcal {A}}_n$ and $h = h_0 + m.$ The number of lattice paths starting from $(v_0,h_0)$ ending with an ith up-step to $(v_i, h_0 + m)$ is given by

$$\begin{aligned} { 2(i-1) - (m-1) \atopwithdelims ()i-1} = \frac{4^{i}}{2^{m+1}\sqrt{\pi i}}\exp \left( -\frac{m^2}{4i} \right) (1+\Delta (i,m)) \end{aligned}$$

where $\Delta (i,m)$ is as defined in Lemma 6.1.

Proof

Let i and d denote the number of up and down steps respectively in a lattice path starting at $v_0$ ending with the ith up-step at position $v_i$. We denote the total number of steps by $v_i=v_0 + i+d$. The change in height for that path is given by $h-h_0 = m= i-d$ so $v_i = v_0+ 2i-m.$ The second to last position of the path is $(v_0 + 2i -m -1,h_0+m-1)$ since the last step is assumed to be an up-step. The total number of lattices paths from $(v_0,h_0)$ to $(v_0+2i-m-1,h_0 +m -1)$ is counted by Lemma 6.1, giving the equation found in Lemma 6.2 $\square$

Now that we can accurately count the number of lattice paths from one point to another we can count the number of non-negative paths between two points. For a pair of points $(v_0,h_0)$ and $(v_i,h_i)$ we let ${\mathcal {E}}_{v_0,h_0}^{v_i,h_i}$ denote the set of non-negative lattice paths ending with an up-step between the two points.

Lemma 6.3

For $(i,m) \in {\mathcal {A}}_n$, let $h_0> n^{0.499},$ $v_i = v_0 + 2i-m,$ and $h_i = h_0+m.$ Then

$$\begin{aligned} \left| {\mathcal {E}}_{v_0,h_0}^{v_i,h_i}\right| = \frac{4^{i}}{2^{m+1}\sqrt{\pi i}}\exp \left( -\frac{m^2}{4i} \right) (1+\Delta '(i,m)) \end{aligned}$$

where $\Delta '(i,m)=o(n^{-0.1})$ uniformly in $i,m \in A_n.$

Proof

We count using standard ballot counting arguments. It is enough to consider only $h_0 < i-m$ in the following approximations.

$$\begin{aligned} \left| {\mathcal {E}}_{v_0,h_0}^{v_0+2i-m,h_0+m}\right| = {2i-m-1 \atopwithdelims ()i-1} - {2i -m -1 \atopwithdelims ()i + h_0 }. \end{aligned}$$

For $h_0 >n^{0.499},$ if $h_i>h_0$ then for $m>0$

$$\begin{aligned} {2i-m-1 \atopwithdelims ()i+h_0 } / {2i-m-1 \atopwithdelims ()i-1}&\le \prod _{r = 0}^{h_0} \left( \frac{ i-m-r}{i+r} \right) \\&\le \prod _{r = 0}^{h_0} \left( 1 - \frac{ r}{4i} \right) \le \exp \left( -(h_0)^2/100i\right) . \end{aligned}$$

If $h_i < h_0$ then $h_0 -|m| > h_0/2$ giving

$$\begin{aligned} {2i+|m|-1 \atopwithdelims ()i+h_0 } \big / {2i+|m|-1 \atopwithdelims ()i-1}&\le \prod _{r = 0}^{ |m|} \left( \frac{ i+|m|-r}{i+r} \right) \prod _{r = |m|+1}^{h_0} \left( \frac{ i+|m|-r}{i+r} \right) \\&\le \prod _{r = |m|+1}^{h_0} \left( 1 - \frac{ 2r-|m|}{i+r} \right) \\&\le \prod _{s=0}^{h_0/2} \left( 1 - \frac{s}{4i} \right) \\&\le \exp \left( -(h_0)^2/100i\right) . \end{aligned}$$

Moreover $(h_0)^2/100i >n^{0.07},$ so by Lemma 6.2

$$\begin{aligned} \left| {\mathcal {E}}_{v_0,h_0}^{v_i,h_i}\right|= & {} {2i-m-1\atopwithdelims ()i-1} \left( 1+ O\left( \exp \left( -n^{0.06}\right) \right) \right) \\= & {} \frac{4^{i}}{2^{m+1}\sqrt{\pi i}}\exp \left( -\frac{m^2}{4i} \right) \left( 1+\Delta '\left( i,m\right) \right) . \end{aligned}$$

