off lattice random walks in 2, and 3 dimensions_rough copy#2

Off-Lattice Random Walks in 2 and 3

DimensionsA. Zimmerman, advised by Dr. M. Nolan

AbstractThe return probabilities and other aspects of lattice random walks are well

known. For a lattice random walk, the return probability to the origin of the walk will be one in both one and two dimensions, proven by Hungarian Mathematician George Pólya. However, less is known about off-lattice random walks. This paper discusses off-lattice random walks with a constant step length of 1. Also, Pólya’s theorem for lattice random walks will be generalized to the off-lattice case with a return area in the shape of a circle or sphere, generally called a ball. The

probability of return to a ball of radius ε< 12 will be found for lattice free walks in

two and three dimensions.

Off-Lattice Random Walks in 2 and 3 Dimensions

Table of ContentsIntroduction: What are Random Walks?..................................................................3

General Process……………………………………………………………………………………………………3

2 Dimensional Case………………………………………………………………………………………………4

3 Dimensional Case………………………………………………………………………………………………6

Conclusion……………………………………………………………………………………………………………8

Future Work…………………………………………………………………………………………………………8

Appendix A: An Explanation of W N(R) and W N ( R⃗ )………………………………………………9

Appendix B: Derivation of W N(R) in 2 and 3 Dimensions…………………………………..10

Appendix C: Analytic forms of W N(R) for 2 Dimensions……………………………………..14

References and Endnotes…………………………………….…………………………………………….18

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Introduction: What are Random Walks?A simple case of a random walk can be defined in the following manner:

A physicist and their significant other (a mathematician) exit a restaurant in an area they are unfamiliar with. This area happens to only have one street (the street that the restaurant is on), and the city blocks are constant, similar to New York City’s grid, but in 1 dimension. They then decide to conduct an experiment and flip a coin. If they flip heads, they go once city block left; if tails, they go one city block right. Once they go to the end of either city block, they repeat the process; heads left, tails right. They continue to repeat the process.

This is an example of a one dimensional lattice random walk. To extend it to two dimensions, think of the New York City street grid with the probability of going north, south, east, or west at each corner; for three dimensions, include the subways and skyscrapers with the probability of going up or down a level. The three dimensional metaphor is somewhat crude, since there is a probability of needing to fly in the air in some cases.

To formalize the idea of a lattice random walk, imagine a particle undergoing a random walk on a Cartesian grid. The particle can only end up at integer junctions (eg (0,1,0)), and can only take steps of length one in one direction at a time.

An interesting question to ask is as follows: What is the probability of returning to the place one started from after a certain number of steps? It turns out, that in one and two dimensional lattice random walks, the return probability to the origin is one as the number of steps becomes very largei. This is known as Pólya’s theorem, and was proved by the Hungarian mathematician George Pólya in 1912ii. Note that the return probabilities for dimensions three or higher is less than one.

These examples are all of lattice random walks; that is, the step length is held to one city block in a finite number of directions. However, in many practical applications of random walks, the walk is not restricted to a Cartesian lattice. This paper will generalize random walks for 2 and 3 dimensions and specifically a generalized version of Polya’s Theorem. Though the step length is kept constant, any direction of the step is allowed. This is easiest to imagine in the two dimensional case; any angle from 0 to 2π can be chosen. In the three dimensional case, any point on a sphere of radius 1 can be chosen. We will call these walks “off-lattice” random walks.

To generalize Pólya’s theorem for an off-lattice random walks, return to a specific point cannot be used; for an off lattice random walk, a return to the origin will in fact be 0. Instead, a circle/sphere/d-dimensional balliii with a radius of ε , with the radius less than the step length (

ε< 12 ) is given as the return area. Note that setting the step length equal to one simplifies the

problem.

The General ProcessFirst, a probability density of the random walk to a single point in space needs to be

derived, labelled as W N ( R⃗ ). It is desirable to generalize this probability density of a random

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walk to a single point to all space for N steps, notated as W N( R⃗). To do this, multiply W N ( R⃗ ) by the surface area of the ball. This gives the probability density for all points in space at a radius of R. For example, in 2 dimensions, W N ( R )=2 πR W N ( R⃗), and in 3 dimensions, W N ( R )=4 π R2W N( R⃗) (See Appendix A for a more detailed explanation).

