Universal Order Statistics for Random Walks & Lévy Flights

Abstract

We consider one-dimensional discrete-time random walks (RWs) of n steps, starting from $x_0=0$, with arbitrary symmetric and continuous jump distributions $f(\eta )$, including the important case of Lévy flights. We study the statistics of the gaps $\varDelta _{k,n}$ between the $k\text {th}$ and $(k+1)\text {th}$ maximum of the set of positions $\{x_1,\ldots ,x_n\}$. We obtain an exact analytical expression for the probability distribution $P_{k,n}(\varDelta )$ valid for any k and n, and jump distribution $f(\eta )$, which we then analyse in the large n limit. For jump distributions whose Fourier transform behaves, for small q, as ${\hat{f}} (q) \sim 1 - |q|^\mu$ with a Lévy index $0< \mu \le 2$, we find that the distribution becomes stationary in the limit of $n\rightarrow \infty$, i.e. $\lim _{n\rightarrow \infty } P_{k,n}(\varDelta )=P_k(\varDelta )$. We obtain an explicit expression for its first moment $\mathbb {E}[\varDelta _{k}]$, valid for any k and jump distribution $f(\eta )$ with $\mu >1$, and show that it exhibits a universal algebraic decay $\mathbb {E}[\varDelta _{k}]\sim k^{1/\mu -1} \varGamma \left( 1-1/\mu \right) /\pi$ for large k. Furthermore, for $\mu >1$, we show that in the limit of $k\rightarrow \infty$ the stationary distribution exhibits a universal scaling form $P_k(\varDelta ) \sim k^{1-1/\mu } \mathcal {P}_\mu (k^{1-1/\mu }\varDelta )$ which depends only on the Lévy index $\mu$, but not on the details of the jump distribution. We compute explicitly the limiting scaling function $\mathcal {P}_\mu (x)$ in terms of Mittag–Leffler functions. For $1< \mu <2$, we show that, while this scaling function captures the distribution of the typical gaps on the scale $k^{1/\mu -1}$, the atypical large gaps are not described by this scaling function since they occur at a larger scale of order $k^{1/\mu }$. This atypical part of the distribution is reminiscent of a “condensation bump” that one often encounters in several mass transport models.

Appendices

Proof of the Schuette–Nesbitt Formula

In this Appendix we provide two different derivations of the Schuette–Nesbitt formula.

1.1 Direct Derivation

In this section, we provide a proof of the Schuette–Nesbitt formula in (33). Let $A_1,\ldots ,A_n$ be a set of events. We introduce the indicator function I(A) which takes the value 1 if A occurred and 0 otherwise. We wish to compute the counting probability $\text {Prob.}(\nu =\sum _{m=1}^n I(A_m))$. To do so, let us first work with indicator functions and denote by $\mathcal {N}=\{1,\ldots ,n\}$ the set of numbers from 1 to n. For exactly $\nu$ events to happen, we sum over all possible combinations of sets of $\nu$ events and require that all the events in the set happened, while requiring that the remaining ones did not happen. This reads

$$\begin{aligned} I\left( \nu =\sum _{m=1}^n I(A_m)\right) = \sum _{\begin{array}{c} J\subset \mathcal {N}\\ |J|=\nu \end{array}} I\left( \bigcap _{j\in J} A_{j}\right) \times I\left( \bigcap _{l\in \mathcal {N}\setminus J}{\bar{A}}_l\right) , \end{aligned}$$

(127)

where |J| denotes the cardinal number of the set J. By using the rule $I(B)=1-I({\bar{B}})$, where $\bar{B}$ is the complementary event of B, we can rewrite the last indicator function as

$$\begin{aligned} I\left( \nu =\sum _{m=1}^n I(A_m)\right) = \sum _{\begin{array}{c} J\subset \mathcal {N}\\ |J|=\nu \end{array}} I\left( \bigcap _{j\in J} A_{j}\right) \times \left[ 1-I\left( \bigcup _{l\in \mathcal {N}\setminus J} A_l\right) \right] . \end{aligned}$$

(128)

Making use of the inclusion-exclusion principle and that $|\mathcal {N}\setminus J|=n-\nu$, we rewrite the last indicator function as

$$\begin{aligned} I\left( \nu =\sum _{m=1}^n I(A_m)\right) = \sum _{\begin{array}{c} J\subset \mathcal {N}\\ |J|=\nu \end{array}} I\left( \bigcap _{j\in J} A_{j}\right) \times \left[ 1-\sum _{p=1}^{n-\nu }(-1)^{p-1}\sum _{\begin{array}{c} K\subset \mathcal {N}\setminus J\\ |K|=p \end{array}}I\left( \bigcap _{k\in K} A_k\right) \right] . \end{aligned}$$

(129)

Distributing the product, we obtain

$$\begin{aligned} I\left( \nu =\sum _{m=1}^n I(A_m)\right) = \sum _{\begin{array}{c} J\subset \mathcal {N}\\ |J|=\nu \end{array}} I\left( \bigcap _{j\in J} A_{j}\right) - \sum _{p=1}^{n-\nu }(-1)^{p-1}\sum _{\begin{array}{c} J\subset \mathcal {N}\\ |J|=\nu \end{array}}\sum _{\begin{array}{c} K\subset \mathcal {N}\setminus J\\ |K|=p \end{array}}I\left( \bigcap _{q\in K\cup J} A_q\right) , \end{aligned}$$

(130)

where we have used that $I\left( \bigcap _{j\in J} A_{j}\right) \times I\left( \bigcap _{k\in K} A_k\right) =I\left( \bigcap _{q\in K\cup J} A_q\right)$. The last double sum enumerates all the sets in $\mathcal {N}$ with $\nu +p$ elements. However, due to the double sum, we over-count them by a factor $\left( {\begin{array}{c}\nu +p\\ p\end{array}}\right) =\left( {\begin{array}{c}\nu +p\\ \nu \end{array}}\right)$, which is the number of ways one can separate a set of size $\nu +p$ into two sets of sizes $\nu$ and p. Therefore, it can be written as

$$\begin{aligned} I\left( \nu =\sum _{m=1}^n I(A_m)\right) = \sum _{\begin{array}{c} J\subset \mathcal {N}\\ |J|=\nu \end{array}} I\left( \bigcap _{j\in J} A_{j}\right) - \sum _{p=1}^{n-\nu }(-1)^{p-1}\left( {\begin{array}{c}p+\nu \\ \nu \end{array}}\right) \sum _{\begin{array}{c} L\subset \mathcal {N}\\ |L|=\nu +p \end{array}}I\left( \bigcap _{q\in L} A_q\right) . \end{aligned}$$

(131)

Finally, noting that the first term on the rhs of (131) can be written as the $p=0$ term of the sum over p, we find

$$\begin{aligned} I\left( \nu =\sum _{m=1}^n I(A_m)\right) = \sum _{p=0}^{n-\nu }(-1)^{p}\left( {\begin{array}{c}p+\nu \\ \nu \end{array}}\right) \sum _{\begin{array}{c} L\subset \mathcal {N}\\ |L|=\nu +p \end{array}}I\left( \bigcap _{q\in L} A_q\right) , \end{aligned}$$

(132)

which, upon taking the expectation value and taking the generating function with respect to $\nu$, we get

$$\begin{aligned} \sum _{\nu =0}^n z^\nu \text {Prob.}\left( \nu =\sum _{m=1}^n I(A_m)\right) = \sum _{k=0}^n (z-1)^k \sum _{1\le i_1<\ldots <i_k\le n}\text {Prob.}\left( \bigcap _{s=1}^k A_{i_s}\right) , \end{aligned}$$

(133)

which recovers the Schuette–Nesbitt formula in (33).

1.2 Alternative Derivation

An alternative proof of this last formula (133) is as follows. The generating function of $P_{\nu , n} = \mathbb {E}[ I\left( \nu =\sum _{m=1}^n I(A_m)\right) ]$—with respect to $\nu =0, \ldots , n$—can also be written as

$$\begin{aligned} \sum _{\nu =0}^n z^\nu P_{\nu ,n} = \mathbb {E}\left[ z^{\sum _{m=1}^n I(A_m)}\right] = \mathbb {E}\left[ \prod _{m=1}^n z^{I(A_m)}\right] . \end{aligned}$$

(134)

Since $I(A_m)$, for $m=1, \ldots , n$, is a binary variable that takes only values 0 or 1, it is easy to check the identity

$$\begin{aligned} z^{I(A_m)} = 1 + I(A_m)(z-1). \end{aligned}$$

(135)

By inserting this identity (135) in (134), one obtains

$$\begin{aligned} \sum _{\nu =0}^n z^\nu P_{\nu ,n} = \mathbb {E}\left[ \prod _{m=1}^n \left( 1 + I(A_m)(z-1) \right) \right] . \end{aligned}$$

(136)

We then expand the product in (136) to obtain

$$\begin{aligned} \sum _{\nu =0}^n z^\nu P_{\nu ,n}= & {} \sum _{k=0}^n (z-1)^k \sum _{1\le i_1< i_2< \cdots< i_k \le n} \mathbb {E}[ I(x_{i_1}) I(x_{i_2}) \cdots I(x_{i_k})]\nonumber \\= & {} \sum _{k=0}^n (z-1)^k \sum _{1\le i_1< i_2< \ldots < i_k \le n} \text {Prob.}\left( \bigcap _{s=1}^k A_{i_s}\right) , \end{aligned}$$

(137)

which is the result given in (133).

