Short Hyperuniform Random Walks

Abstract

Random walks of two steps, with fixed sums of lengths of $1$, taken into uniformly random directions in d-dimensional Euclidean spaces ($d\ge 2$) are investigated to construct continuous step-length distributions which make them hyperuniform. The endpoint positions of hyperuniform walks are spread out in the unit ball as the projections in the walk space of points uniformly distributed on the surface of the unit hypersphere of some k-dimensional Euclidean space ($k>d$). Unique symmetric continuous step-length distributions exist for given d and k, provided that $d<k<2d$. The walk becomes uniform on the unit ball when $k=d+2$. The symmetric densities reduce to simple polynomials for uniform random walks and are mixtures of two pairs of asymmetric beta distributions.

Appendices

Appendix 1: Multivariate Dirichlet Distribution from Gamma Distributions

The probability density function $p_X(x)$ of a gamma random variable $X$ , is [11]

$$\begin{aligned} p_X(x)=\frac{x^{\alpha -1}\exp \left( -x/\theta \right) }{\theta ^\alpha \Gamma (\alpha )} (x>0) \end{aligned}$$

(6.1)

We denote it here as $X\sim \gamma (\alpha , \theta )$ where $\alpha >0$ is the shape parameter and $\theta > 0$ the scale parameter while $\Gamma (\alpha )$ is the Euler gamma function. The characteristic function of $X$ is $\phi _X(t)=E(e^{itX})=\frac{1}{(1-i\theta t)^\alpha }$ [11].

A sum $G=\sum _{i=1}^n G_i$ of $n$ independent gamma random variables, $G_i \sim \gamma (q_i ,\theta )$ $(i=1, \ldots , n)$, with identical scale parameter $\theta$ and a priori different shape parameters $q_i \, (i=1 , \ldots , n)$, is a gamma random variable $G \sim \gamma (n\bar{q} , \theta )$, where $n\bar{q}$ is the sum $\sum _{i=1}^n q_i$. This is for instance deduced from the characteristic function of $G$,

$$\begin{aligned}\phi _G(t)=E(e^{itG})=\frac{1}{\prod _{k=1}^n\left( 1-it\theta \right) ^{q_k}}=\frac{1}{\left( 1-it\theta \right) ^{n\bar{q}}}\end{aligned}$$

From the previous set of $n=m+1$ independent gamma random variables, we define a random vector $\varvec{L}_{(n)}=(L_1,L_2,\ldots ,L_n)$ whose components are $L_i=G_i/G$. The distribution of $\varvec{L}_{(n)}$ is then called a Dirichlet distribution with parameters $\varvec{q_{(n)}}=(q_1, \ldots ,q_n)$, $\varvec{L_{(n)}} \sim D(\varvec{q_{(n)}})$. Its pdf is shown to be ([11] p. 17)

$$\begin{aligned} \left\{ \begin{array}{ll} f(l_1,\ldots ,l_m)=\left( \frac{\Gamma (n\bar{q})}{\prod _{i=1}^n\Gamma (q_i)}\right) \prod _{i=1}^nl_i^{q_i-1} \\ l_n=1-\sum _{i=1}^ml_i, l_i > 0, i=1,\ldots , n \end{array} \right. \end{aligned}$$

(6.2)

When the shape parameters are all equal to $q$, i. e. when $G_i \sim \gamma (q,\theta ) \; (i=1, \ldots ,n)$, the pdf (6.2) becomes invariant under permutation, $f(l_1,\ldots ,l_m)=\left( \frac{\Gamma (nq)}{\Gamma (q)^n}\right) \left\{ \prod _{i=1}^nl_i\right\} ^{q-1}$. Finally, when the $G_i$’s are exponentially distributed, $q=1$, the random vector $\varvec{L}_{(n)}$ is uniformly distributed over the unit $(n-1)$ simplex as $f(l_1,\ldots ,l_m)=m!$.