$\square$

For paths chosen uniformly from ${\mathcal {E}}_{v_0,h_0}^{v_j,h_j}$ for $0<i<j$ we would like to know for various values of i and h how many of these path go through the point $(v_i,h)$ after the ith up-step. Given $\Gamma ^n \in {\mathcal {E}}_{v_0,h_0}^{v_j,h_j}$ chosen uniformly at random what is the probability that $H^n_i = h?$

Lemma 6.4

Fix $v_0, h_0, v_j$ and $h_j$. Let $X^n$ be chosen uniformly from ${\mathcal {E}}_{v_0,h_0}^{v_j,h_j}$. For $h>0$ and $0<i<j$,

$$\begin{aligned} \mathbb {P}( H^n_i = h) = \left| {\mathcal {E}}_{v_0,h_0}^{v_0 + 2i - (h-h_0),h} \right| \left| {\mathcal {E}}_{v_0+2i-(h-h_0),h}^{v_j,h_j}\right| \left| {\mathcal {E}}_{v_0,h_0}^{v_j,h_j}\right| ^{-1}. \end{aligned}$$

Proof

Any path in ${\mathcal {E}}_{v_0,h_0}^{v_j,h_j}$ can be decomposed uniquely into a concatenation of two paths, one in ${\mathcal {E}}_{v_0,h_0}^{v_i,h_i}$ and the other in ${\mathcal {E}}_{v_i,h_i}^{v_j,h_j}$ for some appropriate values of $v_i$ and $h_i$ that satisfy $v_i = v_0 + 2i - (h_i-h_0).$ If $h_i = h,$ then $v_i = v_0 + 2i-(h-h_0).$ The set $\left\{ X^n \in {\mathcal {E}}_{v_0,h_0}^{v_j,h_j} | h_i=h\right\}$ is in bijection with ${\mathcal {E}}_{v_0,h_0}^{v_i,h} \times {\mathcal {E}}_{v_i,h}^{v_j,h_j}.$ Then

$$\begin{aligned} \mathbb {P}( H^n_i = h) =&\left| \left\{ \Gamma ^n \in {\mathcal {E}}_{v_0,h_0}^{v_j,h_j} | H^n_i=h\right\} \right| \left| {\mathcal {E}}_{v_0,h_0}^{v_j,h_j}\right| ^{-1}\\ =&\left| {\mathcal {E}}_{v_0,h_0}^{v_0+2i-(h-h_0),h} \right| \left| {\mathcal {E}}_{v_0 + 2i-(h-h_0),h}^{v_j,h_j}\right| \left| {\mathcal {E}}_{v_0,h_0}^{v_j,h_j}\right| ^{-1} \end{aligned}$$

as desired (see Fig. 5). $\square$

Lemma 6.5

Fix $v_0, h_0, v_j,$ with $h_j$ such that $v_j - v_0 =2j - (h_j-h_0)$ as above. Also fix $i,h> 0$ and $i < j-h.$ For $X^n$ chosen uniformly from ${\mathcal {E}}_{v_0,h_0}^{v_j,h_j},$

$$\begin{aligned} \mathbb {P}(L^n_i/2 = h | H^n_i = h )= C_{h-1} \left| {\mathcal {E}}_{v_0+ 2i-(h-h_0)+2h-1,h-1}^{v_j,h_j}\right| \left| {\mathcal {E}}_{v_0 + 2i-(h-h_0),h}^{v_j,h_j}\right| ^{-1}. \end{aligned}$$

Proof

Let $X^n \in {\mathcal {E}}_{v_0,h_0}^{v_j,h_j}$ also satisfy $H^n_i = h$. From Lemma 6.4 there are precisely

$$\begin{aligned} |{\mathcal {E}}_{v_0,h_0}^{v_0 + 2i-(h-h_0),h}||{\mathcal {E}}_{v_0 + 2i-(h-h_0),h}^{v_j,h_j}| \end{aligned}$$

such paths. Each of these paths that satisfies $L^n_i/2 = h$ has a unique decomposition into three parts:

• A path $X^n_1 \in {\mathcal {E}}_{v_0,h_0}^{v_0 + 2i-(h-h_0),h}$,

• an excursion $X^n_2$ from $(v_0 + 2i-(h-h_0)_i,h)$ to $(v_ 0 + 2i-(h-h_0) + 2h-2,h)$, staying above $h-1,$

• a single down-step from $(v_0 + 2i-(h-h_0)+2h-2,h)$ to $(v_0 + 2i-(h-h_0)+2h-1,h-1)$,