Evaluation of the Integral (49) Using the Pollaczek–Spitzer Formula

In this appendix, we compute the integral in (49), which we decompose in two parts

$$\begin{aligned} \int _{-\infty }^0 da \,{\tilde{p}}_{z,s}(a) = I_1(z,s) - I_2(s), \end{aligned}$$

(138)

where

$$\begin{aligned} I_1(z,s)&=\int _0^\infty da \left[ \frac{1}{1-s} {\bar{S}}_s\left( u\Big |a\right) - {\bar{S}}_0(sz|a)\right] , \end{aligned}$$

(139)

$$\begin{aligned} I_2(s)&= \int _0^\infty da \left[ {\bar{S}}_0(s|-a) -1\right] . \end{aligned}$$

(140)

Below, in Sects. B.1 and B.2, we show that $I_1(z,s)$ and $I_2(s)$ are given by

$$\begin{aligned} I_1(z,s)&= \frac{1}{1-sz}\int _0^\infty \frac{dq}{\pi q^2}\ln \left( \frac{1-s {\hat{f}}(q)}{1-s}\right) , \end{aligned}$$

(141)

$$\begin{aligned} I_2(s)&= \frac{s}{\sqrt{1-s}}\int _{-\infty }^0 da \int _0^\infty dy f(y-a)\int _{\gamma _B} \frac{d\lambda e^{\lambda a}}{2\pi i \lambda }\exp \left( -\lambda \int _0^\infty \frac{dq}{\pi } \frac{\ln (1-s {\hat{f}}(q))}{\lambda ^2+q^2}\right) . \end{aligned}$$

(142)

Inserting these expressions into (138), we recover the expression (52) from the main text.

1.1 Analysis of $I_1(z,s)$

To analyse $I_1(z,s)$ in (139), we introduce an exponential cutoff in the integral which we let go to 0 as

$$\begin{aligned} I_1(z,s) = \lim _{\lambda \rightarrow 0}\int _0^\infty da \,e^{-\lambda a}\left[ \frac{1}{1-s} {\bar{S}}_s\left( u\Big |a\right) - {\bar{S}}_0(sz|a)\right] . \end{aligned}$$

(143)

Using the Pollaczek–Spitzer formula in (50) and the expression of the effective distribution (51), we find

$$\begin{aligned} I_1(z,s)&= \lim _{\lambda \rightarrow 0}\frac{1}{\lambda }\Bigg [\frac{1}{(1-s)\sqrt{1-u}}\exp \left( -\lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln \left[ 1-\frac{s(z-1){\hat{f}}(q)}{1-s{\hat{f}}(q)}\right] }{\lambda ^2+q^2}\right) \nonumber \\&\quad - \frac{1}{\sqrt{1-sz}}\exp \left( -\lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln [1-sz {\hat{f}}(q)]}{\lambda ^2+q^2}\right) \Bigg ]. \end{aligned}$$

(144)

We decompose the logarithms in (144) as

$$\begin{aligned}&\ln \left[ 1-\frac{s(z-1){\hat{f}}(q)}{1-s{\hat{f}}(q)}\right] = \ln \left[ 1-\frac{s(z-1)}{1-s}\right] + \ln \left[ \frac{1-\frac{s(z-1){\hat{f}}(q)}{1-s{\hat{f}}(q)}}{1-\frac{s(z-1)}{1-s}}\right] , \end{aligned}$$

(145)

$$\begin{aligned}&\ln [1-sz {\hat{f}}(q)] = \ln [1-sz ] +\ln \left[ \frac{1-sz {\hat{f}}(q)}{1-sz}\right] , \end{aligned}$$

(146)

to find that (144) becomes

$$\begin{aligned} I_1(z,s)&= \lim _{\lambda \rightarrow 0}\frac{1}{1-sz}\Bigg [\exp \left( -\lambda \int _0^\infty \frac{dq}{\pi }\frac{1}{\lambda ^2+q^2}\ln \left[ \frac{1-\frac{s(z-1){\hat{f}}(q)}{1-s{\hat{f}}(q)}}{1-u}\right] \right) \nonumber \\&\quad - \exp \left( -\lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln \left[ \frac{1-sz {\hat{f}}(q)}{1-sz}\right] }{\lambda ^2+q^2}\right) \Bigg ]. \end{aligned}$$

(147)

Taking the limit $\lambda \rightarrow 0$ and simplifying we get (141).

1.2 Analysis of $I_2(s)$

To analyse $I_2(s)$ in (140), we cannot directly use Pollaczek–Spitzer formula in (50) as the initial position in the generating function of the survival probability is negative. Therefore, we first decompose the generating function of the survival probability over the first step as

$$\begin{aligned} {\bar{S}}_0(s|-a) = 1 + s \int _0^\infty dy f(a+y){\bar{S}}_0(s,y), \end{aligned}$$

(148)

where $f(\eta )$ is the jump distribution. We can then replace ${\bar{S}}_0(s,y)$ by Pollaczek–Spitzer formula (50) in (148) as the initial position is now positive, which gives (142). The limit $s\rightarrow 1$ of $I_2(s)$ can be obtained by rescaling q by $\lambda$ and, a, y and $\lambda$ by $(1-s)^{\frac{1}{\mu }}$, which gives

$$\begin{aligned} I_2(s)&\sim \frac{1}{(1-s)^{\frac{1}{\mu }-\frac{1}{2}}}\int _{-\infty }^0 da \int _0^\infty dy \frac{c_\mu }{(y-a)^{1+\mu }} \nonumber \\&\quad \times \int _{\gamma _B} \frac{d\lambda e^{\lambda a}}{2\pi i \lambda }\exp \left( - \int _0^\infty \frac{dq}{\pi } \frac{\ln \left[ (1-s)(1+\lambda ^\mu q^\mu )\right] }{1+q^2}\right) , \quad s\rightarrow 1, \end{aligned}$$

(149)

where we used the expansion $f(\eta )\sim \frac{c_\mu }{\eta ^{1+\mu }}$ for $\eta \rightarrow \infty$, where $c_\mu =\varGamma (1+\mu )\sin (\pi \mu /2)/\pi$, which can be obtained by inverting the small q expression of the Fourier transform ${\hat{f}}(q)$ in (11). Using that $\int _0^\infty \frac{dq}{\pi }\frac{1}{1+q^2}=\frac{1}{2}$, it gives

$$\begin{aligned} I_2(s)&\sim \frac{1}{(1-s)^\frac{1}{\mu }}\int _{-\infty }^0 da \int _0^\infty dy \frac{c_\mu }{(y-a)^{1+\mu }} \nonumber \\&\quad \times \int _{\gamma _B} \frac{d\lambda e^{\lambda a}}{2\pi i \lambda }\exp \left( - \int _0^\infty \frac{dq}{\pi } \frac{\ln \left( 1+\lambda ^\mu q^\mu \right) }{1+q^2}\right) ,\quad s\rightarrow 1. \end{aligned}$$

(150)

One can easily check that the remaining integrals in (150) are well-behaved for $\mu >1$.

Alternative Derivation of the Expected Gap

As in [33], we start from the celebrated Pollaczek–Wendel identity [47,48,49] which states that the double generating function of the $k\text {th}$ maximum is given by

$$\begin{aligned} \varphi (s,z,\rho )&= \sum _{n=0}^\infty \sum _{k=0}^n s^n z^k \mathbb {E}[ \exp \left( i \rho M_{k+1,n}\right) ] \nonumber \\&= \exp \left( \sum _{n=1}^\infty \frac{s^n}{n} \mathbb {E}[ \exp \left( i\rho x_n^+\right) ] + \frac{(sz)^n}{n} \mathbb {E}[ \exp \left( i\rho x_n^{-}\right) ]\right) . \end{aligned}$$

(151)

Note that the $k\text {th}$ maximum $M_{k,n}$ in the Pollaczek–Wendel identity is the $k\text {th}$ maximum of the set $\{x_0,\ldots ,x_n\}$, which includes the initial position of the random walk. This is different to the definition given in the main text above (2), which does not include the initial position, but should not matter for the analysis of the expected stationary gap $\mathbb {E}[\varDelta _{k}]$. As [33] showed, this formula can be more conveniently written as

$$\begin{aligned} \varphi (s,z,\rho )&= \frac{1}{\sqrt{1-s}\sqrt{1-sz}\sqrt{1-s{\hat{f}}(\rho )}\sqrt{1-sz{\hat{f}}(\rho )}} \nonumber \\&\quad \exp \left( \frac{i}{2\pi }\text {PV}\int _{-\infty }^{\infty } \frac{dq}{q-\rho }\ln (1-s{\hat{f}}(q))\right) \nonumber \\&\quad \times \exp \left( -\frac{i}{2\pi }\text {PV}\int _{-\infty }^\infty \frac{dq}{q-\rho }\ln (1-sz{\hat{f}}(q))\right) , \end{aligned}$$

(152)

where PV refers to the principal value of the integral. From this generating function, we can extract the generating function of the mean value $\mathbb {E}[ M_{k,n}]$ of the $k^{\text {th}}$ maximum:

$$\begin{aligned} \sum _{n=0}^\infty \sum _{k=0}^n z^k \mathbb {E}[ M_{n,k+1}] = \frac{1}{i}\partial _{\rho } \varphi (s,z,\rho )\bigg |_{\rho =0}. \end{aligned}$$

(153)

Next, as in [33], we notice that (152) is not well suited to for an expansion close to $s=0$, and we use trick, similar to theirs but adapted to fat-tailed distributions (11), which consists in writing

$$\begin{aligned} \ln (1-s{\hat{f}}(q)) = \ln \left( 1-s\left( 1-|a q|^\mu \right) \right) + \ln \left( \frac{1-s{\hat{f}}(q)}{1-s\left( 1-|q|^\mu \right) }\right) . \end{aligned}$$

(154)

As in Eq. (45) in [31], we denote

$$\begin{aligned} I_1(s,\rho )&= \frac{\rho }{\pi }\int _{0}^{\infty } \frac{dq}{q^2+\rho ^2}\ln \left( 1-s\left( 1-a^\mu q^\mu \right) \right) \nonumber \\&= \frac{1}{\pi }\int _{0}^{\infty } \frac{dq}{q^2+1}\ln \left( 1-s\left( 1-a^\mu \rho ^\mu q^\mu \right) \right) , \end{aligned}$$

(155)

and

$$\begin{aligned} I_2(s,\rho )&= \frac{\rho }{\pi } \int _0^\infty \frac{dq}{\rho ^2+q^2} \ln \left[ \frac{1-s{\hat{f}}(q)}{1-s\left( 1- q^\mu \right) }\right] . \end{aligned}$$

(156)

Using the decomposition (154) and the definitions (155) and (156), we find that $\varphi (s,z,\rho )$ writes

$$\begin{aligned} \varphi (s,z,\rho )&= \frac{1}{\sqrt{1-s}\sqrt{1-sz}\sqrt{1-s{\hat{f}}(\rho )}\sqrt{1-sz{\hat{f}}(\rho )}}\nonumber \\&\quad \times \exp \left( I_1(s,i\rho )+I_2(s,i\rho )-I_1(sz,i\rho )-I_2(sz,i\rho )\right) . \end{aligned}$$

