3D random walk with complex phase I'm trying to understand the properties of 3D random walks, but with an overall complex phase at each step (the phase is drawn from a uniform distribution between 0 and 2*pi). To be concrete and simplify things, I want to perform a random walk with fixed length in 3-dimensions, but at each step in 3D real space, we also have an overall phase, i.e. the final sum after N steps is
$$\vec{D} = \sum_{i=0}^N \hat{n}_i e^{i \varphi_i}$$
where $\hat{n}_i$ is a 3D unit vector drawn uniformly on a 2-sphere, and $\varphi_i \in [0,2\pi)$ drawn uniformly.
Ultimately, I would like to understand the distribution of the magnitude of the real part of the walk, i.e. $|\text{Re}(\vec{D})|$ as $N \to \infty$ (by symmetry the vector can point in any direction with uniform probability on the 2-sphere). Can this be rewritten as a real random walk somehow?
 A: Here is how I would approach the problem:

*

*The real part of the walk is simply given by
\begin{align}
\mathrm{Re}({D_N}) = \sum_{i=0}^N\hat{n}_i\cos(\varphi_i)
\end{align}
which is already a purely real random walk (with non-unit step sizes)

*We could try to find the pdf of the combined random variable $d:=\hat{n}\cos(\varphi)$, but I don't know if it has a nice form. Luckily, for the $n\to\infty$ limit, thats not neccessary. Obviously, we have $\mathbb{E}(d)=0$, and we can easily compute the variance.
\begin{align}
\mathrm{Var}(d)=\mathbb{E}(dd^T)=\mathbb{E}(\hat{n}\hat{n}^T)\underbrace{\mathbb{E}(\cos^2(\varphi))}_{=\frac{1}{2}} = \frac{1}{6}
\end{align}
Note that $d$ is a 3D vector, so the variance is symmetric $3\times 3$ matrix. In this case the result is proportional to the $3\times 3$ identity matrix. This is due to the spherical symmetry of the problem.

*Now $\mathrm{Re}(D_N)$ is simply the sum of $N$ independent samples of $d$. So for large $N$ it will be approximately a (3-dimensional) normal distribution with mean zero and variance $\frac{N}{6}$. (again, this variance should be interpreted as a diagonal $3\times 3$ matrix).

*The pdf of of this 3-dimensional normal distribution is
\begin{align}
f(x)=\left(\frac{3}{\pi N}\right)^{3/2}\exp\left(-\frac{3}{ N}x^Tx\right)
\end{align}
($x$ is a 3D vector).

*Finally, we can say something about the magnitude of $D_N$:
\begin{align}
\mathbb{E}\left(|\mathrm{Re}(D_N)|\right)
\approx \int_{\mathbb{R}^3} dx\ |x| f(x)
\end{align}
Using spherical coordinates, this integral is easy to do (because $f(x)$ only depends on the magnitude of $x$, not on its direction):
\begin{align}
\mathbb{E}\left(|\mathrm{Re}(D_N)|\right)
&\approx \int_{\mathbb{R}^3} dx\ |x| f(x) \\
&= \left(\frac{3}{\pi N}\right)^{3/2}\int_0^\infty dr\ 4\pi r^3\exp\left(-\frac{3}{N}r^2\right) \\
&= \frac{2\sqrt{N}}{\sqrt{3\pi}}
\end{align}
Note that the calculation is only necessary in order to get the constant factor correct (and please check before using for something important. I did not check it very carefully). The fact that the total magnitude of the walk is proportional to $\sqrt{N}$ is true in general for random walks.