• and a path $X^n_3 \in {\mathcal {E}}_{v_0 + 2i-(h-h_0) + 2h-1,h-1}^{v_j,h_j}.$

Fig. 5

Decomposition of $X^n$ into $X^n_1$, $X^n_2$, and $X^n_3$

The choice of $X^n_1, X^n_2,$ and $X^n_3$ uniquely determines $X^n.$ There are

$$\begin{aligned} \left| {\mathcal {E}}_{v_0,h_0}^{v_0 + 2i-(h-h_0),h} \right| , C_{h-1}, \quad \text {and}\quad \left| {\mathcal {E}}_{v_0 + 2i-(h-h_0)+2h-1,h-1}^{v_j,h_j}\right| \end{aligned}$$

such choices for $X^n_1, X^n_2,$ and $X^n_3$ respectively. Therefore

$$\begin{aligned}&\mathbb {P}(L^n_i/2 = h | H^n_i = h) \\&\quad =\left| {\mathcal {E}}_{v_0,h_0}^{v_0 + 2i-(h-h_0),h} \right| C_{h-1} \left| {\mathcal {E}}_{v_0 + 2i-(h-h_0)+2h-1,h-1}^{v_j,h_j}\right| \\&\qquad \times \left( \left| {\mathcal {E}}_{v_0,h_0}^{v_0 + 2i-(h-h_0),h} \right| \left| {\mathcal {E}}_{v_0 + 2i-(h-h_0),h}^{v_j,h_j}\right| \right) ^{-1}\\&\quad = C_{h-1} \left| {\mathcal {E}}_{v_0 + 2i-(h-h_0)+2h-1,h-1}^{v_j,h_j}\right| \left| {\mathcal {E}}_{v_0 + 2i-(h-h_0),h}^{v_j,h_j}\right| ^{-1} \end{aligned}$$

$\square$

Lemma 6.6

For $X^n \in {\mathcal {E}}_{v_0,h_0}^{v_j,h_j}$ chosen uniformly at random and $0<i<j,$

$$\begin{aligned}&\mathbb {P}(L^n_i/2 = H^n_i ) \\&\quad = \sum _{h=\max (0,h_j-(j-i))}^{h_0 + i} C_{h-1} \left| {\mathcal {E}}_{v_0,h_0}^{v_0 + 2i-(h-h_0),h} \right| \left| {\mathcal {E}}_{v_0 + 2i-(h-h_0)+2h-1,h-1}^{v_j,h_j}\right| \left| {\mathcal {E}}_{v_0,h_0}^{v_j,h_j}\right| ^{-1}. \end{aligned}$$

Proof

If $L^n_i/2 = H^n_i$ then there is some $h \in {\mathbb {N}}$ such that $\{ H^n_i = h\} \cap \{ L^n_i/2 = h\}$ occurs. Therefore

$$\begin{aligned} \{ L^n_i/2 = H^n_i \} = \bigcup _h \left\{ \{H^n_i= h \} \cap \{L^n_i/2 = h \} \right\} . \end{aligned}$$

Luckily the event $\{H^n_i = h \cap L^n_i/2 = h\}$ is disjoint from $\{ H^n_i = h' \cap L^n_i/2 = h' \}$ for $h\ne h'.$ Then

$$\begin{aligned} \mathbb {P}\left( \bigcup _h \{H^n_i= h \} \cap \{L^n_i/2 = h \} \right) = \sum _h \mathbb {P}\left( \{H^n_i= h \} \cap \{L^n_i/2 = h \} \right) . \end{aligned}$$

If $h \notin ( \max (0,H^n_j - (j-i), h_0 + i)$ then $\mathbb {P}( H^n_i = h) = 0.$ Otherwise we have

$$\begin{aligned} \mathbb {P}\left( \{ H^n_i = h \} \cap \{ L^n_i/2 = h\} \right) = \mathbb {P}\left( L^n_i/2 = h | H^n_i = h\right) \mathbb {P}(H^n_i = h ). \end{aligned}$$

Combining Lemmas 6.4 and 6.5 provides the result. $\square$

Lemma 6.7

Let $0<i<j\le 2n^{0.9}$ with $i>2n^{0.6}$ and $j-i > 2n^{0.6},$ and let $h_0 \in (n^{0.499},n^{0.501})$. Define m and $m_j$ such that $h_j = h_0 + m_j$ and $h = h_0 + m$ where $|m_j|< \min (j^{0.55},n^{0.451}).$ Let $m_{max} = \min ( i^{0.55}, m_j+(j-i)^{0.55} )$ and $m_{min} = \max ( -i^{0.55}, m_j - (j-i)^{0.55}).$ For $m_{min}<m<m_{max}$

$$\begin{aligned}&\mathbb {P}( L^n_i/2=h | H^n_i = h) \mathbb {P}(H^n_i = h ) \\&\quad = \frac{ \sqrt{j}}{4\pi \sqrt{i(j-i)(h_0)^3} } \exp \left( - \frac{ ( jm - im_j )^2}{4ij(j-i)}\right) (1+\Delta ) \end{aligned}$$

where $\Delta =o(n^{-0.001} )$ uniformly in $i,j,m,m_j,h_0$ that satisfy the above conditions.