(157)

As in [31], we take a derivative with respect to $\rho$ and evaluate at $\rho =0$, which gives

$$\begin{aligned}&\sum _{n=0}^\infty \sum _{k=0}^n s^n z^k \mathbb {E}[ M_{n,k+1}] = \frac{1}{i}\partial _{\rho } \varphi (s,z,\rho )\bigg |_{\rho =0} \nonumber \\&\quad =\frac{1}{(1-s)(1-sz)}\left[ \partial _\rho I_1(s,0)+\partial _\rho I_2(s,0)-\partial _\rho I_1(sz,0)-\partial _\rho I_2(sz,0) \right] , \end{aligned}$$

(158)

where we used that $I_1(s,0)=I_2(s,0)=0$. The derivative of $I_1(s,\rho )$ at $\rho =0$ is given by

$$\begin{aligned} \partial _\rho I_1(s,0)&=\lim _{\rho \rightarrow 0} \,\frac{\mu s \rho ^{\mu -1}}{\pi }\int _{0}^{\infty } \frac{dq}{q^2+1}\frac{q^\mu }{\left[ 1-s\left( 1- \rho ^\mu q^\mu \right) \right] } \nonumber \\&\quad \times \frac{s^{1/\mu }}{(1-s)^{1/\mu }\sin (\frac{\pi }{\mu })}, \end{aligned}$$

(159)

where we rescaled the integration variable by $\rho ^\mu$. The derivative of $I_2(s,\rho )$ at $\rho =0$ is given by

$$\begin{aligned} \partial _\rho I_2(s,0)&= \frac{1}{\pi } \int _0^\infty \frac{dq}{q^2} \ln \left[ \frac{1-s{\hat{f}}(q)}{1-s\left( 1- q^\mu \right) }\right] . \end{aligned}$$

(160)

Inserting the expressions (159) and (160) in (158), we find

$$\begin{aligned}&\sum _{n=0}^\infty \sum _{k=0}^n s^n z^k \mathbb {E}[ M_{n,k+1}] \nonumber \\&\quad = \frac{1}{(1-s)(1-sz)}\Bigg ( \frac{s^{1/\mu }}{(1-s)^{1/\mu }\sin (\frac{\pi }{\mu })} - \frac{(sz)^{1/\mu }}{(1-sz)^{1/\mu }\sin (\frac{\pi }{\mu })}\nonumber \\&\qquad \times c{1}{\pi } \int _0^\infty \frac{dq}{q^2} \ln \left[ \frac{1-s{\hat{f}}(q)}{1-s\left( 1- q^\mu \right) }\right] - \frac{1}{\pi } \int _0^\infty \frac{dq}{q^2} \ln \left[ \frac{1-sz{\hat{f}}(q)}{1-s\left( 1- q^\mu \right) }\right] \Bigg ). \end{aligned}$$

(161)

Taking the limit $s\rightarrow 0$ gives

$$\begin{aligned}&\sum _{n=0}^\infty \sum _{k=0}^n s^n z^k \mathbb {E}[ M_{n,k+1}] \nonumber \\&\quad \sim \frac{1}{(1-s)^{1+1/\mu }(1-z)\sin (\frac{\pi }{\mu })}+\frac{1}{(1-s)(1-z)}\Bigg (\frac{1}{\pi } \int _0^\infty \frac{dq}{q^2} \ln \left[ \frac{1-{\hat{f}}(q)}{ q^\mu }\right] \nonumber \\&\qquad -\frac{1}{\pi } \int _0^\infty \frac{dq}{q^2} \ln \left[ \frac{1-z{\hat{f}}(q)}{1-z\left( 1- q^\mu \right) }\right] - \frac{a z^{1/\mu }}{(1-z)^{1/\mu }\sin (\frac{\pi }{\mu })}\Bigg ),\quad s\rightarrow 1. \end{aligned}$$

(162)

Using Tauberian theorem, we find

$$\begin{aligned} \sum _{k=0}^n z^k \mathbb {E}[ M_{n,k+1}]&\sim \frac{\mu \, \varGamma \left( 1-\frac{1}{\mu }\right) n^{\frac{1}{\mu }}}{\pi (1-z) }\nonumber \\&\quad +\frac{1}{(1-z)}\Bigg (\frac{1}{\pi } \int _0^\infty \frac{dq}{q^2} \ln \left[ \frac{1-{\hat{f}}(q)}{ q^\mu }\right] \nonumber \\&\quad -\frac{1}{\pi } \int _0^\infty \frac{dq}{q^2} \ln \left[ \frac{1-z{\hat{f}}(q)}{1-z\left( 1- q^\mu \right) }\right] \nonumber \\&\quad - \frac{z^{1/\mu }}{(1-z)^{1/\mu }\sin (\frac{\pi }{\mu })}\Bigg ),\quad n\rightarrow \infty . \end{aligned}$$

(163)

To invert the remaining generating function we use the identity

$$\begin{aligned} \frac{1}{\pi }\int _0^\infty \frac{dq}{q^2}\ln \left( \frac{1-\frac{z}{1+ q^\mu }}{1-z(1- q^\mu )}\right) = \frac{a}{\sin (\frac{\pi }{\mu })} \left( \frac{1}{(1-z)^{1/\mu }}-1-\frac{z^{1/\mu }}{(1-z)^{1/\mu }}\right) , \end{aligned}$$

(164)

in order to rewrite the term $\frac{z^{1/\mu }}{(1-z)^{1/\mu }\sin (\frac{\pi }{\mu })}$, which gives

$$\begin{aligned}&\sum _{k=0}^n z^k \mathbb {E}[ M_{n,k+1}] \sim \frac{1}{1-z}\left( \frac{\mu \, \varGamma \left( 1-\frac{1}{\mu }\right) n^{\frac{1}{\mu }}}{\pi }+\frac{1}{\pi } \int _0^\infty \frac{dq}{q^2} \ln \left[ \frac{1-{\hat{f}}(q)}{ q^\mu }\right] +\frac{1}{\sin (\frac{\pi }{\mu })}\right) \nonumber \\&\quad -\frac{1}{(1-z)^{1+1/\mu }\sin (\frac{\pi }{\mu })}\nonumber \\&\quad +\frac{1}{(1-z)}\left( \frac{1}{\pi }\int _0^\infty \frac{dq}{q^2}\ln \left( 1-\frac{z}{1+ q^\mu }\right) -\frac{1}{\pi } \int _0^\infty \frac{dq}{q^2} \ln \left[ 1-z{\hat{f}}(q)\right] \right) \quad n\rightarrow \infty . \end{aligned}$$

(165)

Using the following generating functions

$$\begin{aligned} \sum _{m=0}^\infty \frac{\varGamma \left( m+\frac{1}{\mu }\right) z^m}{\varGamma \left( \frac{1}{\mu }\right) \varGamma \left( m+1\right) }&= \frac{1}{(1-z)^{1/\mu }}\,,\quad \sum _{m=1}^\infty \frac{z^m}{m} = -\ln (1-z)\,, \end{aligned}$$

(166)

and using the fact that a product of generating functions becomes a convolution, we find

$$\begin{aligned} \mathbb {E}[M_{n,k+1}]&\sim \frac{\mu \, \varGamma \left( 1-\frac{1}{\mu }\right) n^{\frac{1}{\mu }}}{\pi }+\frac{1}{\pi } \int _0^\infty \frac{dq}{q^2} \ln \left[ \frac{1-{\hat{f}}(q)}{ q^\mu }\right] \nonumber \\&\quad - \frac{1}{\pi }\,\varGamma \left( 1-\frac{1}{\mu }\right) \sum _{m=1}^k \frac{\varGamma \left( m+\frac{1}{\mu }\right) }{\varGamma \left( m+1\right) }\nonumber \\&\quad +\frac{1}{\pi } \sum _{m=1}^k \frac{1}{m}\int _0^\infty \frac{dq}{q^2} \left[ {\hat{f}}(q)^m-\frac{1}{(1+ q^\mu )^m}\right] . \end{aligned}$$

(167)

The stationary gap is therefore given by

$$\begin{aligned} \mathbb {E}[\varDelta _{k}]&= \lim _{n\rightarrow \infty } \mathbb {E}[M_{k,n}]- \mathbb {E}[M_{k+1,n}] =\frac{1}{\pi }\,\varGamma \left( 1-\frac{1}{\mu }\right) \frac{\varGamma \left( k+\frac{1}{\mu }\right) }{\varGamma \left( k+1\right) }\nonumber \\&\quad -\frac{1}{\pi } \frac{1}{k}\int _0^\infty \frac{dq}{q^2} \left[ {\hat{f}}(q)^k-\frac{1}{(1+ q^\mu )^k}\right] . \end{aligned}$$

(168)

Upon using the identity (56), we recover the expression (12) given in the introduction.