Appendix 2: Hypergeometrical Differential Equation Characterization

We recall that $F(l) := \ _2F_1(a,b;c;l)$ is the solution of the following differential equation:

$$\begin{aligned} l(1-l)F''(l)+[c-(a+b+1)l]F'(l)-abF(l)=0 \end{aligned}$$

so, letting $_2F_1\left( 2d-1-k,d;\frac{3d-k}{2};l\right) =F(l)$, we have

$$\begin{aligned} l(1-l)F''(l)+\left[ \frac{3d-k}{2}-(3d-k)l\right] F'(l)-d(2d-1-k)F(l)=0 \end{aligned}$$

and similarly

$$\begin{aligned} l(1-l)F''(1-l)+\left[ \frac{3d-k}{2}-(3d-k)(1-l)\right] F'(1-l)-d(2d-1-k)F(1-l)=0 \end{aligned}$$

that is

$$\begin{aligned} l(1-l)F''(1-l)-\left[ \frac{3d-k}{2}-(3d-k)l\right] F'(1-l)-d(2d-1-k)F(1-l)=0 \end{aligned}$$

Let $G(l)=\frac{f_{d,k}(l)}{\left[ l(1-l)\right] ^{d-2}}$, by (22), we get

$$\begin{aligned} G(l)=C\left\{ F(l)+F(1-l)\right\} . \end{aligned}$$

Taking the derivative two times, we obtain

$$\begin{aligned} G'(l)=C\left\{ F'(l)-F'(1-l)\right\} \; \text{ and } \; G''(l)=C\left\{ F''(l)+F''(1-l)\right\} \end{aligned}$$

and substituting

$$\begin{aligned} l(1-l)G''(l)+\left[ \frac{3d-k}{2}-(3d-k)l\right] G'(l)-d(2d-1-k)G(l)=0 \end{aligned}$$

(7.1)

It follows that $G$ is also a solution of a hypergeometrical differential equation. For basic results on these differential equations see [1], Table 15.5.

For $-1< r < 1$ the change of variable $l=\frac{1+r}{2}$ in (7.1) gives

$$\begin{aligned} (1-r^2)F''(r)-(3d-k)rF'(r)-d(2d-1-k)F(r)=0 \end{aligned}$$

where $F(r)=G(\frac{1+r}{2})$. This is the the Gegenbauer equation which has a unique symmetric solution such that: $F(0)=C$ and $F'(0)=0$. It is $F(r)=C\times \ _2F_1(\frac{2d-k-1}{2},\frac{d}{2},\frac{1}{2},r^2)$, see [6], p. 65, the density (15) is easily retrieved.

In the following proposition we prove that the only hyperuniform random walks $HU_d(k)$ which are at the same time symmetric Dirichlet random walks are obtained for $k=2d-1$ and $k=2d-2$ as shown in Eqs. (20) and (21).

Proposition 3

Suppose that $f_L(l)=C\left[ l(1-l)\right] ^{\alpha -1}$, then $\varvec{S}_d(L)$ is $HU_d(k)$ if and only if $k=2d-1$ and $\alpha =d-1$ or $k=2d-2$ and $\alpha =\frac{d}{2}-1$.

Proof

Suppose that $f_{d,k}(l)=C\left[ l(1-l)\right] ^{\alpha -1}$. Then $G(l)=C\left[ l(1-l)\right] ^{\alpha -d+1}$ and we have

$$\begin{aligned} G'(l)=(\alpha -d+1)\left[ l(1-l)\right] ^{\alpha -d}(1-2l) \end{aligned}$$

$$\begin{aligned} G''(l)=(\alpha -d+1)\left[ l(1-l)\right] ^{\alpha -d-1}\left\{ (\alpha -d)(1-2l)^2-2l+2l^2\right\} \end{aligned}$$

So from (7.1) we obtain:

$$\begin{aligned}&(\alpha -d+1)\left\{ (\alpha -d)(1-2l)^2-2l+2l^2\right\} +\frac{3d-k}{2}(\alpha -d+1)(1-2l)^2\\&\quad -d(2d-1-k)(l-l^2)=0 \end{aligned}$$

which is of the form $Bl^2-Bl+A=0$ with

$$\begin{aligned} A=(\alpha -d+1)\left\{ (\alpha -d)+\frac{3d-k}{2}\right\} \end{aligned}$$

and

$$\begin{aligned} B=4(\alpha -d+1)(\alpha -d)+2(\alpha -d+1)+2(3d-k)(\alpha -d+1)+d(2d-1-k) \end{aligned}$$

From the condition $A=B=0$, we get the only solutions

$$\begin{aligned} k=2d-1 ,\;\alpha =d-1 \text { and } k=2d-2 , \;\; \alpha =\frac{d}{2}-1. \end{aligned}$$

$\square$

These solutions were obtained by a different method in [21].