Some Inferences on a 2- and 3-Step Random Walk

Abstract

In 1905, Pearson proposed the following: “A man starts from a point 0 and walks l step in a straight line, then he turns any angle whatever and walks l step in a straight line. He repeats this process n times. I require the probability that after n steps he is at a distance r and r + dr from the starting point o.” In this paper, we will present some basic properties and characterizations of the distribution of the distance for 2 and 3 steps.

1 Introduction

The n-step Pearson’s random walk is a walk in the plane that starts at the origin 0 and consists of n steps of length 1 each taken into a uniformly random direction. Pearson [6] proposed this problem. Let X be the distance traveled in n steps. Kluyer [4] gave the probability density function (pdf) ${p}_{n}\left(x\right)$ of distance X after n steps of unit length. The pdf given by Kluyer [4] is

$$ p_{n} \left( x \right) = \int\nolimits_{0}^{\infty } xtJ_{0} \left( {xt} \right)(J_{0} (t))^{n} {\text{d}}t,\quad 0 \le x < n $$

(1.1)

where ${J}_{0 }(t)$ is the Bessel function of first kind and zeroth order. We will denote PRW(n,x) as the distribution of X whose pdf is given in (1.1).

The solution of (1.1) for n = 2 and 3 is as follows:

$$ \begin{gathered} p_{2} \left( x \right) = \frac{2}{\pi }(4 - x^{2} )^{ - 1/2} ,\quad 0 \le x \le 2, \hfill \\ p_{3} \left( x \right) = \frac{2\sqrt 3 }{\pi }\frac{x}{{3 + x^{2} }}\;_{2} F_{1} \left( {^{{\frac{1}{3}_{1} ,\frac{2}{3}}} \left| {\frac{{x^{2} (9 - x^{2} )^{2} }}{{(3 + x^{2} )^{3} }}} \right.} \right),\quad 0 < x \le 3. \hfill \\ \end{gathered} $$

Rayleigh [4] showed that for $n \ge 5,P_{n} \left( x \right)$ is close to the distribution with pdf ${p}_{nr}(x)$ as

$$ p_{nr} \left( x \right) = \frac{2x}{n}e^{{ - \frac{{x^{2} }}{n}}} ,\quad x \ge 0. $$

(1.2)

In this paper, we will present some distributional properties and characterizations of PRW(n,x) for n = 2 and 3.

2 Basic Properties

Case 1

PRW(2,x).

The cumulative distribution function (cdf) $P_{2} \left( x \right)$ of PRW(2,x) is

$$ \begin{aligned} & P_{2} \left( x \right) = \frac{2}{\pi }\arcsin (x/2)\quad 0 \le x \le 2. \\ P_{2} (1) & = 1/3.\quad {\text{For}}\,{\text{all}}\,n \\ Pn\left( 1 \right) & = \int\nolimits_{0}^{\infty } J_{1} \left( t \right)(J_{0} \left( t \right))^{n} {\text{d}}t \\ & = \left. {\frac{{ - (J_{0} \left( t \right))^{n + 1} }}{n + 1}} \right|_{0}^{ \in } \\ & = \frac{1}{n + 1} \\ \end{aligned} $$

The graph of the cdf for ${P}_{2}$(x) is shown in Fig. 1.

Fig. 1

The cdf for ${P}_{2}$(x)

The percentage point of P₂(x) is given in Table 1.

Table 1 Percentage points of ${P}_{2}(x$)

Let ${\mu }_{2}(m)$ be the mth moment of ${p}_{2}$(x), then we have

$$ \begin{aligned} \mu_{2} \left( m \right) & = \int_{0}^{2} {\frac{2}{\pi }x^{m} (4 - x^{2} )^{ - 1/2} } {\text{dx}} \\ & \quad = - \frac{{2^{m} }}{\surd \pi } \frac{{\Gamma \left( {\frac{m + 1}{2}} \right)}}{{\Gamma \left( {\frac{m + 2}{2}} \right) }} \\ \end{aligned} $$

(2.1)