Analysis of the General Formula in Eq. (98) in the Case of a Double Sided Exponential Jump Distribution

To compute explicitly the generating function of the stationary gap distribution ${\tilde{P}}_z(\varDelta ) = \sum _{k=0} P_k(\varDelta )$ given in Eqs. (97) and (98) for the double exponential jump distribution $f(\eta ) = (1/2)\,e^{-|\eta |}$, it is useful to recall the remark done at the end of Sect. 3.1. Indeed, as done there (see Eqs. (88)–(90)), it is possible to express ${\tilde{p}}_z^{(1)}(\varDelta )$ in (97) in terms of the solution of an integral equation, similar to what was done in Eqs. (88)–(90). One finds indeed

$$\begin{aligned} {\tilde{p}}^{(1)}_{z}(\varDelta ) = \int _{-\varDelta }^0 dx_1 {\bar{S}}_*(z|x_1) {\bar{F}}_*(x_1|\varDelta ), \end{aligned}$$

(169)

where ${\bar{F}}_*(x_1|\varDelta )$ satisfies the integral equation

$$\begin{aligned} \int _{-\varDelta }^0 dx_1 {\tilde{E}}_*(z,x|x_1) {\bar{F}}_*(x_1|\varDelta ) = (1-z) {\bar{F}}_*(x|\varDelta ) + {\bar{S}}_*(z|x). \end{aligned}$$

(170)

In general, it is very difficult to solve this integral equation (170). However, it is possible to solve it explicitly for the double sided exponential distribution $f(\eta )=e^{-|\eta |}/2$, whose Fourier transform is

$$\begin{aligned} {\hat{f}}(k) = \frac{1}{1+k^2}. \end{aligned}$$

(171)

In this case, the effective jump distribution (51) is given by

$$\begin{aligned} {{\hat{f}}_s}(k) = \frac{1-s}{1-s+k^2}, \end{aligned}$$

(172)

which, in real space, reads $\mathrm { f_s}(\eta )= \sqrt{1-s}\exp (-\sqrt{1-s}|\eta |)/2$. Inserting this expression in (92) and (93), and using that in this case $g(\eta )=0$—see Eq. (94)—we find

$$\begin{aligned}&{\bar{S}}_*\left( z|x\right) =1-x\sqrt{1-z}, \end{aligned}$$

(173)

$$\begin{aligned}&{\tilde{E}}_*\left( z-1,y|x\right) =(1-z) \max (x,y)-\sqrt{1-z}, \end{aligned}$$

(174)

where we used the identities

$$\begin{aligned}&-\int _0^\infty \frac{dq}{\pi }\ln \left( 1+\frac{1-z}{q^2}\right) =-\sqrt{1-z}, \end{aligned}$$

(175)

$$\begin{aligned}&-(1-z)\int _0^\infty \frac{dq}{\pi }\frac{1}{q^2}\left( \cos (q(y-x))-1\right) =\frac{|y-x|(1-z)}{2}, \end{aligned}$$

(176)

together with the fact that the Laplace transform of the inverse Laplace transform of a function is the function itself, i.e., $\int _0^\infty dx e^{-x} \int _{\gamma _B} d\lambda \frac{e^{\lambda x}}{2\pi i} h(\lambda )=h(1)$ for any integrable function $h(\lambda )$. Inserting (173) and (174) into the integral Eq. (170) gives

$$\begin{aligned} \int _{-\varDelta }^0 dx_1 \left[ (1-z) \max (x_1,x)-\sqrt{1-z}\right] {\bar{F}}_*(x_1|\varDelta ) = (1-z) {\bar{F}}_*(x|\varDelta ) + 1-x\sqrt{1-z}. \end{aligned}$$

(177)

One can now take two derivatives with respect to x, using $\partial ^2_{x} \max (x_1,x) = \delta (x-x_1)$, to get the simple partial differential equation satisfied by F(x), namely

$$\begin{aligned} {\bar{F}}_*(x|\varDelta ) = \partial ^2_{x}{\bar{F}}_*(x|\varDelta ), \quad x \in [- \varDelta ,0]. \end{aligned}$$

(178)

Therefore ${\bar{F}}_*(x|\varDelta )$ is of the form

$$\begin{aligned} {\bar{F}}_*(x|\varDelta ) = A(\varDelta )\,e^{x} + B(\varDelta )\,e^{-x}, \end{aligned}$$

(179)

where $A(\varDelta )$ and $B(\varDelta )$ are two unknown integration constants, i.e. independent of x. To determine them, we inject this solution (179) in the original integral Eq. (177), which yields a linear system of two independent equations relating $A(\varDelta )$ and $B(\varDelta )$. By solving this linear system, one finds

$$\begin{aligned} A(\varDelta ) =&\frac{\sqrt{1-z}-1}{2 \left( \cosh (\varDelta )+\sqrt{1-z} \sinh (\varDelta )-z \cosh (\varDelta )\right) }, \nonumber \\ B(\varDelta ) =&\frac{\sqrt{1-z}+1}{2 (z-1) \cosh (\varDelta )-2 \sqrt{1-z} \sinh (\varDelta )}. \end{aligned}$$

(180)

Substituting these expressions in the one for ${\bar{F}}_*(x|\varDelta )$, we find

$$\begin{aligned} {\bar{F}}_*(x|\varDelta )= \frac{\sqrt{1-z} \sinh (x)-\cosh (x)}{\cosh (\varDelta )+\sqrt{1-z} \sinh (\varDelta )-z \cosh (\varDelta )}. \end{aligned}$$

(181)

Inserting it in (169) and performing the integration, we find a fairly simple expression, namely

$$\begin{aligned} {\tilde{p}}_z^{(1)}(\varDelta )&= \int _{-\varDelta }^0 dx_1 (1-x_1 \sqrt{1-z})\,\frac{\sqrt{1-z} \sinh (x_1)-\cosh (x_1)}{\cosh (\varDelta )+\sqrt{1-z} \sinh (\varDelta )-z \cosh (\varDelta )}\nonumber \\&= - \frac{1}{\sqrt{1-z}}- \varDelta + \frac{\sqrt{1-z} \sinh \varDelta +\cosh \varDelta }{\sqrt{1-z} \cosh \varDelta +\sinh \varDelta }. \end{aligned}$$

(182)

On the other hand, the term ${\tilde{p}}_{z,s}^{(2)}(\varDelta )$ in (86) in the limit $s \rightarrow 1$ reads

$$\begin{aligned} {\tilde{p}}_{z,s}^{(2)}(\varDelta ) \mathbb {E}({\tilde{\varDelta }}_{z,s}) + \frac{s\varDelta }{1-s}&\sim \frac{1}{1-s} \left( \int _0^\infty \frac{dq}{\pi } \frac{1}{q^2} \ln \left( \frac{1-z{\hat{f}}(q)}{1-z} \right) + \varDelta \right) \end{aligned}$$

(183)

$$\begin{aligned}&\sim \frac{1}{1-s} \left( \frac{1}{\sqrt{1-z}} - 1 + \varDelta \right) \,,\quad s\rightarrow 1\,, \end{aligned}$$

(184)

where we used (54) for the expected gap in the limit $s\rightarrow 1$. Finally, combining the results in (182) and (184) one finds

$$\begin{aligned} {\tilde{p}}_{z,s}(\varDelta ) = {\tilde{p}}_{z,s}^{(1)}(\varDelta ) + {\tilde{p}}_{z,s}^{(2)}(\varDelta ) \underset{s \rightarrow 1}{\sim }\ \frac{1}{1-s} \left( \frac{\sqrt{1-z} \sinh \varDelta +\cosh \varDelta }{\sqrt{1-z} \cosh \varDelta +\sinh \varDelta } - 1 \right) \;. \end{aligned}$$

(185)

Hence we recover the expression obtained in [33] and [39] [see equation (70) in that reference]. Note that there the generating function of $P_{k,n}(\varDelta )$ includes the term $k=0$ (with by convention $P_{0,n} = 1$), which is not the case here: this explains the $-1$ in (185), which is not present in [33, 39]. Note that it is also possible to compute explicitly the expressions of ${\bar{S}}_*(z|x)$ and ${\tilde{E}}_*(z,y|x)$ for the case of the jump distribution $f(\eta ) \propto |\eta |e^{-|\eta |}$ (see Appendix F).

Small s Limit of ${\bar{S}}_s\left( u|x\right)$ and ${\bar{E}}_s\left( u,y|x\right)$ for $x<0$ and $y<0$

In this Appendix, we analyse the small s limit of ${\bar{S}}_s\left( u|x\right)$ and ${\bar{E}}_s\left( u,y|x\right)$, with $u=s(z-1)/(1-s)$ for $x<0$ and $y<0$.

1.1 ${\bar{S}}_s\left( u|x\right)$

From the Pollaczek–Spitzer formula [47, 48], we know that for $x>0$, the generating function of the survival probability is given by (50). In particular, we have the Sparre Andersen result

$$\begin{aligned} {\bar{S}}_s(z,0)&= \frac{1}{\sqrt{1-z}}. \end{aligned}$$

(186)

To obtain an expression for $x<0$, we expand the survival probability over the first jump, which reads

$$\begin{aligned} S_s(n,x) = \int _0^\infty dy f_s(y-x) S_s(n-1,y), \end{aligned}$$

(187)

where $\textrm{f}_s(\eta )=\int _{-\infty }^\infty \frac{dq}{2\pi }e^{-iq\eta }\,{{\hat{f}}_s}(q)$ is the effective jump distribution (51). The expression (187) states that for the random walk to survive from x during n steps, it must jump to $y>0$ at the first step and survive $n-1$ steps from y. Upon taking a generating function of (187) and using that by definition in (41) one has $S_s(0,x)=1$ for $x\in \mathbb {R}$, the generating function of the survival probability reads

$$\begin{aligned} {\bar{S}}_s(z,x)&= 1 + z \int _0^\infty dx' \mathrm {f_s}(x'-x) {\bar{S}}_s(z,x'), \end{aligned}$$

(188)

which is valid for all $x\in \mathbb {R}$. We now analyse (188) in the limit $s\rightarrow 1$. To do so, we found it convenient to add and subtract the following terms

$$\begin{aligned} {\bar{S}}_s(z,x)&= 1 +z \int _0^\infty dx' \mathrm {f_s}(x') {\bar{S}}_s(z,x')+ z \int _0^\infty dx' [\mathrm {f_s}(x'-x)-\mathrm {f_s}(x')] {\bar{S}}_s(z,x'). \end{aligned}$$

(189)

Recognizing that the first two terms correspond to the generating function of the survival probability evaluated at $x=0$, we get

$$\begin{aligned} {\bar{S}}_s(z,x)&= {\bar{S}}_s(z,0) + z \int _0^\infty dx' [\mathrm {f_s}(x'-x)-\mathrm {f_s}(x')] {\bar{S}}_s(z,x'),\nonumber \\&= \frac{1}{\sqrt{1-z}} + z \int _0^\infty dx' [\mathrm {f_s}(x'-x)-\mathrm {f_s}(x')] {\bar{S}}_s(z,x'), \end{aligned}$$

