Record Statistics of Integrated Random Walks and the Random Acceleration Process

Abstract

We address the theory of records for integrated random walks with finite variance. The long-time continuum limit of these walks is a non-Markov process known as the random acceleration process or the integral of Brownian motion. In this limit, the renewal structure of the record process is the cornerstone for the analysis of its statistics. We thus obtain the analytical expressions of several characteristics of the process, notably the distribution of the total duration of record runs (sequences of consecutive records), which is the continuum analogue of the number of records of the integrated random walks. This result is universal, i.e., independent of the details of the parent distribution of the step lengths.

Appendices

Appendix

A A Word on Notations

Asymptotic equivalence

The symbol $\approx$ stands for asymptotic equivalence; the symbol $\sim$ is weaker and means ‘of the order of’.

Probability densities, Laplace transforms, limiting distributions

The probability density function of the continuous random variable X is denoted by $f_X(x)$, with

$$\begin{aligned} f_X(x)=\frac{\mathrm{d}}{\mathrm{d}x}\mathbb {P}(X<x). \end{aligned}$$

In the course of this work, we encounter several positive time-dependent continuous random variables, denoted generically by $Y_{t}$. The probability density function of such a random variable is denoted by $f_{Y_t}(t,y)$ where time t appears as a parameter. The Laplace transform with respect to y of this density is

$$\begin{aligned} \hat{f}_{Y_t}(t,u)=\mathrel {\mathop {{\mathcal {L}}}\limits _{y}^{}}f_{Y}(t,y) =\left\langle {\mathrm{e}}^{-uY_{t}}\right\rangle =\int _{0}^{\infty }\mathrm{d}y\,{\mathrm{e}}^{-uy}\,f_{Y_t}(t,y), \end{aligned}$$

and its double Laplace transform with respect to t and y is denoted by

$$\begin{aligned} \hat{f}_{Y_t}(s,u)=\mathrel {\mathop {{\mathcal {L}}}\limits _{t,y}^{}}f_{Y_t}(t,y)=\mathrel {\mathop {{\mathcal {L}}}\limits _{t}^{}}\left\langle {\mathrm{e}}^{-uY_{t}}\right\rangle =\int _{0}^{\infty }\mathrm{d}t\,{\mathrm{e}}^{-st}\int _{0}^{\infty }\mathrm{d}y\,{\mathrm{e}}^{-uy}\,f_{Y_t}(t,y). \end{aligned}$$

(A.1)

Assume that $Y_{t}$ scales asymptotically as t. As $t\rightarrow \infty$ the density $f_{t^{-1}Y_t}(t,x=y/t)$ of the rescaled variable $Y_{t}/t$ converges to a limit, denoted by

$$\begin{aligned} f_{X}(x)=\lim _{t\rightarrow \infty }f_{t^{-1}Y_t}(t,x=y/t). \end{aligned}$$

(A.2)

B Inversion of the Scaling form of a Double Laplace Transform

For completeness, we reproduce hereafter Appendix B of [21].

Consider the probability density function $f_{Y_t}(t,y)$ of the positive random variable $Y_{t}$, and assume that its double Laplace transform (A.1) with respect to t and y has the scaling behaviour

$$\begin{aligned} \hat{f}_{Y_t}(s,u)=\frac{1}{s}\, g\!\left( \frac{u}{s}\right) \end{aligned}$$

(B.1)

in the regime $s,u\rightarrow 0$, with u/s arbitrary. Then the following properties hold.

• (i) When $t\rightarrow \infty$ the random variable $Y_{t}/t$ possesses a limiting distribution given by (A.2).