Substituting x = 2w in (2.1), we obtain

$$ \begin{aligned} \mu_{2} \left( m \right) & = \int_{0}^{1} {\frac{2}{\pi }} 2^{m} w^{m} \left( {1 - w^{2} } \right)^{ - 1/2} {\text{d}}w \\ & = \int_{0}^{1} {2^{m} (\cos \pi \vartheta /2)^{m} } {\text{d}}\vartheta \\ & = \int_{0}^{1} {2^{m} (1 - \sin^{2} \pi \theta /2 )^{n} } {\text{d}}\vartheta \\ & = \int_{0}^{1} {(4 - 4\sin^{2} \pi \vartheta )^{m/2} } {\text{d}}\vartheta \\ & = \int_{0}^{1} {\left( {2 + 2\cos 2\pi \vartheta } \right)^{m/2} } {\text{d}}\vartheta \\ & = \int_{0}^{1} {|1 + e^{2i\pi \theta } |^{m/2} } \\ \end{aligned} $$

It is known (see [3]) that

$$ \mu_{n} \left( m \right) = \int\nolimits_{0}^{1} \int\nolimits_{0}^{1} ....\int\nolimits_{0}^{1} |1 + e^{{2\pi ix_{1} }} + \cdots + e^{{2\pi ix_{n - 1} }} |^{m} {\text{d}}x_{n - 1} , \ldots ,{\text{d}}x_{2} {\text{d}}x_{1} $$

(2.2)

The first ten moments of PRW(2) are given in Table 2.

Table 2 Moments of PRW(2)

An easier expression of the even moment 2m, m = 1, 2,…. is

$$ \mu_{2} \left( {2m} \right) = \mathop \sum \limits_{{k_{1} + k_{2} = m}}^{m} \left( {\frac{m!}{{k_{1} !k_{2} !}}} \right)^{2} $$

where ${k}_{1}$ and ${k}_{2}$ are nonnegative integers including zero.

The following two characterization theorems are given in Ahsanullah et al. [3].

Theorem 2.1

Suppose the random variable X is absolutely continuous with cdf F(x) such that F(0) = 0, F(x) > 0 for 0 < x, 2, F(x) = 1 for all $x \ge 2$, pdf f(x) and f′(x) exists. Assume that E(X^m) exists for all $m \ge 1.$ Then, $E(X^{m} |X \le x) = g(x)\tau \left( x \right)$, where $\tau \left( x \right) = \frac{f\left( x \right)}{{F\left( x \right)}}$ and $g\left( x \right) = \frac{p(x)}{2}\left( {4 - x^{2} } \right)^{1/2} ,$ $p(x) = \frac{{x^{m + 1} }}{{\pi \left( {m + 1} \right)}} + \frac{1}{\pi }\mathop \sum \limits_{k = 0}^{\infty } \frac{{\left( {2k - 1} \right)!!}}{{\left( {m + 2k + 1} \right)k!!2^{3k} }}x^{m + 2k + 1}$ if and only if $f(x) = \frac{2}{\pi }(4 - x^{2} )^{ - 1/2} ,0 \le x \le 2.$

Theorem 2.2

Theorem 2.1Suppose the random variable X is absolutely continuous with cdf F(x) with F(0) = 0, F(x) > 0 for 0 < x, 2, F(x) = 1 for all $x \ge 2$, pdf f(x) and f′(x) exists. Assume that E(X^m) exists for all $m \ge 1.$ Then, $E(X^{m} {|}X \ge x{)} = h(x)r(x),$ where $\tau \left( x \right)$ = $\frac{f\left( x \right)}{{1 - F\left( x \right)}}$ and $h(x) = \frac{\pi q\left( x \right)}{2}\left( {4 - x^{2} } \right)^{1/2} ,$ $q(x) = E(X) - p(x)$ if and only if $f(x) = \frac{2}{\pi }(4 - x^{2} )^{ - 1/2} ,0 \le x \le 2.$

For characterizations of PRW(2,x), for unequal steps, see Ahsanullah [1].

Case 2

PRW(3,x).