(190)

where we used the Sparre Andersen result (186). As ${\bar{S}}_s(z,x')$ in the integral is only evaluated for positive argument $x'>0$, we can replace it by the Pollaczek–Spitzer formula given in (50). We find

$$\begin{aligned} {\bar{S}}_s(z,x)&= \frac{1}{\sqrt{1-z}} + \frac{z}{\sqrt{1-z}} \int _0^\infty dx' [\mathrm {f_s}(x'-x)-\mathrm {f_s}(x')]\int _{\gamma _B} d\lambda \frac{e^{\lambda x'}}{2\pi i\lambda } \nonumber \\&\quad \times \exp \left( -\lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln (1-z\, {{\hat{f}}_s}(q))}{\lambda ^2+q^2}\right) . \end{aligned}$$

(191)

We now evaluate this expression at $u=s(z-1)/(1-s)$ in the limit $s \rightarrow 1$. It reads explicitly

$$\begin{aligned} {\bar{S}}_s( u,x)&\sim \sqrt{\frac{1-s}{{1-z}}} - \frac{\sqrt{1-z}}{(1-s)^{3/2}} \int _0^\infty dx'[\mathrm {f_s}(x'-x)-\mathrm {f_s}(x')]\int _{\gamma _B} d\lambda \frac{e^{\lambda x'}}{2\pi i\lambda } \nonumber \\&\quad \times \exp \left( -\lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln (1+ \frac{(1-z){\hat{f}}(q)}{1-s {\hat{f}}(q)})}{\lambda ^2+q^2}\right) . \end{aligned}$$

(192)

Naively, it seems that one can take directly the limit $s \rightarrow 1$ in (192) by setting $s=1$ in the integral. This is however too naive, as it can be shown by an explicit calculation for the double exponential case. In fact, this can be seen from the fact that the expansion of the effective distribution as $s \rightarrow 1$ leads, to leading order

$$\begin{aligned} \frac{{{\hat{f}}_s}(q)}{(1-s)} = \frac{{\hat{f}}(q)}{1-s {\hat{f}}(q)} = \frac{{\hat{f}}(q)}{1 - \hat{f}(q)} - (1-s) \frac{{\hat{f}}(q)}{(1 - {\hat{f}}(q))^2} + O((s-1)^2). \end{aligned}$$

(193)

Once inserted in (192), we see that the integral of the term of order $O(1-s)$ is diverging for $q \rightarrow 0$ since $1/(1-{\hat{f}}(q))^2 \propto 1/|q|^{2\mu }$. This signals the fact that one must be careful to take this limit $s \rightarrow 1$. To analyse this limit, it is convenient to use the following trick which amounts to add and subtract ${\hat{f}^\mu _s}(q) = 1/(1+|q|^\mu )$ to ${\hat{f}_s}(q)$

$$\begin{aligned} {\hat{f}_s}(q) = {\hat{f}^\mu _s}(q) + {\hat{F}_s}(q) \quad \textrm{where} \quad {\hat{F}_s}(q) = (1-s)\frac{{\hat{f}}(q)}{1-s\, {\hat{f}}(q)} - (1-s) \frac{1}{(1-s)+|q|^\mu }. \end{aligned}$$

(194)

An interesting property of ${\hat{F}_s}(q)$ is that its expansion for $s \rightarrow 1$, i.e. the analogue of (193) reads, to leading order

$$\begin{aligned} \frac{{{\hat{F}}_s}(q)}{1-s} = \frac{(1+|q|^\mu ){\hat{f}}(q)-1}{(1-{\hat{f}}(q))|q|^\mu } - (1-s) \left( \frac{{\hat{f}}(q)^2}{(1-{\hat{f}}(q))^2} - \frac{1}{|q|^{2\mu }}\right) + O((1-s)^2). \end{aligned}$$

(195)

Since ${\hat{f}}(q) \sim 1 - |q|^\mu$ for small q we see that the divergence $\propto 1/q^{2\mu }$ is thus suppressed in the term of order $O(1-s)$ in (195), while the first term goes to a constant as $q \rightarrow 0$ – assuming that ${\hat{f}}(q) = 1 - q^\mu + O(q^{2\mu })$ as $q \rightarrow 0$. In the limit $s\rightarrow 1$, we find that the Fourier transform of the effective distribution (194) takes the limiting form

$$\begin{aligned} {{\hat{f}}_s}(q)&\sim \mathcal {{\hat{F}}}_\mu \left( \frac{q}{(1-s)^\frac{1}{\mu }}\right) +(1-s){\hat{g}}(q),\quad s\rightarrow 1, \end{aligned}$$

(196a)

which in real space reads

$$\begin{aligned} \mathrm {f_s}(x) = \int _{-\infty }^\infty \frac{dq e^{-iqx}}{2\pi } {{\hat{f}}_s}(q)&\sim (1-s)^{\frac{1}{\mu }} \mathcal {F}_\mu \left( (1-s)^\frac{1}{\mu }x\right) +(1-s)g\left( x\right) \,,\quad s\rightarrow 1, \end{aligned}$$

(196b)

where

$$\begin{aligned} \mathcal {{\hat{F}}}_\mu (u)&= \frac{1}{1+u^\mu }, \end{aligned}$$

(197)

$$\begin{aligned} {\hat{g}}(q)&=\frac{(1+|q|^\mu ){\hat{f}}(q)-1}{(1-{\hat{f}}(q))|q|^\mu }, \end{aligned}$$

(198)

and $\mathcal { F}_\mu (x)=\int _{-\infty }^\infty \frac{dq }{2\pi }e^{-ixu}\mathcal {{\hat{F}}}(u)$, and $g(x)=\int _{-\infty }^\infty \frac{dq }{2\pi }e^{-ixq}{\hat{g}}(q)$. Upon inserting the scaling (196) in (192), and rescaling all integration variables by $(1-s)^\frac{1}{\mu }$, we get upon letting $s\rightarrow 1$,

$$\begin{aligned} {\bar{S}}_s\left( u|x\right)&\sim \frac{\sqrt{1-s}}{\sqrt{1-z}} - \frac{\sqrt{1-z}}{\sqrt{1-s}}\int _0^\infty dx'\left[ \mathcal {F}_\mu \left( x'-(1-s)^\frac{1}{\mu }x\right) -\mathcal {F}_\mu \left( x'\right) \right] \int _{\gamma _B} d\lambda \frac{e^{\lambda x'}}{2\pi i\lambda } \nonumber \\&\quad \times \exp \left( -\lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln (1+\frac{1-z}{1-s}\, \frac{1}{1+q^\mu })}{\lambda ^2+q^2}\right) \nonumber \\&\quad -\sqrt{(1-z)(1-s)}\int _0^\infty dx'\left[ g\left( x'-x\right) -g\left( x'\right) \right] \int _{\gamma _B} d\lambda \frac{e^{\lambda x'}}{2\pi i\lambda } \nonumber \\&\quad \times \exp \left( -\lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln (1+ \frac{(1-z){\hat{f}}(q)}{1- {\hat{f}}(q)})}{\lambda ^2+q^2}\right) . \end{aligned}$$

(199)

Further letting $s\rightarrow 1$ in the logarithm, and using that $\lambda \int _0^\infty \frac{dq}{\pi } \frac{1}{\lambda ^2+q^2}=\frac{1}{2}$, we find

$$\begin{aligned} {\bar{S}}_s\left( u|x\right)&\sim \frac{\sqrt{1-s}}{\sqrt{1-z}} +(1-s)^\frac{1}{\mu }x\int _0^\infty dx'\mathcal {F}'_\mu \left( x'\right) \int _{\gamma _B} d\lambda \frac{e^{\lambda x'}}{2\pi i\lambda } \exp \left( \lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln ( 1+q^\mu )}{\lambda ^2+q^2}\right) \nonumber \\&\quad -\sqrt{1-s}\sqrt{1-z}\int _0^\infty dx'\left[ g\left( x'-x\right) -g\left( x'\right) \right] \int _{\gamma _B} d\lambda \frac{e^{\lambda x'}}{2\pi i\lambda } \nonumber \\&\quad \times \exp \left( -\lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln (1+ \frac{(1-z){\hat{f}}(q)}{1- {\hat{f}}(q)})}{\lambda ^2+q^2}\right) , \end{aligned}$$

(200)

where $\mathcal {F}'_\mu$ is the derivative of $\mathcal {F}_\mu$. Keeping only the highest order terms in (200), which means the first and the last terms for $\mu <2$ and all of them for $\mu =2$, we find the expression (91) given in the main text. For $\mu =2$, we used that

$$\begin{aligned} \int _0^\infty dx'\mathcal {F}_2'\left( x'\right) \int _{\gamma _B} d\lambda \frac{e^{\lambda x'}}{2\pi i\lambda } \exp \left( \lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln ( 1+q^2)}{\lambda ^2+q^2}\right)&= - 1, \end{aligned}$$

(201)

which can be easily shown by noting that $\mathcal {F}_2(x) = e^{-|x|}/2$ and using the fact that the Laplace transform of the inverse Laplace transform of a function is the function itself, i.e. that $\int _0^\infty dx e^{-x} \int _{\gamma _B} d\lambda \frac{e^{\lambda x}}{2\pi i} h(\lambda )=h(1)$ for a function h.