• (ii) The scaling function g is related to $f_{X}$ by

  $$\begin{aligned} g(\xi )=\left\langle \frac{1}{1+\xi X}\right\rangle =\int _{0}^{\infty }\mathrm{d}x\,\frac{f_{X}(x)}{1+\xi x}. \end{aligned}$$

  (B.2)

• (iii) This can be inverted as

  $$\begin{aligned} f_{X}(x)=-\frac{1}{\pi x}\lim _{\epsilon \rightarrow 0}\mathrm{Im}\; g\left( -\frac{1}{x+{\mathrm{i}}\epsilon }\right) . \end{aligned}$$

  (B.3)

• (iv) Finally the moments of X can be obtained, when they exist, by expanding $g(\xi )$ as a Taylor series, since (B.2) implies that

  $$\begin{aligned} g(\xi )=\sum _{k\ge 0}(-\xi )^{k}\left\langle X^{k}\right\rangle . \end{aligned}$$

  (B.4)

These properties can be easily understood as follows.

• (i) First, a direct consequence of the scaling form (B.1) is that $Y_{t}$ scales as t, as can be seen by Taylor expanding the right side of this equation, which generates the moments of $Y_{t}$ in the Laplace space conjugate to t. Therefore (A.2) holds.

• (ii) Then, (B.2) is a simple consequence of (A.2), since

  $$\begin{aligned} {\hat{f}}_{Y_t}(s,u)= \int _0^\infty \mathrm{d}t\,\mathrm{e}^{-st}\langle \mathrm{e}^{-uY_t}\rangle =\int _0^\infty \mathrm{d}t\,\mathrm{e}^{-st}\langle \mathrm{e}^{-u tX}\rangle =\left\langle \frac{1}{s+uX}\right\rangle . \end{aligned}$$

• (iii) Now,

  $$\begin{aligned} f_{X}(x)=\left\langle \delta \left( X-x\right) \right\rangle =-\frac{1}{\pi }\lim _{\epsilon \rightarrow 0}\mathrm{Im}\;\left\langle \frac{1}{x+{\mathrm{i}}\epsilon -X}\right\rangle . \end{aligned}$$

The right side can be rewritten using (B.2), yielding (B.3).

C Some Detailed Derivations

This appendix is devoted to the detailed derivations of a few results used in the body of the paper.

1.1 C.1 Derivation of the Algebraic Expression (3.7) of the Function $h(\xi )$

The function $h(\xi )$ is defined by the integral expression (3.6), where the distribution $f_Z(z)$ is given by (3.3). This reads

$$\begin{aligned} h(\xi )=\frac{12\varGamma (3/4)^2}{\pi ^{3/2}}\, \int _0^\infty \mathrm{d}z\,\frac{z^{1/4}(1+\xi z)^{1/4}}{(1+4z)^{7/4}}. \end{aligned}$$

Setting $z=u/(4(1-u))$ and $\xi =4(1-\zeta )$, we obtain

$$\begin{aligned} h= & {} \frac{3\varGamma (3/4)^2}{\sqrt{2}\,\pi ^{3/2}}\, \int _0^1\mathrm{d}u\,u^{1/4}(1-u)^{-3/4}(1-\zeta u)^{1/4} \nonumber \\= & {} \frac{3}{\sqrt{2}}\,F\!\left( -\frac{1}{4},\frac{5}{4};\frac{3}{2};\zeta \right) . \end{aligned}$$

(C.1)

The hypergeometric function boils down to something more elementary. More precisely, we are facing the first of the 15 entries of the so-called Schwarz Table of all cases where the hypergeometric series reduces to an algebraic function (see, e.g., [16, Vol. I, Sect. 2.7.2]).