After the probability density function is found, the probability of returning to a ball with radius epsilon is easily found by integrating the probability density function from 0 to ε , that is,

PN (ε )=∫0

ε

W N ( R ) dR (1)

A generating function is introduced :

G ( z )=∑N

PN (ε ) zN(2)

Another generating function which describes the probability of return to the ball the first time is notated as G1 (z ). It turns out thativ

G1 (z )=1− 1G ( z )

(3)

If z=1 , if G (1 ) diverges, then the return probability to the ball is 1. If G ( z ) is finite, then the return probability will be less than one.

2 Dimensional CaseThe starting probability function, is given by Chandresekhar and modified for two

dimensionsv: W N ( R⃗ )=W N ( R x , R y )

¿∫−∞

∞

…∫−∞

∞

∏i

N

W j ( x j , y j ) δ ¿¿¿

After writing the Dirac Delta’s as Fourier transform integrals, switching from Cartesian coordinates to polar coordinates, and doing several integrals:

W N ( R⃗ )= 12 π∫0

∞

w J0(Rw )[J 0 (w ) ]N dw , (5 )

where J0 (w ) is a cylindrical Bessel function of the first kind of order 0. A full derivation can be found in Appendix B. Given that W N ( R )=2 πR W N ( R⃗ ) , the probability density becomes:

W N ( R )=∫0

∞

Rw J 0 (Rw ) [ J0 (w ) ]N dw(6)

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At this point, an analytic form of the integral cannot be found. A large N approximation can be found, as well as an analytic form for N=2 (see Appendix C). This is useful for gaining information about what W N(R) could look like. Past that,W N(R) must be left in the integral form. However, it turns out that this is not a problem in finding PN (ε ).

The probability density of returning to a ball of radius ε is found by integrating equation (5):

PN (ε )=∫0

ε

W N ( R ) dR=∫0

ε

∫0

∞

Rw J 0(Rw)[J 0 (w ) ]N dwdR=∫0

∞

ε J1 (εw ) [ J 0 (w ) ] N dw ,(7)

doing the integral with respect to R first. Thus, the generating function defined in equation (2) is:

G ( z )=∑N =0

∞

∫0

∞

ε J1 (εw ) [ J0 (w ) ] N dw (8 )

Note that P0 (ε )=1, and P1 (ε )=0, so the summation starts at 2 and goes to infinity:

G ( z )=1+0+∫0

∞

ε J1 (εw )∑N=2

∞

[ J 0 (w ) z ]N dw (9)

The summation can be re-indexed to N=0 by pulling out [J 0 ( w ) z ]2, and turns into a geometric series, which converges. This makes

G ( z )=1+∫0

∞

ε J 1(εw )(J¿¿0 (w) z)2

1−(J ¿¿0(w)z )dw(10)¿¿

Setting z=1

G (1 )=1+∫0

∞

ε J1(εw)(J¿¿0(w))2

1−J0(w)dw (11)¿

The question is, does the integral diverge at either limit? It turns out that the upper limit of the integral in these types of problems converges. The problem is determining if the integral converges or diverges at w=0. A small w approximation is in order. Making the approximations:

J1 (εw )≈ εw2

−(εw )2

8, J 0 ( w ) ≈1− w2

4,(J ¿¿0(w))2≈1− w2

2¿

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J1 (εw )(J ¿¿0(w))2= εw2

− ε w3

4− (εw )3

8+ ε3w5

16(12)¿

Since w is small, all terms not on the order of w1 can be thrown away. Substituting equation (12) and the small w approximations into equation (11),

G(1)Lower Limit ≈1+∫0

❑

ε2 2w

dw=1+2ε2 limw → 0

lnw (13)

which clearly diverges, since the natural logarithm does not converge at w=0

Thus, since G (1 ) does not exist, G1 (1 )=1− 1

G ( z ) , which makes G1 (1 )=1. This states that a off-

lattice random walk in 2 dimensions to a ball of radius ε<12 will always return to that ball.

3 Dimensional CaseIn 3 dimensions, the starting probability density is given by S. Chandrasekhar asvi:

W N ( R⃗ )=∫−∞

∞

…∫−∞

∞

∏i

N

W j ( x j , y j , z j ) δ ¿¿¿

¿

Again, this requires similar steps to the 2 dimensional case; write the Dirac Delta’s as their Fourier Transform integral, convert to spherical coordinates, and do several integrals. The full steps can be seen in Appendix A. The final form of W N (R) is:

W N ( R )=∫0

∞ 2dwπ

(Rw)2 j0(Rw )¿

where j o(w) is a spherical first order Bessel function of the first kind defined byvii:

j o ( w )= sinww

(16)

