Expectation of random 2d walk

Question

Start at the origin, take $n$ independent steps of length 1 in the direction of $\theta_i$, which is uniformly distributed on $[0,2\pi]$.

If $X,Y$ is the position after $n$ steps and $D = X^2 + Y^2$, what is $E[D]$?

Answer 1

Observe that a move of length $1$ in the direction $\theta$ is a translation by the vector $e^{i\theta}$, where $i^2 = -1$. Thus, if $D: = (X,Y)$ is the position after $n$ moves on the (complex) plane, then $$ D = e^{i\theta_1} + ... + e^{i \theta_n} \in \mathbb{C}, $$ hence $$ |D|^2 = D \overline{D} = \sum\limits_{k,m=1}^n e^{i(\theta_k - \theta_m)} = n + \sum\limits_{1\leq k \neq m\leq n}^n e^{i(\theta_k - \theta_m)}. $$ It follows that $$ \mathbb{E} |D|^2 = n + \sum\limits_{1\leq k \neq m\leq n}^n \mathbb{E} e^{i(\theta_k - \theta_m)} = n + \sum\limits_{1\leq k \neq m\leq n}^n \mathbb{E}e^{i\theta_k} \mathbb{E}e^{-i \theta_m} = n, $$ where we used the independence of $\{\theta_k\}$, and the fact that $\theta_k \sim U[0,2\pi]$ for concluding that $\mathbb{E}(\theta_k) = \frac{1}{2\pi} \int\limits_0^{2\pi} e^{ix} dx = 0$.