This reduction can be shown by elementary means as follows. Starting from the hypergeometric differential equation obeyed by (C.1), i.e.,

$$\begin{aligned} \zeta (1-\zeta )\frac{\mathrm{d}^2h}{\mathrm{d}\zeta ^2} +\left( \frac{3}{2}-2\zeta \right) \frac{\mathrm{d}h}{\mathrm{d}\zeta } +\frac{5}{16}h=0, \end{aligned}$$

and setting $\xi =4\cos ^2\alpha$, i.e., $\zeta =\sin ^2\alpha$, with $0\le \alpha \le \pi /2$ for definiteness, we obtain

$$\begin{aligned} \frac{\mathrm{d}^2h}{\mathrm{d}\alpha ^2} +2\cot \alpha \,\frac{\mathrm{d}h}{\mathrm{d}\alpha } +\frac{5}{16}\,h=0. \end{aligned}$$

Setting $h=v/(\sin \alpha )$, we obtain the simple differential equation

$$\begin{aligned} \frac{\mathrm{d}^2v}{\mathrm{d}\alpha ^2}+\frac{9}{4}\,v=0, \end{aligned}$$

whose solutions are $\sin (3\alpha /2)$ and $\cos (3\alpha /2)$. The regularity of h and its value $h=1$ for $\xi =0$, i.e., $\alpha =\pi /2$, yield

$$\begin{aligned} h=\sqrt{2}\,\frac{\sin (3\alpha /2)}{\sin \alpha }. \end{aligned}$$

Some trigonometric identities finally yield

$$\begin{aligned} h=\frac{1+2\cos \alpha }{\sqrt{1+\cos \alpha }} =\frac{1+\sqrt{\xi }}{\sqrt{1+{\scriptstyle {\scriptstyle 1\over \scriptstyle 2}}\sqrt{\xi }}}. \end{aligned}$$

(C.2)

It can be shown by eliminating radicals that $h(\xi )$ is an algebraic function of degree four, obeying the biquadratic equation

$$\begin{aligned} (\xi -4)h^4+8h^2-4(\xi -1)^2=0. \end{aligned}$$

(C.3)

1.2 C.2 Derivation of the Recursion (5.12) for the Coefficients $a_n$

The gist of 1 and 1 resides in the fact that algebraic functions obey linear differential equations with polynomial coefficients. As a consequence, the coefficients of their power-series expansions obey linear recursions. These properties were known to Abel as early as 1827 (see [4] for an account of historical and algorithmic aspects). In modern times they are only seldom mentioned or used. The present case provides an example of a situation where they are useful.

The function $g_R(\xi )$ obeys the fourth-order algebraic equation (see (5.6))

$$\begin{aligned} P(\xi ,g_R)=4(\xi +1)^3g_R^4-16(\xi +1)^2g_R^3+16(\xi +1)g_R^2-(\xi -2)^2=0. \end{aligned}$$

(C.4)

The linear differential equation obeyed by $g_R(\xi )$ can be derived in three steps.

First, its first derivative reads

$$\begin{aligned} \frac{\mathrm{d}g_R}{\mathrm{d}\xi }= & {} -\frac{\partial P/\partial \xi }{\partial P/\partial g_R} \nonumber \\= & {} -\frac{6(\xi +1)^2g_R^4-16(\xi +1)g_R^3+8g_R^2+2-\xi }{8(\xi +1)g_R((\xi +1)g_R-1)((\xi +1)g_R-2)}. \end{aligned}$$

(C.5)

This expression is a rational function of $g_R$. It can therefore be reduced to the form

$$\begin{aligned} \frac{\mathrm{d}g_R}{\mathrm{d}\xi }=A_3(\xi )g_R^3+A_2(\xi )g_R^2+A_1(\xi )g_R+A_0(\xi ), \end{aligned}$$

(C.6)

where the $A_i$ are rational functions of $\xi$. This can be done by expressing that the difference between (C.5) and (C.6) is a multiple of $P(\xi ,g_R)$. This condition yields coupled linear equations for the $A_i(\xi )$, whose solution yields

$$\begin{aligned} \frac{\mathrm{d}g_R}{\mathrm{d}\xi }=-\frac{N(\xi ,g_R)}{4\xi (\xi +1)(\xi -2)(\xi -3)}, \end{aligned}$$