The cdf P₃(x) is as given follows:

$$ P_{3} \left( x \right) = \int\nolimits_{0}^{\infty } xJ_{1} \left( { xt} \right)(J_{0} (t))^{3} {\text{d}}t $$

Figure 2 shows the cdf of PRW(3,x).

Fig. 2

CDF of PRW(3,x)

Let ${\mu }_{3}$(m) be the mth moment of PRW(3,x}.

We have

$$ |1 + e^{2\pi ix} + e^{2\pi iy} | = (9 - 4\left( {\sin^{2} \pi x + \sin^{2} \pi y + \sin^{2} \left( {x - y} \right)} \right)^{1/2} . $$

(2.3)

Thus, from (2.3) and (2.2), we obtain

$$ \mu_{3} \left( m \right) = (9 - 4\left( {\sin^{2} \pi x + \sin^{2} \pi y + \sin^{2} \left( {x - y} \right)} \right)^{m/2} $$

(2.4)

The first ten moments of PRW(3,x) are given in Table 3.

Table 3 Moments of PRW(3,x)

Some easier expressions of the even moments, ${\mu }_{2}(2m)$, m = 1, 2,…., are

$$ \mu_{2} \left( {2m} \right) = \mathop \sum \limits_{{k_{1} + k_{2} + k_{3} = m}}^{m} \left( {\frac{m!}{{k_{1} !k_{2} !k_{3} }}} \right)^{2} , $$

(2.5)

where ${k}_{1}$, ${k}_{2}$ and ${k}_{3}$ are nonnegative integers including zeros.

$$ \mu_{2} \left( {2m} \right) = \mathop \sum \limits_{j = 0}^{m } \left( {\frac{m!}{{j!\left( {m - j} \right)!}}} \right)^{2} \left( {\frac{{\left( {2j} \right)!}}{j!j!)}} \right). $$

(2.6)

We will present here two characterizations of PRW(3,x).

Theorem 2.3

Suppose the random variable X is absolutely continuous with cdf F(x) such that F(0) = 0, F(x) > 0 for x > 0, F(x) = 1 for all $O \le x \le 3.$ pdf f(x), f′(x) exists$.$ $0\, \le\, x \,\le 3 \,If\,E\left( {X^{m} } \right)$ exists for $m \ge 1$ and $E(X^{m} |X \le x) = g\left( x \right)\tau \left( x \right)$ where $\tau \left( x \right) = \frac{f\left( x \right)}{{F\left( x \right)}}$ and $g\left( x \right)\frac{{\int_{0}^{\infty } x^{m + 1} J_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}$, if and only if $p_{3} { }\left( {\text{x}} \right) = \int\nolimits_{0}^{\infty } xtJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t,0 \le x < 3.$

Proof

We have

$$ \int\nolimits_{0}^{x} u^{m + 1} tJ_{0} \left( {ut} \right){\text{d}}u = \int\nolimits_{0}^{tx} \frac{{w^{m + 1} }}{{t^{m + 1} }}j_{0} \left( w \right){\text{d}}w = x^{m + 1} J_{1} (tx) $$

(2.7)

If ${p}_{3} (x)={\int }_{0}^{\infty }x{J}_{1}( xt)({J}_{0}(t{)}^{3}$dt, then using (2.7), we get

$$ E(X^{m} |X \le x) = \frac{{\int_{0}^{\infty } x^{m + 1} J_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}{{\int_{0}^{\infty } xJ_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}, $$

Thus,

$$ g(x) = \frac{{\int_{0}^{\infty } x^{m + 1} J_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}} $$

Suppose

$$ g(x) = \frac{{\int_{0}^{\infty } x^{m + 1} J_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}, $$

then

$$ \begin{aligned} g^{\prime}(x) & = x^{m} - \frac{{ \int_{0}^{\infty } x^{m + 1} J_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}\frac{B\left( x \right)}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}, \\ & = x^{m} - g\left( x \right)\frac{B\left( x \right)}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}, \\ \end{aligned} $$

where

$$ B(x) = \frac{d}{{{\text{d}}x}}\left( {\int\nolimits_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t} \right). $$

