On Random Flights with Non-uniformly Distributed Directions

Abstract

This paper deals with a new class of random flights in ℝ^d, d≥2, characterized by non-uniform probability distributions on the multidimensional sphere. These random motions differ from similar models appeared in literature where the directions are taken according to the uniform law. The family of angular probability distributions introduced in this paper depends on a parameter ν≥0, which gives the anisotropy of the motion. Furthermore, we assume that the number of changes of direction performed by the random flight is fixed. The time lengths between two consecutive changes of orientation have joint probability distribution given by a Dirichlet density function.

The analysis of the position $\underline{\mathbf{X}}_{d}(t)$, t>0, is not an easy task, because it involves integrals which are not always computable. Therefore, we get the probability distribution of the process $\underline{\mathbf{X}}_{m}^{d}(t)$, t>0, obtained as projection onto the lower space ℝ^m, m<d, of the original random motion in ℝ^d.

In its general framework, the analysis of $\underline{\mathbf{X}}_{d}(t)$, t>0, is very complicated; nevertheless for some values of ν, we provide some explicit results about the process. Indeed, for ν=1 we get the characteristic function of the random flight moving in ℝ^d. By inverting the obtained characteristic function, we derive the analytic form (up to some constants) of the probability distribution of $\underline{\mathbf{X}}_{d}(t)$, t>0.

It is worth to mention that the stochastic processes considered in this paper belong to the class of the non-isotropic random walks, which has several applications in the mechanical statistics.

Appendix

In the random flight problem the integrals involving Bessel functions have a crucial role. For this reason in this section we summarize some important results which are often used in the paper.

In particular we focus our attention on the following integral

$$ \int_0^{2\pi}e^{iz(\alpha\cos\theta+\beta\sin\theta)}\sin^{2\nu}\theta d\theta $$

(4.1)

with α,β∈ℝ and ν≥0. We are able to calculate (4.1) in some particular cases. For example, for ν=0 it is well-known that

$$ \int_0^{2\pi}e^{iz(\alpha\cos\theta+\beta\sin\theta)}d\theta =2\pi J_0\bigl(z\sqrt{\alpha^2+\beta^2}\bigr).$$

(4.2)

Now, we consider (4.1) for ν=n, with n≥1. Following the same approach adopted by De Gregorio [4], we derive an expression of the above quantity given by a linear combination of Bessel functions. The first step consists in expanding the exponential function inside the integral, that is

$$\int_0^{2\pi}e^{iz(\alpha\cos\theta+\beta\sin\theta)}\sin^{2n}\theta d\theta=\sum_{k=0}^{\infty}\frac{(iz)^k}{k!}\sum_{r=0}^k\binom{k}{r}\alpha^{r}\beta^{k-r}\int _0^{2\pi}\cos^{r}\theta \sin^{k-r+2n}\theta d\theta.$$

Hence, we observe that the above integral on (0,π/2), after a change of variable, is equal to zero if k and r are even. Therefore, one has that

Now, we have to handle the quantity

$$\frac{(2(k-r)+2n-1)(2(k-r)+2n-3)\cdots(2(k-r)+1)}{r!(k-r)!}.$$

We observe that, after some suitable manipulations, we can write that

• for n=1, $\frac{2(k-r)+1}{r!(k-r)!}=\frac{1}{r!(k-r)!}+\frac{2}{r!(k-r-1)!}$;

• for n=2, $\frac{(2(k-r)+3)(2(k-r)+1)}{r!(k-r)!}=\frac {3}{r!(k-r)!}+\frac{12}{r!(k-r-1)!}+\frac{4}{r!(k-r-2)!}$;

• for n=3, $\frac {(2(k-r)+5)(2(k-r)+3)(2(k-r)+1)}{r!(k-r)!}=\frac{15}{r!(k-r)!}+\frac {90}{r!(k-r-1)!}+\frac{60}{r!(k-r-2)!}+\frac{8}{r!(k-r-3)!}$.

Then, we can argue that the following equality holds

$$ \frac{(2(k-r)+2n-1)(2(k-r)+2n-3)\cdots(2(k-r)+1)}{r!(k-r)!}=\sum_{j=0}^n \frac{a_{j,n}}{r!(k-r-j)!}$$

(4.3)

where a _j,n s are positive constants. For instance, the previous considerations lead to the following scheme

(4.4)

Unluckily, it is not possible to obtain the value a _j,n for each n and j. Indeed, by performing the same calculations leading to (4.4), it seems that a recursive rule for an arbitrary integer n does not hold. Nevertheless, it is easy to see that

$$a_{0,n}=(2n-1) (2n-3)\cdots3\cdot1,\qquad a_{n,n}=2^n.$$

Finally, we get that

(4.5)

Important results are also the following ones

$$ \int_0^ax^\mu(a-x)^\nu J_\mu(x)J_\nu(a-x)dx=\frac{\Gamma(\mu +\frac{1}{2})\Gamma(\nu+\frac{1}{2})}{\sqrt{2\pi}\Gamma(\mu+\nu +1)}a^{\mu+\nu+\frac{1}{2}}J_{\mu+\nu+\frac{1}{2}}(a),$$

(4.6)

with $\mathop{\mathit{Re}}\mu>-\frac{1}{2}$ and $\mathop{\mathit{Re}}\nu>-\frac{1}{2}$ (see [9], formula 6.581(3)),

$$ \int_0^a\frac{J_\mu(x)}{x}\frac{J_\nu(a-x)}{a-x}dx= \biggl(\frac{1}{\mu}+\frac{1}{\nu}\biggr)\frac{J_{\mu+\nu}(a)}{a},$$

(4.7)

with  Re(μ)>0,  Re(ν)>0 (see [9], formula 6.533(2)),

$$ \int_0^\infty x^{\mu-\nu}J_{\nu+1}(ax)J_\mu(bx)dx=\frac {(a^2-b^2)^{\nu-\mu}b^\mu}{2^{\nu-\mu}a^{\nu+1}\Gamma(\nu-\mu+1)},$$

(4.8)

with a≥b,  Re(ν+1)> Re(μ)>0 (see [9], formula 6.575(1)), and

$$ \int_0^{\pi/2}(\sin x)^{\nu+1}\cos(b\cos x)J_\nu(a\sin x)dx=\sqrt{\frac{\pi}{2}}\frac{a^\nu J_{\nu+\frac{1}{2}}(\sqrt{a^2+b^2} )}{(a^2+b^2)^{\frac{\nu}{2}+\frac{1}{4}}} ,$$

(4.9)

for  Reν>−1 (see [9], p. 743, formula 6.688.(2)).