For $m <m_{min}$ or $m>m_{max},$

$$\begin{aligned} \mathbb {P}(L^n_i/2 = h|H^n_i =h)\mathbb {P}(H^n_i = h ) < \exp \left( -n^{0.001}\right) . \end{aligned}$$

Proof

The summand in Lemma 6.6 is given by

$$\begin{aligned} \mathbb {P}(L^n_i/2= & {} h|H^n_i=h) \mathbb {P}( H^n_i = h ) \\= & {} C_{h-1} \left| {\mathcal {E}}_{v_0,h_0}^{v_0 + 2i-m,h} \right| \left| {\mathcal {E}}_{v_0 + 2i-m+2h-1,h-1}^{v_j,h_j}\right| \left| {\mathcal {E}}_{v_0,h_0}^{v_j,h_j}\right| ^{-1}. \end{aligned}$$

By Lemma 6.3 we can make the following substitutions:

$$\begin{aligned} C_{h-1}&= \frac{4^{h-1}}{\pi ^{1/2}h^{3/2}} (1+\Delta _1(h))),\\ {\mathcal {E}}_{v_0,h_0}^{v_0 + 2i-m,h}&= { 2(i-1) - (m-1) \atopwithdelims ()i-1 } \\&= \frac{ 4^i}{2^{m+1}\pi ^{1/2}i^{1/2}}e^{-m^2/4i}(1+\Delta _2(i-1,m-1)),\\ {\mathcal {E}}_{v_0,h_0}^{v_j,h_j}&= { 2(j-1) - (m_j-1) \atopwithdelims ()j-1 } \\&= \frac{ 4^j}{2^{m_j+1}\pi ^{1/2}j^{1/2}}e^{-m_j^2/4j}(1+\Delta _2(j-1,m_j-1)). \end{aligned}$$

where both $\Delta _1$ and $\Delta _2$ are bounded uniformly by $n^{-0.01}$ over all parameters satisfying the conditions of the lemma. Furthermore,

$$\begin{aligned} {\mathcal {E}}_{v_0 + 2i-m+2h-1,h-1}^{v_j,h_j}&= { 2(j-i-h) - (m_j-m) \atopwithdelims ()j-i-h }\\&= \frac{ 4^{j-i-h}}{2^{m_j-m}\pi ^{1/2}(j-i-h)^{1/2}}e^{-(m_j-m)^2/4(j-i-h)}(1+\Delta _2(j-i\\&\quad -h,mj-m)),\\&= \frac{ 4^{j-i-h}}{2^{m_j-m}\pi ^{1/2}(j-i)^{1/2}}e^{-(m_j-m)^2/4(j-i)}(1+\Delta _3(j-i\\&\quad -h,mj-m)) \end{aligned}$$

where $\Delta _3$ is also uniformly bounded by $n^{-0.01}$ over all parameters satisfying the conditions of the lemma. Combining these equations together proves the first statement of Lemma 6.7. For the second statement we use the second approximation in Lemma 6.1 to bound $\mathbb {P}(H^n_i= h_0 + m)$ using the formula in Lemma 6.4. $\square$

Let’s consider the special case where $j\approx n^{0.9}.$

Lemma 6.8

For $X^n$ chosen uniformly from ${\mathcal {E}}_{v_0,h_0}^{v_j,h_j}$ with $i,j,h_0,$ and $h_j$ satisfying

• $j = n^{0.9}(1+\Delta '), \Delta '\le n^{-0.1}$ uniformly.

• $n^{0.499}<h_0 < n^{0.501}.$

• $i \in ( 2n^{0.6}, n^{0.9}-2n^{0.6}),$

• $h_j = h_0 + m_j$ where $|m_j| \le n^{0.451},$

then

$$\begin{aligned} \mathbb {P}( L^n_i/2 = H^n_i ) = \frac{ 1 }{2 \pi ^{1/2} (h_0)^{3/2} } (1+\Delta ), \end{aligned}$$

where $\Delta = \Delta (i,h_0,h_j) = o(n^{-0.001})$ uniformly in $i,h_0,h_j$ in the ranges above.