Thus

$$ \frac{{x^{m} - g^{\prime}\left( x \right)}}{g\left( x \right)} = \frac{B\left( x \right)}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}. $$

It is easy to show that

$$ \frac{{x^{m} - g^{\prime}\left( x \right)}}{g\left( x \right)} = \frac{f^{\prime}\left( x \right)}{{f(x)}}. $$

Hence,

$$ \frac{f^{\prime}\left( x \right)}{{f(x)}} = \frac{B\left( x \right)}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}} $$

(2.8)

Integrating both sides of (2.8) with respect to x, we obtain

$$ f(x) - c\int\nolimits_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t, $$

where c is a constant. Using the boundary condition ${\int }_{0}^{3}f\left(x\right)\mathrm{d}x =1,$ we obtain

$$ f(x) - \int\nolimits_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t, $$

Theorem 2.4

Suppose the random variable is absolutely continuous with cdf F(x) such that F(0) = 0, F(x) > 0 for x > 0, F(x) = 1 for all $x \cong 3$ and pdf f(x). We assume f'(x) exists for all x. $0 \le x \le 3$/If E(X^m) exists for $m \ge 1$ and $E(X^{m} |X \ge x) = h\left( x \right)r\left( x \right),$ where $r\left( x \right) = \frac{f\left( x \right)}{{1 - F\left( x \right)}}$ and $h\left( x \right) = \frac{{E\left( {X^{m} } \right) - \int_{0}^{\infty } x^{m + 1} J_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}{{E\left( {X^{m} } \right) - \int_{0}^{\infty } xJ_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}$, if and only if $p_{3} { }\left( {\text{x}} \right) = \int\nolimits_{0}^{\infty } xtJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t,\,\,0 \le x < 3$.

Proof

If ${p}_{3} (x)={\int }_{0}^{\infty }x{J}_{1}( xt)({J}_{0}(t{)}^{3}$dt, then using (2.3), we obtain

$$ E(X^{m} |X \ge x) = \frac{{E\left( {X^{m} } \right) - \int_{0}^{\infty } x^{m + 1} J_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}{{E\left( x \right) - \int_{0}^{\infty } xJ_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}} $$

Thus,

$$ h(x) = \frac{{{\text{E}}\left( {X^{m} } \right) - \int_{0}^{\infty } x^{m + 1} J_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}} $$

Suppose

$$ h(x) = \frac{{{\text{E}}\left( {X^{m} } \right) - \int_{0}^{\infty } x^{m + 1} J_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}, $$

then

$$ \begin{aligned} h^{\prime}(x) & = - x^{m} - \frac{{E\left( {X^{m} } \right) - \int_{0}^{\infty } x^{m + 1} J_{1} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}\frac{B\left( x \right)}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}, \\ & = - x^{m} - h\left( x \right)\frac{B\left( x \right)}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}, \\ \end{aligned} $$

where

$$ B(x) = \frac{d}{{{\text{d}}x}}\left( {\int\nolimits_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t} \right). $$

Thus,

$$ - \frac{{x^{m} + h^{\prime}\left( x \right)}}{h\left( x \right)} = \frac{B\left( x \right)}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}}. $$

It is easy to show that

$$ - \frac{{x^{m} + h^{\prime}\left( x \right)}}{h\left( x \right)} = \frac{f^{\prime}\left( x \right)}{{f(x)}}. $$

Hence,

$$ \frac{f^{\prime}\left( x \right)}{{f(x0)}} = \frac{B\left( x \right)}{{\int_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t}} $$

(2.9)

Integrating both sides of (2.4) with respect to x, we obtain

$$ f(x) - c\int\nolimits_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t, $$

where c is a constant. Using the boundary condition ${\int }_{0}^{3}f\left(x\right)\mathrm{d}x =1,$ we obtain

$$ f(x) = \int\nolimits_{0}^{\infty } txJ_{0} \left( {xt} \right)(J_{0} \left( t \right))^{3} {\text{d}}t. $$