Random walk problem in the plane

Let a particle in the plane $R^2$ executes random jumps at discrete times $t= 1, 2, …$. At each step, the particle jumps from the point it is a distance of lenght one. The angle of any new jump (say, with the $x$ axis) is uniformly distributed in $[0,2\pi]$.

Question: If initially ($t=0$) the particle is at the origin, what is the probability that it gets back to the unit disk of the plane for each time $t= 2, 3, 4, …$? In particular, what is the value of this probability for $t=3$?

Thanks for any help!

Solutions Collecting From Web of "Random walk problem in the plane"

$\newcommand{\+}{^{\dagger}}%
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The probability density for the step $n$ is given by
$\ds{{\rm p}\pars{\vec{r}_{n}} \equiv {\delta\pars{r_{n} – 1} \over 2\pi}}$. The probability density $\pp_{N}\pars{\vec{r}}$ of arriving at $\vec{r}$ after $N$ steps is given by:

\begin{align}
\pp_{N}\pars{\vec{r}}
&\equiv
\int\dd^{2}\vec{r}_{1}\,{\rm p}\pars{\vec{r}_{1}}\ldots
\int\dd^{2}\vec{r}_{N}{\rm p}\pars{\vec{r}_{N}}
\delta\pars{\vec{r} – \sum_{\ell = 1}^{N}\vec{r}_{\ell}}
\\[3mm]&=
\int\dd^{2}\vec{r}_{1}\,{\rm p}\pars{\vec{r}_{1}}\ldots
\int\dd^{2}\vec{r}_{N}{\rm p}\pars{\vec{r}_{N}}
\int\exp\pars{\ic\vec{k}\cdot\bracks{\vec{r} – \sum_{\ell = 1}^{N}\vec{r}_{\ell}}}
\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
\\[3mm]&=
\int\expo{\ic\vec{k}\cdot\vec{r}}
\bracks{\int\dd^{2}\vec{R}\,{\rm p}\pars{\vec{R}}\expo{-\ic\vec{k}\cdot\vec{R}}}^{N}
\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
=
\int\expo{\ic\vec{k}\cdot\vec{r}}
\bracks{\int_{0}^{2\pi}\expo{-\ic k\cos\pars{\theta}}\,{\dd\theta \over 2\pi}}^{N}
\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
\end{align}

However
$$
\int_{0}^{2\pi}\expo{-\ic k\cos\pars{\theta}}\,{\dd\theta \over 2\pi}
=
\int_{-\pi}^{\pi}\expo{\ic k\cos\pars{\theta}}\,{\dd\theta \over 2\pi}
=
{1 \over \pi}\int_{0}^{\pi}\expo{\ic k\cos\pars{\theta}}\,\dd\theta
=
{\rm J}_{0}\pars{k}
$$
where ${\rm J}_{\nu}\pars{k}$ is the
$\nu$-$\it\mbox{order Bessel Function of the First Kind}$.

\begin{align}
\pp_{N}\pars{\vec{r}}
&=
\int\expo{\ic\vec{k}\cdot\vec{r}}{\rm J}_{0}^{N}\pars{k}
\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
=
{1 \over 2\pi}\int_{0}^{\infty}\dd k\,k\,{\rm J}_{0}^{N}\pars{k}\int_{0}^{2\pi}
\expo{\ic kr\cos\pars{\theta}}\,{\dd\theta \over 2\pi}
\end{align}
$$
\color{#ff0000}{\pp_{N}\pars{\vec{r}}
=
{1 \over 2\pi}\int_{0}^{\infty}{\rm J}_{0}\pars{kr}{\rm J}_{0}^{N}\pars{k}k
\,\dd k}
$$

The probability ${\rm P}_{N{\Huge\circ}}$ that it returns to the unit circle after $N$ steps is given by:
\begin{align}
\color{#0000ff}{\large{\rm P}_{N{\Huge\circ}}}
&=
\int_{r\ <\ 1}\pp_{N}\pars{\vec{r}}\,\dd^{2}\vec{r}
=
\int_{0}^{\infty}\overbrace{\bracks{\int_{0}^{1}{\rm J}_{0}\pars{kr}r\,\dd r}}
^{{\rm J}_{1}\pars{k}/k}
{\rm J}_{0}^{N}\pars{k}k\,\dd k
\\[3mm]&=
\color{#0000ff}{\large\int_{0}^{\infty}{\rm J}_{1}\pars{k}
{\rm J}_{0}^{N}\pars{k}\,\dd k}
\end{align}
We compute a few values with Wolfram Alpha:
$$
\begin{array}{rclrcl}
{\rm P}_{0{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}
\,\dd k & = & 1\,,\quad
{\rm P}_{1{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}
{\rm J}_{0}\pars{k}\,\dd k & = & \half
\\
{\rm P}_{2{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}{\rm J}_{0}^{2}\pars{k}
\,\dd k & = & {1 \over 3}\,,\quad
{\rm P}_{3{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}
{\rm J}_{0}^{3}\pars{k}\,\dd k & = & {1 \over 4}
\\
{\rm P}_{4{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}{\rm J}_{0}^{4}\pars{k}
\,\dd k & = & {1 \over 5}\,,\quad
{\rm P}_{5{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}
{\rm J}_{0}^{5}\pars{k}\,\dd k & = & {1 \over 6}
\end{array}
$$

It $\tt\large seems$ the exact result is
$\ds{%
{\rm P}_{N{\Huge\circ}}
=
\int_{0}^{\infty}{\rm J}_{1}\pars{k}{\rm J}_{0}^{N}\pars{k} \,\dd k
=
{1 \over N + 1}}$

Mathematica can solve this integral and it yields the $\ul{\mbox{exact result}}$ $\color{#0000ff}{\Large{1 \over N + 1}}$ when $\color{#ff0000}{\large\Re N > -1}$.

You can create a probility dristribution for the location after one step (the unit circle with a certain constant prbability). After two steps, I think that should be doable too. You can also carculate for each point the probability of entering de unit disk in the next step. Then, by combining the probability distribution ($P_1(x,y)$) and the probability of ending in the disk $P_2(x,y)$, you get
$$
\int P_1(x,y)P_2(x,y) \,dx\,dy
$$

This is not a complete answer, but is kind of long for a comment.

For $t=2$, draw a picture. It’s pretty easy to see the answer is $1/3$.

For the $t=3$ case, let the angles be $\theta_1$, $\theta_2$, and $\theta_3$. By symmetry, you can assume without loss of generality that $\theta_1 = 0$. So you need to find the probability that
$\|\langle 1, 0 \rangle + \langle \cos\theta_2, \sin\theta_2 \rangle +\langle \cos\theta_3,\sin\theta_3\rangle \|<1$. A little computation shows that this is equals
$P(1+\cos\theta_2+\cos\theta_3+\cos(\theta_2-\theta_3)<0)$, where $\theta_2$ and $\theta_3$ are chosen independently from the uniform distribution on $[0,2\pi]$. That’s as far as I got.

This is the probability that the particle is in the unit disc at time $t=3$ (not $t$ equals 2 or 3).