Proof

Let $m_{min} = \max ( -i^{0.55}, m_j -(j-i)^{0.55} )$ and $m_{max} = \min ( i^{0.55}, m_j+(j-i)^{0.55})$ and consider the inequality which follows from Lemma 6.6.

$$\begin{aligned}&\sum _{m=m_{min}}^{m_{max}}\mathbb {P}(L^n_i/2=h_0 + m| H^n_i=h_0+m)\mathbb {P}(H^n_i=h_0 + m)\nonumber \\&\quad \le \mathbb {P}(L^n_i/2 =H^n_i )\nonumber \\&\quad \le ne^{-n^{0.001}} + \sum _{m=m_{min}}^{m_{max} }\mathbb {P}(L^n_i/2=h_0 +m| H^n_i=h_0 + m)\mathbb {P}(H^n_i=h_0 +m).\nonumber \\ \end{aligned}$$

(27)

Lemma 6.7 gives

$$\begin{aligned}&\mathbb {P}\left( L^n_i/2 = H^n_i\right) \\&\quad = \left( 1+o\left( n^{-0.01}\right) \right) \sum _{m=m_{min}}^{m_{max}} \frac{ \sqrt{j}}{4\pi \sqrt{i\left( j-i\right) \left( h_0\right) ^3} } \exp \left( - \frac{ \left( jm - im_j \right) ^2}{4ij\left( j-i\right) }\right) \\&\qquad \left( 1+o\left( n^{-0.01}\right) \right) .\\&\quad = \left( 1+o\left( n^{-0.001}\right) \right) \int _{m_{min}}^{m_{max}} \frac{ 1}{4\pi \sqrt{i\left( 1-i/j\right) \left( h_0\right) ^3} } \exp \left( - \frac{ \left( m - im_j/j \right) ^2}{4i\left( 1-i/j\right) }\right) \\&\qquad \left( 1+o\left( n^{-0.01}\right) \right) dm. \end{aligned}$$

By our definition $(m_{min} - \frac{i}{j}m_j ) < -n^{0.01}$ and $(m_{max} - \frac{i}{j}m_j ) > n^{0.01}$. Therefore the integral above is computed in the standard way, with

$$\begin{aligned} \int _{-t}^t \exp \left( -\frac{ (m- m_0)^2 }{4c}\right) dm = 2c^{1/2}\pi ^{1/2} + \delta (t). \end{aligned}$$

where $\delta (t)$ is an error function with exponential decay.

$$\begin{aligned} \mathbb {P}\left( L^n_i/2 =H^n_i \right) = \frac{1}{2\pi ^{1/2}\left( h_0\right) ^{3/2}}\left( 1+o\left( n^{-0.001}\right) \right) . \end{aligned}$$

$\square$

Corollary 6.9

For any $k \in {\mathbb {R}}$ and $\alpha \in (0,.48)$ let $\Gamma ^n$ chosen uniformly from ${\mathcal {E}}_{v_0,h_0}^{v_j,h_j}$ with $i,j,h_0,$ and $h_j$ satisfying

• $j = n^{0.9}(1+ \Delta '), \Delta ' < n^{-0.1}$ uniformly.

• $n^{0.499}<h_0 < n^{0.501}.$

• $i \in ( 2n^{0.6}, n^{0.9}-2n^{0.6}),$

• $h_j = h_0 + m_j$ where $|m_j| \le n^{0.451},$

then

$$\begin{aligned} \mathbb {P}( L^n_i/2 = H^n_i-k(i(n-i)/n)^\alpha ) = \frac{ 1 }{2 \pi ^{1/2} (h_0)^{3/2} } (1+\Delta ), \end{aligned}$$

where $\Delta = \Delta (i,h_0,h_j) = o(n^{-0.001})$ uniformly in $i,h_0,h_j$ in the ranges above.

Proof

The proof goes exactly as in Lemma 6.8 with $L^n_i =H^n_i$ replaced by $L^n_i =H^n_i - k(i(n-i)/n)^\alpha$. The order of $k(i(n-i)/n)^\alpha$ is less than $n^{0.49}$ so it will not affect the approximation. $\square$

Lemma 6.10

Fix $0<a<b<1$ and let $a_k = \lfloor an + nk/K\rfloor$ where $K = \lfloor (b-a) n^{0.1} \rfloor .$ For $\Gamma ^n \in \text {Dyck}^{2n}$ chosen uniformly at random,

$$\begin{aligned} \mathbb {P}\left( \bigcap _{k=0}^{K}\{n^{0.499}< \Gamma ^n({V^n_{a_k}})< n^{0.501}\} \right) > 1 - o(1). \end{aligned}$$