Integrating W N ( R ) to find PN (ε ),

PN (ε )=∫0

ε

W N ( R ) dR=∫0

ε

∫0

∞ 2dwπ

(Rw)2 j0(Rw)¿ ¿

noting that it’s easier to convert the spherical Bessel functions to sines to do the integral, where

j1 (wε )=( sinwε(wε )2

− coswεwε

), a spherical Bessel function of the first kind of order oneviii. Inserting

PN (ε ) into the generating function as in the two dimensional case gives:

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G ( z )=∑N

2π ∫0

∞

ε2 j1 ( εw ) [ j0 (w ) z ]N dw (18)

As before, P0 (ε )=1, and P1 (ε )=0, so the summation starts at 2 and goes to infinity:

G ( z )=1+0+ 2π ∫0

∞

ε2 J 1(εw )∑N=2

∞

[ j0 (w ) z ]N dw(19)

This this sum is the exact same sum in equation (6), with a substitution of spherical Bessel functions instead of the typical cylindrical Bessel functions. It converges to the exact same function as long as spherical Bessel functions are substituted in place of equations (10):

G ( z )=1+ 2π∫0

∞

ε2 j1(εw)( j¿¿ 0(w) z)2

1− j0(w)z(20)¿

Setting z=1,

G (1 )=1+ 2π∫0

∞

ε2 j1(εw )( j¿¿0(w))2

1− j0(w)dw (21)¿

The upper limit of the integral converges, as before. The problem is determining if w=0 limit converges. Plugging in the Taylor series expansion for sines and cosines of small w into their respective spherical Bessel function:

j1 (εw ) ≈ wε3

, j0 ( w ) ≈1− w2

6, ( j¿¿0(w))2≈ w2 ¿

G(1)Lower Limit ≈1+2π

∫0

❑ wε32w2

3¿

w2

3

¿dw=1+ 4 ε3

π ∫0

❑

w dw=1+ limw →0

2 ε3

π w2(22)

Clearly, equation (22) converges at the 0 limit. Thus, since the integral converges, the return

probability for an off-lattice random walk in three dimensions to a radius of ε<12 will be less

than one. A numerical evaluation will give the actual probability.

Conclusion

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It can clearly be seen that the return probability of a random walk to a ball of radius ε will be one if the number of steps is very large in two dimensions. The return probability for three dimensions is less than one, though the exact numerical value of the return probability is uncalculated as of yet.

Future WorkFirst, the return probability of the 3 dimensional case needs to be calculated. Second,

though the formulas for off lattice random walks with a constant step length have been derived, it remains to be seen if the theoretical equations match up with real random walk data. As such, simulations of random walks need to be run and compared to probabilities of returning to different radii. These probabilities of returning to within a certain radii of the origin will also need to be determined from the theoretical equations.

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Appendix A: An Explanation of W N(R) and W N ( R⃗ )

In the notation used in this paper, the differential probability function P, can be written in differential form:

dP=W N (R ) dR( A1),

where W N ( R ) is the probability density over all space. The probability density of a point in that space, W N ( R⃗ ) is generically defined in terms of the probability as:

dP=W N ( R⃗ ) dm+1 R( A2),with m being the number of dimensions. Thus, for two and three dimensions:

dP=W N ( R⃗ ) d3R=2πR W N ( R⃗ ) dR=W N (R ) dR( A3) ,

dP=W N ( R⃗ ) d4 R=4 π R2W N ( R⃗ ) dR=W N ( R ) dR ( A 4 ) ,

respectively. It can also be thought of as multiplying W N ( R⃗ ) by the circumference/surface area/volume/hypervolume of a ball.

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Appendix B: Derivation of W N (R) in 2 and 3 Dimensions2 Dimensional Case:

Starting from equation (4):

W N ( R x , R y )=∫−∞

∞

…∫−∞

∞

∏i

N

W j ( x j , y j ) δ ¿¿¿

Putting the Dirac Delta into it’s Fourier Integral form: δ ¿

and doing the same for y. Thus, using vector notation, plug (B1) into (4)

W N ( R⃗ )= 1(2π )2

∫−∞

∞

∫−∞

∞

e i w⃗ ∙¿ ¿¿¿

Then the equation is converted to polar form, with the following substitutions:

W ( x , y )=W (r )= δ (1−r2)π

=121π

[δ (1−r )+δ (1+r ) ]= δ (1−r )2 π

(B3)

dxdy → rdrdθ ,Change the limits of integration

∓ i w⃗ ∙ R⃗=∓ iwR cosθ' ;∓i w⃗ ∙ r⃗=∓ iwr cosθ

noting that δ (1+r ) can be thrown out, since it is not within the limits of integration. The

function W (r )=δ (1−r )2π

makes logical sense; it’s simply the probability density of being on a

circle of radius r after 1 step. The pi sum generalizes this to N steps. W N ( R⃗ ) becomes

