Consider a random walk on the plane generated as follows:
- $(x_0, y_0)\in\mathbb{R}^2$ is the initial position.
- $\alpha_0\in(-\pi,\pi)$ is the initial direction that the walker is facing.
- The next position is generated by taking a length 1 step towards a random direction in "front" of the walker, that is $(x_{k+1}, y_{k+1}) = (x_k + \cos(\alpha_k), y_k + \sin(\alpha_k))$ and $\alpha_{k+1} = \alpha_k + \beta_k$, where $\beta_k\sim U(-\frac{\pi}{2},\frac{\pi}{2})$.
If $R_N$ is such a random walk with $N$ steps, I would like to know what is the expected value of the square of the final distance from the initial point as a function of N (I'll denote this value by $D_N$).
It is obvious that $E(D_1)=1$ because we take a single step. Also, for $N=2$, since we can assume without loss of generality that $(x_0,y_0)=(0,0)$ and $\alpha_0=0$, one gets that $$E(D_2)=\frac{1}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\|(1+\cos(t), \sin(t))\|_2^2\ dt=\frac{2}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1+\cos(t)\ dt = \frac{2}{\pi}(\pi + 2).$$ If I'm not wrong, in general we get $$E(D_{N+1})=\frac{1}{\pi^N}\int_{[-\frac{\pi}{2},\frac{\pi}{2}]^N}\left(1+\sum_{k=1}^N\cos(t_1+\cdots+t_k)\right)^2+\left(\sum_{k=1}^N\sin(t_1+\cdots+t_k)\right)^2\ d(t_1,\ldots,t_N).$$
I know how to simplify the formula a bit more, or how to even run simulations in a computer to get an approximation of the final result, but would you be able to provide a closed formula for the value of such integral? Thanks!
Let's use complex numbers to represent the position in 2D space. Assuming with out loss of generality that the starting angle is $\alpha_0=0$, then the position after $n$ steps is $$ z_N = \sum_{n=1}^N \exp\left(i\sum_{k=1}^n \alpha_k\right), $$ with $\alpha_k$ being uniformly random on $[-\pi/2,\pi/2]$. We then compute the mean square distance $$ \begin{align} \langle z_N\bar z_N\rangle &= \sum_{n=1}^N\sum_{n'=1}^N \exp\left(i\sum_{k=1}^n \alpha_k-i\sum_{k=1}^{n'} \alpha_k\right)\\ & = N+2(N-1)\langle e^{i\alpha}\rangle+2(N-2)\langle e^{i\alpha}\rangle^2+\ldots+2\langle e^{i\alpha}\rangle^{N-1}\\ &=N+\sum_{k=1}^{N-1}2(N-k)\langle e^{i\alpha}\rangle^k. \end{align} $$ The combinatorics are not too bad and depends on how many $\alpha$s cancel in the square. Also we used that the $\alpha$s are uncorrelated to write $\langle e^{i\alpha_1+i\alpha_2}\rangle=\langle e^{i\alpha_1}\rangle\langle e^{i\alpha_2}\rangle$ etc.
Finally, we can compute $$ \langle e^{\pm i\alpha}\rangle= \frac{1}{\pi}\int_{-\pi/2}^{\pi/2}e^{\pm i\alpha}\, d\alpha= \frac{2}{\pi} $$ and evaluate the sum $$ \langle z_N\bar z_N\rangle = \frac{2^{2+N}+\pi^N(N(\pi^2-4)-4\pi)}{\pi^N(\pi-2)^2}. $$