1.2 ${\tilde{E}}_s\left( u,y|x\right)$

From the Pollaczek–Spitzer formula we know that for $x>0$ and $y>0$, the generating function ${\bar{E}}_s(z,y|x)$ is given by

$$\begin{aligned} {\bar{E}}_s(z,y|x)&= \sum _{n=0}^\infty z^n E_s(n,y|x) = \int _{\gamma _B} d\lambda d\lambda ' \frac{e^{\lambda x}}{2\pi i} \frac{e^{\lambda ' y}}{2\pi i }\frac{1}{\lambda +\lambda '} \nonumber \\&\quad \times \exp \left( -\lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln (1-z\, {{\hat{f}}_s}(q))}{\lambda ^2+q^2}-\lambda ' \int _0^\infty \frac{dq}{\pi }\frac{\ln (1-z\, {{\hat{f}}_s}(q))}{\lambda '^2+q^2}\right) . \end{aligned}$$

(202)

Note that in our case, we are interested in ${\tilde{E}}_s(z,y|x)=\sum _{n=1}^\infty z^n E_s(n,y|x)$, where the index starts from 1 (hence the different notations ${\tilde{E}}_s(z,y|x)$ and ${\bar{E}}_s(z,y|x)$). We therefore need to remove the term $n=0$ in the sum in (202), which gives

$$\begin{aligned} {\tilde{E}}_s(z,y|x) = {\bar{E}}_s(z,y|x) -\delta (y-x), \end{aligned}$$

(203)

where we used that $E_s(0,y|x)=\delta (y-x)$. In particular for $x=y=0$, we have (see Appendix E.3)

$$\begin{aligned} {\tilde{E}}_s(z,0|0)= -\int _{0}^\infty \frac{dq}{\pi }\ln \left( 1-z \,{{\hat{f}}}_s(q)\right) . \end{aligned}$$

(204)

To obtain an expression for $x<0$ and $y<0$, similarly to the previous section, we expand the excursion over the first jump and last jump

$$\begin{aligned} E_s(n,y|x) = \int _0^\infty dx'dy' \mathrm {f_s}(x'-x) E_s(n-2,y'|x') \mathrm {f_s}(y-y'), \end{aligned}$$

(205)

where $\textrm{f}_s(\eta )=\int _{-\infty }^\infty \frac{dq}{2\pi }e^{-iq\eta }\,{{\hat{f}}_s}(q)$ is the effective jump distribution (51). The expression (205) states that for the random walk to make an excursion from x to y during n steps, it must perform a first jump to $x'>0$ at the first step, make an excursion of $n-2$ steps to $y'$, then jump from $y'$ to y. Upon taking a generating function of (205) and using that by definition in (72) $E_s(0,y|x)=\delta (y-x)$ and $E_s(1,y|x)=f_s(y-x)$ for $x,y\in \mathbb {R}$, we get that the generating function of the survival probability reads

$$\begin{aligned} {\bar{E}}_s(z,y|x)&= \sum _{n=0}^\infty z^n E_s(n,y|x)= \delta (y-x) + z f_s(y-x) \nonumber \\&\quad + z^2 \int _0^\infty dx'dy' \mathrm {f_s}(x'-x) {\bar{E}}_s(z,y'|x') \mathrm {f_s}(y-y'). \end{aligned}$$

(206)

Inserting this in (203), we get

$$\begin{aligned} {\tilde{E}}_s(z,y|x) =z f_s(y-x) + z^2 \int _0^\infty dx'dy' \mathrm {f_s}(x'-x) {\bar{E}}_s(z,y'|x') \mathrm {f_s}(y-y'). \end{aligned}$$

(207)

We now analyse the small s limit of (207). As in the previous section, we find it convenient to add and subtract the following terms

$$\begin{aligned} {\tilde{E}}_s(z,y|x)&= z [f_s(y-x)-f_s(0)] + z^2 \int _0^\infty dx'dy' [\mathrm {f_s}(x'-x)\mathrm {f_s}(y-y')\nonumber \\&\quad -\mathrm {f_s}(x')\mathrm {f_s}(-y')] {\bar{E}}_s(z,y'|x') \nonumber \\&\quad +z f_s(0)-z^2 \int _0^\infty dx'dy' \mathrm {f_s}(x') {\bar{E}}_s(z,y'|x') \mathrm {f_s}(-y'). \end{aligned}$$

(208)

Using (207), we recognize that the last line is the generating function evaluated at $x=y=0$, namely

$$\begin{aligned} {\tilde{E}}_s(z,y|x)&= z [f_s(y-x)-f_s(0)] \nonumber \\&\quad + z^2 \int _0^\infty dx'dy'[\mathrm {f_s}(x'-x)\mathrm {f_s}(y-y')-\mathrm {f_s}(x')\mathrm {f_s}(-y')] {\bar{E}}_s(z,y'|x') + {\tilde{E}}_s(z,0|0)\nonumber \\&= z [f_s(y-x)-f_s(0)] \nonumber \\&\quad + z^2 \int _0^\infty dx'dy'[\mathrm {f_s}(x'-x)\mathrm {f_s}(y-y')-\mathrm {f_s}(x')\mathrm {f_s}(-y')] {\bar{E}}_s(z,y'|x') \nonumber \\&\quad -\int _{0}^\infty \frac{dq}{\pi }\ln \left( 1-z \,{{\hat{f}}}_s(q)\right) , \end{aligned}$$

(209)

where we used the expression of ${\tilde{E}}_s(z,0|0)$ in (204). Upon replacing z by $u=s(z-1)/(1-s)$ and using the same trick as explained before in (194) and using the expansion (196), one finds in the limit $s\rightarrow 1$ that

$$\begin{aligned} {\tilde{E}}_s\left( u,0|0\right)&\sim -\frac{1-z}{1-s} [f_s(y-x)-f_s(0)]\nonumber \\&\quad -(1-z)(1-s)^{\frac{2}{\mu }-1}(x+y)\int _0^\infty dx'dy' \mathcal {F}'_\mu \left( x'\right) \mathcal {F}_\mu \left( y'\right) \nonumber \\&\quad \times \int _{\gamma _B} d\lambda d\lambda ' \frac{e^{\lambda x'+\lambda ' y'}}{(2\pi i)^2 (\lambda +\lambda ')}\exp \left( \lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln ( 1+q^\mu )}{\lambda ^2+q^2}\right) \exp \left( \lambda ' \int _0^\infty \frac{dq}{\pi }\frac{\ln ( 1+q^\mu )}{\lambda '^2+q^2}\right) \nonumber \\&\quad + (1-z)^{3/2}(1-s)^{\frac{1}{\mu }-\frac{1}{2}} \int _0^\infty dx' dy' \left[ {{\mathcal {F}}}_\mu (x') g(y-y') - {{\mathcal {F}}}_{\mu }(x') g(-y') \right] \nonumber \\&\quad \times \int _{\gamma _B} d\lambda d\lambda ' \frac{e^{\lambda x'+\lambda ' y'}}{(2\pi i)^2 \lambda '}\exp \left( \lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln ( 1+q^\mu )}{\lambda ^2+q^2}\right) \phi (\lambda ',z)\nonumber \\&\quad + (1-z)^{3/2}(1-s)^{\frac{1}{\mu }-\frac{1}{2}} \int _0^\infty dx' dy' \left[ g(x'-x){{\mathcal {F}}}_{\mu }(-y') - g(x'){{\mathcal {F}}}_{\mu }(-y') \right] \nonumber \\&\quad \times \int _{\gamma _B} d\lambda d\lambda ' \frac{e^{\lambda x'+\lambda ' y'}}{(2\pi i)^2 \lambda }\phi (\lambda ,z)\exp \left( \lambda ' \int _0^\infty \frac{dq}{\pi }\frac{\ln ( 1+q^\mu )}{\lambda '^2+q^2}\right) \nonumber \\&\quad + (1-z)^2 \int _0^\infty dx' dy' \left[ g(x'-x) g(y-y') - g(x') g(-y') \right] \nonumber \\&\quad \times \int _{\gamma _B} d\lambda d\lambda ' \frac{e^{\lambda x'+\lambda ' y'}}{(2\pi i)^2 (\lambda +\lambda ')}\phi (\lambda ,z)\phi (\lambda ',z) \nonumber \\&\quad -\int _{0}^\infty \frac{dq}{\pi }\ln \left( 1+ \frac{(1-z) {\hat{f}}(q)}{1-{\hat{f}}(q)}\right) . \end{aligned}$$

(210)

Next, we note that in the limit $s\rightarrow 1$ with $x=O(1)$ and $y=O(1)$, the first line becomes

$$\begin{aligned} -\frac{1-z}{1-s} [f_s(y-x)-f_s(0)] \sim -(1-z)\int _0^\infty \frac{dq}{\pi }\frac{{\hat{f}}(q)}{1-{\hat{f}}(q)}[\cos (q(y-x))-1]\,,\quad s\rightarrow 1. \end{aligned}$$

(211)

Finally, keeping only the highest order terms in the limit $s\rightarrow 1$, which corresponds to all the lines in (210) for $\mu =2$ and all the lines except the terms of order $O[(1-s)^{\frac{1}{\mu }-\frac{1}{2}}]$, we recover the expression (91) given in the main text. In doing so, we used that, for $\mu =2$, we have

$$\begin{aligned}&\int _0^\infty dx'dy' \left[ \mathcal {F'}_2\left( x'\right) \mathcal {F}_2\left( y'\right) \right] \int _{\gamma _B} d\lambda d\lambda ' \frac{e^{\lambda x'+\lambda ' y'}}{(2\pi i)^2 (\lambda +\lambda ')}\exp \left( \lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln ( 1+q^2)}{\lambda ^2+q^2}\right) \nonumber \\&\quad \times \exp \left( \lambda ' \int _0^\infty \frac{dq}{\pi }\frac{\ln ( 1+q^2)}{\lambda '^2+q^2}\right) =-\frac{1}{2}, \end{aligned}$$

(212)

$$\begin{aligned}&\int _0^\infty dx' \mathcal {F}_2\left( x'\right) \int _{\gamma _B} d\lambda \frac{e^{\lambda x'}}{(2\pi i)^2}\nonumber \\&\quad \times \exp \left( \lambda \int _0^\infty \frac{dq}{\pi }\frac{\ln ( 1+q^2)}{\lambda ^2+q^2}\right) =1, \end{aligned}$$

(213)

which can be easily shown by noting that $\mathcal {F}_2(x) = e^{-|x|}/2$ and using the fact that the Laplace transform of the inverse Laplace transform of a function is the function itself, i.e. that $\int _0^\infty dx e^{-x} \int _{\gamma _B} d\lambda \frac{e^{\lambda x}}{2\pi i} h(\lambda )=h(1)$ for a function h.

1.3 ${\tilde{E}}_s(z,0|0)$

We take the limit $\lambda \rightarrow \infty$ and $\lambda '\rightarrow \infty$ in (202), which gives

$$\begin{aligned} {\bar{E}}_s(z,0|0) = \int _{\gamma _B} d\lambda d\lambda ' \frac{e^{\lambda x}}{2\pi i} \frac{e^{\lambda ' y}}{2\pi i }\frac{1}{\lambda +\lambda '}\left( 1 -\left( \frac{1}{\lambda }+\frac{1}{\lambda '}\right) \int _0^\infty \frac{dq}{\pi } \ln (1-z\,{{\hat{f}}}_s(q))\right) . \end{aligned}$$

(214)

Inverting the Laplace transforms with respect to $\lambda$ and $\lambda '$ and inserting it in (203), we recover (204).

