Probability, Random Walk, Distance from Origin after N steps in 2 and 3 dimensions [duplicate]

Question

I am looking for a formula of the distance from Origin after N equal steps in random directions in a 2 or 3 dimensional spaces. Can someone help me with a reference to a book, article or any publication dealing with this subject? Thanks!

Answer 1

note the question asks for a formula, simulating the thing is trivial..

As a first step, since you only ask for the distance, work out the incremental change in distance with each step assuming a unit step in a random direction. You can then develop a simple 1D numerical approach to find the distance after n steps: (mathematica code, the question was orignally posted to the mathematica site)

md[n_Integer] := (
       Mean@Table[
        dissquared = 0;
        Do[ 
          dissquared += 
               1 + 2 Cos[RandomReal[{0, 2 Pi}]] Sqrt[dissquared ];
         , {n}];
         Sqrt[dissquared], {2000}])


 Show[Plot[ Sqrt[Pi i/4 ], {i, 1, 500}, PlotStyle -> Red], 
      ListPlot[Table[{i, md[i]}, {i, 1, 500, 10}]]]

oops I dont seem to be allowd to post a graphic. Anyway we can emperically work out an expression for the average distance as

d[n] = Sqrt[ n Pi /4 ]

Note obviously d[1] must be 1, but the simulation converges to the above for n larger than 4.. If someone can finish up and show that analytically I'd like to see..

Answer 2

2 apprioriate models for your task could be:

1) Markovian matrix with probabilities between each state.

or 2) Brownian motion

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\large{\mbox{Random steps of length}\ a\ \mbox{in}\ 3D}$:

\begin{align} \pp_{N}\pars{\vec{r}} &\equiv \int\dd^{3}\vec{r}_{1}\,{\delta\pars{r_{1} - a} \over 4\pi a^{2}}\ldots \int\dd^{3}\vec{r}_{N}\,{\delta\pars{r_{N} - a} \over 4\pi a^{2}} \delta\pars{\vec{r} - \vec{r}_{1} - \cdots - \vec{r}_{N}} \\[3mm]&= \int\dd^{3}\vec{r}_{1}\,{\delta\pars{r_{1} - a} \over 4\pi a^{2}}\ldots \int\dd^{3}\vec{r}_{N}\,{\delta\pars{r_{N} - a} \over 4\pi a^{2}} \int\dd^{3}\vec{k}\, \expo{\ic\vec{k}\cdot\pars{\vec{r} - \vec{r}_{1} - \cdots - \vec{r}_{N}}} \\[3mm]&= \int\dd^{3}\vec{k}\,\expo{\ic\vec{k}\cdot\vec{r}} \bracks{\int\dd^{3}\vec{\rho}\,{\delta\pars{\rho - a} \over 4\pi a^{2}} \expo{-\ic\vec{k}\cdot\vec{\rho}}}^{N} \\[3mm]&= \int\dd^{3}\vec{k}\,\expo{\ic\vec{k}\cdot\vec{r}} \bracks{\int_{0}^{\infty}\dd\rho\,4\pi\rho^{2}\, {\delta\pars{\rho - a} \over 4\pi a^{2}} \int\expo{-\ic\vec{k}\cdot\vec{\rho}}{\dd\Omega_{\vec{\rho}} \over 4\pi}}^{N} \\[3mm]&= \int\dd^{3}\vec{k}\,\expo{\ic\vec{k}\cdot\vec{r}} \bracks{\int_{0}^{\infty}\dd\rho\,\delta\pars{\rho - a} \,{\sin\pars{k\rho} \over k\rho}}^{N} = \int\dd^{3}\vec{k}\,\expo{\ic\vec{k}\cdot\vec{r}} \bracks{\sin\pars{ka} \over ka}^{N} \\[3mm]&= \int_{0}^{\infty}\dd k\,4\pi k^{2}\,{\sin\pars{kr} \over kr}\, \bracks{\sin\pars{ka} \over ka}^{N} = {4\pi \over a^{3}}\int_{0}^{\infty}x^{2}{\sin\pars{x\tilde{r}} \over x\tilde{r}}\, \bracks{\sin{x} \over x}^{N}\,\dd x \end{align} where $\tilde{r} \equiv r/a$.