Proof: Mean and Variance of the squared distance of a random walk in n-dimensional space consider a $x$ step random walk starting from origin in $n$-dimensional space where each step is taken into a random direction and has a distance of 1, i.e., each step is a vector on the $n$-dimensional unit sphere.
What is the mean and the variance of the squared euclidean distance from the origin after $x$ steps?
Empirically and with some stochastic juggling, I was able to deduce the following properties:
$\mu = x$
$\sigma^2 = 2x^2/n$
However, I need these properties for a paper and thus have to prove them. My "juggeling" is not a good proof and it is too lengthy to be included in the paper. So is there a short proof for these properties or can you give me a reference where this is proven (then I would just cite it)?
 A: Assume that the $k$th step is $x_k$ in $\mathbb R^n$ with $E[x_k]=0$, $\|x_k\|=1$ almost surely, and that the steps $(x_k)$ are independent. Let $s_N=\|x_1+\cdots+x_N\|^2$. 
Then $s_N=N+\sum\limits_{k\ne\ell} x_k\cdot x_\ell$. By independence, $E[x_k\cdot x_\ell]=E[x_k]\cdot E[x_\ell]=0$ for every $k\ne\ell$ hence 
$$
E[s_N]=N.
$$
Likewise, $s_N^2=N^2+2N\sum\limits_{k\ne\ell} x_k\cdot x_\ell+t_N$ with 
$
t_N=\left(\sum\limits_{k\ne\ell} x_k\cdot x_\ell\right)^2=\sum\limits_{k\ne\ell}\sum\limits_{i\ne j}(x_k\cdot x_\ell)(x_i\cdot x_j).
$
The only terms in the summation whose mean is not zero are such that $\{k,\ell\}=\{i,j\}$. There are $2N(N-1)$ such terms hence $E[t_N]=2N(N-1)\alpha$ with $\alpha=E[(x_1\cdot x_2)^2]$. Finally, 
$$
\mathrm{var}(s_N)=E[t_N]=2N(N-1)\alpha.
$$
This is an exact formula, valid for every $N\geqslant1$. The value of the parameter $\alpha$ depends on the specifics of the distribution of the displacement $x_1$ since, considering $x_1=(x_1^{(i)})_{1\leqslant i\leqslant n}$,
$$
\alpha=\sum\limits_{i=1}^nE[(x_1^{(i)})^2]^2.
$$
If the coordinates of $x_1$ are identically distributed then $E[(x_1^{(i)})^2]=\frac1n$ for every $i$ by symmetry hence $\alpha=\frac1n$ and $\mathrm{var}(s_N)=\frac2nN(N-1)\sim\frac2nN^2$ when $N\to\infty$.