The Special Case of the Distribution $f(\eta ) = \frac{3}{2} |\eta |\,e^{-\sqrt{3}|\eta |}$

In this Appendix, we focus on the special case of

$$\begin{aligned} f(\eta ) = \frac{3}{2} |\eta |\,e^{-\sqrt{3}|\eta |}, \end{aligned}$$

(215)

for which the Fourier transform is given by

$$\begin{aligned} {\hat{f}}(q)=\frac{3 \left( 3-q^2\right) }{\left( q^2+3\right) ^2}. \end{aligned}$$

(216)

In this case, ${\hat{g}}(q)$, which is defined in (198), reads

$$\begin{aligned} {\hat{g}}(q)=-\frac{4}{q^2+9}, \end{aligned}$$

(217)

so that

$$\begin{aligned} g(\eta ) = -\frac{2}{3} e^{-3|\eta |}. \end{aligned}$$

(218)

Inserting this expression (218) in (92) and (93), we find

$$\begin{aligned} S_*(z|x)&= 1 - x \sqrt{1-z} + \frac{2}{3}(z-1) (1-e^{3x}) \frac{\phi (\lambda =3,z)}{3} \quad \;, \end{aligned}$$

(219)

$$\begin{aligned} E_*(z,y|x)&= -\int _{0}^\infty \frac{dq}{\pi }\ln \left( 1+(1-z)\left( \frac{1}{q^2}-\frac{4}{q^2+9}\right) \right) \nonumber \\&\quad - \frac{(1-z)}{6} \left( 4 - 4e^{-3|x-y|} - 3 |x-y| \right) + \frac{(1-z)}{2}(x+y) \end{aligned}$$

(220)

$$\begin{aligned}&\quad + \frac{2(1-z)^2}{27} \phi ^2(3,z) (e^{3x+3y}-1) + \frac{2}{9}(1-z)^{3/2} \phi (3,z) \left( 2- e^{3x}- e^{3y}\right) , \end{aligned}$$

(221)

where the function $\phi (\lambda , z)$ is given in (95) with ${\hat{f}}(k) = 3(3-k^2)/(k^2+3)^2$ and where we used the fact that the Laplace transform of the inverse Laplace transform of a function is the function itself, i.e. that $\int _0^\infty dx e^{-x} \int _{\gamma _B} d\lambda \frac{e^{\lambda x}}{2\pi i} h(\lambda )=h(1)$ for a function h, as well as

$$\begin{aligned} \int _0^\infty \frac{dq}{\pi }\left( \frac{1}{q^2}-\frac{4}{q^2+9}\right) [\cos (q(y-x))-1] = \frac{1}{6} \left( 4 - 4e^{-3|x-y|} - 3 |x-y| \right) . \end{aligned}$$

(222)

In particular, the function $\phi (\lambda ,z)$ admits the small z expansion

$$\begin{aligned} \frac{1}{3}\phi (\lambda =3,z) = (2 - \sqrt{3}) + \frac{z}{4}(2 - \sqrt{3})^2 + O(z^2) \;, \end{aligned}$$

(223)

from which one obtains the small z expansion of $S_*(z|x)$ as

$$\begin{aligned} S_*(z|x)&= \frac{2\sqrt{3}-1}{3} - x + \frac{4-2 \sqrt{3}}{3}\,e^{3x}+ \frac{z}{6} \left( 3x+1-e^{3x} \right) + O(z^2). \end{aligned}$$

(224)

Quite remarkably, one observes from (224), that $S_*(z=0,x)$ coincides exactly with the function $\sqrt{\pi }\, U_0(-x)$ defined in [65]—see (77). Note that this also holds for the double exponential case in (173). Inserting the small z expansion (223) in (221) one can similarly obtain the small z expansion of $E_*(z,y|x)$. These expansions, up to order O(z), can eventually be used in Eq. (97) to obtain the small $\varDelta$ expansion of the distribution of the first gap for this jump distribution $f(\eta ) = \frac{3}{2} |\eta |\,e^{-\sqrt{3}|\eta |}$. We have carefully checked that this small $\varDelta$ expansion exactly coincides (up to order $O(\varDelta ^2)$) with the small $\varDelta$ expansion of the exact result for the distribution of the first gap, obtained from a completely different method in Refs. [34, 35].

Fig. 9

In the limit of $n\rightarrow \infty$ with k fixed, the trajectories that contribute to the gap probability $\int _{-\infty }^\infty da\,p_{k,n}(a,\varDelta )$ are the trajectories that jump an even number of times 2j over the gap. We parametrise these trajectories by letting the random walk start from a distance $x_0$ from the gap, then propagates to a point located a distance $x_1$ during $n_1$ steps while remaining below the gap, after which it jumps over the gap to a point located a distance $x_2$ from the gap, then it propagates to $x_3$ during $k_1$ steps while remaining above the gap, and so on, so that $k=k_1+\cdots +k_j+j$ and that $n-k=n_1+\cdots +n_j+j$ (where the last term j accounts for the first jump in each sequences). We then sum over all possible values of these variables, which corresponds to summing over all possible locations and configurations of the gap

Scaling Function of the Condensed Phase

In this Appendix, we are interested in the stationary probability distribution of the $k\text {th}$ gap $P_k(\varDelta )$ in the limit of large gaps $\varDelta$. To study this limit, we have found convenient to leave aside “Spitzer’s idea” for a moment, and to write out explicitly the random walk path contributions. We argue that all the contributions can be written as (see Fig. 9):

$$\begin{aligned}&\int _{-\infty }^\infty da p_{k,n}(a,\varDelta ) = \sum _{j=1}^\infty \sum _{k_1,\ldots ,k_j=0}^\infty \sum _{n_1,\ldots ,n_j=0}^\infty \nonumber \\&\delta (k_1+\cdots +k_{j}+j-k)\delta (n_1+\ldots +n_{j}+j-[n-k])\nonumber \\&\int _0^\infty dx_0 \cdots dx_{4j+1} E_0(n_1,x_1|x_0)f(x_2+x_1+\varDelta )E_0(k_1,x_3|x_2)f(x_3+x_4+\varDelta )\nonumber \\&\quad E_0(n_2,x_5|x_4)f(x_5+x_6+\varDelta )E_0(k_2,x_7|x_6)f(x_7+x_8+\varDelta ) \nonumber \\&\quad \ldots \nonumber \\&\quad E_0(n_{j-1},x_{4j-3}|x_{4j-4})f(x_{4j-2}+x_{4j-3}+\varDelta )\nonumber \\&\quad E_0(x_{4j-1}|x_{4j-2},k_j)f(x_{4j-1}+x_{4j}+\varDelta )\nonumber \\&\quad E_0(n_j,x_{4j+1}|x_{4j}), \end{aligned}$$

(225)

where $E_0(n,y|x)$ is the propagator of the random walk starting from $x>0$ and reaching $y>0$ after n steps while remaining above the origin. It is given by the Pollaczek–Spitzer formula in (202) with $s=0$. Using the following expansions

$$\begin{aligned}&f(\eta ) \sim c_\mu \,\eta ^{-1-\mu },\quad \eta \rightarrow \infty , \end{aligned}$$

(226)

$$\begin{aligned}&\int _0^\infty dy E_0\left( n,y|x\right) \sim \frac{1}{\sqrt{n}} U_\mu (x),\quad x=O(1), \end{aligned}$$

(227)

$$\begin{aligned}&\sum _{n=0}^\infty E_0\left( n,y|x\right) \sim V_\mu (x,y),\quad x=O(1), y=O(1), \end{aligned}$$

(228)

$$\begin{aligned}&E_0\left( k,y|x\right) \sim \frac{1}{k^\frac{1}{\mu }} \mathcal {J}_\mu \left( u=\frac{y}{k^\frac{1}{\mu }},v=\frac{x}{k^\frac{1}{\mu }}\right) ,\quad x=O(k^\frac{1}{\mu }), y=O(k^\frac{1}{\mu }), \end{aligned}$$

(229)

where

$$\begin{aligned} U_\mu (x)&= \int _{\gamma _B} \frac{d\lambda e^{\lambda x}}{2\pi i} \frac{1}{\lambda \sqrt{\pi }}\exp \left[ -\frac{\lambda }{\pi }\int _0^\infty \frac{dq}{\lambda ^2+q^2}\ln \left( 1-{\hat{f}}(q)\right) \right] , \end{aligned}$$

(230)

$$\begin{aligned} V_\mu (x,y)&= \int _{\gamma _B} \frac{d\lambda e^{\lambda x}}{2\pi i}\int _{\gamma _B} \frac{d\lambda ' e^{\lambda ' y}}{2\pi i} \frac{1}{\lambda +\lambda '}\nonumber \\&\quad \times \exp \left[ -\frac{\lambda }{\pi }\int _0^\infty \frac{dq}{\lambda ^2+q^2}\ln \left( 1-{\hat{f}}(q)\right) -\frac{\lambda '}{\pi }\int _0^\infty \frac{dq}{\lambda '^2+q^2}\ln \left( 1-{\hat{f}}(q)\right) \right] \,, \end{aligned}$$

(231)

$$\begin{aligned} \mathcal {J}_\mu (u,v)&= \int _{\gamma _B} \frac{ds e^s}{2\pi i}\int _{\gamma _B} \frac{d\lambda e^{\lambda u}}{2\pi i} \int _{\gamma _B} \frac{d\lambda 'e^{\lambda ' v}}{2\pi i} \frac{1}{s(\lambda +\lambda ')} \nonumber \\&\quad \exp \left[ -\frac{1}{\pi }\int _0^\infty dq\left( \frac{\ln \left( 1+\frac{\lambda ^\mu q^\mu }{s}\right) }{1+q^2}+\frac{\ln \left( 1+\frac{\lambda '^\mu q^\mu }{s}\right) }{1+q^2}\right) \right] \,, \end{aligned}$$

