Distribution for unit-length random walk on the plane Let $\theta_n \sim \text{uniform}(0, 2\pi)$ and $z = \sum_{n=1}^N \mathrm{e}^{i \theta_n}$. What is $\mathrm{P}(z)$? Alternatively, what is $\mathrm{P}(\lvert z \rvert^2)$? For the latter, I found
\begin{align*}
\lvert z \rvert^2
&= z \overline{z} \\
&= \left( \sum_{n=1}^N \mathrm{e}^{i \theta_n} \right) \left( \sum_{n=1}^N \mathrm{e}^{-i \theta_n} \right) \\
&= \sum_{n=1}^N \mathrm{e}^{i \theta_n} \mathrm{e}^{-i \theta_n} + \sum_{n=1}^N \sum_{m=1}^{n-1} (\mathrm{e}^{i \theta_n} \mathrm{e}^{-i \theta_m} + \mathrm{e}^{-i \theta_n} \mathrm{e}^{i \theta_m}) \\
&= \sum_{n=1}^N 1 + \sum_{n=1}^N \sum_{m=1}^{n-1} (\mathrm{e}^{i \theta_n - i \theta_m} + \mathrm{e}^{i \theta_m - i \theta_n}) \\
&= N + 2 \sum_{n=1}^N \sum_{m=1}^{n-1} \cos(\theta_n - \theta_m) \\
\end{align*}
I know $\cos \theta$ has an arcsine distribution. However, the angle differences are not independent (e.g. they can't all be at right angles to each other), and I'm not sure how to proceed.
 A: If you don't need the exact result for $N$.
Let's consider the projection of the walking point to the $x$ axis. We will see a random walk described by $$\sum_{i=1}^N\cos(\theta_i).$$
The random variables that we sum are independent, have zero expectation and variance $\frac12$. After normalization (dividing the sum by $\sqrt{\frac N2}$), by the central limit theorem, if $N\to \infty$ then the limiting distribution is standard normal. We get the same result if we project the walking particle to the straight $y=mx$, independently of $m$. That is the limiting distribution of the walking pont is two dimensional normal with covariance matrix $\left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$. So, the two coordinates are getting independent. Note that $N$ does not have to be very large for the Gaussian approximation to be suitable.
Since the distribution of the standardized walking point (for a fairly large $N$) is two dimensional standard normal, the distribution of the squared distance from the origin is of $\chi^2$ with parameter $2$.
Note
If the distribution of the $\theta$s  is not uniform over $[0,2\pi]$ then the argumentation above will still be true except that the covariance matrix will not be the unit matrix. That is, the coordinates of the walking point will not be independent even on the long run. In such a case we cannot say that the square of the distance from the origin  is of  $\chi^2$.