W N ( R⃗ )= 1(2 π )2

∫0

∞

∫0

2π

eiwR cosθ'

d θ ' dw 1(2π ) N

∫0

∞

∫0

2 π

…∫0

∞

∫0

2π

∏j

N

δ j (r−1 ) e−iwr j cosθ j r j d r j d θ j(B4)

The portions in the product notation is just the same function, integrated N times. Letting

~W ( w )=∫0

∞

∫0

2π

δ (r−1)e−irwcosθ rdrdθ(B 5)

Inserting (B5) into (B4), W N ( R⃗ ) becomes

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W N ( R⃗ )= 1(2π )2

∫−∞

∞

∫−∞

∞

e iwRcosθ d2w 1(2π )N [~W (w )]N (B6)

Evaluating ~W ( w ),

~W ( w )=∫0

∞

∫0

2π

δ (r−1)e−irw cosθ rdrdθ(B7)

~W ( w )=∫0

2π

e−iw cosθdθ=2 π J0 (w )(B8)

noting that the 2π ’s from ~W ( w ) cancel out with the 1

(2π )N from W N ( R⃗ ) and that J0 (w )is a

cylindrical Bessel function of the first kind of order zero. This can be verified via Mathematica. Thus, W N ( R⃗ ) becomes:

W N ( R⃗ )= 1(2π )2

∫0

2π

∫0

∞

w e i wR cosθ'

[J 0 (w ) ]N dwd θ' (B9)

¿ 12π∫0

∞

w J 0(Rw) [J0 (w ) ]N dw(B10)

Observe that the integral in equation (B10) is the same integral in equation (B8), just with a different variable of integration. Making the substitution W N ( R )=2 πR W N ( R⃗) mentioned in the introduction and Appendix A,

W N ( R )=∫0

∞

Rw J 0 (Rw ) [ J0 (w ) ]N dw (5 )

This is the final form for W N (R) in 2 dimensions that is used in this paper.

3 Dimensional Case

As described in equation (14),

W N ( R⃗ )=∫−∞

∞

…∫−∞

∞

∏i

N

W j ( x j , y j , z j ) δ ¿¿¿

δ ¿

Like the 2 dimensional case, the Dirac Delta is put into it’s Fourier integral form:

δ ¿

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and likewise for y and z. Using vector notation and inserting (A2) into (11)

W N ( R⃗ )= 1(2π )2

∫−∞

∞

∫−∞

∞

e i w⃗ ∙¿ ¿¿¿

The equation is then put into spherical coordinates, with:

W ( x , y , z )=W (r )= δ (1−r 2)2π

=1212 π

[δ (1−r )+δ (1+r ) ]= δ (1−r )4 π

(B12)

noting that δ (1+r ) again is not in the limits of integration. The substitutions made are:

dxdydz → r2sinθ drdθdφ ,Change the limits of integration

∓ i w⃗ ∙ R⃗=∓ iwR cosθ' ;∓i w⃗ ∙ r⃗=∓ iwr cosθ

Thus, W N ( R⃗ ) with the proper substitutions from (B12) is:

W N ( R⃗ )= 1(2 π )3

∫0

∞

∫0

π

∫0

2π

eiwR cosθ'

w2sin θ' dφd θ' dw 1(4 π )N

∫0

∞

∫0

π

∫0

2π

. ..∫0

∞

∫0

π

∫0

2 π

∏j

N

δ j(r−1)e−iw r jcosθ j r j2sin θ j d r j d θ j d φ j(B13)

As before, the portions in the product notations are all the same triple integral, computed N times. Let:

~W ( w )= 14 π ∫0

∞

∫0

π

∫0

2π

δ (r−1 )e−ir cosθ r2sin θ drdθdφ (B14)

such that, plugging (B14) into (B13)

W N ( R⃗ )= 1(2 π )3

∫0

∞

∫0

π

∫0

2π

eiwR cosθ'

w2sin θ' dφd θ' dw ¿

Evaluating ~W ( w ):

~W ( w )= 14 π ∫0

∞

∫0

π

∫0

2π

δ (r−1 )e−ir cosθ r2sin θ drdθdφ (B16)

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¿ 12∫0

π

e iwcosθ sinθ dθ= 12iw

(eiw−e−iw )= sinww

= j0 (w )(B17)

Substituting this back in to (B15),

W N ( R⃗ )= 1(2 π )3

∫0

∞

∫0

π

∫0

2π

eiwR cosθ'

w2sin θ ¿

Doing the polar and azimuthal integrals,

W N ( R⃗ )= 12 π2

∫0

∞ sin RwR

w ¿

Given that W N ( R )=4 π R2W N( R⃗),

W N ( R )= 2π ∫0

∞

Rw sinRw ¿¿¿

which is equation (15).