(232)

and $c_\mu =\varGamma (1+\mu )\sin (\pi \mu /2)/\pi$. we find that (225) becomes $\lim _{n\rightarrow \infty }\int _{-\infty }^\infty da p_{k,n}(a,\varDelta )= \int _{-\infty }^\infty da p_{k}(a,\varDelta )$ where

$$\begin{aligned}&\int _{-\infty }^\infty da p_{k}(a,\varDelta ) \sim \pi \sum _{j=1}^\infty \sum _{k_1,\ldots ,k_j=0}^\infty \delta (k_1+\cdots +k_{j}+j-k)\nonumber \\&\quad \int _0^\infty dx_1 \cdots dx_{4j} U_\mu (x_1)\frac{c_\mu }{(x_1+x_2+\varDelta )^{1+\mu }}\frac{1}{k_1^\frac{1}{\mu }}\mathcal {J}_\mu \left( \frac{x_2}{k_1^\frac{1}{\mu }},\frac{x_3}{k_1^\frac{1}{\mu }}\right) \frac{c_\mu }{(x_3+x_4+\varDelta )^{1+\mu }}\nonumber \\&\quad V_\mu (x_{5},x_{4})\frac{c_\mu }{(x_{5}+x_{6}+\varDelta )^{1+\mu }}\frac{1}{k_2^\frac{1}{\mu }}\mathcal {J}_\mu \left( \frac{x_{6}}{k_2^\frac{1}{\mu }},\frac{x_{7}}{k_2^\frac{1}{\mu }}\right) \frac{c_\mu }{(x_{7}+x_{8}+\varDelta )^{1+\mu }}\nonumber \\&\quad \ldots \nonumber \\&\quad V_\mu (x_{4j-4},x_{4j-3})\frac{c_\mu }{(x_{4j-2}+x_{4j-3}+\varDelta )^{1+\mu }}\frac{1}{k_j^\frac{1}{\mu }}\mathcal {J}_\mu \left( \frac{x_{4j-1}}{k_j^\frac{1}{\mu }},\frac{x_{4j-2}}{k_j^\frac{1}{\mu }}\right) \nonumber \\&\quad \frac{c_\mu }{(x_{4j-1}+x_{4j}+\varDelta )^{1+\mu }}\nonumber \\&\quad U_\mu (x_{4j})\,,\quad k\rightarrow \infty \,, \end{aligned}$$

(233)

where we used that $\sum _{j=1}^{n-1} j^{-1/2}(n-j)^{-1/2}\sim \pi$ for $n\rightarrow \infty$. Rescaling all the integration variables by $\varDelta$ and using the following expansions

$$\begin{aligned}&U_\mu (x) \sim A_\mu x^{\frac{\mu }{2}}\,,\quad x\rightarrow \infty \,, \end{aligned}$$

(234)

$$\begin{aligned}&V_\mu (\alpha \, x,\alpha \, y) \sim C_\mu \,\alpha ^{\mu -1} v_\mu (x,y)\,,\quad \alpha \rightarrow \infty \,, \end{aligned}$$

(235)

where $A_\mu =1/[\sqrt{\pi }\,\varGamma (1+\mu /2)]$, $C_\mu =1/\varGamma (\mu /2)^2$ and $v_\mu (x,y)= \int _0^{\min (x,y)}dz (y-z)^{\frac{\mu }{2}-1}(x-z)^{\frac{\mu }{2}-1}$, gives

$$\begin{aligned}&\int _{-\infty }^\infty da p_{k}(a,\varDelta ) \sim \pi \sum _{j=1}^\infty \sum _{k_1,\ldots ,k_j=0}^\infty \delta (k_1+\cdots +k_{j}+j-k)\,\frac{\varDelta ^{j+1-j\mu }}{k_1^{\frac{1}{\mu }}\ldots k_j^{\frac{1}{\mu }}}\,A_\mu ^2 c_\mu ^{2j} C_\mu ^{j-1}\nonumber \\&\quad \int _0^\infty dx_1 \cdots dx_{4j} x_1^\frac{\mu }{2}\frac{1}{(x_1+x_2+1)^{1+\mu }}\mathcal {J}_\mu \left( \frac{\varDelta x_2}{k_1^\frac{1}{\mu }},\frac{\varDelta x_3}{k_1^\frac{1}{\mu }}\right) \frac{1}{(x_3+x_4+1)^{1+\mu }}\nonumber \\&\quad v_\mu (x_{5},x_{4})\frac{1}{(x_{5}+x_{6}+1)^{1+\mu }}\mathcal {J}_\mu \left( \frac{ \varDelta x_{6}}{k_2^\frac{1}{\mu }},\frac{\varDelta x_{7}}{k_2^\frac{1}{\mu }}\right) \frac{1}{(x_{7}+x_{8}+1)^{1+\mu }}\nonumber \\&\quad \ldots \nonumber \\&\quad v_\mu (x_{4j-4},x_{4j-3})\frac{1}{(x_{4j-2}+x_{4j-3}+1)^{1+\mu }}\mathcal {J}_\mu \left( \frac{\varDelta x_{4j-1}}{k_j^\frac{1}{\mu }},\frac{\varDelta x_{4j-2}}{k_j^\frac{1}{\mu }}\right) \nonumber \\&\quad \frac{1}{(x_{4j-1}+x_{4j}+1)^{1+\mu }}\nonumber \\&\quad x_{4j}^\frac{\mu }{2},\quad k\rightarrow \infty . \end{aligned}$$

(236)

Integrating over $x_1$ and $x_{4j}$ gives

$$\begin{aligned}&\int _{-\infty }^\infty da p_{k}(a,\varDelta ) \sim \pi \sum _{j=1}^\infty \sum _{k_1,\ldots ,k_j=0}^\infty \delta (k_1+\cdots +k_{j}+j-k)\,\frac{\varDelta ^{j+1-j\mu }}{k_1^{\frac{1}{\mu }}\ldots k_j^{\frac{1}{\mu }}}\,A_\mu ^2 c_\mu ^{2j} C_\mu ^{j-1}D_\mu ^2\nonumber \\&\int _0^\infty dx_2 \cdots dx_{4j-1} \frac{1}{(x_2+1)^{\frac{\mu }{2}}}\mathcal {J}_\mu \left( \frac{\varDelta x_2}{k_1^\frac{1}{\mu }},\frac{\varDelta x_3}{k_1^\frac{1}{\mu }}\right) \frac{1}{(x_3+x_4+1)^{1+\mu }}\nonumber \\&\quad v_\mu (x_{5},x_{4})\frac{1}{(x_{5}+x_{6}+1)^{1+\mu }}\mathcal {J}_\mu \left( \frac{\varDelta x_{6}}{k_2^\frac{1}{\mu }},\frac{\varDelta x_{7}}{k_2^\frac{1}{\mu }}\right) \frac{1}{(x_{7}+x_{8}+1)^{1+\mu }}\nonumber \\&\quad \ldots \nonumber \\&\quad v_\mu (x_{4j-4},x_{4j-3})\frac{1}{(x_{4j-2}+x_{4j-3}+1)^{1+\mu }}\mathcal {J}_\mu \left( \frac{\varDelta x_{4j-2}}{k_j^\frac{1}{\mu }},\frac{\varDelta x_{4j-1}}{k_j^\frac{1}{\mu }}\right) \frac{1}{(x_{4j-1}+1)^{\frac{\mu }{2}}}\,,\quad k\rightarrow \infty \,, \end{aligned}$$

(237)

where $D_\mu =\sqrt{\pi } 2^{-\mu } \varGamma \left( \frac{\mu }{2}\right) /\varGamma \left( \frac{\mu +1}{2}\right)$. Formally taking a double derivative with respect to $\varDelta$, we recover the scaling form (22) announced in the introduction. In the limit of large $\varDelta$, we only take the term in $j=1$ in (237) which gives

$$\begin{aligned}&\int _{-\infty }^\infty da p_{k}(a,\varDelta ) \sim \pi \,\frac{\varDelta ^{2-\mu }}{k^{\frac{1}{\mu }}}\,A_\mu ^2 c_\mu ^{2} D_\mu ^2 \int _0^\infty dx_2 dx_{3} \frac{1}{(x_2+1)^{\frac{\mu }{2}}}\mathcal {J}_\mu \left( \frac{\varDelta x_2}{k^\frac{1}{\mu }},\frac{\varDelta x_3}{k^\frac{1}{\mu }}\right) \frac{1}{(x_3+1)^{\frac{\mu }{2}}}\,,\nonumber \\&\quad \varDelta \rightarrow \infty . \end{aligned}$$

(238)

Using that $\mathcal {J}_\mu (x_1,x_2)\rightarrow \mathcal {L}_\mu (x_2-x_1)$ for large $x_1$ and $x_2$ gives

$$\begin{aligned}&\int _{-\infty }^\infty da p_{k}(a,\varDelta )&\sim \pi \,\frac{\varDelta ^{2-\mu }}{k^{\frac{1}{\mu }}}\,A_\mu ^2 c_\mu ^{2} D_\mu ^2 \int _0^\infty dx_2 dx_{3} \frac{1}{(x_2+2)^{\frac{\mu }{2}}}\mathcal {L}_\mu \left( \frac{\varDelta (x_3-x_2)}{k^\frac{1}{\mu }}\right) \frac{1}{(x_3+2)^{\frac{\mu }{2}}},\nonumber \\&\quad \varDelta \rightarrow \infty . \end{aligned}$$

(239)

Changing coordinates $u=(x_3-x_2)/\sqrt{2}$ and $v=(x_3+x_2)/\sqrt{2}$, and rescaling u by $\varDelta k^\frac{1}{\mu }$ it gives

$$\begin{aligned} \int _{-\infty }^\infty da p_{k}(a,\varDelta )&\sim \pi \,\varDelta ^{1-\mu }\,A_\mu ^2 c_\mu ^{2} D_\mu ^2 \int _0^\infty dv\int _{-\infty }^\infty du \frac{1}{(\frac{v}{\sqrt{2}}+1)^{\mu }}\mathcal {L}_\mu \left( u\sqrt{2}\right) \,,\nonumber \\&\sim \pi \,\varDelta ^{1-\mu }\,A_\mu ^2 c_\mu ^{2} D_\mu ^2 E_\mu ,\quad \varDelta \rightarrow \infty , \end{aligned}$$

(240)

where $E_\mu =1/(\mu -1)$. Inserting this result in (29), we find the large argument of $\mathcal {M}_\mu (u)$ in (23).