(C.7)

with

$$\begin{aligned} N(\xi ,g_R) =12(\xi +1)^2g_R^3-36(\xi +1)g_R^2 +(\xi ^3-11\xi ^2+243\xi +12)g_R+3(\xi -2)^2. \end{aligned}$$

Second, higher-order derivatives of the function $g_R(\xi )$ can be readily evaluated by applying iteratively the total derivative operator

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}\xi }=\frac{\partial }{\partial \xi }+\frac{\mathrm{d}g_R}{\mathrm{d}\xi }\frac{\partial }{\partial g_R} \end{aligned}$$

to the expression (C.7). In the present situation, it is sufficient to go up to the second derivative.

Third, eliminating nonlinear terms (those proportional to $g_R^2$ and $g_R^3$) between the resulting expressions of the first and second derivatives, we obtain the desired linear differential equation in the form

$$\begin{aligned} 16(\xi +1)^2(\xi -3)\frac{\mathrm{d}^2g_R}{\mathrm{d}\xi ^2} +16(\xi +1)(3\xi -7)\frac{\mathrm{d}g_R}{\mathrm{d}\xi }+(7\xi -25)g_R+9=0. \end{aligned}$$

(C.8)

Finally, inserting the power-series expansion

$$\begin{aligned} g_R(\xi )=\sum _{n\ge 0}(1-a_n\sqrt{6})(-\xi )^n \end{aligned}$$

(see (5.9), (5.11)) into (C.8), we obtain the following four-term linear recursion for the coefficients $a_n$:

$$\begin{aligned}&(16n^2-9)a_{n-1}+(16n^2+48n+25)a_n -16(n+1)(5n+7)a_{n+1}\nonumber \\&\quad +48(n+1)(n+2)a_{n+2}=0. \end{aligned}$$

(C.9)

1.3 C.3 Derivation of the Recursion (6.11) for the Coefficients $b_n$

The following analysis is in the same vein as the previous section. We start by splitting $g_V(\xi )$ given in (6.8) according to

$$\begin{aligned} g_V(\xi )=\frac{\sqrt{6}}{4}\left( g_U(\xi )-1\right) +g_B(\xi ), \end{aligned}$$

(C.10)

with

$$\begin{aligned} g_B(\xi )=\frac{\sqrt{6}}{4}\frac{1+\sqrt{\xi +1}}{\sqrt{\xi +1+{\scriptstyle {\scriptstyle 1\over \scriptstyle 2}}\sqrt{\xi +1}}}. \end{aligned}$$

It can be shown by eliminating radicals that $g_B$ is an algebraic function of degree four, obeying the biquadratic equation

$$\begin{aligned} 16(\xi +1)(4\xi +3)g_B^4-48(\xi +1)^2g_B^2+9\xi ^2=0. \end{aligned}$$

(C.11)

The linear differential equation obeyed by $g_B(\xi )$ can be derived by means of the three-step procedure presented in 1. We thus obtain

$$\begin{aligned} 16(\xi +1)^2(4\xi +3)\frac{\mathrm{d}^2g_B}{\mathrm{d}\xi ^2}+96(\xi +1)^2\frac{\mathrm{d}g_B}{\mathrm{d}\xi }+5g_B=0. \end{aligned}$$

(C.12)

Inserting the power-series expansion

$$\begin{aligned} g_B(\xi )=\sum _{n\ge 0}b_n(-\xi )^n \end{aligned}$$

(see (6.10), (C.10)) into (C.12), we obtain the following four-term linear recursion for the coefficients $b_n$:

$$\begin{aligned}&32(n-1)(2n-1)b_{n-1}-(176n^2+16n+5)b_n +32(n+1)(5n+3)b_{n+1}\nonumber \\&\quad -48(n+1)(n+2)b_{n+2}=0. \end{aligned}$$

(C.13)

We have $b_0=1$, whereas the $b_n$ enter the expression (6.10) of the moments of V for $n\ge 1$ only.