Proof

This is an immediate consequence of Corollary 6.19 along with the convergence of Dyck paths to Brownian excursion. $\square$

Lemma 6.11

Fix $0<a<b<1.$ For any n large enough and $\Gamma ^n \in \text {Dyck}^{2n}$,

$$\begin{aligned}&\mathbb {P}\left( \bigcup _ {k=0}^{K-1}\left\{ |\Gamma ^n({V^n_{a_k}}) - \Gamma ^n\left( V^n_{a_{k+1}} \right) | > n^{0.451} \right\} \Big | \bigcap _k \left\{ n^{0.499}< \Gamma ^n\left( {V^n_{a_k}} \right)< n^{0.501}\right\} \right) \\&\quad < e^{-n^{0.0001}}. \end{aligned}$$

Proof

This follows by a similar argument to Corollaries 6.19 and 6.20, with slight modifications to the parameters. $\square$

Lemma 6.12

For sufficiently large n, for every $\frac{1}{2}n^{0.9} < j\le 2n^{0.9},$ and $h_0,h_j$ both bounded between $n^{0.499}$ and $n^{0.501}$ with $|h_0-h_j| < n^{0.451}$ we have that if $X^n \in {\mathcal {E}}_{v_0,h_0}^{v_{j},h_j}$ is chosen uniformly at random then

$$\begin{aligned} \mathbb {P}\left( \sup _{t\in [0,1]} \left| X^n( v_0 + tv_j ) - h_0\right| > 2 n^{0.452} \right) < e^{-n^{0.001}}. \end{aligned}$$

Proof

First note that the maximum fluctuation of $|X^n(v_0 + tv_j) - h_0|$ is within 1 of the maximum fluctuation of $|H^n_i- h_0|$ for $i \le j$. Suppose $i > j/2$, $H^n_i=h$, and $|h -h_0| > n^{0.452}$. Lemma 6.3 gives

$$\begin{aligned} |{\mathcal {E}}_{v_0,h_0}^{V^n_i,h}| \le \frac{4^{i}}{2^{h-h_0+1}}e^{-n^{0.904}/4j}, \end{aligned}$$

for sufficiently large n, independent of $j,h_0$, and $h_j$ satisfying the hypotheses of the lemma. There are $2(j-i)-(h_j-h)$ steps remaining to go from $(v_i,h)$ to $(v_j,h_j).$ So

$$\begin{aligned} |{\mathcal {E}}_{V^n_i,h}^{v_j,h_j}| \le \frac{ 4^{j-i}}{2^{h_j-h}}. \end{aligned}$$

Again by Lemma 6.3,

$$\begin{aligned} \left| {\mathcal {E}}_{v_0,h_0}^{v_j,h_j}\right| \ge \frac{4^j}{2^{h_j-h_0+2}\sqrt{\pi j}}\exp \left( -\frac{(h_j-h_0)^2}{4j}\right) , \end{aligned}$$

for sufficiently large n, independent of $j,h_0$, and $h_j$ satisfying the hypotheses of the lemma. Then we can conclude that

$$\begin{aligned} \mathbb {P}\left( H_i^n =h \right) \le 4\sqrt{\pi j} \exp \left( -\frac{n^{0.904}}{4j} + \frac{(n^{0.451})^2}{4j}\right) \le 8\sqrt{\pi } n^{0.9} e^{-n^{0.004}/8}. \end{aligned}$$

A similar bound can be used for $i < j/2$ with a little more work. Note that if $|h - h_0| > n^{0.452}$ and $|h_j-h_0|<n^{0.451}$ then $|h-h_j| > \frac{1}{2}n^{0.452}.$ The same argument now applies. Using the union bound and summing over the possible values of h and i now gives the result. $\square$

Lemma 6.13

Let $I \subset [an,bn]$ denote an interval of length at most $n^{\alpha }.$ For $\gamma \in \text {Dyck}^{2n},$ if $h_i > n^{0.49}$ for all $i\in I$, then

$$\begin{aligned} \theta _I \le 1 + n^{\alpha - 0.49}. \end{aligned}$$

Proof

If the jth excursion is contained in ith excursion, then both $h_j > h_i$ and $l_j/2 < l_i/2.$ If $\theta _i=1$ then for $i<j< l_i/2$, $\theta _j=0.$

