Two-dimensional random walk, but with steps in $n$ directions (rather than just 4)

Question

At each step, choose a (uniform) random number $i=1\ldots n$, and walk one unit (i.e., constant stepsize) in the direction $\theta=i\frac{2\pi}n$. Then $n=4$ reduces to the usual case, with steps parallel to the $x,y$-axes. And $n=2$ "collapses" to the one-dimensional case.

Now, the expected distance for a $d$-dimensional walk after $N$ steps is given by Reference Request, Random Walk as $\sqrt{\frac{2N}d}\frac{\Gamma\left(\frac{d+1}2\right)}{\Gamma(d/2)}$. And for $d=2$, I'd expect that answer to be correct for the usual $n=4$ directions. But not necessarily correct for (the unusual) $n\ne4$.

And I've programmed it, and that answer's indeed correct for $n=4$, e.g., ...

rwalk2d> ntrials=100000, nsteps=256, ndirections=4
         avg_distance=14.183 (expected=14.180,stddev=7.4119)

where the program runs 100,000 independent trials of 256 steps each, with ndirections$\equiv n$=4. Output is avg_distance, the average of all 100,000 random walks, and expected distance from the preceding formula. And stddev is just the standard deviation of those 100,000 numerical trials.

Okay, so here's the goofy-to-me question. What happens when $n\ne4$??? Absolutely nothing!!! ...

rwalk2d> ntrials=100000, nsteps=256, ndirections=16
         avg_distance=14.180 (expected=14.180,stddev=7.4036)

rwalk2d> ntrials=100000, nsteps=256, ndirections=32
         avg_distance=14.182 (expected=14.180,stddev=7.3986)

rwalk2d> ntrials=100000, nsteps=256, ndirections=99
         avg_distance=14.185 (expected=14.180,stddev=7.4203)

Not even the standard deviation changes, which I calculated after seeing no change in the avg_distance.
  So my question: Is there some closed-form explanation why not?
I'd have intuitively expected some kind of $n$-dependence.

Moreover, note my remark above that for $n=2$ the situation "collapses" to the one-dimensional $d=1$ case. And >>that<< indeed works...

rwalk2d> ntrials=100000, nsteps=256, ndirections=2
         avg_distance=12.757 (expected=12.766,stddev=9.6640)

(And just in case you're wondering, running the program with $n=1$ gives avg_distance=256 and stddev=0:) So I'm quite surprised at this $n$-behavior (or lack of it), and trying to understand the underlying reason.

Answer 1

We will assume that $n \geq 2$. Let $X_k=(X_k^{(1)}, X_k^{(2)})$ denote the $k$-th increment. Then $\mathbf{E}[X_k] = \mathbf{0}$ and the covariance matrix $\Sigma_n$ satisfies

$$ \Sigma_2 = \begin{pmatrix}1&0\\0&0\end{pmatrix} \qquad \text{and} \qquad \Sigma_n = \begin{pmatrix}1/2&0\\0&1/2\end{pmatrix} \quad \text{for} \quad n \geq 3. $$

Now, if $S_N=\sum_{k=1}^{N}X_k$ denotes the $N$-th step of the random walk, then CLT tells that

$$ \mathbf{E}[|S_N|] \sim \sqrt{N} \, \mathbf{E}[|Z_n|] $$

as $N\to\infty$, where $Z_n \sim \mathcal{N}(\mathbf{0}, \Sigma_n)$. Now the conjectured behavior is explained by the fact that

$$ \mathbf{E}[|Z_n|] = \begin{cases}\sqrt{\frac{2}{\pi}}, &n = 2 \\ \frac{\sqrt{\pi}}{2}, & n \geq 3 \end{cases} $$