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Appendix C: Analytic forms of W N (R) for 2 Dimensions2 Dimensions: A case for N=2

Letting N=2 in equation (5)

W 2 ( R )=R∫0

∞

w J 0 ( Rw ) [J 0 (w ) ]2dw (C 1)

According to Gradshteyn and Ryzhikix,

∫0

∞

x J 12(v+n )

(ax )J 12(v−n )

(ax)J v (bx )dx=¿

2π b

1

√(4a2−b2 )T n( b

2a ) (C 2 ) ,

where T n is the Chebyshev Polynomial. In our case, a=1 , b=R , v=0 , n=0, T 0 ( g )=1, so

W 2 ( R )= 2π

1

√(4−R2 )(C3)

with R<∓2 (since we can’t divide by 0 and since our step length is one, we can’t go past a radius of 2)

We check to see if this probability density is properly normalized by integrating from 0 to 2, and setting the integral equal to 1:

∫0

2 2π

1

√(4−R2 )dR=1(C 4)

This means that this form is properly normalized and valid. A graph and contour plot can be seen in Figure 1.

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2 Dimensions: A Large N Case

An analytical form of W N(R)=∫0

∞

Rw J 0(Rw )[J 0 ( w ) ]N dw is currently unknown.

However, by making a few approximations, a general shape of the function as N → ∞ can be found. First, let

[J 0 ( w ) ]N=eNln (J0 (w ))(C5)

Insert the Bessel function summation form into (C5), keeping the first two terms, and approximating the logarithm:

J0 (w )=∑l=0

∞ (−1 )l

22l (l ! )2w2l

≅ 1− w2

4 (1! )2

ln (1−w2

4 )≅−w2

4(C6)

Thus inserting (C6) into (6) and evaluating the integral

W N ( R⃗ )= 1(2π )∫0

∞

w J 0(Rw )e−N w2

4 dw

W N ( R⃗ )= 1Nπ

e−R2

N

W N ( R )=2 πR W N ( R⃗)

W N ( R )=2RN

e−R 2

N (C 7)

Is the form normalized properly? If so, then W N ( R ) does in fact give an accurate function for the probability density for N being very large:

∫0

∞2RN

e−R 2

N =1(C8)

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Figure 1: A Probability density plot of W 2 ( R ).

15


Graphs of the function can be seen in Figures 2 and 3.

As it can be seen, the probability density of returning to the origin is in fact 0, as is

stated in the introduction section.

A. Zimmerman

Figure 2: A graph of W N ( R )=2RN

e−R 2

N for N=100 over 2π radians.

Figure 3: A graph of the cross section W N ( R )=2RN

e−R 2

N for N=100.

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References and Endnotes

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i Heldt, Karl. “Return Probabilities of Two Dimensional Lattice Random Walks.” 2013. Unpublished.ii Piaskowski, Kevin. “One and Two-Dimensional Random Walks with One-Step Memory.” 2016. Unpublished.iii From now on, the term “ball” will be exclusively used to describe the area/volume/hypervolume of return for this problem.iv Piaskowski, Kevin. “One and Two-Dimensional Random Walks with One-Step Memory.” 2016. Unpublished.v Ibid.vi Chandrasekhar, S. “Stochastic Problems in Physics and Astronomy.” Reviews of Modern Physics, Vol 15, Number 1. 1943. Note that this is modified for 2 dimensions (the original is in 3 dimensions) and put into modern notation. Apparently, the modern Dirac delta notation used today was not commonly in use at the time of Chandrasekhar published the paper cited. An alternative approach used by Chandrasekhar is called the “Method of Markov”, which gets around the diverging nature of the Dirac Delta.vii Ibid. In this case, the function is unmodified except in the case of the Dirac delta , as mentioned in footnote VI.viii Oliver, Frank W.J., Lozier, Daniel W., Boisvert, Ronald F., Clark, Charles W. “NIST Handbook of Mathematical Functions.” Cambridge University Press, 2010, New York, NY.ix Ibid.

off lattice random walks in 2, and 3 dimensions_rough copy#2

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