For an interval I with $|I|< n^{0.49}$ suppose at least one fixed point exists. Let $i^*$ denote the first excursion that satisfies $\theta _{i*}=1.$ Since $i^*$ corresponds to a fixed point,

$$\begin{aligned} l_{i^*} = h_{i^*} > n^{0.49}. \end{aligned}$$

Therefore, for all $j\in I$ such that $j>i^*$ the jth excursion is contained in the $i^*$th excursion and $\theta _j=0$. Therefore either $\theta _{I}$ is either 0 or 1. For an interval of size greater than or equal to $n^{\alpha }$, it can be covered by $n^{\alpha -0.49}$ intervals of size $n^{0.49}$ each of which has at most one fixed point so the total number of fixed points will be bounded by $n^{\alpha -0.49}.$ $\square$

Corollary 6.14

Let $I \subset [an,bn]$ denote an interval of length at most $n^{\alpha }.$ For $\gamma \in \text {Dyck}^{2n},$ if $h_i > n^{0.49}$ for all $i\in I$, then

$$\begin{aligned} \theta ^{K,\alpha }_I \le 2n^{\alpha - 0.49}. \end{aligned}$$

Proof

The function $f(i)=(i(n-i)/n)^{\alpha }$ can change by at most 1 over any interval of length $n^{.49}$. This implies that over any interval $I'$ of length $n^{.49}$ has $\theta ^{K,\alpha }_{I'}\le 2$. Then the result follows as in Lemma 6.13. $\square$

Lemma 6.15

There exists constant $C>0$ such that or every n large enough and i, j, w that satisfy the following:

• (a) $2n^{0.6}<i<j<n^{0.9}-2n^{0.6},$

• (b) $|i-j| > 2n^{0.6},$

• (c) and $w < n^{0.451}$

  $$\begin{aligned} \Psi (i,j,w,n)= & {} \sum _{m' = -\infty }^{\infty } \sum _{ m = -\infty }^{\infty } \exp \left( - \frac{1}{4} \left( \frac{m^2}{i} + \frac{(m-m')^2}{j-i} + \frac{(w-m')^2}{n^{0.9}-j} - \frac{w^2}{n^{0.9}} \right) \right) \\< & {} C \sqrt{ \frac{i (j-i) (n^{0.9} - j) }{n^{0.9}}}. \end{aligned}$$

Proof

Our goal will be to convert this double sum into a recognizable form.

$$\begin{aligned} \Psi (i,j,w,n) \le&\sum _{m'=-\infty }^{\infty }\sum _{m = -\infty }^{\infty } \exp \left( - \frac{1}{4} \left( \frac{m^2}{i} + \frac{(m-m')^2}{j-i} +\frac{(w-m')^2}{n^{0.9}-j} - \frac{w^2}{n^{0.9}} \right) \right) \\ \le&\sum _{m' =-\infty }^{\infty } \left[ \exp \left( -\frac{1}{4} \left( \frac{(w-m')^2}{n^{0.9}-j} - \frac{w^2}{n^{0.9}} \right) \right) G\left( i,j,m'\right) \right] \end{aligned}$$

where

$$\begin{aligned} G( i,j,m' ) = \sum _{m= -\infty }^{\infty }\exp \left( -\frac{1}{4} \left( \frac{m^2}{i} + \frac{(m-m')^2}{j-i} \right) \right) . \end{aligned}$$

With some algebra we see

$$\begin{aligned} G(i,j,m')= & {} \exp \left( -\frac{1}{4} \frac{ {m'}^2}{j} \right) \sum _{m=-\infty }^{\infty } \exp \left( -\frac{1}{4} \frac{j(m-m'i/j)^2}{i(j-i)} \right) \\\le & {} C_1 \exp \left( -\frac{1}{4} \frac{ {m'}^2}{j} \right) \int _{-\infty }^{\infty } \exp \left( -\frac{1}{4} \frac{j}{i(j-i)} (t-m'i/j)^2\right) dt\\\le & {} C_1 \exp \left( -\frac{1}{4} \frac{ {m'}^2}{j} \right) \sqrt{ \frac{i(j-i)}{j} } \end{aligned}$$

for some positive constant $C_1$ that does not depend on i, j or $m'$. Inserting this into the upper bound for $\Psi (i,j,w,n)$ gives

$$\begin{aligned} \Psi (i,j,w,n)\le & {} C_1\sqrt{ \frac{i(j-i)}{j} } \sum _{m'=-\infty }^{\infty }\exp \left( -\frac{1}{4}\left( \frac{{m'}^2}{j} + \frac{(m'-w)^2}{n^{0.9}-j} - \frac{w^2}{n^{0.9}} \right) \right) \\\le & {} C_1\sqrt{ \frac{i(j-i)}{j} }\sum _{m' = -\infty }^{\infty } \exp \left( -\frac{1}{4}\left( \frac{n^{0.9} ( m' - wj/n^{0.9} )^2}{j(n^{0.9}-j)}\right) \right) \\\le & {} C \sqrt{ \frac{i(j-i)(n^{0.9}-j)}{n^{0.9}} } \end{aligned}$$

where $C>0$ and does not depend on i, j, w,  and n. $\square$

The next two results are special cases of [24, Theorem III.12, Theorem III.15] respectively (see also [21, Lemma A1, Lemma A2]).

