Ant in a circle

Calling $\mathbb{E}(d)$ the expected time for leaving the circle if you are at distance $d$, $\mathbb{E}(d)$ should have the following property: $$\mathbb{E}(d)=1+\frac{1}{2\pi}\int_0^{2\pi}\mathbb{E}(\sqrt{d^2+1-2d\cos\alpha})d\alpha$$ with the constraint $\mathbb{E}(d)=0$ if $d>3$.

This formula say that the expected leaving time is 1 (the cost of doing another step) + the mean of the expected time over each point at distance 1 from the current point ($\sqrt{d^2+1-2d\cos\alpha}$ is the cosine law applied at the triangle between the origin, the current point and next point).

I don't know how to explicit solve this equation (nor if is feasible to be done), but it can be used for iteratively approximate the solution.

edit: I've tried to solve with R, seem that the solution is $\mathbb{E}(0)\simeq11.42$, and this is the plot of the approximation of the function:

The ant's position at each time $n\in\mathbb{N}$ is a random variable which we could model in the complex plane as $Z_0=0$, $Z_n=Z_{n-1}+\Delta Z_n=\sum_{k-1}^n\Delta Z_k$ for moves $\Delta Z_n=e^{i\Theta_n}$ which are i.i.d. and which depend on uniform i.i.d. random variates $\Theta_n=2\pi\,U_n\sim\operatorname{Unif}\left(0,2\pi\right)$ . If we also let $$ R_n=|Z_n|\,, \qquad S_n=\sup\limits_{1\le k\le n}R_n \qquad \text{and} \qquad T=\inf\left\{n\mid R_n>R\right\} $$ be the first time $n$ for which $R_n>R=3$, then we are asking for $\mathbb{E}[T]$, which can be calculated as $$ \mathbb{E}[T]=\sum_{n=R}^{\infty}n\cdot\mathbb{P}[R_{n-1} &lt R &lt R_n]\,. $$ Now if we let $F_n(r)=\mathbb{P}[R_n &lt r]$ and let $f_n(r)=\mathbb{P}[R_n = r]$ be the PDF and CDF of $R_n$, we can see that $F_n(n)=1$, and we may be able to determine these functions inductively, with a recursion: $$ \eqalign{ F_n(r)&= \mathbb{P}[R_n &lt r]\\&= \mathbb{P}[R_{n-1} &lt r-1]+ \mathbb{P}[R_n&ltr\,|\,r-1&lt R_{n-1}&lt r+1] \\&= F_{n-1}(r-1)+ \int_{r-1}^{r+1} f_{n-1}(s)\cdot \mathbb{P}[R_n&ltr|R_{n-1}=s] \,ds } $$ The latter integral can certainly be split into the two cases $s&ltr$ and $s>r$, for which the indeterminate probability inside the integrand can be evaluated using geometry by subtracting the area of sectors of circles (and triangles for $s>r$):

$$ \eqalign{ \mathbb{P}[R_n&ltr|R_{n-1}=s]&= \left\{ \begin{align} 1-\frac{\beta -r^2 \alpha}{2\pi} && s \le r\\ \frac{A_\gamma+r^2A_\alpha}{2\pi} && s \gt r\\ \end{align} \right. } $$ where the angles $\alpha,\beta,\gamma$ can be expressed using the law of cosines: $$ \eqalign{ \cos\alpha &=\frac{r^2+s^2-1}{2rs}\\ \cos\beta &=\frac{r^2-s^2-1}{2 s} \qquad\text{or}\qquad\sin\beta=r\,\sin\alpha\\ \cos\gamma &=\frac{-r^2+s^2+1}{2 s}=-\cos\beta\\ } $$ (or solved from the equations $re^{i\alpha}=s+e^{i\beta}$ & $\beta+\gamma=\pi$, from which one can infer the diagrams for each case) and $A_\alpha=2\alpha-\sin2\alpha$ is twice the area of $\{z\in\mathbb{C}:\,|z|>r,\,|z-s|&lt1\}$.

Well, that's a start. There is surely a more elegant recursion; perhaps @carlop has found it.

Just for fun and simplicity, here's the results if the ant were constrained to 1 dimension (can only move left or right). Also note that I considered the ant out when he reaches 3 from the center:

1E:  1 from edge
  Move Out: Exit and stop considering ant
  Move In: Goto 2E
2E:  2 from edge
  Move Out: Goto 1E
  Move In: Goto C
C:  Center
  Move Right or Left: Goto 2E

After Move 1: 1:1 at 2E
After Move 2: 1:2 at 1E, 1:2 at C
After Move 3: 1:4 out, 3:4 at 2E
After Move 4: 1:4 out, 3:8 at 1E, 3:8 at C
After Move 5: 7:16 out, 9:16 at 2E

Chance of being at 2E after (1 + 2*n) moves: $$(3/4)^n$$ Chance of being out at (1 + 2*n) moves: $$1 - (3/4)^n$$ Chance of an ant getting out on the (1 + 2*n)th move: $$(1 - (3/4)^n) - (1 - (3/4)^{n-1}) = (3/4)^{n-1}-(3/4)^n$$

Thus, the average time it takes an ant to get out is: $$\sum_{n=1}^{\infty}{(1 + 2n)*((3/4)^{n-1}-(3/4)^n)}$$

Whose final value turns out to be 9, which will clearly be lower than the average time for the 2-dimensional problem.

It would also be interesting to consider a fly in a sphere with similar constraints :)

I developed code using Monte Carlo Method as show below. I used Java because that is what I had available, but it could be translated to Matlab or other code:

import java.util.Random;
import static java.lang.Math.* ;

public class RandomWalk {
static int N = 100000 ; // Number of iterations
static double R = 3.0 ; // Radius of circle
static int L = 200 ;

public static void main(String[] args) {
    int countArr[] = new int[L] ;
    for (int ci : countArr) countArr[ci]=0 ;
    Random rGen = new Random();   
    rGen.setSeed(856) ;

    for (int i=0 ; i<N ; i++)
    {
        int c = 0 ; // Count number of legs
        double d=0 ; // distance from center ;
        double xt=0.0, yt=0.0 ; // Total x,y distance from 0

        while (d <= R)
        {
            double ang = 2*PI*rGen.nextDouble() ;
            xt += cos(ang) ;
            yt += sin(ang) ;
            d = sqrt(xt*xt + yt*yt) ;
            c++ ;
        }
        countArr[c]++ ;
    }
    for (int i=0 ; i<L ; i++)
        System.out.println(i + ", " + countArr[i]/(double)N) ;
 }
}

The PDF of the ant leaving the circle on leg $i$ looks like this:

Note that the plot has modes at $i=4$ and $i=6$. The expected value is 11.4.