Lemma 6.16

Let $X_1,X_2,\dots$ be i.i.d with ${\mathbb {E}}X_1=0$ and let $S_n =X_1+\cdots + X_n$. Suppose that $\sigma ^2= {\mathbb {E}}(X_1^2) < \infty$. For all x and n we have

$$\begin{aligned} {\mathbb {P}}\left( \max _{1\le k\le n} S_k \ge x\right) \le 2 {\mathbb {P}}\left( S_n\ge x - \sqrt{2n\sigma ^2}\right) . \end{aligned}$$

Lemma 6.17

Let $X_1,X_2,\dots$ be i.i.d with ${\mathbb {E}}X_1=0$ and let $S_n =X_1+\cdots + X_n$. Suppose that there exists $a>0$ such that ${\mathbb {E}}(e^{t |X_1|}) < \infty$. Then there exist constants $g, T>0$, independent of n, such that

$$\begin{aligned} {\mathbb {P}}(S_n\ge x) \le {\left\{ \begin{array}{ll} \exp \left( -\frac{x^2}{2gn}\right) &{} \text {if } 0\le x\le ngT \\ \exp \left( -\frac{Tx}{2}\right) &{} \text {if } x\ge ngT. \end{array}\right. } \end{aligned}$$

These lemmas lead immediately to the following corollary.

Corollary 6.18

Maintaining the hypotheses of Lemma 6.17, fix $\epsilon , c>0$ and $0< \alpha < 2\beta$ and let $\nu = \min (\beta , 2\beta -\alpha )$. There exist constants $A,B >0$ such that

$$\begin{aligned} {\mathbb {P}}\left( \max _{1\le i\le n} \max _{|i-j|\le cn^{\alpha }} |S_j -S_i| \ge \epsilon n^\beta \right) \le A \exp \left( -B n^{\nu }\right) \end{aligned}$$

Let $S=(S_m,m\ge 0)$ be a simple symmetric random walk on ${\mathbb {Z}}$ with $S_0=0$. Define $V_0=0$ and for $m\ge 1$ let $V_m = \inf \{k>V_{m-1} : S_k-S_{k-1} = 1\}$. Let $\eta (S) = \inf \{ k : S_k=-1\}$. Observe that $(V_m-V_{m-1})_{m\ge 1}$ is an i.i.d sequence of geometric random variables with parameter 1 / 2.

Corollary 6.19

Fix $\epsilon >0$ and $1/2 < \alpha \le 1$. There exist constants $A, B>0$ such that

$$\begin{aligned} {\mathbb {P}}\left( \max _{1\le i\le n} |V_i - 2i| \ge \epsilon n^\alpha \right) \le A \exp \left( -B n^{2 \alpha -1}\right) . \end{aligned}$$

Corollary 6.20

Let $\Gamma ^n$ be a uniformly random Dyck path of length 2n. For any $\delta >0$, there exist constants $A,B,\nu >0$ such that for all $n\ge 1$

$$\begin{aligned} {\mathbb {P}}\left( \max _{1\le i\le 2n} \Gamma ^n_i \ge 0.4 n^{0.5 +\delta }\right) \le A\exp \left( -Bn^\nu \right) \end{aligned}$$

and

$$\begin{aligned} {\mathbb {P}}\left( \max _{1\le i\le 2n} \max _{|i-j|\le 2n^{0.5+\delta }} |\Gamma ^n_j -\Gamma ^n_i| \ge 0.5 n^{0.25 + \delta } \right) \le A \exp \left( -B n^{\nu }\right) \end{aligned}$$

Proof

Noting that

$$\begin{aligned} \Gamma ^n \overset{d}{=} (S_k, 0\le k \le 2n) \text { given } \eta (S) =2n+1, \end{aligned}$$

the first claim is an immediate consequence of Lemmas 6.16 and 6.17 combined with the fact that ${\mathbb {P}}(\eta (S) =2n+1) \sim cn^{-3/2}$ for some $c>0$. The second claim follows similarly from Corollary 6.18. $\square$